Wastewater operator certification study guide : a guide to preparing for wastewater treatment

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certification Study Guide

Wastewater operator Related Titles From the American Water Works Association

Operator Certification Study Guide: A Guide to Preparing for Water Treatment and Distribution Operator Certification Exams This is the best-selling companion volume to Wastewater Operator Certification Study Guide. It includes 560 sample questions and answers found on water operator certification tests, organized by Grades 1, 2, 3, and 4. No. 20517

Math for Wastewater Treatment Operators Grades 1 and 2Math for Wastewater Treatment Operators Grades 3 and 4These are comprehensive wastewater math study guides—separate volumes are available for Grades 1 and 2 and for

Grades 3 and 4. Each provides sample questions and answers similar to those found in the math sections of the exams, plus they are excellent on-the-job references for math-related questions and problems. Grades 1 and 2, No. 20662 Grades 3 and 4, No. 20663

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Wastewater Microbiology: A Handbook for OperatorsThis book explains how to control the microbiological process from the viewpoint

of the wastewater treatment operator. It provides an understanding of basic microbiology and how microorganisms are used in the treatment of wastewater. No. 20563

The Wastewater Operator Certification Study Guide provides practice questions to thoroughly prepare for certification exams. It includes a range of sample questions and answers for Class I, II, III, and IV wastewater-operator certification exams on all relevant topics, including multiple levels of wastewater treatment, equipment, sampling, reporting, safety, solids handling, and more.

20683 Cover-ca.indd 1 6/30/2009 10:31:06 AM

A Guide to Preparing for Wastewater Treatment Certification Exams

First Edition

John Giorgi

CertifiCation Study Guide

Wastewater operator

WW OpCert SG FINAL01.indb 1 6/26/2009 12:23:09 PM

iii

Wastewater Operator Certification Study Guide: A Guide to Preparing for Wastewater Treatment Certification ExamsCopyright © 2009 American Water Works Association

AWWA Publications Manager: Gay Porter De NileonProject Manager: Martha Ripley GrayCopy Editor: Deborah J. LynesProduction: Carrie MattoxCover Art: Cheryl Armstrong

All rights reserved. No part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopy, recording, or any information or retrieval system, except in the form of brief excerpts or quotations for review purposes, without the written permission of the publisher.

DisclaimerAlthough this study guide has been extensively reviewed for accuracy, there may be an occasion to dispute an answer, either factually or in the interpretation of the question. AWWA has made every effort to correct or eliminate any questions that may be confusing or ambiguous. If you do find a question that you feel is confusing or incorrect, please contact the AWWA Publishing Group.

Additionally, it is important to understand the purpose of this study guide. It does not guarantee cer-tification. It is intended to provide the operator with an understanding of the types of questions he or she will be presented with on a certification exam and the areas of knowledge that will be covered. AWWA highly recommends that you make use of the additional resources listed at the end of this study guide in preparing for your exam.

Library of Congress Cataloging-in-Publication DataGiorgi, John.Wastewater operator certification study guide : a guide to preparing for wastewater treatment certi-fication exams / by John Giorgi. -- 1st ed. p. cm. Includes bibliographical references. ISBN 978-1-58321-728-3

1. Sewage--Purification--Examinations--Study guides. 2. Sewage disposal plants--Examinations --Study guides. 3. Sanitary engineers--Certification--United States--Examinations--Study guides. I. Title.

TD745.G565 2009628.3076--dc22

2009011271

6666 West Quincy AvenueDenver, CO 80235-3098303.794.7711www.awwa.org


iii

Contents

List of Figures, viiPreface, ixAcknowledgments, xi

Grade 1 Questions, 1Wastewater Sources and Characteristics, 1Basic Wastewater Treatment, 2Wastewater Treatment Facilities, 3Preliminary Treatment, 3Racks, Screens, Comminutors, and Grit Removal, 4Primary Sedimentation, 4Sedimentation and Flotation, 4Secondary Treatment Processes, 5Activated Sludge, 5Waste Treatment Ponds, 6Odor Control, 6Phosphorus Removal, 7Disinfection, 7Wastewater Solids Processing, 7Sampling, Testing, and Reporting, 8Maintenance, 10Safety, 10Basic Chemistry, 12

Grade 1 Answers and References, 15

Grade 2 Questions, 25General Wastewater Treatment Principles, 25Wastewater Treatment Facilities, 25Wastewater Sources and Characteristics, 27Racks, Screens, Comminutors, and Grit Removal, 28Sedimentation and Flotation, 28Secondary Treatment Processes, 29Activated Sludge, 30


iviv vv

Trickling Filters, 33Rotating Biological Contactors, 34Waste Treatment Ponds, 35Sludge Digestion and Solids Handling, 36Odor Control, 40Enhanced Biological (Nutrient) Control, 41Disinfection, 41Solids Removal from Secondary Effluents, 42Effluent Disposal, 43Wastewater Solids Processing, 43Residual Solids Management, 46Water and Wastewater Microbiology, 47Maintenance, 47Safety, 48Basic Chemistry, 49


Grade 3 Questions, 67General Wastewater Treatment Principles, 67Wastewater Treatment Facilities, 68Wastewater Sources and Characteristics, 69Racks, Screens, Comminutors, and Grit Removal, 69Sedimentation and Flotation, 70Activated Sludge, 71Trickling Filters, 74Rotating Biological Contactors, 76Waste Treatment Ponds, 77Sludge Digestion and Solids Handling, 79Odor Control, 81Phosphorus Removal, 82Enhanced Biological (Nutrient) Control, 83Disinfection, 84Solids Removal from Secondary Effluents, 85Effluent Disposal, 86Residual Solids Management, 86Wastewater Reclamation and Reuse, 88Water and Wastewater Microbiology, 90Basic Chemistry, 91



iviv vv

Grade 4 Questions, 111General Wastewater Treatment Principles, 111Sedimentation and Flotation, 112Activated Sludge, 112Trickling Filters, 120Rotating Biological Contactors, 121Waste Treatment Ponds, 122Sludge Digestion and Solids Handling, 123Odor Control, 124Phosphorus Removal, 124Enhanced Biological (Nutrient) Control, 125Disinfection, 126Solids Removal from Secondary Effluents, 127Residual Solids Management, 129Water and Wastewater Microbiology, 133Basic Chemistry, 134


Grades 1–4 Pumping Questions, 155Grade 1 Pumping Questions, 155

Basic Water Hydraulics, 155Introduction to Centrifugal Pumps, 155Centrifugal Pump Components, 156Centrifugal Pumps—Theory, 157Pump Nomenclature, 157Mechanical Seals, 158

Grades 2–4 Pumping Questions, 158Basic Water Hydraulics, 158Centrifugal Pumps—Theory, 159Centrifugal Pumps—Types, 160Pump Nomenclature, 160Replacement of Pump Components on End Suction Centrifugal Pumps, 161Selection and Replacement of Packing, 162Pump Piping System, 163End Suction and Split-Case Centrifugal Pumps, 163Line Shaft Turbine Operating Conditions, 164Special Pumping Units, 164Introduction to Centrifugal Pumps, 165Centrifugal Pump Components, 166Centrifugal Pumps: Operational Procedures, 168Centrifugal Pump: Maintenance Procedures, 168Centrifugal Pump: Lubrication, 168


vivi vii

Centrifugal Pump: Troubleshooting, 169Centrifugal Pump: Modifications, 170Positive-Displacement Pumps, 170

Grades 1–4 Pumping Answers and References, 171Math Questions, 181Answers to Math Questions, 195Math Questions with Solutions, 197Appendix A—Common Conversion Factors, 229Appendix B—Summary of Wastewater Treatment Equations (All Grades), 233Appendix C—Wastewater Flowchart Diagrams, 255Appendix D—Abbreviations, 269References, 273


vivi vii

Figures

Figure 1 Flowchart of Typical Wastewater Treatment Processes, 256Figure 2 Flowchart of Conventional Activated Sludge Process, 257Figure 3 Flowchart of Contact Stabilization Process, 258Figure 4 Flowchart of Activated Sequencing Batch Reactor, 259Figure 5 Flowchart of Typical Wastewater Treatment Using Ponds after Secondary Treatment, 260Figure 6 Flowchart of Typical Wastewater Treatment Using Polishing Ponds in Series Treatment, 261Figure 7 Flowchart of Wastewater Treatment Using Rotating Biological Contactor Process, 262Figure 8 Flowchart of Typical Wastewater Treatment Using an Oxidation Ditch, 263Figure 9 Flowchart of Typical Wastewater Treatment Using Electrodialysis or Nanofiltration, 264Figure 10 Flowchart of Typical Wastewater Treatment Using a Membrane Bioreactor, 265Figure 11 Flowchart of Wastewater Treatment Nitrification Process, 266Figure 12 Flowchart of Wastewater Treatment Plant Processing Sludge, 267


viiiviii ix

This work is dedicated to my wife, Flora Zhou Giorgi; my children, Sara, Stephanie, and Steve; my mother, Thelma Giorgi; and my father, Albert Peter Giorgi.


viiiviii ix

Preface

This guide is intended to give operators practice answering questions that are similar in format and content to the questions that appear on certification exams. The questions in this study guide are not the same questions that will appear on the certification exam. However, the questions allow operators to experience the types of questions that may be on a certifi-cation exam. If you have difficulty answering any of the questions in this study guide, you should consult the reference source provided following the answer to the question.

American Water Works AssociationAWWA is an international nonprofit and educational society and the largest and oldest organization of water professionals in the world. Its more than 55,000 members represent the full spectrum of the water community: treatment plant operators and managers, scientists, environmentalists, manufacturers, academicians, regulators, and others who hold genuine interest in water supply and public health. Membership includes more than 4,600 utilities that supply water to roughly 180 million people in North America. Through our collective strength, we become better stewards of water for the greatest good of the people and the environment.

Founded in 1881, AWWA provides knowledge, information, and advocacy to improve the quality and supply of water in North America and beyond. AWWA advances public health, safety, and welfare by uniting the efforts of the full spectrum of the water community.


xi


xi

Acknowledgments

John Giorgi would like to thank the staff and editors of the American Water Works Associa-tion; Gay Porter De Nileon, Publications Manager; and Dean Bugher for his assistance and recommendations.

The following people served as reviewers for this edition of the Wastewater Operator Certifi-cation Study Guide. Their assistance is greatly appreciated.

E.E. Arasmith, ACR, Albany, Ore. (AWWA)K.D. Kerri, California State University, Sacramento, Calif. (AWWA)S.M. Passarelli, AWWA, Denver, Colo. (AWWA)D.S. Sarai, Asheville, N.C. (AWWA)


1

Grade 1 Questions

Wastewater Sources and Characteristics1. What is the most common reason for a wastewater to contain intermittent large

amounts of sand, gravel, and grit?a. groundwater infiltrationb. storm eventsc. cement company dischargesd. refinery discharges

2. Combined wastewater is defined asa. domestic wastewater and industrial wastewaterb. sanitary wastewaterc. domestic wastewater and stormwater runoffd. sanitary wastewater and stormwater runoff

3. In general, the solids content in wastewater is abouta. 0.1%b. 0.6%c. 1%d. 2.5%

4. In general, what is the percentage of suspended solids in the total amount of solids?a. 30%b. 35%c. 45%d. 55%

5. Generally, the percentage of dissolved solids in the amount of total solids is abouta. 45%b. 55%c. 65%d. 70%

6. Which type of organisms would most probably be associated with poor treatment or young biomass?a. amoebasb. free-swimming ciliatesc. rotifersd. stalked ciliates

7. Which type of organisms would most probably be associated with very clear plant effluent, low biochemical oxygen demand (BOD), and low amounts of suspended solids?a. free-swimming ciliatesb. stalked ciliatesc. amoebasd. rotifers


2 WASTEWATER OPERATOR CERTIFICATION STUDY GUIDE

Basic Wastewater Treatment8. Aerobic bacteria will reproduce and live in an environment that contains only

a. hydrogen sulfideb. oxygenc. waterd. methane

9. Anaerobic bacteria will reproduce and live only in an environment that completely lacksa. free or dissolved nitratesb. dissolved methanec. dissolved hydrogen sulfided. free or dissolved oxygen (DO)

10. The biochemical oxygen demand (BOD) test usually has a specific time ofa. 2 daysb. 3 daysc. 5 daysd. 10 days

11. An Imhoff cone measuresa. volume of settleable solidsb. how fast sludge water will filter through a membranec. depth of transparency in a digesterd. sludge carry-over from a digester

12. Septic conditions are caused bya. fungib. moldsc. anaerobic bacteriad. hydrogen sulfide accumulation

13. Scum is usually removed in the _______ treatment stage.a. primaryb. secondaryc. biologicald. chemical

14. When domestic wastewater enters natural waters, what species deplete most of the oxygen?a. anaerobic bacteriab. aerobic bacteriac. protozoansd. rotifers

15. Once the oxygen is depleted in a natural body of water, which organisms take over?a. flagellatesb. sporozoansc. anaerobic bacteriad. coliform bacteria


GRADE 1 QUESTIONS 3

16. The major gas produced by anaerobic decomposition isa. methaneb. carbon dioxidec. nitrogend. hydrogen sulfide

Wastewater Treatment Facilities17. Another name for primary treatment is

a. chemical treatmentb. sedimentationc. grit removald. biological treatment

18. Digester gas is _______ when mixed with air in the proper proportions.a. extremely toxic and moderately explosiveb. extremely explosivec. slightly toxic and explosived. slightly explosive when there is an ignition source and

19. A pond that is used to treat raw wastewater is calleda. an oxidation pondb. a facultative pondc. a waste stabilization pondd. a decomposition pond

20. A pond that has an aerobic top layer and an anaerobic bottom layer is calleda. an oxidation pondb. a facultative pondc. a waste stabilization pondd. a decomposition pond

Preliminary Treatment21. What is the angle of the bars on a manually cleaned bar screen?

a. 25°b. 30°c. 33°d. 45°

22. What is the angle of the bars on a mechanically cleaned bar screen?a. 30 to 45°b. 40 to 45°c. 45 to 60°d. 50 to 60°

23. What is the ideal wastewater velocity in feet per second (ft/s) through a screen channel?a. 0.5 ft/sb. 1.0 ft/sc. 1.5 ft/sd. 2.0 ft/s



24. What is the ideal wastewater velocity in feet per second (ft/s) for grit removal in a grit chamber?a. 0.7 ft/sb. 1.0 ft/sc. 1.2 ft/sd. 1.4 ft/s

25. In general, what is the water content of screened material?a. 50%b. 65%c. 80%d. 89%

26. What is the wastewater velocity range in feet per second (ft/s) for grit removal in a grit chamber?a. 0.5 to 1.0 ft/sb. 0.7 to 1.4 ft/sc. 0.8 to 1.5 ft/sd. 1.0 to 1.5 ft/s

Racks, Screens, Comminutors, and Grit Removal27. Which one of the following chemicals gives water its greatest capacity to neutralize

acids?a. hydroxideb. silicatec. borated. phosphate

28. The bar rack is useda. when the bar screen is out of serviceb. before the bar screenc. after the bar screend. either before or after the bar screen

Primary Sedimentation29. Typically, what is the average detention time for wastewater in a primary clarifier?

a. 2 hrb. 3 hrc. 3.5 hrd. 4.5 hr

Sedimentation and Flotation30. Which one of the following gases is irritating to the eyes?

a. carbon monoxideb. chlorinec. methaned. hydrogen sulfide


GRADE 1 QUESTIONS 5

31. Activated sludge clarifiers accept material to be settled froma. an aeration tankb. a primary clarifierc. a trickling filterd. a primary digester tank

Secondary Treatment Processes32. According to the Clean Water Act, secondary treatment is defined as a wastewater

plant producing an effluent that has no more than _______ total suspended solids and no more than _______ BOD5.a. 20 mg/L, 40 mg/Lb. 30 mg/L, 30 mg/Lc. 40 mg/L, 40 mg/Ld. 40 mg/L, 20 mg/L

33. The color of the biofilm on a rotating biological contactor would appear _______ under normal operating conditions.a. dark greenb. greenish grayc. grayd. brown

Activated Sludge34. A primary effluent and returned activated sludge mixture is called

a. mixed liquorb. mixed liquor suspended solidsc. activated sludge liquord. waste activated sludge

35. The mixture of organisms and solids in water that is removed from the settling tank is calleda. mixed liquorb. mixed liquor suspended solidsc. activated sludge liquord. mixed activated sludge

36. Under normal operating conditions, what color is the activated sludge?a. dark chocolate brownb. light to medium brownc. greenish brownd. dark green

37. What test(s) could an operator use to make process changes immediately if a problem develops?a. BOD testb. chemical oxygen demand (COD) testc. sludge volume index (SVI) testd. food-to-microorganism (F/M) ratio and percent volatile matter



38. The main “workers” in the activated sludge process area. bacteriab. protozoansc. ciliatesd. rotifers

39. A yellow gas flame from a digester’s waste gas burner may indicate a poor quality gas with aa. high CO2 contentb. high H2S contentc. high H2S and CO2 contentd. high H2S and skatole content

Waste Treatment Ponds40. Ponds used in series after primary wastewater treatment are sometimes called

a. secondary treatment pondsb. tertiary treatment pondsc. oxidation pondsd. polishing ponds

41. The most common type of pond isa. an anaerobic pondb. an aerobic pondc. a facultative pondd. an oxidation pond

42. Which one of the following nuisance organisms is controlled simply by the water surface being disturbed by wind or currents?a. mosquitoesb. shrimp-like animalsc. dragonfly larvaed. chironomid midges

43. When is the dissolved oxygen in a waste pond the lowest?a. sunriseb. sunsetc. around noond. around 4:00 pm

Odor Control44. What would be the most likely solution, if the influent to a wastewater treatment plant

had an H2S odor?a. airstream treatmentb. add treatment chemicalsc. remove sludge fasterd. improve housekeeping


GRADE 1 QUESTIONS 7

Phosphorus Removal45. What is the main reason why wastewater treatment plants remove phosphorus from

their effluent before discharging to a lake or river?a. because it seriously depletes oxygenb. because if forms phosphoric acid that is detrimental to fishc. to reduce bacteria that can kill aquatic wildlifed. to reduce the growth of algae

Disinfection46. What is the color of liquid chlorine?

a. yellowb. yellowish greenc. greend. amber

47. How many times heavier than air is chlorine gas?a. 1.95 timesb. 2.0 timesc. 2.5 timesd. 2.7 times

48. Free available chlorine consists ofa. hypochlorous acid and hypochlorite ionb. hypochlorous acid, dissolved chlorine gas, and dichloraminec. dissolved chlorine gas and hypochlorite acidd. dissolved chlorine gas, hypochlorous acid, and hypochlorite ion

49. The fusible plug on a chlorine cylinder will start to melt at approximatelya. 145°Fb. 150°Fc. 155°Fd. 160°F

Wastewater Solids Processing50. This solids treatment method usually has 20 days of retention time:

a. lime stabilizationb. aerobic digestionc. gravity thickeningd. thermal treatment

51. This solids treatment method processes solids in a sealed unit:a. gravity thickeningb. solids concentratorsc. lime stabilizationd. anaerobic digestion



52. This solids treatment method produces methane gas that can be used in other plant processes:a. aerobic digestionb. anaerobic digestionc. chlorine oxidationd. thermal treatment

53. This solids treatment method cures the final product and then it may be applied to farmlands:a. solids concentratorsb. compostingc. anaerobic digestiond. flotation thickening

54. This solids treatment method uses evaporation but also requires a great deal of labor:a. sand drying bedsb. thermal treatmentc. gravity thickeningd. incineration

55. This solids treatment method uses a partially submerged rotating drum:a. centrifugeb. vacuum filterc. solids concentratorsd. aerobic digestion

56. These solid treatment methods use a coagulant polymer and pressure:a. solids concentratorsb. centrifugesc. vacuum filtersd. belt filter and plate

Sampling, Testing, and Reporting57. Tests such as pH, temperature, and dissolved oxygen should always be performed

within _______ after collection.a. 15 minb. 30 minc. 1 hrd. 2 hr

58. When calibrating a pH meter, the buffers used should bracket the expected pH anda. be no more than 2 pH units apartb. be at least 2 pH units apartc. be at least 3 pH units apart, such as 4, 7, and 10d. be no more than 3 pH units apart

59. A test method for determining dissolved oxygen is thea. iodometric direct titrationb. azide modification of the Winkler titration methodc. amperometric direct titrationd. Odzakovic titration method


GRADE 1 QUESTIONS 9

60. How long can a dissolved oxygen sample wait for analysis using the modified Winkler method, if it is preserved with sulfuric acid?a. 2 hrb. 4 hrc. 6 hrd. 8 hr

61. BOD5 samples can be preserved bya. freezingb. storing them in a refrigerator above freezing and up to 4°Cc. adding 3 mL of 0.2N HNO3d. adding 2 mL of 0.2N H2SO4

62. The maximum holding time for preserved BOD5 samples isa. 2 daysb. 3 daysc. 4 daysd. 5 days

63. Which one of the following tests is least likely to effectively help an operator control a waste plant’s process?a. DOb. pHc. temperatured. BOD5

64. The maximum holding time for a preserved total suspended solids sample isa. 2 daysb. 3 daysc. 5 daysd. 7 days

65. Generally, the state regulatory body must be notified of all unusual or extraordinary discharges by telephone withina. 1 hrb. 2 hrc. 24 hrd. 48 hr

66. Generally, the state regulatory body must be notified for any noncompliance that could adversely affect that state’s waters or that could endanger public health withina. 1 hrb. 2 hrc. 8 hrd. 24 hr



Maintenance67. Electromotive force (emf) is also known as

a. voltageb. ohmsc. wattsd. amps

68. Electron flow is measured ina. wattsb. ohmsc. ampsd. voltage

69. What type of fuses would most likely be used to protect a meter?a. current-limiting fusesb. dual-element fusesc. phase fusesd. voltage sensitive fuses

Safety70. What class of fire involves electrical equipment?

a. Class Ab. Class Bc. Class Cd. Class D

71. When a permit is required to enter a confined space, who may sign the permit?a. attending personb. entrantc. entry supervisord. Occupational Safety and Health Administration (OSHA) representative

72. Which one of the following is a measure of a material’s opposition to the flow of electric current?a. voltsb. ohmsc. amperesd. watts

73. What class of fire involves oil and grease?a. Class Ab. Class Bc. Class Cd. Class D

74. Why is it essential to ventilate a valve vault before entry?a. equalize temperature and pressureb. eliminate foul odorsc. remove excessive moistured. remove dangerous gases


GRADE 1 QUESTIONS 11

75. The primary purpose of check valves is to preventa. loss of primeb. excessive pump pressurec. water from flowing in two directionsd. water hammer

76. Which one of the following infectious diseases is not transmitted by wastewater?a. tetanusb. tuberculosisc. the acquired immune deficiency syndrome (AIDS) virusd. polio

77. Which one of the following gases is least likely to collect in a belowground vault?a. hydrogen sulfideb. gasoline vaporsc. carbon dioxided. carbon monoxide

78. When working in a manhole, your vehicle, whenever possible, should be placeda. on the right side of the manholeb. on the left side of the manholec. ahead of the manhole and oncoming trafficd. between the manhole and oncoming traffic

79. OSHA requires that adequate protection, such as shoring, be placed in any trench that isa. 3.5 ft or greater in depthb. 4 ft or greater in depthc. 4.5 ft or greater in depthd. 5 ft or greater in depth

80. How far back from the opening of a trench should soil excavated from that trench be placed?a. 1 ft at leastb. 2 ft at leastc. 3 ft at leastd. 4 ft at least

81. In a trench, ladders are required everya. 10 ftb. 20 ftc. 25 ftd. 30 ft

82. How many ladders are required to be in a 5-ft deep trench when six people are working down in the trench?a. one ladderb. two laddersc. three laddersd. six ladders



83. Which is not an essential ingredient for all ordinary fires?a. ignition sourceb. fuelc. heatd. oxygen

84. If some clothes are on fire, what class of fire is it?a. Class Ab. Class Bc. Class Cd. Class D

85. If some magnesium is on fire, what class of fire is it?a. Class Ab. Class Bc. Class Cd. Class D

86. What is the minimum number of people that should be on-site working at a confined space site, where one of them is entering a manhole?a. twob. threec. fourd. five

87. What type of fire extinguishers should be in pump stations?a. foamb. carbon dioxidec. A-B-C chemical typed. water and carbon dioxide

Basic Chemistry88. What piece of laboratory glassware is primarily used to mix chemicals and measure

approximate volumes?a. pipetteb. beakerc. buretted. graduated cylinder

89. The electric potential between two points is measured ina. amperesb. voltsc. ohmsd. watts

90. What is the most common use today for a positive-displacement pump?a. lift or booster pumpb. chemical feed pumpc. filter feed pumpd. wastewater water intake pump



91. Which laboratory glassware will give the most accurate measurements of liquids?a. graduated cylindersb. beakersc. flasksd. burettes

92. Which laboratory glassware is used for measuring volumes?a. flasksb. beakersc. graduated cylindersd. burettes

93. People working in and around wastewater must be immunized fora. tetanusb. cholerac. amebiasisd. dysentery

94. Which one of the following laboratory tests has a zero holding time?a. nitrateb. chlorine residualc. turbidityc. BOD5

95. Which one of the following statements concerning colloids is false?a. solutes have a greater effect on freezing and boiling points of liquids than do

colloidsb. only very small colloids will pass through a semipermeable membranec. the average wavelength of white light is smaller than the dimensions of colloidal

particlesd. dialysis is the separation of colloids from solutes through a semipermeable

membrane96. Which one of the following statements concerning colloids is true?

a. an ion opposite in charge to a colloidal particle will force its way into the colloidal particle

b. electrolytes will stabilize colloidsc. van der Waal’s forces help stabilize colloidal particlesd. electrophoresis is a term used to describe the settling of a colloid in an electrical

field97. Water is used as a standard because it has unique properties and is readily available

in its pure state. Which standard is not correct?a. Celsius scaleb. the caloriec. the kilogramd. specific gravity



98. What acids are formed when water reacts with nonmetal oxides?a. acid hydroxidesb. acid anhydridesc. acid oxidesd. acid hydrates

99. Which one of the following is not a hydrate?a. alumb. blue stonec. sodium nitrated. ferrous sulfate


15

Grade 1 Answers and References

Wastewater Sources and Characteristics1. Answer: b. storm events

Reference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 3, Page 92.

2. Answer: d. sanitary wastewater and stormwater runoffReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 3, Page 93.

3. Answer: a. 0.1%Reference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 3, Page 95.

4. Answer: a. 30%Reference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 3, Page 95.

5. Answer: d. 70%Reference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 3, Page 95.

6. Answer: a. amoebasReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 3, Page 99.

7. Answer: b. stalked ciliatesReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 3, Page 100.

Basic Wastewater Treatment8. Answer: b. oxygen

Reference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 2, Page 15.

9. Answer: d. free or dissolved oxygen (DO)Reference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 2, Page 15.

10. Answer: c. 5 daysReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 2, Page 15.



11. Answer: a. volume of settleable solidsReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 2, Page 15.

12. Answer: c. anaerobic bacteriaReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 2, Page 17.

13. Answer: a. primaryReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 2, Page 19.

14. Answer: b. aerobic bacteriaReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 2, Page 19.

15. Answer: c. anaerobic bacteriaReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 2, Page 19.

16. Answer: a. methaneReference: Water and Wastewater Treatment Plant Operations, Frank R. Spellman, Chapter 11, Page 328.

Wastewater Treatment Facilities17. Answer: b. sedimentation


18. Answer: b. extremely explosiveReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 3, Page 49.

19. Answer: c. a waste stabilization pondReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 3, Page 54.

20. Answer: b. a facultative pondReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 3, Page 54.

Preliminary Treatment21. Answer: b. 30°


22. Answer: c. 45 to 60°Reference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 4, Page 111.


GRADE 1 ANSWERS AND REFERENCES 17

23. Answer: c. 1.5 ft/sReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 4, Page 111.

24. Answer: b. 1.0 ft/sReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 4, Page 112.

25. Answer: c. 80%Reference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 4, Page 112.

26. Answer: b. 0.7 to 1.4 ft/sReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 4, Page 114.

Racks, Screens, Comminutors, and Grit Removal27. Answer: a. hydroxide


28. Answer: a. when the bar screen is out of serviceReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 4, Page 70.

Primary Sedimentation29. Answer: a. 2 hr


Sedimentation and Flotation30. Answer: b. chlorine


31. Answer: a. an aeration tankReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 5, Page 138.

Secondary Treatment Processes32. Answer: b. 30 mg/L, 30 mg/L


33. Answer: c. grayReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 6, Page 135.



Activated Sludge34. Answer: d. waste activated sludge


35. Answer: a. mixed liquorReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 7, Page 141.

36. Answer: a. dark chocolate brownReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 7, Page 144.

37. Answer: b. chemical oxygen demand (COD) testReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 64.

38. Answer: a. bacteriaReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 116.

39. Answer: a. high CO2 contentReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 209.

Waste Treatment Ponds40. Answer: c. oxidation ponds


41. Answer: c. a facultative pondReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 9, Page 297.

42. Answer: a. mosquitoesReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 9, Page 305.

43. Answer: a. sunriseReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 9, 5.

Odor Control44. Answer: b. add treatment chemicals

Reference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 1, Page 36.



Phosphorus Removal45. Answer: d. to reduce the growth of algae


Disinfection46. Answer: d. amber


47. Answer: c. 2.5 timesReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 8, Page 152.

48. Answer: d. dissolved chlorine gas, hypochlorous acid, and hypochlorite ionReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 10, Page 350.

49. Answer: d. 160°FReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 10, Page 379.

Wastewater Solids Processing50. Answer: b. aerobic digestion


51. Answer: d. anaerobic digestionReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 9, Page 162.

52. Answer: b. anaerobic digestionReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 9, Page 162.

53. Answer: b. compostingReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 9, Page 162.

54. Answer: a. sand drying bedsReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 9, Page 163.

55. Answer: b. vacuum filterReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 9, Page 163.

56. Answer: d. belt filter and plateReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 9, Page 164.



Sampling, Testing, and Reporting57. Answer: a. 15 min


58. Answer: b. be at least 2 pH units apartReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 10, Page 177.

59. Answer: b. azide modification of the Winkler titration methodReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 10, Page 185.

60. Answer: d. 8 hrReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 10, Page 186.

61. Answer: b. storing them in a refrigerator above freezing and up to 4°CReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 10, Page 191.

62. Answer: a. 2 daysReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 10, Page 191.

63. Answer: d. BOD5


64. Answer: d. 7 daysReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 10, Page 195.

65. Answer: c. 24 hrReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 10, Page 199.

66. Answer: d. 24 hrReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 10, Page 199.

Maintenance67. Answer: a. voltage

Reference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 15, Page 375.

68. Answer: c. ampsReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 15, Page 375.



69. Answer: b. dual-element fusesReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 15, Page 382.

Safety70. Answer: c. Class C

Reference: Title 29, Code of Federal Regulations, Part 155 (c) (10).71. Answer: c. entry supervisor

Reference: Title 29, Code of Federal Regulations, Part 1910.146(e) (2).72. Answer: b. ohms

Reference: Basic Science Concepts and Applications, AWWA, Principles and Practices of Water Supply Operations Series, Second Edition, Page 541.

73. Answer: b. Class BReference: Small Water System Operation and Maintenance, Kenneth D. Kerri, Third Edition, Chapter 7, Page 401, Section 7.162.

74. Answer: d. remove dangerous gasesReference: Small Water System Operation and Maintenance, Kenneth D. Kerri, Third Edition, Chapter 6, Section 6.20.

75. Answer: c. water from flowing in two directionsReference: Water Transmission and Distribution, AWWA, Principles and Practices of Water Supply Operations Series, Second Edition, Chapter 3.

76. Answer: c. the acquired immune deficiency syndrome (AIDS) virusReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 14, Page 266.

77. Answer: d. carbon monoxideReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 14, Page 268.

78. Answer: d. between the manhole and oncoming trafficReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 14, Page 278.

79. Answer: d. 5 ft or greater in depthReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 14, Page 285.

80. Answer: b. 2 ft at leastReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 14, Page 285.

81. Answer: c. 25 ftReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 14, Page 285.



82. Answer: d. six laddersReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 14, Page 285.

83. Answer: a. ignition sourceReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 14, Page 303.

84. Answer: a. Class AReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 14, Page 303.

85. Answer: d. Class DReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 14, Page 304.

86. Answer: b. threeReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 14, Pages 266, 267, and 270.

87. Answer: c. A-B-C chemical typeReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 14, Page 320.

Basic Chemistry for Wastewater Operators88. Answer: b. beaker

Reference: Water Quality, AWWA, Principles and Practices of Water Supply Operations Series, Second Edition, Chapter 3.

89. Answer: b. voltsReference: Basic Science Concepts and Applications, AWWA, Principles and Practices of Water Supply Operations Series, Third Edition, Electricity 1, Page 533.

90. Answer: b. chemical feed pumpReference: Water Transmission and Distribution, AWWA, Principles and Practices of Water Supply Operations Series, Second Edition, Chapter 12, Page 369.

91. Answer: d. burettesReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 16, Pages 485 and 486.

92. Answer: c. graduated cylindersReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 16, Page 485.

93. Answer: a. tetanusReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 16, Page 493.

94. Answer: b. chlorine residualReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 16, Page 502.



95. Answer: d. dialysis is the separation of colloids from solutes through a semipermeable membraneReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 1, Pages 70 and 71.

96. Answer: a. an ion opposite in charge to a colloidal particle will force its way into the colloidal particleReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 1, Page 72.

97. Answer: c. the kilogramReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 9, Pages 78 and 79.

98. Answer: b. acid anhydridesReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 9, Page 79.

99. Answer: c. sodium nitrateReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 9, Page 80.


25

Grade 2 Questions

General Wastewater Treatment Principles1. What is pressure head caused by?

a. water flowb. water elevationc. potential energy plus water velocityd. potential energy plus kinetic energy

2. What color should the piping be that contains reclaimed water?a. orangeb. purplec. yellowd. red

3. What is the term for the height to which a column of water will rise in a well?a. potentiometric surfaceb. unconfined levelc. piezometric surfaced. groundwater level

4. In general, what is the maximum practical lift of pumps?a. 10 to 15 ftb. 15 to 25 ftc. 25 to 30 ftd. 30 to 35 ft

5. Which one of the following valves is best to use in throttling flow situations?a. gateb. butterflyc. globed. ball

6. Which one of the following devices is used to measure water depths in storage facilities?a. transducerb. Venturi meterc. magnetic sensord. thermistor

Wastewater Treatment Facilities7. The velocity in sanitary sewers is approximately

a. 1 foot per second (ft/s)b. 2 ft/sc. 3 ft/sd. 3.5 ft/s



8. The distance between manholes ranges froma. 300 to 500 ftb. 500 to 600 ftc. 750 to 1,000 ftd. 900 to 1,200 ft

9. What should the flow velocity be through a grit channel?a. 0.5 ft/sb. 1.0 ft/sc. 1.5 ft/sd. 2.0 ft/s

10. Grit washers are used toa. remove some of the organic matterb. remove large pieces of woodc. remove rags and eggshellsd. remove the odors

11. The most common flow-measuring device for wastewater is aa. Parshall flumeb. magnetic flowmeterc. weird. Venturi meter

12. The detention time for primary clarifiers usually has a range ofa. 1.5 to 2 hrb. 4 to 5 hrc. 3 to 7 hrd. 5 to 12 hr

13. The effluent from which process is called mixed liquor?a. secondary clarificationb. primary clarificationc. sludge drying bed or pondd. activated sludge

14. Aerobic ponds usually range in depth froma. 2 to 4 ftb. 3 to 6 ftc. 3 to 8 ftd. 4 to 8 ft

15. Anaerobic ponds usually range in depth froma. 4 to 8 ftb. 6 to 12 ftc. 8 to 12 ftd. 10 to 14 ft



16. Another name for a waste stabilization pond isa. raw wastewater lagoonb. facultative pondc. anaerobic pondd. decomposition pond

Wastewater Sources and Characteristics17. Typically, what average percent of total solids in domestic wastewater, both dissolved

and suspended, are organics?a. 20%b. 25%c. 38%d. 50%

18. Typically, what average percent of total solids in domestic wastewater, both dissolved and suspended, are inorganics?a. 25%b. 35%c. 45%d. 50%

19. Approximately what percent of the dissolved solids in wastewater are inorganics?a. 33%b. 50%c. 75%d. 87%

20. Approximately what percent of the dissolved solids in wastewater are organics?a. 12%b. 20%c. 25%d. 36%

21. Which type of organisms would most probably be associated with moderate sludge age and effluent water quality?a. stalked ciliatesb. free-swimming ciliatesc. amoebasd. rotifers

22. Which type of organisms would most probably be associated with high suspended solids and biochemical oxygen demand (BOD)?a. amoebasb. stalked ciliatesc. free-swimming ciliatesd. rotifers



23. High suspended solids and BOD are most probably associated with which type of organisms?a. stalked ciliatesb. rotifersc. free-swimming ciliatesd. flagellates

24. Which type of organism would most likely enhance the penetration of oxygen into the activated sludge?a. rotifersb. flagellatesc. amoebasd. free-swimming ciliates

Racks, Screens, Comminutors, and Grit Removal25. Chlorine gas has a specific gravity of

a. 1.5 sp grb. 2.0 sp grc. 2.5 sp grd. 3.0 sp gr

26. Proper flow velocity through a grit channel is indicated when the grit being removed containsa. no organicsb. a little organic materialc. about half the volume as the grit being removedd. about the same volume as the grit being removed

Sedimentation and Flotation27. Surface loading and detention time in a clarifier are directly related to

a. the amount of settleable solidsb. size of the sludge hopperc. weir lengthd. flow through the clarifier

28. What is the expected range for the removal of settleable solids from a primary clarifier?a. 80 to 90%b. 85 to 90%c. 90 to 95%d. 95 to 99%

29. What condition will occur if sludge is not pumped out of primary clarifiers at regular intervals?a. broken or mechanical failure of the flightsb. clogged pump(s) when the pumps come onc. clarifier will become septicd. sludge will clog the sludge removal pipe



30. Secondary clarifiers are usually located aftera. the primary clarifierb. an anaerobic digesterc. a biological processd. preaeration

31. Filter sloughings are a product ofa. biological actionb. chemical actionc. physical processesd. chemical and physical processes

32. When one liquid is dissolved in another liquid it is calleda. a colloidb. an emulsionc. a compoundd. floc

33. What is the optimum temperature for digester substances?a. 86°Fb. 88°Fc. 90°Fd. 95°F

Secondary Treatment Processes34. Which one of the following types of wastewater ponds listed below is not classified like

the other three?a. polishing pondb. oxidation pondc. aerobic pondd. raw sewage stabilization pond

35. What type of pond would receive wastewater from a secondary treatment system?a. aerated pondb. oxidation pondc. stabilization pondd. polishing pond

36. The color of the biofilm on a rotating biological contactor undergoing nitrification would appeara. grayish greenb. grayish brownc. greenish brownd. reddish brown to golden

37. The color of the biofilm on a rotating biological contactor that has high sulfur concentrations would appeara. chalky whiteb. yellowish greenc. yellowd. golden brown



38. Rotating biological contactors have all the following advantages except:a. high sludge productionb. they can handle a large range of flowsc. power requirements are lowd. low operating costs, as they are easy to run

39. A wastewater plant has a train of three rotating biological contactor (RBC) units. The color of the first RBC is _______; the color of the second RBC is _______; and the color of the third RBC is _______.a. (1) yellowish green; (2) green; (3) very dark green to grayb. (1) green; (2) green; (3) golden greenc. (1) gray; (2) reddish brown; (3) reddish brownd. (1) yellowish green; (2) green; (3) golden brown

Activated Sludge40. Which one of the following activated sludge process modifications is very sensitive to

bulking?a. conventional modificationb. extended aeration activated sludgec. complete mix aerationd. pure oxygen

41. Which modified activated sludge process does not require primary treatment?a. complete mix aerationb. contact stabilizationc. tapered aerationd. step aeration

42. Primary treatment is not required in which modified activated sludge process?a. step aerationb. complete mix aerationc. tapered aerationd. extended aeration activated sludge

43. Mixed liquor is a combination of the following mixture:a. activated sludge with primary effluent mixed in a pond or clarifierb. raw water with primary effluent and return sludge mixed in an aeration tankc. activated sludge in an aeration tank mixed with primary effluent and return

sludge or raw wastewaterd. activated sludge mixed with raw wastewater in a clarifier or aeration tank

44. In a small package wastewater treatment plant, the operator should try to keep the dissolved oxygen in the aeration tank at approximatelya. 1.0 mg/Lb. 2.0 mg/Lc. 3.0 mg/Ld. 4.0 mg/L



45. How much of the solids should be wasted each week in the summer from a package wastewater treatment plant for the best water quality?a. 5%b. 10%c. 20%d. 25%

46. At what concentration level should dissolved oxygen be maintained during the initial startup of an oxidation ditch?a. 0.5 mg/Lb. 1.0 mg/Lc. 1.5 mg/Ld. 2.0 mg/L

47. What should be the color of an oxidation ditch that is working properly?a. light yellow to tanb. tan to light brownc. light brown to medium brownd. medium brown to dark brown

48. Which one of the following is the least important in the successful operation of an activated sludge plant?a. BODb. food entering the process (the solids)c. sludge aged. organisms available to treat the wastes

49. What is the critical factor in any method to control the aeration tank?a. BODb. F/M (food-to-microorganism) ratio c. COD (chemical oxygen demand)d. DO (dissolved oxygen)

50. What is the most accurate measure of the organisms present in an aerator for treating incoming wastes?a. BODb. MLSS (mixed liquor suspended solids)c. MLVSS (mixed liquor volatile suspended solids)d. COD

51. Most algae are in thea. plant kingdomb. animal kingdomc. Monera kingdomd. Protista kingdom

52. Bacteria are in thea. plant kingdomb. animal kingdomc. Monera kingdomd. Protista kingdom



53. Grease should be cleaned from walkways with a stiff brush anda. detergentb. bleachc. trisodium phosphated. phosphate-free surfactant

54. Which one of the following types of substances is not soluble in water and has the potential of creating explosive conditions?a. oils and greasesb. toxic gasesc. flammable oilsd. organic toxicants

55. Which one of the following types of substances can cause obstructions in the sewer system and may cause anaerobic decomposition where the substances accumulate?a. heavy metalsb. oils and greasesc. flammable oilsd. settleable solids

56. Which one of the following types of substances is corrosive and may interfere with biological treatment?a. organic toxicantsb. cyanidesc. acids and alkaliesd. heavy metals

57. Which one of the following types of substances can be toxic to wastewater plants using biological systems and may adversely affect potable water supplies downstream?a. heavy metalsb. toxic gasesc. cyanidesd. acids and alkalies

58. Where would toxicants least likely cause a problem for a wastewater treatment plant?a. clarifierb. trickling filterc. sludge digesterd. activated sludge

59. If an activated sludge plant is hit with a high organic load, what will most likely occur in the aeration tank?a. BOD will increaseb. there will be a significant decrease in DOc. a more turbid effluentd. effluent will exhibit floc carry-over



60. In the following list, which is the least important nutrient in wastewater treatment for microorganisms to reproduce?a. potassiumb. nitrogenc. carbon in organic matterd. phosphorus

61. Adequate BOD/nitrogen/phosphorus ratio for microorganisms to grow is usuallya. 100:2:1, respectivelyb. 100:5:1, respectivelyc. 100:10:3, respectivelyd. 100:12:5, respectively

Trickling Filters62. Humus sludge is biomass from

a. trickling filtersb. primary clarifiersc. secondary clarifiersd. primary digesters

63. Excessive slime growths on a trickling filter may cause a condition on the filter calleda. overloadb. pondingc. cloggingd. sloughing

64. Another name for slime growth isa. slagb. zoogleal filmc. algal matd. humus

65. It is not recommended to start up trickling filters in the _______ because bacteria are most often dormant at this time.a. late fall, winter, and early springb. late fall and winterc. winter and early springd. winter

66. When the influent load to a wastewater treatment plant changes rapidly, the best test for control purposes isa. total solidsb. DOc. BODd. COD

67. A problem with odors from trickling filters may be solved bya. reducing the number of weedsb. using masking agentsc. applying chlorined. flooding the filter



68. What can be done to the trickling filter to solve high effluent suspended solids?a. flush out distributor armsb. decrease recirculationc. reduce the number of weedsd. flood the filter

Rotating Biological Contactors69. Approximately how much surface area of a rotating biological contactor is immersed in

the wastewater?a. 40%b. 45%c. 50%d. 60%

70. How many stages are usually present in the rotating biological contactor process?a. three stagesb. four stagesc. six stagesd. eight stages

71. Rotating biological contactors are adversely affected by temperatures belowa. 40°Fb. 43°Fc. 48°Fd. 55°F

72. How often should the oil be changed in the speed reducer of a rotating biological contactor?a. every 2 monthsb. every 3 monthsc. every 6 monthsd. every 12 months

73. How often should the motor bearings be greased in a rotating biological contactor unit?a. every 2 monthsb. every 3 monthsc. every 6 monthsd. every 12 months

74. How often should the mainshaft bearings be greased in a rotating biological contactor unit?a. every weekb. every monthc. every 3 monthsd. every 6 months

75. How often should the drive bearings be greased in a rotating biological contactor unit?a. every weekb. every monthc. every 3 monthsd. every 6 months



76. How often should the oil in the chain casing be changed?a. every 2 monthsb. every 3 monthsc. every 6 monthsd. every 12 months

Waste Treatment Ponds77. Ponds used in series after a trickling filter are sometimes called

a. polishing pondsb. oxidation pondsc. secondary treatment pondsd. tertiary treatment ponds

78. A pond that contains algae and requires the use of compressors is classified asa. an anaerobic pondb. an aerobic pondc. a facultative pondd. an oxidation pond

79. A pond that depends on fermentation, is odorous, and is used mainly for processing industrial wastes is classified asa. an anaerobic pondb. an aerobic pondc. a facultative pondd. a oxidation pond

80. A pond that contains oxygen in its upper layers and lacks oxygen in the bottom layer is classified asa. an anaerobic pondb. an aerobic pondc. a facultative pondd. an oxidation pond

81. When is the pH in a waste pond the lowest?a. sunriseb. sunsetc. around noond. around 4:00 pm

82. When is the pH in a waste pond the highest?a. sunriseb. sunsetc. around noond. around 4:00 pm

83. When is the dissolved oxygen in a waste pond the highest?a. sunriseb. sunsetc. around noond. around 4:00 pm



84. Regarding waste pond lab sampling, which one of the following parameters is sampled at the influent of the pond?a. pHb. BODc. DOd. temperature

85. How far out from the water’s edge should waste pond samples be collected?a. 6 ftb. 8 ftc. 10 ftd. 15 ft

86. At what depth should samples be collected from a waste pond?a. 1 ftb. 1.5 ftc. 2.0 ftd. 2.5 ft

87. The recommended minimum depth for a waste pond isa. 2.0 ftb. 2.5 ftc. 3.0 ftd. 3.5 ft

Sludge Digestion and Solids Handling88. Saprophytic microorganisms are also called

a. acid formersb. methane fermentersc. alcohol fermentersd. facultative fermenters

89. Microorganisms that are methane fermenters will only produce in the pH range ofa. 5.8 to 6.5b. 6.1 to 7.0c. 6.6 to 7.6d. 6.8 to 8.3

90. Digestion is working properly if the volatile solids content of the raw sludge has been reduced bya. 40 to 60%b. 50 to 60%c. 55 to 65%d. 60 to 75%

91. An anaerobic sludge digester should be operated such thata. acid formation is greater than methane formationb. acid formation is equal to methane formationc. acid formation is less than methane formationd. acid formation is less than alkaline formation



92. The most common imbalance in the operation of an anaerobic sludge digester isa. excess acid fermentationb. excess methane fermentationc. insufficient mixingd. pH becomes too alkaline

93. The most important factor in the high-rate process of anaerobic digestion isa. DOb. CODc. F/M ratiod. mixing

94. Operators should, as a rule, never change the temperature in an anaerobic digester by more thana. 1°F/dayb. 2°F/dayc. 3°F/dayd. 4°F/day

95. What is the optimum temperature range of mesophilic bacteria in anaerobic digestion?a. 68 to 80°Fb. 80 to 100°Fc. 85 to 100°Fd. 88 to 105°F

96. What is the purpose of a draft tube in an anaerobic sludge digester?a. removes methane gas from the digester tankb. directs raw and digested sludge mixture to a heating areac. removes sludge to be used as seed sludged. collects supernatant for moving to the primary clarifier or headworks

97. How often should the liquid level be checked on a digester to prevent overfilling?a. once/dayb. two times/dayc. three times/dayd. four times/day

98. Typically, an anaerobic digester will produce about 30% to 35% by volume of which particular gas from the following list?a. CO2b. H2c. CH4d. N2

99. How often should flame arresters be serviced on anaerobic digesters?a. every 3 monthsb. every 6 monthsc. once/yeard. every 2 years



100. How often should the thermal valves be dismantled for service?a. every 3 monthsb. every 6 monthsc. once/yeard. every 2 years

101. Typically, how often should sediment and drip traps be drained in cold weather?a. twice a dayb. dailyc. once/weekd. once/month

102. Draft tube propeller mixers are subject to corrosion froma. nitric acidb. phosphoric acidc. sulfuric acidd. hydrogen sulfide

103. What do corbels provide in an anaerobic digester?a. stability and proper cover buoyancyb. they prevent the cover from scraping the sidewallc. support for the floating cover when it is at its lowest pointd. they prevent foaming in the annular space

104. What do roller guides provide in an anaerobic digester?a. stability and proper cover buoyancyb. they prevent the cover from scraping the sidewallc. support for the floating cover when it is at its lowest pointd. they prevent foaming in the annular space

105. What do ballast blocks provide in anaerobic digesters?a. stability and proper cover buoyancyb. they prevent the cover from scraping the sidewallc. support for the floating cover when it is at its lowest pointd. prevents the cover from sinking into the digested sludge

106. What do skirts provide in anaerobic digesters?a. stability and proper cover buoyancyb. they prevent the cover from scraping the sidewallc. prevents the cover from sinking into the digested sludged. they prevent foaming in the annular space

107. The floating cover on an anaerobic digester should be inspecteda. dailyb. weeklyc. bimonthlyd. monthly



108. Bacteria in an anaerobic digester will effectively utilizea. hairb. mineral oilsc. petroleum productsd. fats

109. What would happen if too much sludge was added to an anaerobic digester?a. methane fermenters would predominate and pH would increaseb. methane fermenters would predominate and pH would decreasec. acid fermenters would predominate and pH would increased. acid fermenters would predominate and pH would decrease

110. The best operation performance for an anaerobic digester is when it is feda. several times a dayb. twice per dayc. once per dayd. once every other day

111. The least desirable control method for an anaerobic digester is the use of digestera. pHb. CO2c. volatile-acid-to-alkalinity ratiod. temperature

112. What would be the best solution if the volatile-acid-to-alkalinity ratio increased to 0.3 in a digester due to pumping thin sludge?a. thicken the raw sludgeb. reduce mixing timec. increase sludge withdrawald. suspend the addition of seed sludge

113. If a single stage or primary tank has foam in the supernatant due to an organic overloaded in the digester, what is the best solution?a. reduce digester temperatureb. increase digester temperature and mixingc. reduce the feeding rated. increase supernatant withdrawal

114. What would be the most probable cause for gas leaking through the pressure relief valve on the digester’s roof?a. plugged gas line or closed valveb. not properly lubricated gasket or gasket crackedc. waste gas burner valve not closed all the way or leak past the valved. excessive gas production

115. If the manometer indicates the gas pressure in a digester is below normal, what would be the most probable cause?a. the digester’s pressure relief valve is stuck in the open positionb. there is water or an obstruction in the main gas linec. too much lime has been added, reducing gas productiond. pressure control valve on the waste gas burner line is closed



116. If the gas pressure in a digester is lower than normal, what is the most probable cause?a. gas line is frozenb. pressure relief valve is stuck in the closed positionc. excessive sludge withdrawald. gas line or hose is leaking

117. If a wastewater treatment plant has four aerobic digesters operated in series, which tank will exert the highest oxygen demand?a. first tankb. second tankc. third tankd. second and third tanks

118. Where is the best place to collect scum from wastewater plants that use aerobic digesters?a. primary clarifierb. secondary clarifierc. first aerobic digester tankd. last aerobic digester tank

119. What should the initial depth of sludge from a digester be when it is placed in a sludge drying bed?a. 6 to 12 in.b. 12 to 18 in.c. 18 to 24 in.d. 24 to 30 in.

120. Operators should use special precautions when placing anaerobically digested sludge on a drying bed because sludge containsa. H2S gasb. carbon monoxidec. methaned. a high concentration of pathogens

121. What does the wastewater industry call sludge that is only partially digested?a. grayb. yellowc. greend. brown

Odor Control122. Hydrogen sulfide will exist in a system depending on the presence or absence of

a. iron bacteriab. oxygenc. organic acidsd. inorganic sulfides



123. To what pH should wastewater be raised to halt sulfide production and inhibit sludge growth and biological slimes?a. above 8.5b. above 8.8c. above 9.0d. above 9.5

124. If the headworks in a wastewater treatment plant had an H2S odor, what would be the most likely solution?a. airstream treatmentb. correct faulty plant operationc. add treatment chemicalsd. remove sludge from grit channel

125. If the primary sedimentation unit to a wastewater treatment plant had an H2S odor, what would be the most likely solution?a. improve housekeepingb. reduce turbulencec. increase mixingd. remove sludge faster

126. If the sludge drying beds had a decaying organic odor, what would be the most likely solution?a. increase air in aeration systemb. improve operations before digesterc. improve housekeeping around drying bedsd. chemical addition with countermasking or masking agents

127. If the ponds or lagoons had an ammonia odor, what would be the most likely solution?a. improve housekeeping around ponds or lagoonsb. remove sludge along sides of ponds or lagoonsc. add chemical treatmentd. add air or check diffusers in ponds or lagoons for proper function

Enhanced Biological (Nutrient) Control128. In the activated sludge system for the process control of effluent suspended solids,

what is the primary controlling variable for the desired microbial population?a. the sludge volume index (SVI) and sludge density index (SDI)b. the settled sludge blanketc. the concentration of MLSSd. the concentration of MLVSS

Disinfection129. The process that uses heat to selectively destroy undesirable organisms is called

a. ultraviolet radiationb. thermal-pulsec. autoclavingd. pasteurization



130. Which chemical in the following list is the strongest disinfectant?a. chlorineb. hypochlorous acidc. hypochlorite iond. hydrochloric acid

131. What is the recommended withdrawal of chlorine in pounds from a 100- or 150-lb cylinder in a 24-hr period such that freezing does not occur?a. 25 lbb. 30 lbc. 35 lbd. 40 lb

132. If there is a leak from a valve on a sulfur dioxide tank and the leak is found using ammonia, what color will the vapor fumes be?a. whiteb. yellowc. greend. tan

133. What is the acceptable sulfur dioxide level in mg/L in wastewater plant effluent?a. 0.5 mg/L or lessb. 2.0 mg/L or lessc. 5.0 mg/L or lessd. 10.0 mg/L or less

134. What is the ultimate measure of effective chlorination?a. bacteriological resultb. chlorine residual before dechlorinationc. tests indicate the absence of pathogensd. receiving waters remain viable around waste plant’s discharge line

Solids Removal from Secondary Effluents135. Which one of the following methods would not be used to feed dry chemicals?

a. screw feederb. vibrating troughc. rotating feederd. plunger feeder

136. Which one of the following methods would not be used to feed liquid chemicals?a. screw pumpb. plunger pumpc. gear pumpd. diaphragm pump

137. Which type of feeder would most likely feed clear solutions?a. rotameterb. decanterc. orificed. positive displacement



138. Which type of feeder would always feed a gas?a. rotameterb. direct feedc. ribbon feederd. loss of weight feeder

139. Which one of the following is a major factor in predicting the performance of wastewater clarifiers?a. DOb. BOD5c. CODd. solids loading

Effluent Disposal140. What is the most probable cause for excessive algae in the effluent of a wastewater

pond?a. pond has too much nitrogenb. pond has too much nitrate and not enough surface mixingc. detention time is excessived. temperature favoring the growth of a particular species of algae

141. If a secondary clarifier has floatable debris in the effluent, what is the most likely cause?a. solids detention time is excessiveb. the clarifier is hydraulically overloadedc. outlet baffle is not in the proper positiond. denitrification is occurring in the bottom of the clarifier

142. Because water samples sometimes require repeat or backup tests, the amount of sample that should be collected for samples that are preserved and transported for any particular analyses should be at least _______ times the amount required for those analyses.a. twob. threec. fourd. five

143. Which one of the following parameters can be preserved for later laboratory analysis?a. nitrateb. pHc. temperatured. dissolved oxygen

Wastewater Solids Processing144. Which one of the following is not a major solids treatment method?

a. thickeningb. concentrationc. dewateringd. stabilization



145. A solids treatment thickening process method would bea. gravityb. filter pressesc. compostingd. chemical oxidation

146. One solids treatment thickening process would bea. heat treatmentb. flotationc. compostingd. belt press

147. Another solids treatment thickening process would bea. sand drying bedsb. vacuum filtersc. solids concentratorsd. composting

148. A solids treatment stabilization process method would bea. flotationb. gravityc. vacuum filtersd. incineration

149. A solids treatment dewatering process method would bea. gravityb. incinerationc. compostingd. flotation

150. Which one of the following solids treatment methods can be classified as both stabilization and dewatering?a. incinerationb. wet air oxidationc. lime stabilizationd. chemical oxidation

151. This solids treatment method has a solids buildup that forms a blanket on the bottom that forces the water out due to compression:a. solids concentratorb. vacuum filterc. gravity thickeningd. centrifuge

152. This solids treatment method aerates the recycled water from the unit under pressure:a. flotation thickeningb. centrifugec. aerobic digestiond. solids concentrators



153. Typically, in the flotation thickening method, how much performance improvement in the amount of solids for waste activated sludge is realized when polymer is added?a. 1%b. 5%c. 15%d. 25%

154. This solids treatment method chemically treats the residual flow, which is then spread evenly over a porous belt:a. lime stabilizationb. vacuum filterc. chlorine oxidationd. solids concentrators

155. This solids treatment method usually dewaters solids by mixing them with a bulking agent and then waiting for biological stabilization to occur:a. solids concentratorsb. compostingc. anaerobic digestiond. flotation thickening

156. This solids treatment method maintains a certain pH for at least 2 hr and the final product can be applied directly to land:a. anaerobic digestionb. lime stabilizationc. aerobic digestiond. composting

157. This solids treatment method produces a waste of very high strength, and this waste must be returned to the wastewater treatment process:a. thermal treatmentb. anaerobic digestionc. incinerationd. solids concentrators

158. This solids treatment method uses a chemical and mixes the recycled flow with the process residual flow in a reactor:a. lime stabilizationb. compostingc. chlorine oxidationd. solids concentrators

159. This solids treatment method first chemically conditions the solids and then later in the process “throws” the solids out of the water:a. vacuum filterb. belt filter and platec. centrifuged. solids concentrators



160. This solids treatment method maximizes the reduction of moisture and solids:a. incinerationb. centrifugec. thermal treatmentd. vacuum filter

161. Which solids treatment method removes 100% of the pathogens, water, and organics?a. centrifugeb. chlorine oxidationc. thermal treatmentd. incineration

Residual Solids Management162. In general, secondary sludges

a. contain 20 to 25% nonvolatile inorganic matterb. are more fibrousc. have less flocculant than primary sludgesd. have a density significantly greater than water

163. Which type of centrifuge uses a knife or plow insertion to scrap off the cake from its walls?a. scroll centrifugeb. basket centrifugec. pressure-cone centrifuged. disc-nozzle centrifuge

164. Which type of centrifuge uses an outer screw conveyor to move the solids to the discharge point?a. scroll centrifugeb. basket centrifugec. pressure-cone centrifuged. disc-nozzle centrifuge

165. What is the ideal temperature range for methane-forming bacteria in anaerobic digesters?a. 90 to 98°Fb. 92 to 102°Fc. 94 to 97°Fd. 95 to 101°F

166. What type of wastewater solids would most likely be sent to a rendering reclamation?a. screeningsb. gritc. secondary sludged. scum



167. Which one of the following wastewater solids would most likely never end up on a dedicated land disposal (DLD) site?a. incinerated materialb. screeningsc. gritd. compost

168. If a sludge cake consists of 60% solids, where did it most likely originate?a. drying lagoonb. vacuum filterc. filter pressd. centrifuge

Water and Wastewater Microbiology169. What type of protozoan is typically associated with a poor treatment process and a

young biosolids age?a. sporozoansb. ciliatesc. flagellatesd. the Sarcodina

170. The most common pathogen identified in waterborne disease outbreaks in the United States isa. Cryptosporidiumb. Giardiac. Entamoebad. Cyclospora

171. The major gas produced by aerobic decomposition isa. nitrogenb. hydrogen sulfidec. carbon dioxided. methane

Maintenance172. If an induction motor is rated at 40 amps for a full-load current, what will be the most

probable starting sequence amps required?a. 60 to 80 ampsb. 80 to 120 ampsc. 120 to 200 ampsd. 200 to 320 amps

173. Motor insulation material comes in four classes and they are based ona. temperature limitationb. thickness of insulationc. composition of insulationd. type of motor for which the insulation is being used



174. The current imbalance between legs of polyphase motors should never exceeda. 1%b. 2%c. 5%d. 10%

175. The allowable voltage above or below the value stamped on a motor’s nameplate isa. 1%b. 3%c. 5%d. 10%

176. The allowable frequency deviation above or below the value stamped on a motor’s nameplate isa. 1%b. 3%c. 5%d. 10%

177. What would be the cause of a shorted winding in a motor?a. overloadedb. stalledc. foreign material in motord. incorrect power

178. What should be the air gap separation between a potable water pipe and the rim of a wastewater tank?a. at least one pipe diameter above the rimb. at least one and a half pipe diameters above the rimc. at least two pipe diameters above the rimd. at least two and a half pipe diameters above the rim

Safety179. What is the only acceptable breathing device to wear while handling chlorine leaks?

a. oxygen supply breathing apparatusb. self-contained breathing apparatusc. potassium tetraoxide canister typed. activated carbon canister type

180. Which of the following types of fire extinguishers should be used on electrical fires?a. fire-resistant blanketb. sodium bicarbonatec. carbon dioxided. soda–acid

181. Which one of the following provides a profile of hazardous substances or mixtures?a. Occupational Safety and Health Administration (OSHA)b. Comprehensive Environmental Response, Compensation, and Liability Act

(CERCLA)c. Code of Federal Regulations (CFR)d. material safety data sheet (MSDS)



182. An atmosphere is defined as oxygen deficient if it contains less than what percent oxygen by volume?a. 16.5%b. 18.5%c. 19.5%d. 19.8%

183. What class of fire involves sodium and magnesium?a. Class Ab. Class Bc. Class Cd. Class D

184. What is the proper emergency chlorine kit for a 1-ton chlorine cylinder?a. emergency kit Ab. emergency kit Bc. emergency kit Cd. emergency kit T

Basic Chemistry185. The specific gravity standard for gases is

a. water vapor at 100°Cb. airc. oxygend. nitrogen

186. The standard for gases has a density ofa. 1.00 g/Lb. 1.13 g/Lc. 1.23 g/Ld. 1.29 g/L

187. How many steps are there in coagulation?a. three stepsb. four stepsc. five stepsd. six steps

188. Low-temperature water can be compensated for when using alum bya. increasing the pHb. decreasing the pHc. increasing the alum dosaged. decreasing the alum dosage

189. When an electrolyte is dissolved in water to form an ion, the ion is said to bea. dissociatedb. ionizedc. hydratedd. hydrolyzed



190. Which one of the following species is ozone?a. O2+3

b. NO2c. O3d. N2O

191. Which laboratory glassware can be used to heat liquids in an open flame?a. flasksb. graduated cylindersc. volumetric flasksd. burettes

192. When inserting glass tubing into rubber hose or stoppers, never use _______ for lubrication.a. oilb. flame polishingc. waterd. lubricating jelly

193. Electrical units operating in areas exposed to flammable vapors shoulda. be removed and reinstalled in a separate roomb. be housed in a tightly sealed containerc. be double-insulatedd. be explosion-proof

194. Usually, what is the percent accuracy of a graduated cylinder and burette?a. 1%b. 1.5%c. 0.5 and 1%, respectivelyd. 1% and 0.5%, respectively

195. The clarity of a trickling filter’s effluent is considered good when the Secchi disk can be read at a level ofa. 1 ftb. 2 ftc. 3 ftd. 4 ft

196. The clarity of the effluent from an activated sludge process is considered good when the Secchi disk can be read at a level ofa. 3 ftb. 4 ftc. 5 ftd. 6 ft

197. The clarity of a chlorine contact basin’s effluent is considered good when the Secchi disk can be read at a level ofa. 4 ftb. 5 ftc. 6 ftd. 7 ft


51


General Wastewater Treatment Principles1. Answer: b. water elevation

Reference: Basic Science Concepts and Applications, AWWA, Principles and Practices of Water Supply Operations Series, Second Edition, Hydraulics 4, Page 246.

2. Answer: b. purpleReference: No reference source specified.

3. Answer: c. piezometric surfaceReference: Water Distribution Operator Training Handbook, AWWA, Second Edition, Chapter 16, Page 182.

4. Answer: b. 15 to 25 ftReference: Water Distribution Operator Training Handbook, AWWA, Second Edition, Chapter 17, Page 194.

5. Answer: c. globeReference: Water Distribution Operator Training Handbook, AWWA, Second Edition, Chapter 7, Page 62.

6. Answer: a. transducerReference: Water Distribution Operator Training Handbook, AWWA, Second Edition, Chapter 20, Page 221.

Wastewater Treatment Facilities7. Answer: b. 2 ft/s


8. Answer: a. 300 to 500 ftReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 3, Page 34.

9. Answer: b. 1.0 ft/sReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 3, Page 42.

10. Answer: a. remove some of the organic matterReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 3, Page 42.

11. Answer: a. Parshall flumeReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 3, Page 43.



12. Answer: a. 1.5 to 2 hrReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 3, Page 43.

13. Answer: d. activated sludgeReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 3, Page 47.

14. Answer: b. 3 to 6 ftReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 3, Page 53.

15. Answer: c. 8 to 12 ftReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 3, Page 53.

16. Answer: a. raw wastewater lagoonReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 3, Page 54.

Wastewater Sources and Characteristics17. Answer: d. 50%





21. Answer: b. free-swimming ciliatesReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 3, Page 100.

22. Answer: a. amoebasReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 3, Page 100.

23. Answer: d. flagellatesReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 3, Page 100.

24. Answer: a. rotifersReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 3, Page 101.



Racks, Screens, Comminutors, and Grit Removal25. Answer: c. 2.5 sp gr


26. Answer: b. a little organic materialReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 4, Page 87.

Sedimentation and Flotation27. Answer: d. flow through the clarifier


28. Answer: d. 95 to 99%Reference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 5, Page 129.

29. Answer: c. clarifier will become septicReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 5, Page 130.

30. Answer: c. a biological processReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 5, Page 137.

31. Answer: a. biological actionReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 5, Page 138.

32. Answer: b. an emulsionReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 5, Page 145.

33. Answer: c. 90°FReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 5, Page 153.

Secondary Treatment Processes34. Answer: c. aerobic pond

Reference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 6, Pages 124 and 127.

35. Answer: d. polishing pondReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 6, Page 125.

36. Answer: d. reddish brown to goldenReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 6, Page 135.



37. Answer: a. chalky whiteReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 6, Page 135.

38. Answer: a. high sludge productionReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 6, Page 135.

39. Answer: c. (1) gray; (2) reddish brown; (3) reddish brownReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 6, Page 137.

Activated Sludge40. Answer: a. conventional modification


41. Answer: b. contact stabilizationReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 7, Page 143.

42. Answer: d. extended aeration activation sludgeReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 7, Page 143.

43. Answer: c. activated sludge in an aeration tank mixed with primary effluent and return sludge or raw wastewaterReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 8, Page 246.

44. Answer: b. 2.0 mg/LReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 8, Page 264.

45. Answer: a. 5%Reference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 8, Page 264.

46. Answer: d. 2.0 mg/LReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 8, Page 273.

47. Answer: d. medium brown to dark brownReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 8, Page 274.

48. Answer: a. BODReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 54.



49. Answer: b. F/M (food-to-microorganism) ratioReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 58.

50. Answer: c. MLVSS (mixed liquor volatile suspended solids)Reference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 58.

51. Answer: d. Protista kingdomReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 116.

52. Answer: c. Monera kingdomReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 116.

53. Answer: c. trisodium phosphateReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 133.

54. Answer: c. flammable oilsReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 2, Page 98.

55. Answer: d. settleable solidsReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 2, Page 98.

56. Answer: c. acids and alkaliesReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 2, Page 98.

57. Answer: a. heavy metalsReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 2, Page 98.

58. Answer: a. clarifierReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 2, Page 98.

59. Answer: b. there will be a significant decrease in DOReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 2, Page 99.

60. Answer: a. potassiumReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 2, Page 101.

61. Answer: b. 100:5:1, respectivelyReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 2, Page 101.



Trickling Filters62. Answer: a. trickling filters


63. Answer: b. pondingReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 6, Page 168.

64. Answer: b. zoogleal filmReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 6, Page 168.

65. Answer: d. winterReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 6, Page 179.

66. Answer: d. CODReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 6, Page 182.

67. Answer: b. using masking agentsReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 6, Page 191.

68. Answer: b. decrease recirculationReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 6, Page 191.

Rotating Biological Contactors69. Answer: a. 40%


70. Answer: b. four stagesReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 7, Page 214.


72. Answer: c. every 6 monthsReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 7, Page 231.

73. Answer: d. every 12 monthsReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 7, Page 231.



74. Answer: a. every weekReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 7, Page 231.

75. Answer: a. every weekReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 7, Page 231.

76. Answer: b. every 3 monthsReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 7, Page 231.

Waste Treatment Ponds77. Answer: a. polishing ponds


78. Answer: b. an aerobic pondReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 9, Page 297.

79. Answer: a. an anaerobic pondReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 9, Page 297.

80. Answer: c. a facultative pondReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 9, Page 297.

81. Answer: a. sunriseReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 9, Page 315.

82. Answer: d. around 4:00 pmReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 9, Page 315.

83. Answer: d. around 4:00 pmReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 9, Page 315.

84. Answer: b. BODReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 9, Page 315.

85. Answer: b. 8 ftReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 9, Page 315.

86. Answer: a. 1 ftReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 9, Page 315.



87. Answer: c. 3.0 ftReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Fourth Edition, Chapter 9, Page 331.

Sludge Digestion and Solids Handling88. Answer: a. acid formers


89. Answer: c. 6.6 to 7.6Reference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 148.

90. Answer: b. 50 to 60%Reference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 148.

91. Answer: b. acid formation is equal to methane formationReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 148.

92. Answer: a. excess acid fermentationReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 148.

93. Answer: d. mixingReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 151.

94. Answer: a. 1°F/dayReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 151.

95. Answer: c. 85 to 100°FReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 151.

96. Answer: b. directs raw and digested sludge mixture to a heating areaReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 153.

97. Answer: a. once/dayReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 153.

98. Answer: a. CO2


99. Answer: a. every 3 monthsReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 162.



100. Answer: c. once/yearReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 162.

101. Answer: a. twice a dayReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 166.

102. Answer: d. hydrogen sulfideReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 179.

103. Answer: c. support for the floating cover when it is at its lowest pointReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 179.

104. Answer: b. they prevent the cover from scraping the sidewallReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 181.

105. Answer: a. stability and proper cover buoyancyReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 181.

106. Answer: d. they prevent foaming in the annular spaceReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 181.

107. Answer: a. dailyReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 181.

108. Answer: d. fatsReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 183.

109. Answer: d. acid fermenters would predominate and pH would decreaseReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 184.

110. Answer: a. several times a dayReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 185.

111. Answer: a. pHReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 192.

112. Answer: a. thicken the raw sludgeReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 206.



113. Answer: c. reduce the feeding rateReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 207.

114. Answer: a. plugged gas line or closed valveReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 209.

115. Answer: c. too much lime has been added, reducing gas productionReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 209.

116. Answer: d. gas line or hose is leakingReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 210.

117. Answer: a. first tankReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 218.

118. Answer: a. primary clarifierReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 219.

119. Answer: b. 12 to 18 in.Reference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 221.

120. Answer: c. methaneReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 221.

121. Answer: c. greenReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 222.

Odor Control122. Answer: b. oxygen


123. Answer: c. above 9.0Reference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 1, Page 17.

124. Answer: c. add treatment chemicalsReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 1, Page 36.

125. Answer: d. remove sludge fasterReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 1, Page 36.



126. Answer: d. chemical addition with countermasking or masking agentsReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 1, Page 37.

127. Answer: c. add chemical treatmentReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 1, Page 37.

Enhanced Biological (Nutrient) Control128. Answer: c. the concentration of MLSS


Disinfection129. Answer: d. pasteurization


130. Answer: b. hypochlorous acidReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 10, Page 348.

131. Answer: d. 40 lbReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 10, Page 398.

132. Answer: a. whiteReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 10, Page 405.

133. Answer: a. 0.5 mg/L or lessReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 10, Page 410.

134. Answer: a. bacteriological resultReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 10, Page 418.

Solids Removal from Secondary Effluents135. Answer: d. plunger feeder


136. Answer: a. screw pumpReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 4, Page 361.

137. Answer: a. rotameterReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 4, Page 373.



138. Answer: b. direct feedReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 4, Page 373.

139. Answer: d. solids loadingReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 4, Page 377.

Effluent Disposal140. Answer: d. temperature favoring the growth of a particular species of algae


141. Answer: b. the clarifier is hydraulically overloadedReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 13, Page 246.

142. Answer: a. twoReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 13, Page 253.

143. Answer: a. nitrateReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 13, Page 252.

Wastewater Solids Processing144. Answer: b. concentration


145. Answer: a. gravityReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 9, Page 159.

146. Answer: b. flotationReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 9, Page 159.

147. Answer: c. solids concentratorsReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 9, Page 159.

148. Answer: d. incinerationReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 9, Page 159.

149. Answer: b. incinerationReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 9, Page 159.



150. Answer: a. incinerationReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 9, Page 159.

151. Answer: c. gravity thickeningReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 9, Page 160.

152. Answer: a. flotation thickeningReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 9, Page 160.


154. Answer: d. solids concentratorsReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 9, Page 161.

155. Answer: b. compostingReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 9, Page 162.

156. Answer: b. lime stabilizationReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 9, Page 162.

157. Answer: a. thermal treatmentReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 9, Page 163.

158. Answer: c. chlorine oxidationReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 9, Page 163.

159. Answer: c. centrifugeReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 9, Page 164.

160. Answer: a. incinerationReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 9, Page 164.

161. Answer: d. incinerationReference: Wastewater Treatment Plant Operations Made Easy, Frank R. Spellman and Joanne Drinan, Chapter 10, Page 220.

Residual Solids Management162. Answer: a. contain 20 to 25% nonvolatile inorganic matter




163. Answer: b. basket centrifugeReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 189.

164. Answer: a. scroll centrifugeReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 189.

165. Answer: c. 94 to 97°FReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 206.

166. Answer: d. scumReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 290.

167. Answer: d. compostReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 290.

168. Answer: a. drying lagoonReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 294.

Water and Wastewater Microbiology169. Answer: c. flagellates

Reference: Water and Wastewater Treatment Plant Operations, Frank R. Spellman, Chapter 11, Page 315.

170. Answer: b. GiardiaReference: Water and Wastewater Treatment Plant Operations, Frank R. Spellman, Chapter 11, Page 319.

171. Answer: c. carbon dioxideReference: Water and Wastewater Treatment Plant Operations, Frank R. Spellman, Chapter 11, Page 328.

Maintenance172. Answer: d. 200 to 320 amps


173. Answer: a. temperature limitationReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 15, Page 391.

174. Answer: c. 5%Reference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 15, Page 400.



175. Answer: d. 10%Reference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 15, Page 400.

176. Answer: c. 5%Reference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 15, Page 400.

177. Answer: c. foreign material in motorReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 15, Page 406.

178. Answer: c. at least two pipe diameters above the rimReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 15, Page 446.

Safety179. Answer: b. self-contained breathing apparatus

Reference: Water Treatment, AWWA, Principles and Practices of Water Supply Operations Series, Second Edition, Chapter 7, Page 225.

180. Answer: c. carbon dioxideReference: Transmission and Distribution, AWWA, Principles and Practices of Water Supply Operations Series, Second Edition, Chapter 13, Page 428.

181. Answer: d. material safety data sheet (MSDS)Reference: Small Water System Operation and Maintenance, Kenneth D. Kerri, Third Edition, Chapter 6, Section 6.18.

182. Answer: c. 19.5%Reference: Code of Federal Regulations, Title 29 Part 1910.146(b) Definitions.

183. Answer: d. Class DReference: Small Water System Operation and Maintenance, Kenneth D. Kerri, Third Edition, Chapter 7, Page 401, Section 7.162.

184. Answer: b. emergency kit BReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 10, Pages 389 and 406.

Basic Chemistry185. Answer: b. air

Reference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 1, Page 3.

186. Answer: d. 1.29 g/LReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 1, Page 3.



187. Answer: a. three stepsReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 8, Page 73.

188. Answer: c. increasing the alum dosageReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 8, Page 73.

189. Answer: a. dissociatedReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 10, Page 84.

190. Answer: c. O3


191. Answer: a. flasksReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 16, Pages 485 and 486.

192. Answer: a. oilReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 16, Page 495.

193. Answer: d. be explosion-proofReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 16, Page 495.

194. Answer: a. 1%Reference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 16, Page 498.


196. Answer: d. 6 ftReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 16, Page 504.



67

Grade 3 Questions

General Wastewater Treatment Principles1. Which of the following best defines adsorption?

a. converting small particles of suspended solids into larger particles by the use of chemicals

b. assimilation of one substance into the body of another by molecular and chemical action

c. chemical complexing of metallic cations with certain inorganic compoundsd. adhesion of a gas, liquid, or dissolved substance onto the surface or interface zone

of another substance2. Autoclaving will sterilize

a. with steam at 121°C and 15 psib. at high temperatures near 400°Cc. with ultraviolet lightd. with ozone and 50 psi

3. During a confined space entry, how often must the confined space be monitored for hazardous atmospheres?a. continuouslyb. every 5 minc. before entry onlyd. before entry and once every 15 min

4. Under any soil conditions, cave-in protection is required for trenches or excavations that are how many feet deep?a. 3.5 ftb. 4.0 ftc. 4.5 ftd. 5.0 ft

5. Which one of the following gases is colorless, odorless, lighter than air, flammable, and sometimes called swamp gas?a. hydrogen sulfideb. methanec. radond. carbon dioxide

6. Material safety data sheets (MSDSs) are required fora. all chemicals used in the workplace regardless of hazardb. only chemicals with known health hazardsc. only flammable or explosive chemicalsd. only chemicals with suspected health hazards



7. What health risk is associated with nitrate in water?a. kidney damageb. liver damagec. nervous system damaged. methemoglobinemia

8. If only two rings of packing are used in the stuffing box, how many degrees should the joints be staggered?a. 15 to 45 degreesb. 45 to 90 degreesc. 90 to 180 degreesd. 180 to 225 degrees

9. Algae will grow by assimilatinga. ureab. ammoniac. amino acidsd. nitrate

10. Permits that regulate discharges into navigable waters must meet requirements froma. Department of Health and Human Services (HHS)b. National Pollutant Discharge Elimination System (NPDES)c. US Environmental Protection Agency (USEPA)d. NSF International

11. An NPDES discharge permit does not requirea. monthly average of settleable solidsb. biochemical oxygen demand (BOD)c. most probable number (MPN) of coliform group bacteriad. dissolved solids

Wastewater Treatment Facilities12. In secondary aerobic treatment, living organisms partially stabilize organic matter by

the process ofa. putrefactionb. fermentationc. hydrolysisd. oxidation

13. Most trickling filters will remove BOD-causing wastes and suspended solids from the influent at percentages ranging from:a. 30 to 50%b. 40 to 60%c. 70 to 85%d. 90 to 98%

14. Acid formers producea. carbon dioxide and hydrogen sulfideb. carbon monoxide and methanec. methane and oxygend. methane and hydrogen sulfide



15. The material that sloughs off a trickling filter from the buildup of organisms is calleda. scumb. nonvolatile organicsc. sludged. humus

16. The activated sludge is recirculated back to thea. primary clarifierb. secondary clarifierc. aeration tankd. before flowmeter and after comminutor

17. Which group of bacteria best describes the breakdown of organics in an anaerobic digester?a. fermentersb. acid formersc. saprophytic bacteriad. autotrophs

18. A properly digested sludge has _______ of the organics destroyed and converted to gas.a. 50%b. 60%c. 74%d. 90%

19. Digester gas consists ofa. 30% methane and 70% carbon dioxideb. 40% methane and 60% carbon dioxidec. 50% methane and 50% carbon dioxided. 60% methane and 40% carbon dioxide

Wastewater Sources and Characteristics20. The total Kjeldahl nitrogen is a measure of

a. nitrates and nitritesb. organic and inorganic nitrogenc. all organic nitrogend. organic nitrogen and ammonia nitrogen

Racks, Screens, Comminutors, and Grit Removal21. During normal flows, grit should be removed from the grit channel

a. every 6 hrb. every 8 hrc. every 12 hrd. on a daily basis



22. Aerated grit chambers are found most often at wastewater plants havinga. an activated sludge processb. rotating biological contactorsc. anaerobic digestiond. trickling filters

23. In a cyclone grit separator the water and lighter particles are carried out in thea. apex channelb. orifice holec. primary vortexd. secondary vortex

Sedimentation and Flotation24. In the activated sludge process, bulking is caused by

a. slime moldsb. filamentous fungic. filamentous bacteriad. filamentous algae and slime molds

25. In a large wastewater facility with several clarifiers, what detention time should be followed to determine when to take a clarifier off-line because the detention time is too long?a. 2 hrb. 2.5 hrc. 4 hrd. 6 hr

26. In a small package wastewater treatment plant on startup, digester gas will typically start being produced in the digester after approximatelya. 2 weeksb. 3 weeksc. 4 weeksd. 5 weeks

27. On startup of a small package wastewater treatment plant, when should lime be added to the digester unit?a. on the very first dayb. after about 3 daysc. after about 2 weeksd. after gas production

28. What is the best temperature range to maintain digester contents?a. 80 to 90°Fb. 80 to 95°Fc. 85 to 100°Fd. 90 to 100°F



Activated Sludge29. When nitrification is required in the activated sludge process, what are the organisms

doing the nitrification called?a. facultative bacteriab. obligate aerobesc. aerobesd. anaerobes

30. If the dissolved oxygen (DO) is too low in the aeration tank at an activated sludge plant, what will occur?a. rotifers will thriveb. the flocculated sludge will settlec. filamentous bacteria will proliferated. pinpoint floc will develop

31. The outlet to the receiving waters should be submerged toa. prevent public scrutinyb. facilitate mixingc. increase the dissolved oxygen levelsd. reduce foam and scum

32. When an environment is low in oxygen, which of the following organisms would do best?a. obligate aerobesb. facultative organismsc. anaerobic bacteriad. protozoans

33. Which one of the following organisms can cause sludge bulking in the activated sludge process because it is filamentous?a. Serratiab. Balamuthiac. Thiothrixd. Acinetobacter

34. What is the typical age of sludge at a high rate activated sludge plant?a. 0.5 to 2.0 daysb. 1.5 to 3 daysc. 2 to 4 daysd. 2 to 5 days

35. What would be the most likely BOD range for the final effluent of a conventional activated sludge plant?a. 10 to 20 mg/Lb. 15 to 35 mg/Lc. 20 to 40 mg/Ld. 30 to 80 mg/L



36. The preferred way to waste activated sludge isa. continuous wastingb. a set time of dayc. whenever it is neededd. variable depending on time, flow, and amount of influent sludge, but never to

change more than 15% from one day to the next37. What would be the most probable cause for sludge floating to the surface of a secondary

clarifier?a. toxic shock loadb. sludge return rate is inadequatec. mixed liquor is predominately filamentous organismsd. excessive turbulence in clarifier

38. What would be the most probable cause for the sludge blanket in the secondary clarifier to overflow the weirs uniformly?a. overloaded clarifiers due to peak flowsb. too many filamentous organisms causing sludge to floatc. aeration tank is too alkalined. large influx of cold water causing warmer sludge to float

39. What would be the most probable cause for a dark tan foam in the aeration tank that sprays cannot dissipate?a. inadequate aerationb. septic sludgec. mean cell residence time (MCRT) is too longd. solids loading too high

40. What would be the most probable cause for the aerator contents to turn dark?a. aeration is inadequateb. mixed liquor suspended solids (MLSS) are too lowc. MCRT too longd. nitrification occurring

41. What would be the most probable cause for large air bubbles in the aeration tank?a. dead zone causing an anaerobic condition in that areab. denitrification in that areac. broken diffuser(s) in that aread. nitrification occurring and low alkalinity in that area

42. What would be the most probable cause for the return sludge to have too low a sludge concentration?a. sludge is being overoxidizedb. sludge is being underoxidizedc. filamentous growthd. nitrification of sludge is occurring



43. What is the usual dissolved oxygen level in a sequencing batch reactor during the reaction cycle?a. 0.5 to 2.0 mg/Lb. 1.0 to 3.0 mg/Lc. 2.0 to 4.0 mg/Ld. 3.0 to 5.0 mg/L

44. What is the usual range of sludge age in a sequencing batch reactor?a. 8 to 30 daysb. 20 to 40 daysc. 25 to 45 daysd. 30 to 65 days

45. Microorganism samples should be prepared for analyses withina. 15 minb. 60 minc. 2 hrd. 4 hr

46. Bacteria in the lag phase area. losing population, as more are dying than multiplyingb. growing in sizec. growing in numbersd. remaining stationary in population size because cells multiplying equals cells

dying47. What group of protozoans do the flagellates comprise?

a. ciliatesb. suctoriac. Mastigophorad. amoebas

48. What should be done if too much nitrification is occurring in an aeration tank at an activated sludge plant? Assume the process is not moving out of nitrification.a. reduce sludge influentb. reduce DOc. increase sludge influentd. reduce return activated sludge

49. In the step-feed and contact stabilization modes, what would most likely happen if the waste activated sludge (WAS) flow was increased?a. increased nitrificationb. reduced nitrificationc. decrease the food-to-microorganism (F/M) ratiod. increase the F/M ratio

50. The last stage of pure oxygen reactors will usually vent gas to the atmosphere when the concentration of oxygen drops belowa. 70%b. 65%c. 60%d. 50%



51. In regard to cryogenic air separation, what component of the air is the least volatile?a. oxygenb. carbon dioxidec. nitrogend. methane

52. Which one of the following types of substances can be of animal or vegetable origin (which are biodegradable and can be successfully treated) and can also be of mineral origin, which will cause treatment problems and thus generally require pretreatment?a. flammable oilsb. oils and greasesc. acids and alkaliesd. organic toxicants

53. If an activated sludge plant is hit with a toxic load, what will most likely occur in the aeration tank?a. a more turbid effluentb. the BOD will increasec. a DO increase without an increase in air inputd. effluent will exhibit floc carry-over

54. The first process in the treatment of industrial wastes is (are)a. grit removalb. scum and grit removalc. screening coarse materialsd. pH adjustments

55. What is the COD:N:P ratio that is usually added when an individual waste treatment plant is started up?a. 100:5:1b. 100:10:2c. 150:5:1d. 150:10:5

56. Ammonia nitrogen is removed from wastewater by a process called desorption or stripping. What chemical is added to the wastewater to accomplish this removal?a. soda ashb. limec. caustic sodad. muriatic acid

Trickling Filters57. The first stage of biochemical oxygen demand (BOD) can also be called _______ BOD.

a. hydrolysisb. nitrificationc. oxidationd. carbonaceous



58. The second stage of biochemical oxygen demand (BOD) can also be called _______ BOD.a. oxidativeb. carbonaceousc. nitrogenousd. hydrolysis

59. The media depth in trickling filters that use rock ranges froma. 1 to 4 ftb. 2 to 5 ftc. 3 to 8 ftd. 6 to 12 ft

60. The purpose of the splinter box on a trickling filter is toa. divide the flow for recirculationb. provide adjustment for the distributor armc. provide level control for wastewater applicationd. provide flow control to the distributor arm

61. Which one of the following trickling filter types has more uniform sloughing of filter growth?a. low-rate filterb. high-rate filterc. standard-rate filterd. roughing filter

62. Although the startup of trickling filters depends on local conditions, it is usually best to start them ina. late April to early Juneb. early March to late Septemberc. the summer onlyd. late May to late September

63. What is the typical amount of BOD from the final effluent of a trickling filter plant?a. 5 to 15 mg/Lb. 15 to 40 mg/Lc. 20 to 50 mg/Ld. 25 to 60 mg/L

64. One of the trickling filter plant’s main functions is to stabilize thea. oxygen demandb. BODc. total solidsd. dissolved solids

65. A trickling filter plant should use the _______ recirculation rate possible to produce the best results and to also meet the _______ requirements.a. lowest, NPDES permitb. highest, NPDES permitc. lowest, EPA discharged. highest, EPA discharge



66. How effective a trickling filter plant is operating is mainly indicated bya. total solidsb. plant effluentc. effluent coliform countsd. settleable solids

67. Which two plant effluent factors indicate a trickling filter plant is working properly?a. suspended solids and BODb. dissolved solids and DOc. DO and BODd. total solids and DO

68. Ponding in a trickling filter may be solved bya. reducing sprayb. increasing recirculation ratec. checking plant influentd. partially opening the end gates

69. A problem with filter flies on the trickling filter may be solved bya. drying out the filter for a short timeb. operating filters in parallelc. high-pressure spraying the filter surfaced. increasing the rate of recirculation

70. Problems with odors from trickling filters may be solved bya. flushing out distributor armsb. decreasing recirculationc. drying out filter for a short period of timed. preventing splashing out of filters

71. Problems with icing on a trickling filter may be solved bya. pumping sludge out of clarifierb. high-pressure spraying the filter surfacec. reducing sprayd. checking ventilation

72. If the secondary clarifier has a problem with low dissolved oxygen, what can be done to the trickling filter to solve this problem?a. reduce the number of weedsb. operate the plant in parallelc. increase recirculation rated. partially open the end gates

Rotating Biological Contactors73. What usually starts to happen as BOD levels decrease from stage to stage in a four-

stage rotating biological contactor?a. carbonaceous cycle beginsb. methane gas starts to be producedc. large increase in slime growths sloughing off contactord. nitrification starts



74. Typically, how long will an even biomass develop on a rotating biological contactor if the wastewater is of normal strength and the water is not too cold?a. 1 to 2 weeksb. 2 to 4 weeksc. 3 to 5 weeksd. 4 to 6 weeks

75. Rotating biological contactors designed to remove suspended solids and BOD will usually have dissolved oxygen levels after the first stage ofa. 0.5 to 1.0 mg/Lb. 1.0 to 3.0 mg/Lc. 2.0 to 4.0 mg/Ld. 2.5 to 4.0 mg/L

76. If the appearance of a rotating biological contactor becomes black, what is most probably the cause?a. death of the slimesb. COD too lowc. BOD overloadingd. too slow a rotation causing high dissolved oxygen

77. If the appearance of a rotating biological contactor becomes white, what is most probably the cause?a. revolutions per minute (rpm) too fastb. a type of bacteria that feeds on sulfur compounds is thrivingc. BOD too lowd. mass dying off of the slime biomass

78. What is a possible course of action if a rotating biological contactor (RBC) turns black?a. increase the rpmb. put another RBC unit onlinec. shut off the black unit; put another unit online; then clean the black unitd. pressure-wash the RBC unit to remove the black material; the sloughings will be

removed later from the secondary clarifier79. What is a possible course of action if a rotating biological contactor (RBC) turns white?

a. pressure-wash the RBC unit to remove the white material; the sloughings will be removed later from the secondary clarifier

b. shut off the white unit; put another unit online; then clean the white unitc. put another RBC unit onlined. increase the rpm

Waste Treatment Ponds80. Methane-fermenting bacteria in an anaerobic waste treatment pond will produce

a. carbon dioxideb. nitrogenc. organic acidsd. alkalinity



81. Methane-producing bacteria require a pH level within the sludge ranging froma. 6.0 to 7.0b. 6.5 to 7.5c. 6.5 to 8.0d. 7.0 to 8.5

82. Typically, a thriving biological community in a wastewater treatment pond will take at least _______ to develop.a. 1 weekb. 2 weeksc. 1 monthd. 2 months

83. A waste pond should be cleaned when the depth of the wet sludge is abovea. 0.5 ftb. 1.0 ftc. 1.5 ftd. 2.0 ft

84. A pond that freezes over in the winter and thaws in the spring would most likely follow which sequence below during the course of one year (starting in winter)?a. anaerobic to facultative to aerobicb. anaerobic to oxidative to facultativec. anaerobic to facultative to oxidative to aerobicd. oxidative to anaerobic to facultative to aerobic

85. Which one of the following reasons would most likely cause a waste treatment pond to have offensive odors?a. excessive evaporationb. poor circulationc. excessive aerationd. anaerobic conditions

86. Regarding waste pond lab sampling, which one of the following parameters is sampled at the influent and the effluent of the pond?a. dissolved oxygenb. dissolved solidsc. BODd. pH

87. Waste ponds that provide the best physical storage of dissolved oxygen are _______ in depth.a. about 2.5 ftb. about 3.0 ftc. about 3.5 ftd. 4 ft or more



Sludge Digestion and Solids Handling88. Mesophilic bacteria thrive between the temperature range of

a. 58 to 103°Fb. 60 to 107°Fc. 64 to 110°Fd. 68 to 113°F

89. What is the temperature at which thermophilic bacteria begin to thrive in anaerobic digestion?a. above 98°Fb. above 105°Fc. above 113°Fd. above 118°F

90. Typically, how many cubic feet of gas will an anaerobic digester produce for every pound of volatile matter destroyed?a. 4 to 7 ft3

b. 6 to 10 ft3

c. 8 to 12 ft3

d. 12 to 18 ft3

91. What type of corrosion can occur in an anaerobic digester if air is constantly in contact with the gas?a. hydrochloric acid corrosionb. sulfuric acid corrosionc. galvanic corrosiond. nitric acid corrosion

92. What is the lower explosive limit in percent for a mixture of anaerobic digester gas and air?a. 5.3% digester gasb. 5.8% digester gasc. 7.1% digester gasd. 7.7% digester gas

93. Bacteria in an anaerobic digester cannot effectively utilizea. proteinsb. fatsc. petroleum productsd. enzymes

94. Sludge pumped to an anaerobic digester is considered thin if it contains solids that are less thana. 2.5%b. 3.5%c. 4%d. 4.7%



95. What is the best method to prevent foaming from occurring in an anaerobic digester?a. keep the F/M ratio highb. by adding enzymes slowly but continuously each dayc. add soda ash at a dose ranging from 1.0 mg/L to 1.5 mg/Ld. adequate mixing from top to bottom of the tank

96. If the volatile-acid-to-alkalinity ratio was increasing in an anaerobic digester, what would be the most probable cause?a. heat exchanger temperature set too lowb. not enough settling timec. raw sludge feed point is too close to the supernatant draw-off lined. withdrawing too much sludge

97. If the odor of the supernatant is sour from either the primary or secondary anaerobic digester, what is the most probable cause?a. hydraulic overloadb. pH is too highc. excessive withdrawal of sludged. toxic load entered digester

98. If the bottom sludge from a digester is too watery, what is the most probable cause?a. excessive mixingb. toxic load killed most of microorganismsc. digester being heated too muchd. digester needs cleaning

99. If the sludge at the point of disposal is too thin, what is the most probable cause?a. hydraulic overloading has occurredb. excessive gas in digesterc. heat exchanger not working properly or brokend. excessive mixing

100. If the pH in a digester drops, the CO2 becomes so high that no burnable gas is produced, and the volatile-acid-to-alkalinity ratio has increased to 0.8, what would be the best solution?a. add more sludgeb. add limec. decrease sludge withdrawald. add seed sludge from another digester

101. If the shaft seal is leaking on a mechanical mixer, what would be the most probable cause?a. poor lubricationb. poor alignment of equipmentc. grit has infiltrated packing or seald. packing or seal worn or dried out

102. What is the most probable cause for the scum blanket being too thick?a. too much gas productionb. digester temperature too lowc. volatile-acid-to-alkalinity ratio too lowd. volatile-acid-to-alkalinity ratio too high



103. The most probable reason for an aerobic digester to have an odor problem isa. pH is too low due to excess acid fermentersb. sludge being added too fastc. scum blanket not being removed often enoughd. the supply of oxygen is insufficient

104. What type of diffusers should be installed for aerobic digesters that use diffused air?a. open orifice diffusersb. dome diffusersc. plate diffusersd. fiberglass diffusers

105. Before digested sludge is fed to a centrifuge, it is usually conditioned witha. hydroxidesb. phosphatesc. polymersd. phyllosilicates

Odor Control106. What substance has the odor characteristic of a skunk?

a. crotyl mercaptanb. ethyl sulfidec. methyl sulfided. pyridine

107. If a wastewater plant’s biological filter has an H2S odor, what would be the most likely solution?a. add more lime to pretreatment systemb. improve housekeepingc. check operations of digesterd. add H2O2 to liquid before the filter

108. If in a biological system the anaerobic digestion unit has a decaying organic odor, what would be the most likely solution?a. reduce mixing in the digesterb. check waste gas burner and light if it is offc. remove solids and reseedd. add H2O2 or increase the addition of H2O2 by increments of 5% until odor ceases

109. If the vacuum filters or filter presses have an organic odor, what would be the most likely solution?a. apply minimal dosage of lime to vacuum filters or filter pressesb. remove solids from the areac. increase air to aeration systemd. correct design problem



Phosphorus Removal110. Besides a microbial population, what other element do all biological treatment systems

share?a. a liquid–solids separationb. a reactorc. a digesterd. inert media to which the bacteria can attach

111. After luxury uptake of phosphorus by microorganisms, to what environment are the microorganisms transferred in order for them to release this phosphorus?a. aerobicb. anaerobicc. anoxicd. facultative

112. What should the pH be for the most efficient removal of phosphorus to very low levels?a. above 9.0b. above 9.5c. above 10d. above 11

113. What type of microorganisms are Nitrosomonas and Nitrobacter?a. aerobesb. heterotrophsc. prototrophsd. autotrophs

114. This nitrification process facility has generally too short a contact time for complete nitrification.a. complete mix activated sludge processb. extended aerationc. conventional aeration systemd. step-feed aeration

115. Which one of the following determinations is the best means for monitoring the day-to-day operations of the nitrification process?a. pHb. alkalinityc. dissolved oxygend. CO2 levels

116. In the nitrification process, too many alkaline chemicals were added to the wastewater causing a very high pH. What was most probably produced that inhibited the nitrifying organisms?a. calcium oxideb. calcium carbonatec. ammoniad. nitrite and nitrate



117. If a denitrification unit shows the effluent COD has a sudden increase, what is the most probable cause?a. excessive DOb. methanol has been overdosedc. filter has an accumulation of nitrogen gasd. filter has an excessive accumulation of solids

118. If a denitrification unit shows the effluent has a sudden increase in nitrate, what is the most probable cause?a. DO levels are lowb. filter surface is blinded by solids floating to the top of the bedc. filter has an excessive accumulation of solidsd. the pH has drifted outside the 7.0 to 7.5 range

119. If a denitrification unit shows the effluent has a sudden increase in nitrate, what is the most plausible solution?a. decrease sludge returnb. add lime to raise the pHc. increase methanold. increase DO level

120. If a fluidized bed denitrifier is started up and immediately the filter is blinded, what is the most probable cause?a. too much lime was added before unit was stopped on previous runb. unit has excess bacterial slimec. solids floated to the top of the bedd. there was a loss of solids from the denitrifier

121. If an ammonia stripping process loses its efficiency for removing ammonia, what is the most probable cause?a. insufficient backwashing of unitb. pH of tower is too lowc. hydraulic loading is too lowd. weather is too cold

122. If an ammonia stripping process loses its efficiency for removing ammonia, what is the most plausible solution?a. increase hydraulic flow to unitb. increase pH by adding limec. backwash the unitd. add methanol to the unit

Enhanced Biological (Nutrient) Control123. In the enhanced biological activated sludge bulking control system for the process

control of activated sludge bulking, what is the primary controlling variable for the desired microbial population?a. DOb. F/M ratioc. soluble BOD5d. MLSS concentration



124. In the enhanced biological nutrient control activated sludge system for the process control of nitrogen oxidation, what is the primary controlling variable for the desired microbial population?a. BOD5b. DOc. alkalinityd. MLSS concentration

Disinfection125. The use of hypochlorite increases the pH because it will form _______ when added to

water.a. hydroxyl ionsb. carbonate ionsc. bicarbonate ionsd. calcium or sodium carbonates

126. Typically, chlorine dioxide does not react witha. base metalsb. ammoniac. sulfide iond. nitrite

127. Chlorine dioxide is an effective disinfectant when the pH isa. below 7.0b. above 7.6c. below 8.0d. above 8.5

128. What would be the maximum chlorine dosage at a waste plant for control of filter flies?a. 0.1 mg/Lb. 0.5 mg/Lc. 0.8 mg/Ld. 1.1 mg/L

129. How many turns should the valve be opened that supplies chlorine to the evaporator?a. 1 turnb. 1.5 to 2 turnsc. 2 to 3 turnsd. about 5 turns

130. A plant that uses liquid chlorine should set the high temperature alarm for the evaporator water bath ata. 185°Fb. 190°Fc. 195°Fd. 200°F



131. It is recommended that chlorine cylinders be stored in rooms where the temperature never exceedsa. 100°Fb. 110°Fc. 120°Fd. 125°F

132. Which one of the following chemicals is most often used at wastewater plants to de-chlorinate water before it enters the receiving waters?a. sodium bisulfiteb. sodium dioxidec. sodium metabisulfited. sodium thiosulfate

133. If a chlorine rotameter was used to measure the flow of sulfur dioxide gas, what would you multiply the reading by to determine lb/day of sulfur dioxide?a. 0.9b. 0.95c. 1.03d. 1.05

134. How can sulfuric acid damage to water treatment plant structures be reduced or eliminated?a. apply oxygenation to destroy hydrogen sulfideb. apply sodium bicarbonate, which combines with sulfuric acidc. apply calcium carbonate, which combines with sulfuric acidd. apply aeration to the wastewater to remove hydrogen sulfide

Solids Removal from Secondary Effluents135. Usually, the maximum solution strength for a polymer is _______ so it can be pumped

easily.a. 0.5%b. 1.0%c. 2.0%d. 5.0%

136. If a dual-media filter is improperly backwashed continually over a period of time, what will most likely result?a. bacterial slime will develop on filter mediab. air bindingc. mudball formationd. pathogen breakthrough

137. If the air-relief valve on a pressure filter discharges water, what is the most probable cause?a. debris lodged between valve and seatb. the effluent valve is blocked or closedc. underdrain system has an air pocketd. filter is overloaded



138. If a continuous backwash, upflow, deep-bed silica sand media filter has greatly reduced filter flow capacity, what is the most probable cause?a. sand has backflowed into plenum areab. screen in airlift is restricted with solidsc. sand level is too highd. airlift is not properly functioning

Effluent Disposal139. If a secondary clarifier has effluent that is high in BOD, what is the most likely cause?

a. the outlet baffle is not in the proper positionb. too low a DO content on the clarifierc. the biological treatment process has been organically overloadedd. anaerobic reactions on the bottom of the tank due to excessive detention time

Residual Solids Management140. If sludge is rising in a gravity thickener and solids are being carried over in the

effluent due to the sludge gassing off, what would be the most likely solution?a. increase chemical feed rateb. increase sludge collector speedc. increase the pumping of sludge from the clarifierd. increase sludge withdrawal rate

141. A gravity thickener has the following characteristics: thin, dilute underflow sludge and a clear liquid level, but sludge is rising and being carried over with the effluent. What is the most likely solution?a. increase collector speedb. lower influent sludgec. lower sludge blanketd. increase chemical feed rate

142. If a gravity thickener has thin, dilute underflow sludge, abundant solids on the liquid surface, and solids carry-over, what is the most likely solution?a. lower influent sludgeb. increase sludge withdrawal ratec. decrease chemical feed rated. lower collector speed

143. If a dissolved air flotation unit has solids being carried over with the effluent but has a good thickened sludge float concentration, what is the most probable cause?a. air to solids lowb. pressure too lowc. chemical addition inadequated. float blanket too thick

144. If a dissolved air flotation unit has good quality effluent but the float is thin (dilute), what is the most probable cause?a. air to solids lowb. air to solids highc. float blanket too thind. chemical addition is inadequate



145. If a basket centrifuge thickener has good quality centrate but the discharge solids are dilute, what is the most probable cause?a. feed rate too highb. feed rate too lowc. incorrect chemical dosed. bowl speed too low

146. If a basket centrifuge thickener has poor quality centrate and the solids discharged are dilute, what is the most probable cause?a. mechanical malfunctionb. bowl speed too lowc. feed time too shortd. high loadings

147. If a basket centrifuge thickener has poor quality centrate and the solids discharged are dilute, what is the most likely solution?a. lower pool depthb. decrease chemical dosagec. raise flow rated. lower flow rate

148. If a scroll centrifuge thickener has poor quality centrate but the discharge solids are good, what is the most probable cause?a. feed time too shortb. scroll speed too fastc. hydraulic load too highd. pool depth too high

149. If a scroll centrifuge thickener has poor quality centrate and the solids discharged are dilute, what is the most likely solution?a. increase chemical dosageb. feed time too longc. decrease bowl speedd. increase flow rate

150. If a disc-nozzle centrifuge thickener has good centrate but the solids discharged are dilute, what is the most likely solution?a. decrease flow rateb. unplug nozzles as requiredc. decrease number and size of nozzlesd. increase number and size of nozzles

151. If a gravity belt thickener has cake solids that are too wet, what is the most probable cause?a. belt bindingb. belt speed too highc. hydraulic load too highd. polymer dose too low



152. When microorganisms enter the endogenous phase, they area. multiplyingb. oxidizing their own cellular massc. going into their cyst formd. coming out of their cyst form

153. If an aerobic digester has a high residual DO and a low uptake rate, what is the most probable cause?a. filamentous growth too highb. toxic material in digesterc. air rates too highd. high digester pH

154. If an aerobic digester has reduced volatile suspended solids (VSS) destruction, what is the most likely solution?a. increase air rateb. decrease air ratec. lower temperatured. increase pH

155. If a windrow composting process develops anaerobic conditions, what is the most probable cause?a. ballingb. stack moisture is too lowc. lack of nitrogend. lack of nitrogen and phosphorus

Wastewater Reclamation and Reuse156. Which treatment process will give the required results for using treated wastewater

for agricultural forage crops irrigation?a. extended aerationb. two-stage nitrificationc. trickling filtersd. selective ion exchange

157. Which treatment process will give the required results for using treated wastewater for livestock and wildlife watering?a. tertiary lime treatmentb. activated sludgec. carbon tertiary lime treatmentd. alum added to aeration basin

158. Which treatment process will give the required results for using treated wastewater for power plant and industrial cooling?a. rotating biological contactorsb. extended aerationc. nitrification-denitrificationd. ferric chloride added to primary clarifier



159. Which treatment process will give the required results for using treated wastewater for industrial water supply primary metals?a. two-stage nitrificationb. carbon adsorption or filtered secondary effluentc. tertiary lime plus ion exchanged. trickling filter

160. Which treatment process will give the required results for using treated wastewater for direct potable water supply?a. carbon adsorptionb. activated sludgec. rotating biological contactorsd. reverse osmosis of advanced waste treatment (AWT) effluent

161. If a wastewater reclamation system has floatables in the pond’s effluent, what is the most probable cause?a. detention time too longb. temperature of pond too highc. mixing inadequated. improper location of outlet baffle

162. If a wastewater reclamation plant system has excessive algae in the pond’s effluent, what is the most probable cause?a. temperature has encouraged a particular algal species to proliferateb. detention time too longc. improper location of outlet baffled. pond is hydraulically overloaded

163. If a wastewater reclamation plant system has excessive algae in the pond’s effluent, what is the most plausible solution?a. reduce the pond’s level by decreasing influent flowb. install algae screen at pond effluentc. draw off effluent from below surface of pondd. add 0.25 mg/L dosage of chlorine to reduce algae numbers

164. If a wastewater reclamation system has excessive BOD in the pond’s effluent, what is most probable cause?a. organic overloadb. detention time too longc. insufficient polymer additiond. insufficient lime addition

165. If a wastewater reclamation system has high suspended solids in the effluent of the rotating biological contactor, what is the most probable cause?a. excessive mixingb. improper skimmer operationc. excessive contactor speedd. organically overloaded



166. If a wastewater reclamation system has high suspended solids in the effluent of the rotating biological contactor (RBC), what is the most plausible solution?a. adjust outlet baffleb. reduce contactor speedc. take RBC off-line and cleand. after RBC add 0.25 mg/L chlorine to oxidize suspended solids

167. Which one of the following parameters is collected as a daily grab sample in wastewater reclamation water quality monitoring?a. phosphorusb. alkalinityc. ammonia nitrogend. nitrate nitrogen

168. Which type of irrigation is not suited for crop growth?a. furrowb. sprinklersc. floodingd. infiltration–percolation

169. Which type of land disposal of wastewater requires no vegetation?a. overland flowb. infiltration–percolationc. furrowd. irrigation

Water and Wastewater Microbiology170. Under optimal conditions bacteria can double their numbers approximately every

a. 20 to 30 minb. 1 to 3 hrc. 3 to 4 hrd. 6 to 8 hr

171. Large numbers of nematodes would most likely be found ina. aeration tanksb. secondary clarifiersc. trickling filtersd. conventional activated sludge systems

172. The major gas produced by facultative bacteria (anaerobic decomposition) isa. nitrogenb. hydrogen sulfidec. methaned. carbon dioxide

173. The major gas produced by anoxic decomposition isa. hydrogen sulfideb. methanec. carbon dioxided. nitrogen



Basic Chemistry174. Atoms that have the same atomic number but a different mass number are called

a. isotopesb. radioactivec. radicalsd. isomers

175. What element is used as a standard for atomic weights?a. hydrogenb. heliumc. oxygend. carbon

176. The oxidation number of an element is also called its _______ number.a. ionb. valencec. radicald. isotope

177. When reduction occurs,a. electrons are lost by the species being reducedb. electrons are gained by the species being reducedc. electrons are shared with the oxygen atomd. oxygen is always lost to the species being reduced

178. An example of ionic bonding would bea. HClb. NaClc. H2d. CH4

179. What is the oxidation number of ferric iron?a. +1b. +2c. +3d. +4

180. What is the quantitative relationship of chemical reactants and their products known as?a. mass balanceb. mass ratioc. stoichiometryd. a chemical reaction

181. An acid is a substance that willa. donate a proton to another substanceb. accept an electron from another substancec. donate an electron from another substanced. accept a proton from another substance



182. Which element is found in all acids?a. hydrogenb. carbonc. oxygend. chlorine

183. A salt isa. a covalent compoundb. an ionic compoundc. a precipitated compoundd. an electron accepter

184. A heterogeneous mixture isa. composed of a solvent and a soluteb. two or more dissolved substances in waterc. called a solution if composed of two or more substancesd. called a suspension if composed of two or more substances

185. From the following list, which one will show Brownian motion?a. soluteb. colloidsc. solventd. electrolyte

186. A solution used to determine the concentration of another solution is called aa. saturated solutionb. standard solutionc. concentrated solutiond. dilute solution

187. Which one of the following does not counteract the zeta potential?a. covalent forcesb. Brownian motionsc. van der Waals forcesd. gravitational forces

188. The maximum density of water occurs ata. – 0.2°Cb. 0°Cc. 0.2°Cd. 4°C

189. When metal reacts with water, they forma. acidsb. hydroxides and hydrogen gasc. basesd. hydroxides and acids



190. Crystalline compounds that contain water are calleda. hygroscopic crystalsb. hydroxidesc. hydrous crystalsd. hydrates

191. What is it called when ions retain a specific number of water molecules when forming crystals?a. water of crystallizationb. hydration of ionsc. crystal hydrationd. ionic hydration

192. What is the process called, when a solid is separated from a solution?a. crystallizationb. precipitationc. distillationd. exsolution

193. Which gas law states that the rate of diffusion of the gas is inversely proportional to the square root of the gas’s molecular weight?a. Graham’s lawb. Henry’s lawc. Dalton’s lawd. Charles’s law

194. Which one of the following hydrocarbon series has one or more rings in their structure?a. alkyne seriesb. aromatic seriesc. alkane seriesd. alkene series

195. If a hydrocarbon has one or more of its hydrogen atoms replaced with a hydroxyl (OH–) group, it is calleda. a ketoneb. an esterc. an aldehyded. an alcohol

196. Saponification is the hydrolysis of aa. proteinb. polysaccharidec. carbohydrated. lipid

197. Temporary hardness is caused bya. magnesium sulfateb. calcium nitratec. magnesium chlorided. calcium bicarbonate



198. Soft waters that have alkalinity levels below 50 mg/L are most corrosivea. in the summerb. in the winter and springc. in the summer and springd. in the winter

199. When ozone with an oxygen or air mixture is dispersed by diffusers at the bottom of a contact chamber, it is calleda. diffusionb. contactionc. dispersiond. disinfection


95


General Wastewater Treatment Principles1. Answer: d. adhesion of a gas, liquid, or dissolved substance onto the surface or

interface zone of another substanceReference: Water Treatment, AWWA, Principles and Practices of Water Supply Operations Series, Second Edition, Chapter 13, Page 375.

2. Answer: a. with steam at 121°C and 15 psiReference: Water Quality, AWWA, Principles and Practices of Water Supply Operations Series, Second Edition, Chapter 3, Page 79.

3. Answer: a. continuouslyReference: Code of Federal Regulations, Title 29, Part 1910.146(c) (5) (ii) (E) and Appendix C, Example 1.

4. Answer: d. 5.0 ftReference: Water Transmission and Distribution, AWWA, Principles and Practices of Water Supply Operations Series, Second Edition, Page 147.

5. Answer: b. methaneReference: Water Transmission and Distribution, AWWA, Principles and Practices of Water Supply Operations Series, Second Edition, Page 496; and Small Water System Operation and Maintenance, Kenneth D. Kerri, Third Edition, Chapter 6, Section 6.18.

6. Answer: a. all chemicals used in the workplace regardless of hazardReference: Code of Federal Regulations, Title 29, Part 1910.1200.

7. Answer: d. methemoglobinemiaReference: Water Quality, AWWA, Principles and Practices of Water Supply Operations Series, Second Edition, Page 205.

8. Answer: c. 90 to 180 degreesReference: Water Transmission and Distribution, AWWA, Principles and Practices of Water Supply Operations Series, Second Edition, Appendix E, Page 565.

9. Answer: d. nitrateReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 2, Page 23.

10. Answer: b. National Pollutant Discharge Elimination System (NPDES)Reference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 2, Page 23.

11. Answer: d. dissolved solidsReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 2, Page 23.



Wastewater Treatment Facilities12. Answer: d. oxidation


13. Answer: c. 70 to 85%Reference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 3, Page 46.

14. Answer: a. carbon dioxide and hydrogen sulfideReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 3, Page 49.

15. Answer: d. humusReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 3, Page 47.

16. Answer: c. aeration tankReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 3, Page 47.

17. Answer: b. acid formersReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 3, Page 49.

18. Answer: a. 50%Reference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 3, Page 49.

19. Answer: a. 30% methane and 70% carbon dioxideReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 3, Page 49.

Wastewater Sources and Characteristics20. Answer: d. organic nitrogen and ammonia nitrogen


Racks, Screens, Comminutors, and Grit Removal21. Answer: d. on a daily basis


22. Answer: a. an activated sludge processReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 4, Page 88.

23. Answer: d. secondary vortexReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 4, Page 91.



Sedimentation and Flotation24. Answer: c. filamentous bacteria


25. Answer: c. 4 hrReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 5, Page 119.

26. Answer: b. 3 weeksReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 5, Page 149.

27. Answer: a. on the very first dayReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 5, Page 149.

28. Answer: b. 80 to 95°FReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 5, Page 153.

Activated Sludge29. Answer: b. obligate aerobes


30. Answer: c. filamentous bacteria will proliferateReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 8, Page 254.

31. Answer: d. reduce foam and scumReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 8, Page 283.

32. Answer: b. facultative organismsReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 9.

33. Answer: c. ThiothrixReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 10.

34. Answer: a. 0.5 to 2.0 daysReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 21.

35. Answer: a. 10 to 20 mg/LReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 54.



36. Answer: a. continuous wastingReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 55.

37. Answer: c. mixed liquor is predominately filamentous organismsReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Pages 74 and 75.

38. Answer: a. overloaded clarifiers due to peak flowsReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Pages 74 and 75.

39. Answer: c. mean cell residence time (MCRT) is too longReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 75.

40. Answer: a. aeration is inadequateReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Pages 74 and 75.

41. Answer: c. broken diffuser(s) in that areaReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Pages 74 and 75.

42. Answer: c. filamentous growthReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Pages 74 through 76.

43. Answer: b. 1.0 to 3.0 mg/LReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 105.

44. Answer: c. 25 to 45 daysReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 105.

45. Answer: a. 15 minReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 112.

46. Answer: b. growing in sizeReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 116.

47. Answer: c. MastigophoraReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 119.

48. Answer: b. reduce DOReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 126.



49. Answer: d. increase the F/M ratioReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 127.

50. Answer: d. 50%Reference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 2, Page 60.

51. Answer: a. oxygenReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 2, Page 63.

52. Answer: b. oils and greasesReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 2, Page 98.

53. Answer: c. a DO increase without an increase in air inputReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 2, Page 99.

54. Answer: d. pH adjustmentsReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 2, Page 104.

55. Answer: c. 150:5:1Reference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 2, Page 105.

56. Answer: b. limeReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 2, Page 139.

Trickling Filters57. Answer: d. carbonaceous


58. Answer: c. nitrogenousReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 6, Page 167.

59. Answer: c. 3 to 8 ftReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 6, Page 169.

60. Answer: a. divide the flow for recirculationReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 6, Page 173.

61. Answer: b. high-rate filterReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 6, Page 177.



62. Answer: a. late April to early JuneReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 6, Page 179.

63. Answer: b. 15 to 40 mg/LReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 6, Page 181.

64. Answer: a. oxygen demandReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 6, Page 182.

65. Answer: a. lowest, NPDES permitReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 6, Page 184.

66. Answer: b. plant effluentReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 6, Page 184.

67. Answer: a. suspended solids and BODReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 6, Page 184.

68. Answer: b. increasing recirculation rateReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 6, Page 191.

69. Answer: d. increasing the rate of recirculationReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 6, Page 191.

70. Answer: d. preventing splashing out of filtersReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 6, Page 191.

71. Answer: c. reducing sprayReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 6, Page 191.

72. Answer: c. increase recirculation rateReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 6, Page 191.

Rotating Biological Contactors73. Answer: d. nitrification starts


74. Answer: a. 1 to 2 weeksReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 7, Page 225.



75. Answer: a. 0.5 to 1.0 mg/LReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 7, Page 226.

76. Answer: c. BOD overloadingReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 7, Page 228.

77. Answer: b. a type of bacteria that feeds on sulfur compounds is thrivingReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 7, Page 228.

78. Answer: b. put another RBC unit onlineReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 7, Page 228.

79. Answer: c. put another RBC unit onlineReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 7, Page 228.

Waste Treatment Ponds80. Answer: d. alkalinity

Reference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 9, Pages 299 and 301.

81. Answer: b. 6.5 to 7.5Reference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 9, Page 301.

82. Answer: d. 2 monthsReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 9, Page 303.

83. Answer: b. 1.0 ftReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 9, Page 307.

84. Answer: a. anaerobic to facultative to aerobicReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 9, Page 308.

85. Answer: d. anaerobic conditionsReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 9, Page 311.

86. Answer: c. BODReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 9, Page 315.

87. Answer: d. 4 ft or moreReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 9, Page 321.



Sludge Digestion and Solids Handling88. Answer: d. 68 to 113°F


89. Answer: c. above 113°FReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 151.

90. Answer: d. 12 to 18 ft3


91. Answer: b. sulfuric acid corrosionReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 158.

92. Answer: a. 5.3% digester gasReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 166.

93. Answer: c. petroleum productsReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 183.

94. Answer: b. 3.5%Reference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 185.

95. Answer: d. adequate mixing from top to bottom of the tankReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 187.

96. Answer: d. withdrawing too much sludgeReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Pages 206 and 207.

97. Answer: d. toxic load entered digesterReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Pages 206 and 207.

98. Answer: a. excessive mixingReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Pages 206 and 207.

99. Answer: d. excessive mixingReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Pages 206 and 207.

100. Answer: b. add limeReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 206.



101. Answer: d. packing or seal worn or dried outReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 208.

102. Answer: b. digester temperature too lowReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 209.

103. Answer: d. the supply of oxygen is insufficientReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 219.

104. Answer: a. open orifice diffusersReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 219.

105. Answer: c. polymersReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 224.

Odor Control106. Answer: a. crotyl mercaptan


107. Answer: d. add H2O2 to liquid before the filterReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 1, Page 36.

108. Answer: b. check waste gas burner and light if it is offReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 1, Page 37.

109. Answer: b. remove solids from the areaReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 1, Page 37.

Phosphorus Removal110. Answer: a. a liquid–solids separation


111. Answer: b. anaerobicReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 5, Page 478.

112. Answer: d. above 11Reference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 5, Page 485.



113. Answer: d. autotrophsReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 5, Page 518.

114. Answer: d. step-feed aerationReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 5, Page 520.

115. Answer: b. alkalinityReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 5, Page 521.

116. Answer: c. ammoniaReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 5, Page 522.

117. Answer: b. methanol has been overdosedReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 5, Page 536.

118. Answer: d. the pH has drifted outside the 7.0 to 7.5 rangeReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 5, Page 536.

119. Answer: b. add lime to raise the pHReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 5, Page 536.

120. Answer: c. solids floated to the top of the bedReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 5, Page 536.

121. Answer: b. pH of tower is too lowReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 5, Page 539.

122. Answer: b. increase pH by adding limeReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 5, Page 539.

Enhanced Biological (Nutrient) Control123. Answer: a. DO


124. Answer: b. DOReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 7, Page 554.



Disinfection125. Answer: a. hydroxyl ions


126. Answer: b. ammoniaReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 10, Page 349.

127. Answer: d. above 8.5Reference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 10, Page 349.

128. Answer: b. 0.5 mg/LReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 10, Page 352.

129. Answer: b. 1.5 to 2 turnsReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 10, Page 364.


131. Answer: a. 100°FReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 10, Page 398.

132. Answer: b. sodium dioxideReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 10, Page 403.

133. Answer: b. 0.95Reference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 10, Page 403.

134. Answer: a. apply oxygenation to destroy hydrogen sulfideReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 10, Page 420.

Solids Removal from Secondary Effluents135. Answer: b. 1.0%


136. Answer: c. mudball formationReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 4, Page 398.



137. Answer: c. underdrain system has an air pocketReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 4, Page 423.

138. Answer: a. sand has backflowed into plenum areaReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 4, Page 438.

Effluent Disposal139. Answer: c. the biological treatment process has been organically overloaded


Residual Solids Management140. Answer: d. increase sludge withdrawal rate


141. Answer: a. increase collector speedReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 181.

142. Answer: a. lower influent sludgeReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 181.

143. Answer: d. float blanket too thickReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 188.

144. Answer: c. float blanket too thinReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 188.

145. Answer: b. feed rate too lowReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 200.

146. Answer: d. high loadingsReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 200.

147. Answer: d. lower flow rateReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 200.

148. Answer: c. hydraulic load too highReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 200.



149. Answer: a. increase chemical dosageReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 200.

150. Answer: c. decrease number and size of nozzlesReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 200.

151. Answer: b. belt speed too highReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 205.

152. Answer: b. oxidizing their own cellular massReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 208.

153. Answer: b. toxic material in digesterReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 216.

154. Answer: a. increase air rateReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 216.

155. Answer: a. ballingReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 266.

Wastewater Reclamation and Reuse156. Answer: c. trickling filters

Reference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 8, Pages 581 and 582.

157. Answer: b. activated sludgeReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 8, Pages 581 and 582.

158. Answer: a. rotating biological contactorsReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 8, Pages 581 and 582.

159. Answer: d. trickling filterReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 8, Pages 581 and 582.

160. Answer: d. reverse osmosis of advanced waste treatment (AWT) effluentReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 8, Pages 581 and 582.

161. Answer: d. improper location of outlet baffleReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 8, Page 593.



162. Answer: a. temperature has encouraged a particular algal species to proliferateReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 8, Page 593.

163. Answer: c. draw off effluent from below surface of pondReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 8, Page 593.

164. Answer: a. organic overloadReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 8, Page 593.

165. Answer: d. organically overloadedReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 8, Page 593.

166. Answer: a. adjust outlet baffleReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 8, Page 593.

167. Answer: b. alkalinityReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 8, Page 594.

168. Answer: d. infiltration–percolationReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 8, Page 599.

169. Answer: b. infiltration–percolationReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 8, Page 599.

Water and Wastewater Microbiology170. Answer: a. 20 to 30 min


171. Answer: c. trickling filtersReference: Water and Wastewater Treatment Plant Operations, Frank R. Spellman, Chapter 11, Page 318.

172. Answer: b. hydrogen sulfideReference: Water and Wastewater Treatment Plant Operations, Frank R. Spellman, Chapter 11, Page 328.

173. Answer: d. nitrogenReference: Water and Wastewater Treatment Plant Operations, Frank R. Spellman, Chapter 11, Page 328.



Basic Chemistry174. Answer: a. isotopes


175. Answer: d. carbonReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 3, Page 16.

176. Answer: b. valenceReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 4, Page 26.

177. Answer: b. electrons are gained by the species being reducedReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 4, Page 28.

178. Answer: b. NaClReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 4, Page 30.

179. Answer: c. +3Reference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 4, Page 31.

180. Answer: c. stoichiometryReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 5, Page 39.

181. Answer: a. donate a proton to another substanceReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 6, Page 47.

182. Answer: a. hydrogenReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 6, Page 47.

183. Answer: b. an ionic compoundReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 6, Page 54.

184. Answer: d. called a suspension if composed of two or more substancesReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 7, Page 57.

185. Answer: b. colloidsReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 7, Page 57.

186. Answer: b. standard solutionReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 7, Page 63.



187. Answer: a. covalent forcesReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 8, Page 71.

188. Answer: d. 4°CReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 9, Page 77.

189. Answer: b. hydroxides and hydrogen gasReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 9, Page 79.

190. Answer: d. hydratesReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 9, Pages 79 and 80.

191. Answer: a. water of crystallizationReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 10, Page 85.

192. Answer: b. precipitationReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 10, Page 86.

193. Answer: a. Graham’s lawReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 11, Page 93.

194. Answer: b. aromatic seriesReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 12, Pages 102 and 103.

195. Answer: d. an alcoholReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 12, Page 105.

196. Answer: d. lipidReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 12, Page 114.

197. Answer: d. calcium bicarbonateReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 13, Page 120.

198. Answer: a. in the summerReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 14, Page 130.

199. Answer: b. contactionReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 15, Page 152.


111

Grade 4 Questions

General Wastewater Treatment Principles1. What is the odor detection limit of chlorine gas?

a. 0.1 ppmb. 0.3 ppmc. 0.5 ppmd. 1.0 ppm

2. What is a cathode?a. negative pole of an electrolyteb. positive pole of an electrolytec. negative pole of an electrolytic cell or systemd. positive pole of an electrolytic cell or system

3. The least reactive metals are calleda. anodic metalsb. galvanic metalsc. cathodic metalsd. tempered metals

4. Unless water cooled, the operating temperature of a mechanical seal in a pump should never exceeda. 95°F (35°C)b. 120°F (49°C)c. 140°F (60°C)d. 160°F (71°C)

5. What term describes the condition that exists when the source of the water supply is below the centerline of the pump?a. total discharge headb. suction liftc. velocity headd. pressure head

6. All occupied trenches 4 or more ft deep must provide exits ata. 15-ft intervalsb. 20-ft intervalsc. 25-ft intervalsd. 30-ft intervals

7. What is the difference between a weak acid and a strong acid?a. amount of buffering releasedb. amount of hydroxyl ions releasedc. amount of carbonate ions releasedd. amount of hydrogen ions released



8. Which one of the following is the best type of valve to use to dampen a water hammer?a. needleb. pressure-reliefc. pressure-reducingd. pinch

9. Which one of the following best defines the term stray-current corrosion?a. decomposition of a metal by its reaction with an acidic waterb. decomposition of a material caused by an outside electric currentc. reaction between a metal and waterd. reaction between two different metals with water acting as an electrolyte

Sedimentation and Flotation10. A shock load of toxic wastes coming into a plant can be treated or controlled with

a. coagulants onlyb. coagulants and limec. coagulants and coagulant aidd. coagulants or chlorine

11. Toxic solids should be disposed of ina. a digester to destabilize and break down the toxinsb. an approved sanitary landfillc. an incineratord. a sludge pond for further aerobic breakdown

12. On startup of a small package wastewater treatment plant, lime should be added to the digestion unit when the pH goes belowa. 6.8b. 7.0c. 7.2d. 7.5

Activated Sludge13. If the effluent to a small package wastewater treatment plant using the activated

sludge process appears turbid, what is most probably the cause?a. low concentration of dissolved oxygen (DO) in the aeration tankb. pumping rate for the return of activated sludge is out of balancec. excessive aeration rates in the aeration tankd. alkalinity is too high

14. If a small package treatment plant utilizing the activated sludge process has a light-colored floating sludge on the surface of the clarifier, which one of the following would most likely eliminate the floating sludge?a. increase the sludge wasting rateb. add a pH stabilizer, as pH is too lowc. reduce the aeration ratesd. increase the return sludge rate



15. What usually causes frothing in the activated sludge process?a. too low a solids levelb. too high a solids leavingc. too short a retention timed. too long a detention time

16. Which one of the following suggestions would most likely control foaming?a. increase the detention timeb. water spraysc. increase the dissolved oxygend. increase solids wasting

17. What is the ideal level of solids in a jar test sample collected from an aeration tank if it is allowed to settle for 30 min?a. ¼ of the jar depthb. ⅓ of the jar depthc. ⅖ of the jar depthd. ½ of the jar depth

18. In a small wastewater treatment plant using the activated sludge process, solids in the settling compartment should not be allowed to remain longer thana. 2 hrb. 3 hrc. 4 hrd. 5 hr

19. In an oxidation ditch, what would cause a dark gray color to develop in the mixed liquor?a. the buildup of facultative bacteriab. the buildup of aerobic bacteriac. the buildup of anaerobic bacteriad. the lack of bacteria buildup

20. During the startup of the oxidation sludge process in the oxidation ditch, the operator should maintain the highesta. recommended level of dissolved oxygenb. inflow of raw wastewater possible in order to fill the ditchc. possible rate of returned activated sludged. possible brush rotor speed advisable

21. What is the best range of dissolved oxygen in an oxidation ditch for optimum operation?a. 0.5 to 2.0 mg/Lb. 1.0 to 2.0 mg/Lc. 1.0 to 2.5 mg/Ld. 1.0 to 3.0 mg/L

22. Overoxidation of an oxidation ditch will causea. an algae bloomb. suspended solids to floatc. oxygen toxicity for protozoans and other organisms, especially rotifersd. the formation of pinpoint floc



23. In the oxidation ditch process, what should be done if dark solids appear in the settling tank?a. increase disinfection to reduce septic conditions and wasteb. increase the return sludge ratec. add oxygen to the settling tank through the clarifier’s diffusersd. waste all the solids and start the settling process over

24. A thick dark foam on the surface of the oxidation ditch indicates that thea. process is performing wellb. rotor revolutions per minute (rpm) are too fastc. wasting rate should be increasedd. flow through the ditch is too slow

25. What should the operator do if the oxidation ditch becomes hydraulically overloaded due to a heavy rainstorm?a. temporarily store the excess rainwater in the primary clarifierb. increase the dissolved oxygen levelsc. shut down one or more rotor assembliesd. send excess wastewater to the settling tank

26. The blower system at a wastewater plant has to be shut off for cleaning the activated sludge process. The process will not be adversely affected, if the system is off for onlya. 5 to 10 minb. 15 to 30 minc. 30 to 60 mind. 1 to 2 hr

27. When the pressure difference indicated on a manometer reads _______ , the air filters on a blower for an aeration tank should be cleaned or changed.a. 3 in.b. 5 in.c. 7 in.d. 10 in.

28. The waste activated sludge wasting rate should not be changed by more than _______ from one day to the next.a. 10 to 15%b. 20%c. 20 to 25%d. 25%

29. What is the possible number of pounds of sludge volatile matter produced per pound of biochemical oxygen demand (BOD) removed from a conventional plant running the activated sludge process?a. 0.15 lbb. 0.25 lbc. 0.55 lbd. 0.85 lb



30. If an operator calculates the wasting rate for activated sludge to be 45 gpm over the following 24 hr, the operator should wastea. the theoretical wasting rate calculated (45 gpm)b. a little less than the theoretical wasting ratec. a little more than the theoretical wasting rated. at least 15% more, since the amount of activated sludge will increase by about

that much over the next 24 hr31. In an activated sludge plant, what would a major storm event do to the percent volatile

matter in the aerator?a. decrease the percent volatile matterb. increase the percent volatile matter by a few percentc. increase the volatile matter significantlyd. the volatile matter would remain the same

32. What is the 5-day sludge age equivalent to?a. 1 lb of food per 5 lb of organismsb. 1 lb of organisms per 5 lb of foodc. 5 days of BODd. 5 days by weight of mixed liquor suspended solids

33. A significant temperature change for an activated sludge plant is usually plus or minusa. 4°Fb. 5°Fc. 7°Fd. 10°F

34. Rising sludge is caused by _______ and _______ .a. bulking; septicityb. septicity; denitrificationc. too high a flow; bulkingd. too high a flow; septicity

35. A settleability test is performed on a secondary clarifier at an activated sludge plant. If the sludge floats too soon, then the sludge age should be reduced. What else may be done?a. increase the DOb. decrease the BODc. increase the F/M (food-to-microorganism) ratiod. decrease the F/M ratio

36. If the mixed liquor suspended solids (MLSS) concentrations of one aeration tank are substantially different from another aeration tank at an activated sludge plant, what would be the most probable cause?a. alkalinity too low in one of the tanksb. diffusers plugged or broken in one of the aeration tanksc. denitrification has occurred in one of the tanksd. aeration tanks have unequal flow distribution



37. What would be the most probable cause for the return sludge to have too low a sludge concentration?a. the mean cell retention time is too shortb. sludge is being overoxidizedc. sludge is being underoxidizedd. the sludge consists mainly of actinomycetes

38. If filamentous organisms are the cause of sludge floating on the surface of secondary clarifiers, what would be the best solution?a. decrease DO in aeration tankb. increase sludge return ratec. reduce mean cell residence time (MCRT)d. decrease pH to 6.5

39. If a toxic shock load caused pin floc in the overflow of the secondary clarifier, what would be the best course of action?a. increase the MCRTb. get the sludge return rate as high as possiblec. decrease the sludge return rate as soon as possibled. decrease the DO in the aeration tank(s)

40. If the mean cell retention time was too long and caused dark tan foam on the aeration tank, what would be the best course of action?a. moderately increase sludge wasting to reduce the MCRTb. quickly increase sludge wasting to reduce the MCRTc. slowly but significantly decrease sludge wasting to reduce the MCRTd. quickly decrease sludge wasting to reduce the MCRT

41. If a low concentration of MLSS caused thick sudsy foam on the aeration tank, what would be the best solution?a. monitor industrial dischargesb. increase DO to the aeration tankc. increase MCRT and MLSS by decreasing sludge wastingd. decrease MLSS by increasing sludge wasting or decrease sludge return rate

42. Sphaerotilus is usually associated witha. high DOb. low F/M ratioc. low DO or low F/M ratiod. low DO or high F/M ratio

43. Nocardia growth is most associated witha. white scum or frothb. low DO or low F/M ratioc. low DO or high F/M ratiod. high F/M ratio



44. What type of organisms predominate in the activated sludge, when the F/M ratio is high?a. small Mastigophorab. rotifersc. stalked ciliatesd. free-swimming ciliates

45. Which types of organisms are usually present in old activated sludge that has a high MCRT?a. stalked ciliatesb. free-swimming ciliatesc. rotifersd. Mastigophora

46. If an activated sludge plant is not experiencing sludge bulking or high sludge volume index (SVI) problems, then there is no need to counta. stalked ciliatesb. free-swimming ciliatesc. filamentsd. large Mastigophora

47. When looking through a microscope, an operator sees the presence of nematodes and a predominance of rotifers from a sample collected from the activated sludge plant. What is the most likely cause?a. DO is too lowb. anaerobic conditions existc. F/M ratio is too lowd. MCRT is too short

48. If the activated sludge sample contains predominately nematodes and rotifers, then the activated sludge process most probablya. has high F/M ratio and high MCRTb. has high F/M ratio and low MCRTc. has low F/M ratio and high MCRTd. has low F/M ratio and low MCRT

49. In the step-feed and contact stabilization modes, what would most likely happen if the return activated sludge (RAS) flow was increased?a. it would decrease the SVIb. it would increase the SVIc. reaeration detention time would be reducedd. it would decrease the F/M ratio

50. Why does a sequencing batch reactor (SBR) usually produce a better effluent than a conventional activated sludge plant?a. because only the top 50% of the supernatant is decantedb. because only the top 70% of the supernatant is decantedc. because the water is motionless just before it is decantedd. because the SBR concentrates the suspended solids more efficiently



51. If a sequencing batch reactor normally removed BOD and suspended solids, what would an operator have to do to also remove nitrogen and phosphorus?a. increase the mixing timeb. increase the mixing and aeration timec. increase the reaction and settling timed. increase the aeration and settling time

52. An operator starting up a new sequencing batch reactor plant obtains 2,000 gal of seed sludge from another activated sludge plant. How long will it take before this seed sludge is capable of treating wastewater?a. 3 to 5 daysb. 1 to 2 weeksc. 1 monthd. 2 months

53. What is the purpose of aeration besides providing dissolved oxygen to organisms?a. mixing of the mixed liquorb. oxidation and reduction reactionsc. reduction and reduce volatilesd. reduce volatiles and purification

54. What type of activated sludge process has a maximum return activated sludge rate of 200%?a. carbonaceous stage of separate stage nitrificationb. nitrification stage of separate stage nitrificationc. step-feed aerationd. contact stabilization

55. What is the advantage of using the constant percentage return activated sludge (RAS) flow rate control in the activated sludge process versus the constant RAS flow rate?a. required operational time is lessb. simplicityc. the F/M ratio has less variabilityd. maximum solids loading on the clarifier will result during peak flow periods

56. An activated sludge plant has wastewater with a temperature of 20°C. In general, in days, what MCRT is needed for the plant to produce a nitrified effluent?a. 10 daysb. 15 daysc. 20 daysd. 30 days

57. In general, a wastewater activated sludge plant will lose activated sludge in the effluent at a rate of _______ of the total solids required to be wasted.a. < 2.5%b. < 5%c. < 7.5%d. < 10%



58. Respiration rate values for activated sludge of less than 4 mg O2/hr/gm MLSS indicatesa. typical values of most conventional activated sludge processesb. understabilized organic matterc. biological organisms are not healthy or stabled. organic matter is undergoing rapid biodegradation

59. Respiration rate values for activated sludge of more than 20 mg O2/hr/gm MLSS indicatesa. slow biodegradable organic matterb. typical values of most conventional activated sludge processesc. biological organisms are not healthy or stabled. organic matter is undergoing rapid biodegradation

60. Which one of the following is the most common cause of wastewater toxicity to microorganisms due to its excessive amounts?a. heavy metalsb. acidsc. petroleum productsd. detergents

61. Which one of the following is the least common cause of wastewater toxicity to microorganisms?a. residual chlorineb. free ammoniac. chlorinated hydrocarbonsd. solvents

62. In the startup of an industrial waste treatment plant, chemicals are added to the aeration basin until the microorganisms reproduce and approach an MLSS level ofa. 2,000 mg/Lb. 2,500 mg/Lc. 3,000 mg/Ld. 3,500 mg/L

63. In the activated sludge process, the pounds of nitrogen required is equal to what percent of dry weight volatile solids produced each day?a. 3%b. 7%c. 10%d. 18%

64. If a wastewater plant receives highly variable flows, it can store the wastewater during periods of high flow and release the wastewater during periods of low flow. What is the recommended storage volume in percent of total daily flow for such a wastewater plant?a. 10 to 20%b. 25 to 33%c. 30 to 40%d. at least 50%



65. Why do petroleum waste treatment plants use the extended aeration mode?a. to maximize petroleum waste degradation by oxidationb. to increase the mixed liquor volatile suspended solids (MLVSS) and thus

maximize oxidation of petroleum wastesc. to maintain the nitrification population essential for ammonia oxidationd. to maintain a low F/M ratio, which causes the microorganisms to use sulfides as

an energy source66. What could be added to the effluent if phenols are close to the National Pollutant

Discharge Elimination System (NPDES) limit at a petroleum waste plant?a. soda ashb. caustic sodac. nitric acidd. hydrogen peroxide

67. What is the optimal pH range for the biological nitrification process?a. 7.0 to 7.3b. 7.3 to 7.9c. 7.3 to 8.0d. 7.5 to 8.5

68. What is the temperature range for good nitrification to take place?a. 60 to 95°Fb. 70 to 80°Fc. 75 to 85°Fd. 75 to 95°F

69. In the nitrification process, phosphorus is added to the aeration system and adjusted according to _______ and the concentration of total Kjedahl nitrogen (TKN) in the wastewater.a. MLSSb. MLVSSc. BOD5 leveld. MCRT

Trickling Filters70. Adequate amounts of nitrogen and phosphorous are especially lacking most often from

which one of the following wastes?a. industrialb. domesticc. fermentation industriesd. food processing

71. When treating industrial wastes, for every 100 lb of incoming BOD there should be how many pounds of ammonia nitrogen?a. 1 lbb. 5 lbc. 10 lbd. 12 lb



72. What is the primary nuisance insect associated with the operation of a trickling filter?a. psychodab. tachinid fliesc. syrphid fliesd. thrips

73. Which one of the following high flow rates into a plant would probably have high dissolved oxygen content higher than normal (assume domestic wastes)?a. broken collection system pipeb. stormwaterc. clearance of a sewer stoppaged. industrial discharge

74. Which one of the following chemicals is used to neutralize wastes entering a wastewater treatment plant that contains cyanide?a. perchloric acidb. chlorinec. limed. sulfuric acid

75. Which one of the following chemicals is used to decrease the pH entering a wastewater treatment plant if the pH entering the plant is greater than 9.0?a. acetic acidb. HNO3c. NH3d. H3PO4

76. Which one of the following chemicals is used to increase the pH entering a wastewater treatment plant if the pH entering the plant is less than 6.0?a. Na2CO3b. CaCO3c. NaOHd. CaSO4

77. What can be done to the trickling filter if there is floating sludge on the secondary clarifier?a. apply chlorine to trickling filterb. decrease recirculationc. partially open end gatesd. check plant influent and make adjustments

Rotating Biological Contactors78. At what level should the alkalinity level be maintained in a rotating biological

contactor wastewater plant when nitrification is occurring?a. 3.4 times the influent ammoniab. 4.2 times the influent ammoniac. 5.8 times the influent ammoniad. 7.1 times the influent ammonia



79. What is the possible cause of excessive drive noise in rotating biological contactor units?a. excessive heatb. belt too narrowc. flange installed incorrectlyd. misalignment

Waste Treatment Ponds80. A high pH in a waste treatment pond will balance

a. aerobic fermentationb. anaerobic fermentationc. CO2 and O2 production by algae and other microorganismsd. algae and bacterial communities

81. If a hard scum develops on a wastewater treatment pond, what organisms may grow and cause an odor problem?a. green algaeb. blue-green algaec. aerobic bacteriad. anaerobic bacteria

82. Which one of the following nuisance organisms is sufficiently controlled only by using a pesticide?a. mosquitoesb. dragonfly larvaec. chironomid midgesd. shrimp-like animals

83. An operator working at a waste treatment pond knows that a surface aerator is working properly whena. no foam is on the surfaceb. very little foam is on the surfacec. some scum and foam is on the surfaced. very little foam is on the surface so aeration time is backed off 10%

84. A waste pond that is overloaded will usually bea. light greenb. dull greenc. grayish greend. gray

85. Why are dairy products hard to treat in a waste pond?a. because most bacteria in a waste pond lack the enzyme lactase necessary to break

down lactoseb. because the pH is too highc. because the pH is too lowd. because the dairy products kill the algae



Sludge Digestion and Solids Handling86. The addition of soda ash to a sour digester will

a. increase acid fermentationb. not add anything extra to the system that would not be added laterc. add less solids loading than limed. add more solids loading than lime

87. The addition of this chemical to a sour digester may cause a dangerous vacuum to develop.a. limeb. soda ashc. alumd. anhydrous ammonia

88. Neutralization of a sour digester with sodium bicarbonate is considered a good substitution for which one of the following chemicals?a. limeb. soda ashc. alumd. anhydrous ammonia

89. Biocatalysts are calleda. proteinsb. mitochondriac. enzymesd. catabolites

90. What is the beneficial concentration range of ammonia nitrogen in an anaerobic digester?a. 50 to 150 mg/Lb. 50 to 200 mg/Lc. 75 to 225 mg/Ld. 100 to 250 mg/L

91. What is the beneficial concentration range of magnesium in an anaerobic digester?a. 50 to 100 mg/Lb. 75 to 150 mg/Lc. 100 to 200 mg/Ld. 150 to 250 mg/L

92. If the supernatant from an anaerobic digester that is returning to the plant process is of poor quality and is causing plant upsets, what is the most probable cause?a. organic overload has occurredb. a toxic load has entered the digesterc. there is excessive mixing and settling time is insufficientd. sludge recirculation line is partially or completely plugged



93. If the odor of the supernatant is sour from either the primary or secondary anaerobic digester, what is the most probable cause?a. digester being overheatedb. recirculation of gas is excessivec. digester pH is too lowd. volatile-acid-to-alkalinity ratio has increased to 0.8

94. What is the most probable cause of foam in the supernatant from either the single stage or the primary tank?a. digester too stagnant as mixing is inadequateb. filamentous organisms are dying off by the billionsc. excessive gas recirculationd. digester being overheated

Odor Control95. At what pH will sodium hydroxide control sulfide generation for periods of about a

month or more?a. 9.5b. 10.5c. 11.5d. 12.5

96. What is the recommended organic loading rate in lb BOD/d/yd3 of media for an odor removal tower?a. 0.2b. 0.5c. 1.0d. 1.4

97. The best temperatures reported for combusting odorous gases from a wastewater plant begin at temperatures greater thana. 1,000°Fb. 1,200°Fc. 1,500°Fd. 1,800°F

98. How many times can activated carbon be regenerated before it is usually replaced?a. 3 timesb. 6 timesc. 9 timesd. 11 times

Phosphorus Removal99. A modified version of the activated sludge process for removing phosphorus and

nitrogen would have the following sequence of tank environments:a. anaerobic, anoxic, and aerobicb. aerobic, nitrification, and anaerobicc. anaerobic, anoxic, and denitrificationd. anaerobic, nitrification, aerobic, and denitrification



100. To what should the pH be raised for converting NH4+ to gaseous ammonia (NH3)?a. 9.4 to 9.8b. 9.5 to 10.5c. 10.5 to 11.5d. 11.0 to 11.3

101. The clinoptilolite used in ion exchange for nitrogen removal is regenerated witha. HClb. NaOHc. KMnO4d. NaCl

Enhanced Biological (Nutrient) Control102. In the enhanced-biological-nutrient-control activated sludge system for the process

control of phosphorus removal, what are the primary controlling variables for the desired microbial population?a. volatile fatty acidsb. soluble BOD5 and phosphatec. pH and alkalinityd. oxidized nitrogen and DO

103. In the four-stage Bardenpho process, ammonia from the influent is converted to nitrate in which area?a. anoxic zoneb. aeration basinc. anaerobic zoned. aerobic zone

104. What is the theoretical amount of oxygen in milligrams required to convert 1 mg of ammonia nitrogen to nitrate nitrogen?a. 2.5 mgb. 3.3 mgc. 4.6 mgd. 7.1 mg

105. What is the operating guideline in the four-stage Bardenpho process for the mean cell residence time?a. 15 to 20 daysb. 20 to 25 daysc. 20 to 30 daysd. 20 to 40 days

106. The optimum pH range in the enhanced chemical nutrient control reactions for phosphorus removed by Fe3+ addition isa. 4.0 to 6.0b. 4.5 to 6.0c. 4.5 to 6.5d. 5.0 to 6.5



107. What is the desired DO level for an anaerobic selector?a. 0.05 mg/Lb. 0.10 mg/Lc. 0.15 mg/Ld. 0.20 mg/L

108. Which one of the following types of foam is the most persistent and most difficult to control?a. solids from the dewatering processb. filamentous or Nocardia organismsc. polymer overdosingd. nutrient deficiencies

Disinfection109. In the reaction of hydrogen sulfide with chlorine to form sulfuric acid, it takes one part

hydrogen sulfide for every _______ parts of chlorine.a. 2.5b. 4.5c. 6.5d. 8.5

110. Dichloramine will be present in wastewaters at pH values below what?a. 5.5b. 5.0c. 4.7d. 4.3

111. What is the recommended withdrawal of chlorine from a 100- or 150-lb cylinder for each degree Fahrenheit ambient temperature such that freezing does not occur?a. 1 lbb. 2 lbc. 3 lbd. 4 lb

112. What chlorine emergency repair kit is used to stop leaks only at the valves of tank cars?a. emergency repair kit Ab. emergency repair kit Bc. emergency repair kit Cd. emergency repair kit T

113. If a liquid sulfonator was misting, what would most likely be the cause?a. loss of water pressureb. wrong size injector valvec. dirty injector valved. defective evaporator



114. To prevent reliquefaction of sulfur dioxide gas in a conducting pipe, what should the temperature range of the room be?a. 60 to 80°Fb. 70 to 85°Fc. 80 to 90°Fd. 90 to 100°F

Solids Removal from Secondary Effluents115. If the floc is too small in the solids removal physical–chemical treatment process, what

is the most probable cause?a. paddle speed of flocculators too slowb. not enough dilution waterc. improper or misaligned bafflingd. change in pH

116. If the floc is too small in the solids removal physical–chemical treatment process, what is the most plausible solution?a. increase paddle speed of first flocculatorb. add a coagulant aidc. check dosage with jar testd. increase detention time by pumping out sludge

117. If the floc is too large and thus settles too soon in the solids removal physical–chemical treatment process, what is the most probable cause?a. too much dilution waterb. chemical makeup is too strongc. excessive addition of coagulant aidd. cross-collectors not functioning

118. If the floc is too large and thus settles too soon in the solids removal physical–chemical treatment process, what is the most plausible solution?a. inspect spiral screwb. neutralize the pHc. decrease paddle speed in rapid mix unitd. optimize location of coagulant aid addition

119. In the solids removal physical–chemical treatment process, if there is floating sludge, what is the most probable cause?a. floc is not settlingb. change in the influent wastewater characteristicsc. cross-collectors not functioningd. coagulant aid dosage too low

120. In the solids removal physical–chemical treatment process, if there is floating sludge, what is the most plausible solution?a. do a jar test and adjust chemicalsb. adjust bafflingc. check scum collecting deviced. neutralize pH of wastewater



121. In the solids removal physical–chemical treatment process, if there is a loss of solids over the effluent weir, what is the most probable cause?a. sludge collectors malfunctioningb. improper chemical dosagec. cross-collectors not functioningd. sludge blanket too deep

122. In the solids removal physical–chemical treatment process, if the sludge is thin with a deep sludge blanket, what is the most probable cause?a. improper or misaligned bafflingb. cross-collectors not functioning properlyc. not enough coagulantd. excessive coagulant dosage

123. In the solids removal physical–chemical treatment process, if the sludge collector has a jerky operation, what is the most probable cause?a. misaligned baffleb. broken bafflec. sludge blanket too deepd. board on cross-collector has broken off and fallen into area of sludge collector

124. In the solids removal physical–chemical treatment process, if the sludge collector has a jerky operation, what is the most plausible solution?a. align baffling to proper depthb. pump out excess sludgec. repair broken bafflesd. repair cross-collector

125. It is recommended that filter system alarms should be tested everya. 30 daysb. 60 daysc. 90 daysd. 6 months

126. What should the head loss on a filter be after a proper backwash?a. 0.5 ftb. 1.0 ftc. 2.0 ftd. 2.5 ft

127. If a wastewater plant with a filtration system has low suspended solids in the applied water, but solids are still passing through the filter, what is the most plausible solution?a. backwash at a lower head lossb. add chlorine to applied water to oxidize suspended solidsc. place more filters in serviced. backwash more frequently



128. If a wastewater plant with a filtration system has air binding, what is the most plausible solution?a. backwash more frequentlyb. backwash at a lower head lossc. backwash less frequentlyd. place more filters in service to reduce hydraulic load on problem filter(s)

129. If a wastewater plant with a filtration system has high BOD and COD, what is the most plausible solution?a. change type of filter aidb. increase coagulant dosagec. increase lime at secondary clarifierd. chlorinate applied water

130. Media loss from a filter in any four-month period is considered to be excessive if the amount of loss is greater thana. 2 cmb. 4 cmc. 5 cmd. 10 cm

Residual Solids Management131. If an aerobic digester has poor sludge settling, what is the most probable cause?

a. excessive turbulenceb. filamentous growthc. air rates too highd. digestion time too high or too low

132. What percentage of fecal coliforms and streptococci will be destroyed when sludge is stabilized with lime?a. 90%b. 95%c. 96%d. 99%

133. What is the approximate curing or mixing time for lime used in residual solids management?a. 30 minb. 45 minc. 60 mind. 90 min

134. If a reactor in a thermal conditioning process cannot maintain its temperature, what is the most probable cause?a. feed pump is inoperativeb. inadequate makeup water supplyc. scaling too thick in reactord. operation of decant thickener poor



135. If a thermal conditioning process has a reduction in sludge dewatering ability, what is the most probable cause?a. operation of decant thickener is poorb. fuel exhaustedc. fuel pump inoperatived. temperature sensor inoperative

136. The heat exchanger for a thermal conditioning process has excessive pressure drop (∆p). What is the most likely solution?a. clean or repair or replace actuatorb. increase reactor temperaturec. acid flush the heat exchangerd. open suction valve or throttle discharge valve

137. If the inner portions of the cake of a plate and frame filter press are wet and the cake is messy on discharge, what is the most probable cause?a. precoat inefficienciesb. influent characteristics have changedc. low filtration timed. torn filter cloths

138. If a plate and frame press has cakes discharged that are wet throughout, what is the most probable cause?a. precoat inefficienciesb. chemical inefficienciesc. torn filter clothsd. plugged filter drains

139. If a belt filter press has poor filtrate quality, what is the most probable cause?a. sludge squeezed from beltb. polymer dosage is insufficientc. belt speed too highd. influent characteristics have changed

140. If a belt filter press has cake solids that are too wet, what is the most probable cause?a. belt tension too lowb. belt tension too highc. polymer dosage insufficientd. hydraulic load too high

141. What is the typical percent solids recovery from vacuum filters?a. 70 to 80%b. 80 to 87%c. 85 to 95%d. 94 to 98%

142. If a vacuum filter experiences a loss in vacuum, what is the most probable cause?a. trough is fullb. tear in filter mediac. clogged mediad. drum speed too low



143. If a vacuum filter has poor quality filtrate, what is the most probable cause?a. insufficient conditioningb. influent characteristics have changedc. vacuum pump failedd. clogged media

144. If a vacuum filter has wet cake and a poor discharge, what is the most probable cause?a. drum speed too lowb. trough too fullc. trough empty or too lowd. insufficient conditioning

145. If a vacuum filter has poor quality mat formation, what is the most probable cause?a. clogged mediab. vacuum on filter too lowc. trough too fulld. drum speed too low

146. What is the typical loading rate of dry sludge solids for open sand drying beds?a. 10 to 25 lb/yr/ft2

b. 10 to 35 lb/yr/ft2

c. 25 to 40 lb/yr/ft2

d. 30 to 45 lb/yr/ft2

147. In general, percent solids recovery from sand drying beds for primary, secondary, and combined digested sludge ranges froma. 80 to 88%b. 88 to 93%c. 93 to 97%d. 95 to 99%

148. If the dried product from a rotary dryer is too wet, what is the most probable cause?a. vacuum too lowb. overdosing of conditioning agentc. drum speed is too fastd. plugged feed port

149. If the dried product from a rotary dryer is too wet, what is the most likely solution?a. increase drum speedb. drum needs alignmentc. reduce polymer dosage and increase conditioning agentd. increase polymer dosage and reduce conditioning agent

150. If sludge balling is occurring in a rotary dryer, what is the most probable cause?a. low gas flowb. detention time too longc. drum speed too highd. polymer dosage too high



151. If sludge balling is occurring in a rotary dryer, what is the most likely solution?a. reduce the polymer doseb. increase vacuum applied to drumc. unplug outlet portsd. decrease drum speed

152. If a rotary dryer is experiencing vibrations, what is the most probable cause?a. incoming sludge too wetb. vacuum too highc. feed or outlet ports pluggedd. blending insufficient

153. What is the temperature range for a multiple-hearth furnace?a. 800 to 1,000°Fb. 1,000 to 1,200°Fc. 1,200 to 1,500°Fd. 1,500 to 1,700°F

154. The process used to describe the moving material inside a multiple-hearth furnace is calleda. plowingb. amalgamatingc. rabblingd. homogenizing

155. Ideally, the moisture content of sludge fed to a furnace should not exceeda. 50%b. 60%c. 70%d. 75%

156. Why would the flame inside a multiple-hearth furnace be bright and sharp?a. lack of air supplyb. there is an excess of airc. the air supply is in balance with the systemd. there is an adequate excess air supply, but not too much

157. What would a short and blue flame inside a multiple-hearth furnace indicate?a. lack of air supplyb. there is an excess of airc. the air supply is in balance with the systemd. there is an adequate excess air supply, but not too much

158. The shaft air temperature in a multiple-hearth furnace should not exceeda. 451°Fb. 550°Fc. 625°Fd. 735°F



159. Since the last sludge to burn in a multiple-hearth furnace can become very hot, it is important to control this burnout so as not to damage the furnace. What temperature increase in a hearth is ideal during a burnout?a. 75°Fb. 100°Fc. 150°Fd. 200°F

160. Once the hearth temperature on a multiple-hearth furnace reaches 1,000°F, how fast can the furnace be heated in °F/hr?a. 25°F/hrb. 50°F/hrc. 100°F/hrd. 125°F/hr

161. What type of landfill for wastes usually limits the annual application of the wastes to a particular area based on nitrogen requirement?a. on-site dedicated land disposal (DLD)b. agricultural reclamationc. sanitary landfilld. high-rate dedicated land disposal

Water and Wastewater Microbiology162. Where is the surface charge site of bacteria?

a. outer cell wallb. inner cell wallc. plasma membraned. capsule

163. What is the name of the site where protein synthesis in bacterial cells occurs?a. mesosomesb. deoxyribonucleic acid (DNA)c. mitochondriad. ribosomes

164. Microscopic crustaceans would most likely be used to clarify effluents froma. oxidation pondsb. aeration tanksc. secondary clarifiersd. aerobic digesters

165. Along with nutrients anaerobic bacteria need mostlya. carbon dioxideb. organic matterc. hydrogen sulfided. volatile acids



Basic Chemistry166. What chemical is used to prevent the interference of sulfide, sulfite ions, and

thiosulfate with the chloride test?a. 1 mL of 20% sodium bicarbonateb. 2 mL of 0.5N sodium thioiodatec. 1 mL of 30% H2O2d. 0.5 mL of 10% aluminum hydroxide

167. How do you convert degrees Celsius to degrees absolute?a. multiply degrees Celsius by 3.26b. add 200°Cc. add 273°Cd. multiply degrees Celsius by 1.8

168. What is the maximum number of electrons in the first energy shell (“k” shell) of an atom?a. one electronb. two electronsc. four electronsd. eight electrons

169. Ionic compounds are formed whena. an acid is added to a mixtureb. an acid is added to a compoundc. a transfer of electrons occursd. electrons are shared

170. A double replacement reaction is the same asa. a composition reactionb. a decomposition reactionc. an ionic reactiond. a redox reaction

171. All replacement reactions area. redox reactionsb. ionic reactionsc. composition reactionsd. decomposition reactions

172. All acids area. hydrophilic compoundsb. polar covalent compoundsc. amphiprotic compoundsd. dipolar ionic compounds

173. When a base reacts with an oxide of a nonmetal, it will producea. a saltb. an acidc. a hydroxided. binary acid



174. Amphiprotic compounds can best be described as compounds thata. behave as electron or proton acceptors, but never both at the same time when

reacting with another substanceb. accept two or more protons when reacting with another substancec. behave as proton acceptors or proton donators depending on the substances with

which they are reactingd. accept at least one electron and one proton when reacting with another substance

175. When anhydrides react with water, they producea. bases and oxidesb. acids and hydroxidesc. oxides and acidsd. acids or bases

176. What are basic anhydrides?a. oxides of metalsb. hydroxide of metalsc. oxides of nonmetalsd. hydroxide of nonmetals

177. Soluble oxides of metals in water willa. cause acidityb. cause alkalinityc. produce amphiprotic compoundsd. produce neutral compounds

178. Gram equivalent weight of an element is thea. gram equivalent weight of solute per liter of solutionb. gram atomic weight per oxidation numberc. moles of solute per kilogram of solventd. moles of solute per liter of solution

179. Which one of the following would increase in solubility as the temperature decreases?a. magnesium carbonateb. chlorine gasc. iron hydroxided. sulfuric acid

180. What will the attractive forces between solute particles cause?a. precipitationb. the formation of a colloidal suspensionc. lower solubilityd. a nonelectrolyte solution

181. Colloids range in size froma. 0.01 to 25 nmb. 1 to 100 nmc. 50 to 250 nmd. 200 to 1,000 nm



182. What is the usual effective pH range of iron salt coagulants?a. 3.5 to 9.0b. 6.5 to 8.8c. 3.0 to 9.5d. 4.2 to 9.0

183. What type of polar bond is there between the hydrogen atoms and the oxygen atom in water?a. ionicb. covalentc. van der Waal’sd. hydrogen bonding

184. When oxides of nonmetals react with water, they forma. acidsb. hydroxides and hydrogen gasc. hydroxidesd. a base and an acid

185. The number of water molecules needed for an ion to be dissociated depends on thea. pH and temperature of the waterb. pH and the type of ion in questionc. charge and size of the iond. pH of the water and charge on the ion

186. Which gas law states that the mass of a dissolved gas in a liquid is directly proportional to the pressure it exerts in the liquid at a given temperature?a. Graham’s lawb. Henry’s lawc. Charles’ lawd. Boyle’s law

187. If one of the hydrogen atoms of a hydrocarbon is replaced with a carboxylic (COOH) group, it is calleda. an esterb. an acidc. an etherd. a ketone

188. The ratio of hydrogen to oxygen atoms in all carbohydrates is always:a. 2 to 1, respectivelyb. 3 to 1, respectivelyc. 4 to 1, respectivelyd. there is no set ratio

189. Simple proteins are made up ofa. amino acidsb. carbohydrates and aminesc. deoxyribonucleic acidd. monosaccharides



190. Permanent hardness isa. magnesium and calcium carbonate and sulfates of magnesium and calciumb. the same as noncarbonated hardnessc. magnesium and calcium carbonated. the same as carbonate hardness

191. Natural zeolites that have become exhausted with use are regenerated by immersing them in a strong solution ofa. NaClb. NaOHc. HCld. H2SO4

192. Ferrous iron commonly found in well water can be removed with lime by increasing the pH to at leasta. 8.4b. 8.7c. 9.2d. 9.4

193. With what should zeolites be pretreated for the removal of iron?a. MnSO4b. NaOHc. KMnO4d. H2SO4

194. What should zeolites be pretreated with for the removal of manganese?a. MnSO4b. NaOHc. KMnO4d. H2SO4

195. What is the chemical makeup of a tubercle?a. Fe(OH)2b. Fe(OH3)c. Fe(OH)2 on the inside and Fe(OH3) on the outsided. Fe(OH)2 on the outside and Fe(OH3) on the inside

196. Which one of the following types of organisms is hardest to kill?a. cholerab. typhoidc. Cryptosporidiumd. infectious hepatitis

197. One of the disadvantages of using chloramines is that chloramines can inducea. blood clots in the kidneysb. hemolytic anemiac. amyloidosis kidney diseased. renal dysplasia



198. If air is used to generate ozone, what percentage of the air is usually converted?a. 1 to 3%b. about 4%c. 2 to 6%d. 9 to 11%

199. How much more soluble is ozone in water than oxygen in water?a. 7.6 timesb. about 10 timesc. 12 timesd. 20 times


139


General Wastewater Treatment Principles1. Answer: b. 0.3 ppm

Reference: Water Treatment, AWWA, Principles and Practices of Water Supply Operations Series, Second Edition, Chapter 7, Page 182.

2. Answer: c. negative pole of an electrolytic cell or systemReference: Water Treatment, AWWA, Principles and Practices of Water Supply Operations Series, Second Edition, Chapter 9, Page 266, and Glossary, Page 487.

3. Answer: c. cathodic metalsReference: Water Treatment, AWWA, Principles and Practices of Water Supply Operations Series, Second Edition, Chapter 9, Table 9-2, Page 269.

4. Answer: d. 160°F (71°C)Reference: Water Transmission and Distribution, AWWA, Principles and Practices of Water Supply Operations Series, Second Edition, Chapter 12, Page 388.

5. Answer: b. suction liftReference: Basic Science Concepts and Applications, AWWA, Principles and Practices of Water Supply Operations Series, Second Edition, Hydraulics 6, Figure H6-8B, Page 282.

6. Answer: c. 25-ft intervalsReference: Title 29 CFR, Part 1926, Subpart P.

7. Answer: d. amount of hydrogen ions releasedReference: Basic Science Concepts and Applications, AWWA, Principles and Practices of Water Supply Operations Series, Second Edition, Page 423.

8. Answer: b. pressure-reliefReference: Water Distribution Operator Training Handbook, AWWA, Second Edition, Page 64.

9. Answer: b. decomposition of a material caused by an outside electric currentReference: Water Transmission and Distribution, AWWA, Principles and Practices of Water Supply Operations Series, Second Edition, Page 248.

Sedimentation and Flotation10. Answer: d. coagulants or chlorine




11. Answer: b. an approved sanitary landfillReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 5, Page 120.

12. Answer: b. 7.0Reference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 5, Page 149.

Activated Sludge13. Answer: b. pumping rate for the return of activated sludge is out of balance


14. Answer: c. reduce the aeration ratesReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 8, Page 265.

15. Answer: d. too long a detention timeReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 8, Page 266.

16. Answer: b. water spraysReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 8, Page 266.

17. Answer: d. ½ of the jar depthReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 8, Page 266.

18. Answer: a. 2 hrReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 8, Page 267.

19. Answer: d. the lack of bacteria buildupReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 8, Page 273.

20. Answer: c. possible rate of returned activated sludgeReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 8, Page 273.

21. Answer: a. 0.5 to 2.0 mg/LReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 8, Page 274.

22. Answer: d. the formation of pinpoint flocReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 8, Page 274.

23. Answer: b. increase the return sludge rateReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 8, Page 274.



24. Answer: c. wasting rate should be increasedReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 8, Page 274.

25. Answer: c. shut down one or more rotor assembliesReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 8, Page 277.

26. Answer: c. 30 to 60 minReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 40.

27. Answer: a. 3 in.Reference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 45.

28. Answer: a. 10 to 15%Reference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 55.

29. Answer: c. 0.55 lbReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 57.

30. Answer: b. a little less than the theoretical wasting rateReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 57.

31. Answer: a. decrease the percent volatile matterReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 61.

32. Answer: a. 1 lb of food per 5 lb of organismsReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 70.

33. Answer: d. 10°FReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 71.

34. Answer: b. septicity; denitrificationReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 73.

35. Answer: c. increase the F/M (food-to-microorganism) ratioReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 73.

36. Answer: d. aeration tanks have unequal flow distributionReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Pages 74 and 75.



37. Answer: d. the sludge consists mainly of actinomycetesReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Pages 74 through 76.

38. Answer: b. increase sludge return rateReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 74.

39. Answer: b. get the sludge return rate as high as possibleReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 74.

40. Answer: a. moderately increase sludge wasting to reduce the MCRTReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 75.

41. Answer: c. increase MCRT and MLSS by decreasing sludge wastingReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 75.

42. Answer: d. low DO or high F/M ratioReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 117.

43. Answer: b. low DO or low F/M ratioReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 117.

44. Answer: a. small MastigophoraReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 119.

45. Answer: c. rotifersReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 120.

46. Answer: c. filamentsReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 120.

47. Answer: c. F/M ratio is too lowReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 120.

48. Answer: c. has low F/M ratio and high MCRTReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 123.

49. Answer: c. reaeration detention time would be reducedReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 127.



50. Answer: c. because the water is motionless just before it is decantedReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 103.

51. Answer: c. increase the reaction and settling timeReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 104.

52. Answer: b. 1 to 2 weeksReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 106.

53. Answer: a. mixing of the mixed liquorReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 11, Page 130.

54. Answer: b. nitrification stage of separate stage nitrificationReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 2, Page 68.

55. Answer: c. the F/M ratio has less variabilityReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 2, Page 70.

56. Answer: b. 15 daysReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 2, Page 80.

57. Answer: b. < 5%Reference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 2, Page 80.

58. Answer: c. biological organisms are not healthy or stableReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 2, Page 87.

59. Answer: d. organic matter is undergoing rapid biodegradationReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 2, Page 88.

60. Answer: d. detergentsReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 2, Page 101.

61. Answer: c. chlorinated hydrocarbonsReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 2, Page 101.

62. Answer: a. 2,000 mg/LReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 2, Page 105.



63. Answer: c. 10%Reference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 2, Page 106.

64. Answer: a. 10 to 20%Reference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 2, Page 102.

65. Answer: c. to maintain the nitrification population essential for ammonia oxidationReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 2, Pages 136 and 137.

66. Answer: d. hydrogen peroxideReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 2, Page 138.

67. Answer: d. 7.5 to 8.5Reference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 2, Page 141.

68. Answer: a. 60 to 95°FReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 2, Page 142.

69. Answer: c. BOD5 levelReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 2, Page 142.

Trickling Filters70. Answer: d. food processing


71. Answer: b. 5 lbReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 6, Page 183.

72. Answer: a. psycodaReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 6, Page 187.

73. Answer: b. stormwaterReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 6, Page 189.

74. Answer: b. chlorineReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 6, Page 190.

75. Answer: a. acetic acidReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 6, Page 190.



76. Answer: c. NaOHReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 6, Page 190.

77. Answer: d. check plant influent and make adjustmentsReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 6, Page 191.

Rotating Biological Contactors78. Answer: d. 7.1 times the influent ammonia


79. Answer: d. misalignmentReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 7, Page 233.

Waste Treatment Ponds80. Answer: b. anaerobic fermentation


81. Answer: b. blue-green algaeReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 9, Pages 303 and 304.

82. Answer: c. chironomid midgesReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 9, Page 306.

83. Answer: b. very little foam is on the surfaceReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 9, Page 312.

84. Answer: d. grayReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 9, Page 315.

85. Answer: c. because the pH is too lowReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 9, Page 319.

Sludge Digestion and Solids Handling86. Answer: c. add less solids loading than lime


87. Answer: a. limeReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 186.



88. Answer: a. limeReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 186.

89. Answer: c. enzymesReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 187.

90. Answer: b. 50 to 200 mg/LReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 205.

91. Answer: c. 100 to 200 mg/LReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Page 205.

92. Answer: c. there is excessive mixing and settling time is insufficientReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Pages 206 and 207.

93. Answer: c. digester pH is too lowReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Pages 206 and 207.

94. Answer: c. excessive gas recirculationReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 2, Seventh Edition, Chapter 12, Pages 206 and 207.

Odor Control95. Answer: d. 12.5


96. Answer: b. 0.5Reference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 1, Page 18.

97. Answer: c. 1,500°FReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 1, Page 21.

98. Answer: a. 3 timesReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 1, Page 34.

Phosphorus Removal99. Answer: a. anaerobic, anoxic, and aerobic




100. Answer: c. 10.5 to 11.5Reference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 5, Page 519.

101. Answer: d. NaClReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 5, Page 519.

Enhanced Biological (Nutrient) Control102. Answer: d. oxidized nitrogen and DO


103. Answer: b. aeration basinReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 7, Page 556.

104. Answer: c. 4.6 mgReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 7, Page 558.

105. Answer: c. 20 to 30 daysReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 7, Page 560.

106. Answer: a. 4.0 to 6.0Reference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 7, Page 561.

107. Answer: b. 0.10 mg/LReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 7, Page 563.

108. Answer: b. filamentous or Nocardia organismsReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 7, Page 563.

Disinfection109. Answer: d. 8.5


110. Answer: a. 5.5Reference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 10, Page 349.

111. Answer: a. 1 lbReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Fourth Edition, Chapter 10, Page 396.



112. Answer: c. emergency repair kit CReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 10, Page 406.

113. Answer: d. defective evaporatorReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 10, Page 414.

114. Answer: c. 80 to 90°FReference: Operation of Wastewater Treatment Plants, Kenneth D. Kerri, Volume 1, Sixth Edition, Chapter 10, Page 421.

Solids Removal from Secondary Effluents115. Answer: d. change in pH


116. Answer: c. check dosage with jar testReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 4, Page 390.

117. Answer: b. chemical makeup is too strongReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 4, Page 390.

118. Answer: d. optimize location of coagulant aid additionReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 4, Page 390.

119. Answer: b. change in the influent wastewater characteristicsReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 4, Page 390.

120. Answer: a. do a jar test and adjust chemicalsReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 4, Page 390.

121. Answer: b. improper chemical dosageReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 4, Page 390.

122. Answer: b. cross-collectors not functioning properlyReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 4, Page 390.

123. Answer: c. sludge blanket too deepReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 4, Page 390.

124. Answer: b. pump out excess sludgeReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 4, Page 390.



125. Answer: b. 60 daysReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 4, Page 402.

126. Answer: a. 0.5 ftReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 4, Page 403.

127. Answer: c. place more filters in serviceReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 4, Page 405.

128. Answer: b. backwash at a lower head lossReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 4, Page 405.

129. Answer: d. chlorinate applied waterReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 4, Page 406.

130. Answer: c. 5 cmReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 4, Page 406.

Residual Solids Management131. Answer: d. digestion time too high or too low



133. Answer: a. 30 minReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 227.

134. Answer: b. inadequate makeup water supplyReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 232.

135. Answer: a. operation of decant thickener is poorReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 232.

136. Answer: c. acid flush the heat exchangerReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 232.

137. Answer: c. low filtration timeReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 241.



138. Answer: b. chemical inefficienciesReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 241.

139. Answer: a. sludge squeezed from beltReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 245.

140. Answer: a. belt tension too lowReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 245.

141. Answer: c. 85 to 95%Reference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 251.

142. Answer: b. tear in filter mediaReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 252.

143. Answer: a. insufficient conditioningReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 252.

144. Answer: d. insufficient conditioningReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 252.

145. Answer: a. clogged mediaReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 252.

146. Answer: b. 10 to 35 lb/yr/ft2


147. Answer: d. 95 to 99%Reference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 255.

148. Answer: a. vacuum too lowReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 270.

149. Answer: a. increase drum speedReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 270.

150. Answer: d. polymer dosage too highReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 270.



151. Answer: a. reduce the polymer doseReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 270.

152. Answer: c. feed or outlet ports pluggedReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 270.

153. Answer: d. 1,500 to 1,700°FReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 271.

154. Answer: c. rabblingReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 271.


156. Answer: b. there is an excess of airReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 282.

157. Answer: d. there is an adequate excess air supply, but not too muchReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 282.

158. Answer: b. 550°FReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 283.

159. Answer: b. 100°FReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 284.

160. Answer: c. 100°F/hrReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 285.

161. Answer: b. agricultural reclamationReference: Advanced Waste Treatment, Kenneth D. Kerri, Fifth Edition, Chapter 3, Page 296.

Water and Wastewater Microbiology162. Answer: c. plasma membrane


163. Answer: d. ribosomesReference: Water and Wastewater Treatment Plant Operations, Frank R. Spellman, Chapter 11, Page 312.



164. Answer: a. oxidation pondsReference: Water and Wastewater Treatment Plant Operations, Frank R. Spellman, Chapter 11, Page 316.

165. Answer: d. volatile acidsReference: Water and Wastewater Treatment Plant Operations, Frank R. Spellman, Chapter 11, Page 328.

Basic Chemistry166. Answer: c. 1 mL of 30% H2O2


167. Answer: c. add 273°CReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 2, Page 10.

168. Answer: b. two electronsReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 3, Page 17.

169. Answer: c. a transfer of electrons occursReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 4, Page 26.

170. Answer: c. an ionic reactionReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 5, Page 42.

171. Answer: a. redox reactionsReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 5, Page 44.

172. Answer: b. polar covalent compoundsReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 6, Page 47.

173. Answer: a. a saltReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 6, Page 50.

174. Answer: c. behave as proton acceptors or proton donators depending on the substances with which they are reactingReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 6, Page 50.

175. Answer: b. acids and hydroxidesReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 6, Page 51.



176. Answer: a. oxides of metalsReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 6, Page 51.

177. Answer: b. cause alkalinityReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 6, Page 51.

178. Answer: b. gram atomic weight per oxidation numberReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 7, Page 59.

179. Answer: b. chlorine gasReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 7, Page 65.

180. Answer: c. lower solubilityReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 7, Page 65.

181. Answer: b. 1 to 100 nmReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 8, Page 67.

182. Answer: a. 3.5 to 9.0Reference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 8, Page 74.

183. Answer: b. covalentReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 9, Page 78.

184. Answer: a. acidsReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 9, Page 79.

185. Answer: c. charge and size of the ionReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 10, Page 84.

186. Answer: b. Henry’s lawReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 11, Page 93.

187. Answer: b. an acidReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 12, Page 108.

188. Answer: a. 2 to 1, respectivelyReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 12, Page 109.



189. Answer: a. amino acidsReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 12, Page 115.

190. Answer: b. the same as noncarbonated hardnessReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 13, Page 120.

191. Answer: a. NaClReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 13, Page 123.

192. Answer: d. 9.4Reference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 13, Page 125.

193. Answer: a. MnSO4


194. Answer: c. KMnO4


195. Answer: c. Fe(OH)2 on the inside and Fe(OH3) on the outsideReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 14, Page 128.

196. Answer: c. CryptosporidiumReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 15, Page 137.

197. Answer: b. hemolytic anemiaReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 15, Page 148.

198. Answer: a. 1 to 3%Reference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 15, Page 151.

199. Answer: d. 20 timesReference: Basic Chemistry for Water and Wastewater Operators, Darshan Singh Sarai, Revised Edition, Chapter 15, Page 151.


155

Grades 1–4 Pumping Questions

Grade 1 Pumping QuestionsBasic Water Hydraulics1. The amount of energy in feet required to overcome resistance to flow in a pipe is called

a. pump headb. cut-off headc. pressure headd. friction head

2. The amount a liquid can be raised vertically by a given pressure is calleda. pressure headb. total headc. velocity headd. pump head

3. The amount of energy in feet that a pump supplies to a fluid is calleda. velocity headb. pump headc. total headd. pressure head

4. How many inches is a total vacuum at sea level?a. 14.7 in.b. 30 in.c. 34 in.d. 39.6 in.

5. How many feet of water will equal the atmospheric pressure at sea level?a. 28 ftb. 30 ftc. 32 ftd. 34 ft

Introduction to Centrifugal Pumps6. The most widely used pump type in wastewater is

a. centrifugalb. rotaryc. diaphragmd. peripheral



7. What is used to stop air leakage into the casing around a pump shaft?a. packing glandb. lantern ringc. sealsd. shaft sleeves

8. What assembly holds the lantern ring and packing?a. shaft assemblyb. casing ring assemblyc. packing gland casingd. stuffing box

9. The lantern ring is also known as thea. packing ringb. seal cagec. packing seald. mechanical seal

10. What device applies an even pressure to the packing such that it compresses tight around the pump shaft?a. lantern ringb. mechanical sealc. packing glandd. seal cage

Centrifugal Pump Components11. Which centrifugal pump impeller type would usually be used for pumping medium-

sized solids?a. openb. closedc. semiopend. radial

12. Which centrifugal pump impeller type would be used for pumping large solids?a. openb. closedc. semiopend. radial

13. Which centrifugal pump impeller type would typically be used for pumping clear wastewater?a. openb. closedc. semiopend. radial


GRADES 1–4 PUMPING QUESTIONS 157

14. What part of a centrifugal pump has the function of separating the high-pressure zone in the volute from the low-pressure zone at the impeller’s eye?a. wear ringsb. packing glandc. shaft sleevesd. slinger ring

15. What is at the top of a stuffing box?a. packing glandb. lantern ringc. mechanical seald. seal cage

Centrifugal Pumps—Theory16. What part of a pump houses the packing or mechanical seal?

a. the shroudb. the stuffing boxc. the voluted. the casing head

17. Velocity (V) head is expressed mathematically asa. V/gb. V2/gc. V2/2gd. V2/32.2 ft/s2

18. When water from a dam flows through turbines, it converts elevated head intoa. pressure headb. mechanical energyc. electrical energyd. potential energy

19. What is the only type of pump that can be operated against a closed valve?a. vertical turbine pumpb. centrifugal pumpc. axial-flow pumpd. mixed-flow pump

Pump Nomenclature20. The purpose of the lantern ring is to

a. distribute lubrication to the packingb. prevent water from getting into the stuffing boxc. seal the low-pressure zone from the high-pressure zoned. prevent water from getting between the shaft and shaft sleeves



21. What device serves the same function as the packing?a. in-line suction glandb. packing glandc. mechanical seald. lantern seal

22. The bearings are protected from water leaking out of the stuffing box by thea. shaft sleeveb. lantern ringc. slinger ringd. packing gland

23. The “heart” of a pump is called thea. volute caseb. impellerc. motord. pump

Mechanical Seals24. The main purpose of mechanical seals is to

a. keep lubrication in and dirt and other foreign materials outb. control water leakage from the stuffing boxc. keep contamination from entering or leavingd. save on costs, as mechanical seals last longer than packing

Grades 2–4 Pumping QuestionsBasic Water Hydraulics1. Hydraulics is the study of

a. fluid pressure in pipes or conduitsb. the force of fluids in motionc. the pressure of fluids in motiond. fluids in motion and at rest

2. What is it called if the energy supplied to a pump is equal to the energy required for the liquid to be moved to a specific point?a. static headb. pump headc. cut-off headd. total head

3. The arrangement of pumps in series is primarily used to increase thea. flow, usually by 70–80%b. discharge headc. flow but keep discharge head the samed. pump efficiency



4. The arrangement of pumps in parallel is typically useda. when head is not sufficient and more flow is requiredb. to increase discharge head without increasing flowc. to increase flow but keep discharge head the samed. to increase pump efficiency

5. What type of centrifugal pump should be used when pumping wastewater with large solids?a. single-suction with closed impellersb. double-suction with closed impellersc. single-suction with open or semiopen impellersd. double-suction with open impellers

6. What type of pump would be most likely used for chemical solutions?a. centrifugalb. diaphragmc. peripherald. pneumatic ejector

Centrifugal Pumps—Theory7. An object doing work by virtue of its motion is called

a. kinetic energyb. mechanical energyc. potential energyd. velocity energy

8. Water behind a dam and above a water treatment plant has energy by virtue of its elevation. This difference in elevation is called elevation head ora. kinetic energyb. velocity headc. potential energyd. pressure head

9. Open impellers are usually used to pump water that containsa. any chemicals except acidsb. alkaline chemicals onlyc. viscous polymersd. large solids

10. Which type of centrifugal pump can have water enter from two sides (double suction)?a. end suction pumpb. split-case pumpc. vertical turbine pumpd. close coupled pump

11. Which type of impeller is best to use for pumping water with low volumes of solids?a. closedb. openc. semiopend. radial



Centrifugal Pumps—Types12. What type of pump has the ability to rotate its discharge 360°?

a. vertical pumpsb. split-case pumpc. end suction pumpd. jet pumps

13. A split-case pump has three impellers. What type of multistage pump is this?a. one stageb. two stagec. three staged. six stage

14. A split-case pump has two equal smaller impellers placed on either side of two equal-size large impellers. How many stages does this pump have?a. one stageb. two stagesc. four stagesd. eight stages

Pump Nomenclature15. The lowest pressure point in the pump is

a. the center of the impellerb. the outermost part of the impellerc. suction side of the pumpd. discharge side of the pump

16. Sleeve bearings that are used in a vertical pump’s column may be lubricated bya. lithiumb. unsealed grease-packed bearingsc. sealed grease-packed bearingsd. oil drip system

17. The shaft’s main function is to transmit _______ from the motor to the impeller.a. centrifugal forceb. torquec. kinetic energyd. thrust

18. The purpose of roller bearings is toa. increase torqueb. increase thrustc. reduce shaft erosion corrosiond. reduce friction and heat

19. A horizontal load parallel to a pump shaft is called aa. shear loadb. thrust loadc. radial loadd. stress load



20. A load, perpendicular to a pump shaft, applied to a bearing is calleda. a shear loadb. a thrust loadc. a radial loadd. axial

21. Which one of the following is not a component of a bearing?a. rolling pinb. inner and outer racesc. rolling elementd. cage

22. All of the following cause bearing failure, but which one is the major cause of bearing failure?a. poor lubrication practicesb. contaminationc. vibrationd. faulty mounting

Replacement of Pump Components on End Suction Centrifugal Pumps23. When removing the volute case at the gasket from a pump, you should use

a. hammer and screwdriversb. chiselsc. soft-faced hammerd. screwdrivers or chisels

24. The impeller of a pump is prevented from turning on the shaft by aa. lock nut on threaded shaftb. keyc. steel pind. caliper pin

25. How should a pump’s shaft be stored when not in use?a. stored vertically, on endb. stored horizontally in foam or other soft materialc. stored in a shaft box and coated with a light oil to prevent corrosiond. stored horizontally in a long box on foam that is saturated with a light oil

26. The purpose of the shaft sleeve is to preventa. erosion of the shaftb. wear at the stuffing boxc. corrosion of the shaftd. wear at the eye of the impeller

27. Before installing casing wear rings, they should bea. lubricated with a light oilb. greased with lithiumc. cooledd. heated



28. To ease installation of impeller wear rings, they can bea. lubricated with a light oilb. greased with lithiumc. heatedd. cooled

29. Before reinstalling a shaft sleeve or impeller, it should be cleaned witha. alcoholb. acetonec. mild detergentd. kerosene

30. What should be used to clean an old bearing?a. solventb. mild detergentc. kerosened. ethyl alcohol

Selection and Replacement of Packing31. The purpose of packing is to

a. keep oil or graphite on the shaftb. control water leakage along the pump’s shaftc. prevent water leakage from the pump’s shaftd. help prevent shaft from warping

32. Most pump packing is composed of _______ with _______ lubrication.a. asbestos base, graphiteb. flax, polytetrafluoroethylene (PTFE)c. PTFE, inert oild. aluminum or copper, inert oil

33. The packing used for medium pressures and temperatures is most likelya. aluminumb. plant fiberc. asbestosd. PTFE

34. Packing used in high-speed pumps and temperatures greater than 450°F should be made ofa. asbestosb. metal braided with asbestosc. 100% PTFEd. synthetics

35. New packing rings should be staggered ata. 45 degreesb. 90 degreesc. 120 degreesd. 180 degrees



36. Packing is designed toa. add lubricant to the shaftb. expand and deteriorate with normal usec. protect the shaftd. wear and deteriorate with normal use

37. Packing replacement is usually performed whena. water leakage sprays out of the pump housingb. no further tightening can be done on the packing glandc. the packing gland bolts are exposed by more than 2½ in. above the nutd. the packing has completely disintegrated

Pump Piping System38. Seal water used to cool and flush mechanical seals should originate from an external

source for all the following circumstances except:a. when chemicals and sewage, muddy water, or other dirty water are being pumpedb. when discharge pressure is below 10 psic. when suction lift is greater than 15 psid. when potable water is being pumped

39. When external seal water is used as a water source, what should its pressure be in relation to the volute pressure?a. 25 to 30 psi higherb. 20 to 25 psi higherc. 10 to 15 psi higherd. 3 to 10 psi higher

End Suction and Split-Case Centrifugal Pumps40. What type of coupling is used when the coupled shafts are of two different sizes?

a. split couplingb. flexible disk couplingc. flange couplingd. jaw coupling

41. This type of rigid coupling is easy to install and remove.a. split couplingb. jaw couplingc. gear couplingd. flexible disk coupling

42. Which one of the following statements is false?a. most pumps are self-primingb. bearings can be overlubricatedc. fire point is defined as the point at which a lubricant will burnd. oil is used in high rpm conditions while grease is used in low speeds and heavy

loads



43. How often should amperage and voltage tests be performed on three-phase motors?a. monthlyb. at least every 3 monthsc. at least every 6 monthsd. at least once per year

44. The purpose of lubrication is toa. remove heat and reduce frictionb. prevent corrosionc. reduce weard. increase torque

45. Grease that is water resistant is made witha. graphiteb. animal fatc. calciumd. potassium phosphate

46. What is the approximate reduction in oil’s “life” for each rise in temperature of 20°F over 140°F?a. 15%b. 33%c. 50%d. 60%

Line Shaft Turbine Operating Conditions47. Lubrication used on bearings of line shaft turbines can be

a. oil or waterb. grease or oilc. lithium or greased. graphite or grease

48. Excessive upthrust on line shaft turbines is caused bya. misalignment of shaftb. pumping too much waterc. impeller that is too small or wornd. vertical clearance between bottom of pump bowl and impeller that is too small

Special Pumping Units49. Under what conditions are regenerative turbine pumps used?

a. high head conditionsb. low head conditionsc. high-volume output conditionsd. low suction conditions

50. The screw pump is most effective for lifts that are less thana. 30 ftb. 32 ftc. 34 ftd. 40 ft



51. Approximately how much flow increase in percent occurs when a two-flight screw pump is changed to a three-flight screw pump, all other things being equal?a. 10%b. 20%c. 33%d. 40%

Introduction to Centrifugal Pumps52. What type of centrifugal pump is the most common?

a. axial-flow impellerb. radial-flow impellerc. mixed-flow impellerd. vertical-flow impeller

53. The capacity of a centrifugal pump is related to total heada. directlyb. indirectlyc. linearlyd. disproportionately

54. As the total head on a system increases, the volume that a centrifugal pump delivers is reduceda. directlyb. indirectlyc. proportionatelyd. disproportionately

55. In wastewater treatment, nonclogging pumps are usually designed to pass solids that have a diameter of at leasta. 3 in.b. 5 in.c. 6 in.d. 8 in.

56. What type of centrifugal pump is used to ensure that a wet well’s pump maintains sufficient prime?a. service water pumpb. submersible pumpc. dry pit pumpd. recirculation pump

57. Typically, pumps used to return activated sludge, recycle trickling filter effluent, or return digester supernatant would bea. recirculation pumpsb. submersible pumpsc. service water pumpsd. wet well pumps



Centrifugal Pump Components58. Besides providing the enclosure for the impeller, the volute case is also cast and

machined to provide a seat for thea. lantern ringb. impeller ringsc. mechanical seald. slinger ring

59. If a centrifugal pump with a single volute is operated less than _______ of its design capacity, an excess radial load will result, causing premature bearing failure.a. 20%b. 25%c. 30%d. 35%

60. If a centrifugal pump with a single volute is operated more than _______ of its design capacity, an excess radial load will result, causing premature bearing failure.a. 110%b. 120%c. 125%d. 135%

61. What material is usually used for centrifugal pump sleeves?a. steelb. copperc. aluminumd. brass

62. Rigid couplings are most common ona. horizontal mounted pumpsb. vertical mounted pumpsc. positive-displacement pumpsd. rotary pumps

63. A type of rigid coupling would bea. chain couplingb. split couplingc. gear couplingd. jaw coupling

64. How much will a flexible disk coupling compensate for in angular movement?a. 1 degreeb. 2 degreesc. 3.2 degreesd. 3.7 degrees

65. How much will a flexible disk coupling compensate for in parallel misalignment?a. ¹⁄₃₂ in.b. ¹⁄₁₆ in.c. ⅛ in.d. ¼ in.



66. What will a flexible diaphragm coupling compensate for in angular movement?a. 1 degreeb. 2 degreesc. 3.7 degreesd. 4 degrees

67. What will a flexible diaphragm coupling compensate for in parallel misalignment?a. ¹⁄₃₂ in.b. ¹⁄₁₆ in.c. ⅛ in.d. ¼ in.

68. Which one of the following statements is false concerning lantern rings?a. the lantern ring is placed above the stuffing boxb. lantern rings have holes in the sidesc. lantern rings have an I-beam constructiond. lantern rings are made of brass or plastic

69. When should a mechanical seal be replaced?a. manufacturer’s recommended time and before any leakage occursb. when seal first starts to leakc. when seal leaks a small amountd. when leakage becomes excessive

70. What are the bearings called that maintain the radial positioning of a shaft?a. radial bearingsb. rolling bearingsc. sleeve bearingsd. thrust bearings

71. What are the bearings called that maintain the axial positioning of a shaft?a. sleeve bearingsb. line bearingsc. thrust bearingsd. rolling bearings

72. What are the bearings of choice for centrifugal pumps today?a. sleeve bearingsb. plain bearingsc. roller bearingsd. sliding contact bearings

73. What type of bearing can operate well against radial loads but can withstand only very low thrust loads?a. self-aligning spherical roller bearingsb. single-row tapered roller bearingsc. self-aligning, double-row ball bearingsd. angular contact bearings



Centrifugal Pumps: Operational Procedures74. Pump bases with four anchor bolts usually require _______ for each bolt.

a. a set of shimsb. a lock washerc. a large washer, then a lock washerd. a set of wedges

75. Centrifugal pump misalignment causes between _______ of all bearing failures.a. 5 to 10%b. 15 to 25%c. 50 to 70%d. 80 to 90%

Centrifugal Pump: Maintenance Procedures76. How much seal water does a mechanical seal require?

a. 5 drops/minb. 3 drops/minc. 1 or 2 drops every few minutesd. none—mechanical seals run dry

77. Centrifugal pump motor bearings should have a temperature of abouta. 160°Fb. 180°Fc. 195°Fd. 200°F

78. How often should the temperature of centrifugal pump motor bearings be checked with a thermometer?a. every dayb. once a weekc. twice a monthd. once a month

79. Inspection of the shaft and coupling alignment should be performed every _______ or immediately if problem signs develop.a. monthb. quarterc. 6 monthsd. year

Centrifugal Pump: Lubrication80. Which one of the following statements concerning lubricants is false?

a. lubricants work to transfer heatb. all lubricants have the ability to separate surfacesc. lubricants eliminate corrosiond. lubricants can provide a seal from contamination



81. The best lubricant for cooling is most probablya. oilb. greasec. graphited. water

82. The best lubricant that would dampen shocks to the pump isa. greaseb. oilc. waterd. graphite

83. What type of lubrication is best used on pump bearings for light to moderate, high-speed, horizontally shafted pumps?a. waterb. oilc. vegetable-based greased. animal-based grease approved by NSF International

84. Bearings should not be lubricated with this type of oil because as this oil begins to breakdown it forms acids.a. nondetergent mineral oilb. animal oilc. SAE 10d. SAE 20

85. Pump bearings are usually designed to use grease lubricants that provide oil lubricant ina. a soap-like baseb. an animal oil base approved by NSF Internationalc. a graphite based. a vegetable oil base approved by NSF International

Centrifugal Pump: Troubleshooting86. If a pump fails to prime because the suction lift is too high, what is the best solution?

a. priming unit should be inspected and cleaned or repairedb. suction piping air bleed off valve should be openedc. check external water seal unitd. reevaluate the pump’s requirements and correct the condition

87. If a pump loses its prime because the suction line has an air pocket, what is the best solution?a. check pump’s amperage and be sure pump’s strainer is cleanb. clean or repair priming unitc. suction piping air bleed off valves should be opend. check external water seal unit



Centrifugal Pump: Modifications88. What type of pump has little chance of losing its prime but has the disadvantage of

having a lack of access for maintenance or repair?a. recessed impeller or vortex pumpb. turbine pumpc. submersible pumpd. positive-displacement pump

89. What type of pump can handle materials that would typically clog or damage the pump’s impeller but has the disadvantage of being less efficient?a. recessed impeller or vortex pumpb. turbine pumpc. submersible pumpd. positive-displacement pump

90. What type of pump can produce a high head but requires high maintenance if large amounts of solids are being pumped?a. recessed impeller or vortex pumpb. turbine pumpc. submersible pumpd. positive-displacement pump

Positive-Displacement Pumps91. What type of pump adds energy to the fluid flow intermittently?

a. recessed impeller or vortex pumpb. turbine pumpc. submersible pumpd. positive-displacement pump

92. Diaphragm pumps do not work very well if they have to lift liquids more than abouta. 4 ftb. 8 ftc. 12 ftd. 20 ft


171

Grades 1–4 Pumping Answers and References

Grade 1 Pumping Answers and ReferencesBasic Water Hydraulics1. Answer: d. friction head

Reference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 2, Page 13.2. Answer: a. pressure head

Reference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 2, Page 13.3. Answer: b. pump head

Reference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 2, Page 13.4. Answer: b. 30 in.

Reference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 2, Page 18.5. Answer: d. 34 ft

Reference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 2, Page 20.

Introduction to Centrifugal Pumps6. Answer: a. centrifugal

Reference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 3, Page 51.7. Answer: c. seals

Reference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 3, Page 53.8. Answer: d. stuffing box

Reference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 3, Page 53.9. Answer: b. seal cage

Reference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 3, Page 53.10. Answer: c. packing gland


Centrifugal Pump Components11. Answer: c. semiopen

Reference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 4, Page 83.12. Answer: a. open

Reference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 4, Page 84.13. Answer: b. closed




14. Answer: a. wear ringsReference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 4, Page 86.

15. Answer: a. packing glandReference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 4, Page 96.

Centrifugal Pumps—Theory16. Answer: b. the stuffing box

Reference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 1, Page 1.

17. Answer: c. V2/2gReference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 1, Page 6.

18. Answer: b. mechanical energyReference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 1, Page 6.

19. Answer: b. centrifugal pumpReference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 1, Page 7.

Pump Nomenclature20. Answer: a. distribute lubrication to the packing


21. Answer: c. mechanical sealReference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 3, Page 32.

22. Answer: c. slinger ringReference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 3, Page 33.

23. Answer: b. impellerReference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 3, Page 36.

Mechanical Seals24. Answer: b. control water leakage from the stuffing box



GRADES 1–4 PUMPING ANSWERS AND REFERENCES 173

Grades 2–4 Pumping Answers and ReferencesBasic Water Hydraulics1. Answer: d. fluids in motion and at rest

Reference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 2, Page 9.2. Answer: c. cut-off head

Reference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 2, Page 13.3. Answer: b. discharge head

Reference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 2, Page 41.4. Answer: a. when head is not sufficient and more flow is required

Reference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 2, Page 41.5. Answer: c. single-suction with open or semiopen impellers

Reference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 2, Page 41.6. Answer: b. diaphragm


Centrifugal Pumps—Theory7. Answer: a. kinetic energy


8. Answer: c. potential energyReference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 1, Page 1.

9. Answer: d. large solidsReference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 1, Page 8.

10. Answer: b. split-case pumpReference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 1, Page 13.

11. Answer: a. closedReference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 1, Page 14.

Centrifugal Pumps—Types12. Answer: c. end suction pump


13. Answer: b. two stageReference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 2, Page 23.



14. Answer: b. two stagesReference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 2, Page 23.

Pump Nomenclature15. Answer: a. the center of the impeller


16. Answer: d. oil drip systemReference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 3, Page 39.

17. Answer: b. torqueReference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 3, Page 41.

18. Answer: d. reduce friction and heatReference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 3, Page 58.

19. Answer: b. thrust loadReference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 3, Page 58.

20. Answer: c. a radial loadReference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 3, Page 58.

21. Answer: a. rolling pinReference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 3, Page 58.

22. Answer: a. poor lubrication practicesReference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 3, Page 62.

Replacement of Pump Components on End Suction Centrifugal Pumps23. Answer: c. soft-faced hammer


24. Answer: b. keyReference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 4, Page 69.

25. Answer: a. stored vertically, on endReference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 4, Page 71.



26. Answer: d. wear at the eye of the impellerReference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 4, Page 71.

27. Answer: c. cooledReference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 4, Page 74.

28. Answer: c. heatedReference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 4, Page 74.

29. Answer: d. keroseneReference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 4, Page 76.

30. Answer: a. solventReference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 4, Page 77.

Selection and Replacement of Packing31. Answer: b. control water leakage along the pump’s shaft


32. Answer: a. asbestos base, graphiteReference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 5, Page 80.

33. Answer: c. asbestosReference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 5, Page 81.

34. Answer: b. metal braided with asbestosReference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 5, Page 81.

35. Answer: b. 90 degreesReference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 5, Page 82.

36. Answer: d. wear and deteriorate with normal useReference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 5, Page 83.

37. Answer: b. no further tightening can be done on the packing glandReference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 5, Page 84.



Pump Piping System38. Answer: d. when potable water is being pumped


39. Answer: c. 10 to 15 psi higherReference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 7, Page 108.

End Suction and Split-Case Centrifugal Pumps40. Answer: c. flange coupling

Reference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 9, Pages 153 and 154.

41. Answer: a. split couplingReference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 9, Pages 154 through 156.

42. Answer: a. most pumps are self-primingReference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 9, Pages 149, 169, and 170.

43. Answer: b. at least every 3 monthsReference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 9, Page 160.

44. Answer: d. increase torqueReference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 9, Page 169.

45. Answer: c. calciumReference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 9, Page 170.

46. Answer: c. 50%Reference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 9, Page 171.

Line Shaft Turbine Operating Conditions47. Answer: a. oil or water


48. Answer: b. pumping too much waterReference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 10, Page 183.



Special Pumping Units49. Answer: a. high head conditions


50. Answer: a. 30 ftReference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 11, Page 199.

51. Answer: b. 20%Reference: Pumps and Pumping, E.E. “Skeet” Arasmith, Fourth Edition, Lesson 11, Page 200.

Introduction to Centrifugal Pumps52. Answer: b. radial-flow impeller

Reference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 3, Page 59.53. Answer: a. directly

Reference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 3, Page 63.54. Answer: c. proportionately

Reference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 3, Pages 63 and 64.55. Answer: a. 3 in.

Reference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 3, Page 73.56. Answer: c. dry pit pump

Reference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 3, Page 74.57. Answer: a. recirculation pumps


Centrifugal Pump Components58. Answer: b. impeller rings

Reference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 4, Page 78.59. Answer: c. 30%

Reference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 4, Page 79.60. Answer: b. 120%

Reference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 4, Page 79.61. Answer: d. brass

Reference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 4, Page 91.62. Answer: b. vertical mounted pumps

Reference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 4, Page 92.63. Answer: b. split coupling

Reference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 4, Pages 92 through 95.



64. Answer: b. 2 degreesReference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 4, Page 95.

65. Answer: a. ¹⁄₃₂ in.Reference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 4, Page 95.

66. Answer: d. 4 degreesReference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 4, Page 95.

67. Answer: c. ⅛ in.Reference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 4, Page 95.

68. Answer: a. the lantern ring is placed above the stuffing boxReference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 4, Page 98.

69. Answer: d. when leakage becomes excessiveReference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 4, Page 101.

70. Answer: a. radial bearingsReference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 4, Page 102.

71. Answer: c. thrust bearingsReference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 4, Page 102.

72. Answer: c. roller bearingsReference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 4, Page 103.

73. Answer: c. self-aligning, double-row ball bearingsReference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 4, Page 103.

Centrifugal Pumps: Operational Procedures74. Answer: a. a set of shims

Reference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 5, Page 114.75. Answer: c. 50 to 70%


Centrifugal Pump: Maintenance Procedures76. Answer: c. 1 or 2 drops every few minutes

Reference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 6, Page 129.77. Answer: b. 180°F

Reference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 6, Page 130.78. Answer: d. once a month

Reference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 6, Page 130.79. Answer: c. 6 months




Centrifugal Pump: Lubrication80. Answer: c. lubricants eliminate corrosion

Reference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 8, Pages 142 through 144.

81. Answer: d. waterReference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 8, Page 143.

82. Answer: a. greaseReference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 8, Page 143.

83. Answer: b. oilReference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 8, Page 147.

84. Answer: b. animal oilReference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 8, Page 148.

85. Answer: a. soap-like baseReference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 8, Page 149.

Centrifugal Pump: Troubleshooting86. Answer: d. reevaluate the pump’s requirements and correct the condition

Reference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 9, Pages 160 and 161.

87. Answer: c. suction piping air bleed off valves should be openReference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 9, Pages 160 and 161.

Centrifugal Pump: Modifications88. Answer: c. submersible pump

Reference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 10, Page 180.89. Answer: a. recessed impeller or vortex pump

Reference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 10, Page 181.90. Answer: b. turbine pump


Positive-Displacement Pumps91. Answer: d. positive-displacement pump

Reference: Pumping, Frank R. Spellman and Joanne Drinan, Chapter 11, Page 185.92. Answer: a. 4 ft



181

Math Questions

1. Convert 45 degrees Fahrenheit to degrees Celsius.a. 7.2°Cb. 13°Cc. 23.4°Cd. 81°C

2. If 157 is 37.4%, what is 74%?a. 57.5b. 58.7c. 116d. 310

3. If 62 grams (g) of lime are dissolved in 1.75 liters (L) of water, what is the percent strength of the lime solution?a. 3.4% lime solutionb. 3.5% lime solutionc. 34% lime solutiond. 35% lime solution

4. If a stabilization pond is 318 ft long, 68.3 ft wide, and 7.65 ft deep, what is the number of cubic feet in the stabilization pond?a. 21,719 ft3

b. 94,500 ft3

c. 166,000 ft3

d. 216,000 ft3

5. If a substance weighs 9.55 lb/gal, what is the density of the solution in g/cm3?a. 0.137 g/cm3

b. 1.15 g/cm3

c. 1.75 g/cm3

d. 2.08 g/cm3

6. An average of 61 gallons per day of screenings is removed from a screening pit that has a capacity of 8.75 yd3. How long will it take to fill the pit in days, if the screenings are not removed?a. 23 daysb. 26 daysc. 29 daysd. 31.6 days

7. What is the velocity in feet per second (ft/s) for water flowing through a channel that is 9.8 ft wide and 4.15 ft deep, and the flow is 18.5 ft3/s?a. 0.27 ft/sb. 0.45 ft/sc. 2.2 ft/sd. 7.8 ft/s



8. What is the hydraulic loading rate on a trickling filter in gallons per day per square foot (gpd/ft2), given the following data?Flow = 2.63 mgdDiameter of tricking filter = 100.1 ftClarifier recirculation rate = 0.22 mgda. 90.6 gpd/ft2

b. 138 gpd/ft2

c. 293 gpd/ft2

d. 360 gpd/ft2

9. What is the estimated biosolids pumping rate for the following wastewater system?Assume sludge = 8.34 lb/galPlant flow = 1.27 mgdRemoved biosolids = 1.08%Influent total suspended solids (TSS) = 317 mg/LEffluent TSS = 105 mg/La. 8.57 gpmb. 17.3 gpmc. 34.5 gpmd. 144 gpm

10. How many gpd of a 10.4% sodium hypochlorite solution are needed to disinfect a flow of 375,000 gallons, if the dosage required is 9.15 mg/L? Assume the solution weighs 8.34 lb/gal.a. 33.0 gpdb. 35.7 gpdc. 275 gpdd. 298 gpd

11. A wastewater plant is treating 1.81 mgd at a chlorine dosage of 9.75 mg/L. If the sodium hypochlorite being used is 11.5% available chlorine, what is the chlorine usage in lb/d?a. 16.6 lb/d sodium hypochloriteb. 24.3 lb/d sodium hypochloritec. 153 lb/d sodium hypochlorited. 1,280 lb/d sodium hypochlorite

12. A wastewater treatment pond receives a flow of 243,000 gpd. What is the organic loading rate in pounds of biochemical oxygen demand per day per acre (lb BOD5/d/acre), if the pond has a surface area of 4.12 acre-ft and the influent BOD5 concentration is 216 mg/L?a. 25.9 lb BOD5/d/acreb. 54.6 lb BOD5/d/acrec. 106 lb BOD5/d/acred. 1,804 lb BOD5/d/acre

13. Calculate the influent flow to a trickling filter in mgd, if the BOD5 loading is 1,470 lb/d and the BOD5 is 212 mg/L.a. 0.8 mgdb. 0.831 mgdc. 1.20 mgdd. 1.42 mgd


MATH QUESTIONS 183

14. What is the soluble BOD5 if the total BOD5 is 209 mg/L, the K factor is 0.58, and the suspended solids (SS) are 198 mg/L?a. 76.8 mg/L soluble BOD5b. 77 mg/L soluble BOD5c. 94 mg/L soluble BOD5d. 323.84 mg/L soluble BOD5

15. What is the hydraulic digestion time for a 49.8-ft diameter digester with a level of 10.85 ft and sludge flow of 9,105 gallons per day (gpd)?a. 2.32 daysb. 6.85 daysc. 12.6 daysd. 17.4 days

16. Determine the loading rate on a digester in pounds of volatile solids added per day per cubic foot (lb VSA/d/ft3), if the volume of sludge in the digester is 200,100 gal and the digester has an influent of 7,630 lb/d of volatile solids (VS).a. 0.0381 lb VSA/d/ft3

b. 0.285 lb VSA/d/ft3

c. 0.642 lb VSA/d/ft3

d. 0.957 lb VSA/d/ft3

17. Given the following data, calculate the volatile solids (VS) loading ratio on a digester.Sludge weight in digester = 164,740 lbVS loading = 1,305 lb/dTotal solids (TS) percentage = 4.33%VS percentage = 67.1%a. 0.123 VS ratiob. 0.273 VS ratioc. 8.46 VS ratiod. 3.67 VS ratio

18. What must have been the gas production by a digester in ft3/d, given the following data?Volatile solids destroyed = 428 lb/dGas produced in ft3/lb VS destroyed = 11.8 ft3/lba. 36.3 ft3/db. 3,630 ft3/dc. 4,606 ft3/dd. 5,050 ft3/d

19. Given the following data, calculate the mean cell residence time (MCRT) for this activated sludge system.Aeration tank and final clarifier volume = 0.677 mil galMixed liquor suspended solids (MLSS) = 3,580 mg/LSuspended solids (SS) wasted = 1,910 lb/dSecondary effluent SS = 368 lb/da. 6.25 daysb. 7.49 daysc. 8.87 daysd. 10.6 days



20. If a sour digester has a volume of 188,000 gallons and a volatile acid concentration of 2,605 mg/L, how many pounds of lime will it take to neutralize the volatile acids?a. 489.7 lb of limeb. 3,663 lb of limec. 4,084 lb of limed. 4,897 lb of lime

21. The capacity of an aeration tank is 510,000 gallons. How many pounds of MLSS are being aerated, if the aeration tank is 78% full and the concentration of MLSS is 2,640 mg/L?a. 6,400 lb MLSSb. 7,700 lb MLSSc. 8,800 lb MLSSd. 11,000 lb MLSS

22. What is the food-to-microorganism (F/M) ratio for an aeration tank with a volume of 419,500 gallons, if the primary effluent flow is 5.77 mgd, the mixed liquor volatile suspended solids (MLVSS) is 3,095 mg/L, and the BOD5 is 238 mg/L?a. 1.06 F/M ratiob. 1.09 F/M ratioc. 1.10 F/M ratiod. 1.12 F/M ratio

23. Given the following data, determine the seed sludge required in gallons.Digester has a radius of 24.8 ftLiquid level in digester is 16.5 ftRequires 15.0% seed sludgea. 4,780 gal of seed sludgeb. 23,100 gal of seed sludgec. 31,800 gal of seed sludged. 35,800 gal of seed sludge

24. A gravity thickener with a radius of 29.7 ft receives a flow of 0.0752 mgd. If the percent solids are 4.50%, what are the solids loading on the gravity thickener in lb/d/ft2?a. 1.22 lb/d/ft2

b. 1.23 lb/d/ft2

c. 1.27 lb/d/ft2

d. 1.30 lb/d/ft2

25. What is the solids loading for a dissolved air flotation (DAF) unit in lb/hr/ft2 that is 59.4 ft by 18.0 ft, with a sludge flow of 0.201 mgd and a waste-activated sludge (WAS) concentration of 7,830 mg/L, and the sludge weighs 8.71 lb/gal?a. 0.534 lb/hr/ft2

b. 4.05 lb/hr/ft2

c. 8.10 lb/hr/ft2

d. 12.8 lb/hr/ft2


MATH QUESTIONS 185

26. If a DAF unit receives air at an average rate of 9.11 ft3/min, how many lb/d of air does it receive? Know: air = 0.0807 lb/ft3 at standard temperature, pressure, and average composition.a. 1,060 lb/d of airb. 2,120 lb/d of airc. 5,240 lb/d of aird. 5,420 lb/d of air

27. Given the following data, determine the removal efficiency of sludge solids on a disk centrifuge.Influent sludge solids = 5,917 mg/LEffluent sludge solids = 536 mg/La. 88.2%b. 90.3%c. 90.6%d. 90.9%

28. Calculate the filter loading in lb/hr/ft2 on a vacuum filter that has a surface area of 275 ft2, a digested biosolids flow rate of 47 gpm, and a solids concentration of 4.33%.a. 0.37 lb/hr/ft2

b. 3.7 lb/hr/ft2

c. 37 lb/hr/ft2

d. 370 lb/hr/ft2

29. A settleability test result shows that 219 mL of activated biosolids settled in a 1-L graduated cylinder. If the concentration of mixed liquor suspended solids (MLSS) is 2,440 mg/L, what is the biosolids density index (BDI)?a. 1.11 BDIb. 1.14 BDIc. 1.83 BDId. 5.34 BDI

30. Calculate the unseeded BOD5 in mg/L, given the following data.Start of test bottle dissolved oxygen (DO) = 9.3 mg/LBottle was incubated for 5 days in the dark at 20°CAfter 5 days DO = 3.8 mg/LSample size = 100 mLTotal volume = 300 mLGive answer to three significant figures.a. 1.83 mg/L unseeded BOD5b. 3.10 mg/L unseeded BOD5c. 16.5 mg/L unseeded BOD5d. 27.9 mg/L unseeded BOD5

31. If 125 lb of magnesium hydroxide are dissolved 15.0 gallons of water, what is the percent strength of the solution?a. 45.0%b. 50.0%c. 55.0%d. 60.0%



32. If the sludge entering a digester has a volatile solids (VS) content of 53.8% and the digester effluent sludge has a VS content of 40.1%, calculate the percent VS reduction.a. 25.5%b. 42.5%c. 51.0%d. 74.5%

33. What is the percent moisture reduction for a digester, if the raw biosolids is 7.18% solids and the digested biosolids solids is 14.9%?a. 45.7% moisture reductionb. 48.2% moisture reductionc. 51.8% moisture reductiond. 55.8% moisture reduction

34. Given the following data, how many lb/d of volatile solids (VS) are pumped to a digester?Pumping rate = 2,180 gpdSolids content = 5.51%Volatile solids = 60.1%Specific gravity of sludge = 1.06a. 72.2 lb/d VSb. 638 lb/d VSc. 866 lb/d VSd. 7,220 lb/d VS

35. A solution containing exactly 150 gallons of 9.5% hypochlorite is required. How many gallons of a 12.5% solution must be mixed with a 3.5% solution to make the required solution?Solve the problem using the dilution triangle.a. 40 gallons of the 3.5% solution and 110 gallons of the 12.5% solutionb. 50 gallons of the 3.5% solution and 100 gallons of the 12.5% solutionc. 60 gallons of the 3.5% solution and 90 gallons of the 12.5% solutiond. 70 gallons of the 3.5% solution and 80 gallons of the 12.5% solution

36. A soda ash tank is conical at the bottom and cylindrical at the top. If the diameter of the cylinder is 17.5 ft, with a depth of 34.5 ft, and the cone depth is 14.0 ft, what is the volume of the tank in ft3?a. 9,420 ft3

b. 9,720 ft3

c. 9,840 ft3

d. 9,980 ft3

37. If a 10.0-ft diameter chemical tank drops 4.31 in. in exactly 3 hr, what is the pumping rate for the chemical in gpm?a. 1.17 gpmb. 4.89 gpmc. 16.8 gpmd. 28.3 gpm


MATH QUESTIONS 187

38. What is the solids loading rate for a secondary clarifier, given the following information?Radius of clarifier = 65.4 ftPrimary effluent flow = 3,779,000 gpdReturn sludge flow = 725,000 gpdMixed liquor suspended solids (MLSS) = 2,110 mg/La. 0.71 lb MLSS/d/ft2

b. 2.46 lb MLSS/d/ft2

c. 4.95 lb MLSS/d/ft2

d. 5.90 lb MLSS/d/ft2

39. Determine the waste activated sludge (WAS) pumping rate in gpm, given the following data.Amount of WAS to be wasted = 4,380 lb/dWAS suspended solids concentration = 3,170 mg/La. 115 gpmb. 275 gpmc. 485 gpmd. 960 gpm

40. Find the motor horsepower (mhp) for a pump with the following parameters.Motor efficiency (ME): 88.7%Pump efficiency (PE): 76.5%Total head (TH): 215 ftFlow: 3.81 mgda. 97.5 mhpb. 120 mhpc. 200 mhpd. 212 mhp

41. What is the chlorine dosage in mg/L for a wastewater plant, if the chlorinator setting is 135 lb/d and it treats 2,360,000 gpd?a. 5.75 mg/Lb. 6.28 mg/Lc. 6.86 mg/Ld. 9.28 mg/L

42. How many gpd of a 11.8% sodium hypochlorite solution are needed to disinfect a flow of 1,744,000 gallons, if the dosage required is 9.65 mg/L and the specific gravity of the hypochlorite is 1.04?a. 118 gpd sodium hypochloriteb. 137 gpd sodium hypochloritec. 148 gpd sodium hypochlorited. 157 gpd sodium hypochlorite



43. Given the following data, calculate the feed rate of a polymer solution in mL/min.Influent flow = 875 gpmPolymer dose = 6.80 mg/LPolymer solution specific gravity = 1.29Polymer percent purity = 40.5%a. 5.32 mL/min polymerb. 6.85 mL/min polymerc. 7.07 mL/min polymerd. 11.8 mL/min polymer

44. Given the following data, determine the amount of dry solids produced in lb/d.Flow = 1,248,000 gpdInfluent suspended solids = 295 mg/LPrimary effluent suspended solids = 106 mg/La. 1,100 lb/db. 1,590 lb/dc. 1,970 lb/dd. 3,070 lb/d

45. Determine the age of the sludge in an aeration tank, given the following data.Primary effluent flow = 1.27 mgdVolume of aeration tank = 435,000 galMLSS = 2,040 mg/LSuspended solids = 73 mg/La. 9.2 daysb. 9.6 daysc. 10.7 daysd. 15.4 days

46. What is the organic loading rate in lb BOD5/d/1,000 ft3 for a trickling filter that is 68.5 ft in diameter and 5.05 ft deep, if the primary effluent flow is 2.46 mgd and the BOD5 is 149 mg/L?a. 122 lb BOD5/d/1,000 ft3

b. 138 lb BOD5/d/1,000 ft3

c. 156 lb BOD5/d/1,000 ft3

d. 164 lb BOD5/d/1,000 ft3

47. What is the hydraulic digestion time for a digester that is 25.0 ft in radius, has a level of 13.7 ft, and has a sludge flow of 10,840 gallons per day (gpd)?a. 2.48 daysb. 7.45 daysc. 13.8 daysd. 18.6 days

48. What is the ratio of volatile solids loading on a digester, if the digester has 59,980 kg of volatile solids (VS), 1,330 lb/d are pumped into it, percentage total solids (TS) are 4.89%, and percentage volatile solids (VS) are 70.8%?a. 0.290b. 0.845c. 1.66d. 3.66


MATH QUESTIONS 189

49. Given the following data, calculate the population equivalent.Average wastewater flow for the day = 3.06 ft3/sBOD5/person = 0.22 lb/dBOD5 concentration in the wastewater = 2,525 mg/La. 48,000 peopleb. 96,000 peoplec. 190,00 peopled. 275,000 people

50. How many pounds of MLSS are being aerated, if the aeration tank is 49.8 ft in diameter, with a sludge height of 13.5 ft and a concentration of MLSS of 1,905 mg/L, and the specific gravity of the MLSS is 1.03?a. 3,123 lb MLSSb. 3,220 lb MLSSc. 3,370 lb MLSSd. 3,410 lb MLSS

51. A digester tank has a level capacity of 28.5 ft. Currently there is 18.2 ft of sludge water in the tank. What would the SCADA reading be on the board in mA for a 4 mA to 20 mA signal?a. 11.9 mAb. 13.3 mAc. 14.2 mAd. 16.8 mA

52. A total of 2,860 gpd of sludge (primary sludge) with 5.64% solids content and weighing 8.43 lb/gal is pumped to a wastewater thickener tank. Determine the amount of sludge that should flow from the thickener tank in gpd, if the sludge (secondary sludge) is further concentrated to 7.18% solids and weighs 8.52 lb/gal. Note: For convenience, primary sludge is abbreviated to 1° sludge and secondary sludge to 2° sludge.a. 2,220 gpdb. 2,267 gpdc. 2,845 gpdd. 3,610 gpd

53. What is the weir overflow rate in gpd/ft, if the flow is 0.266 mgd and the radius of the clarifier is 39.9 ft?a. 1,060 gpd/ftb. 1,090 gpd/ftc. 1,140 gpd/ftd. 1,190 gpd/ft

54. Given the following data, how many lb/d of volatile solids (VS) are pumped to a digester?Influent pumping rate = 2,850 gpdSolids content = 5.15%Volatile solids (VS) = 59.6%Specific gravity of sludge = 1.05a. 515 lb/d VSb. 628 lb/d VSc. 766 lb/d VSd. 1,280 lb/d VS



55. What percent hypochlorite solution would result, if 825 gallons of a 12.5% solution were mixed with exactly 175 gallons of a 3.5% solution? Assume both solutions have the same density and give answer to three significant numbers.a. 8.94% final solutionb. 9.37% final solutionc. 10.1% final solutiond. 10.9% final solution

56. Find the number of pounds of MLSS being aerated, if the aeration tank is 50.0 ft in diameter, with a sludge height of 14.2 ft and a concentration of MLSS of 2,280 mg/L, and the specific gravity of the MLSS is 1.05.a. 556 lb MLSSb. 3,960 lb MLSSc. 3,964 lb MLSSd. 4,160 lb MLSS

57. What is the flow velocity in feet per second (ft/s) for a trapezoidal channel, given the following data?Bottom width, w1 = 5.11 ftWater surface width, w2 = 8.73 ftDepth = 3.91 ftFlow = 22.5 ft3/sa. 0.715 ft/sb. 0.832 ft/sc. 1.76 ft/sd. 3.25 ft/s

58. What is the biosolids retention time (BRT) in days for a digester that is 70.1 ft in diameter, has a working level of 17.55 ft, and receives an average flow of 10.5 gpm?a. 4.48 daysb. 12.9 daysc. 27.1 daysd. 33.5 days

59. Given the following data, calculate the cost of running a pump in dollars and cents per day.Flow = 1.58 mgdTDH = 167 ftMotor efficiency = 89.9%Pump efficiency = 71.4%Cost in kilowatt hours = $0.076a. $72.09/db. $92.23/dc. $98.09/dd. $99.93/d


MATH QUESTIONS 191

60. A wastewater treatment plant is treating 2,080,000 gpd with a 12.5% sodium hypochlorite solution. If the dosage required is 9.75 mg/L and the specific gravity of the hypochlorite is 1.03, how many gpd of sodium hypochlorite are required?a. 158 gpd sodium hypochloriteb. 161 gpd sodium hypochloritec. 167 gpd sodium hypochlorited. 172 gpd sodium hypochlorite

61. What should the chemical feed pump be set on in gpd, given the following data?Plant’s treatment flow = 2,610,000 gpdAlum dose = 11.05 mg/LAlum = 5.37 lb/gal (assuming 48.5% purity of aluminum sulfate)a. 44.8 gpd of liquid alum solutionb. 45.1 gpd of liquid alum solutionc. 48.2 gpd of liquid alum solutiond. 49.8 gpd of liquid alum solution

62. Given the following data on a wastewater treatment pond, calculate the organic loading rate in lb BOD5/d/acre.Influent flow = 278,000 gpdSurface area of pond = 4.11 acre-ftInfluent BOD5 concentration = 231 mg/La. 31.6 lb BOD5/d/acreb. 78.1 lb BOD5/d/acrec. 99.6 lb BOD5/d/acred. 130 lb BOD5/d/acre

63. Given the following data, calculate the digester loading in lb volatile solids (VS)/d/1,000 ft3.Digester diameter = 49.9 ftSludge level = 14.8 ftInfluent sludge flow = 10,500 gpdPercent sludge solids = 4.88%Percent volatile solids = 70.1%Specific gravity of sludge = 1.04a. 104 lb VS/d/1,000 ft3

b. 108 lb VS/d/1,000 ft3

c. 112 lb VS/d/1,000 ft3

d. 115 lb VS/d/1,000 ft3

64. Calculate the required waste rate from an aeration tank in mgd and gpm, given the following data.Aeration tank volume = 0.585 mil galDesired COD lb/MLVSS lb = 0.18Primary effluent flow = 2.39 mgdPrimary effluent COD = 134 mg/LMixed liquor volatile suspended solids (MLVSS) = 3,588 mg/LWaste volatile solids (WVS) concentration = 4,075 mg/La. 49 gpmb. 52 gpmc. 54 gpmd. 57 gpm



65. What is the pressure 5.25 ft from the bottom of a tank at exactly 48 hr after the pumping started, given the following data?Tank radius = 35.5 ftInitial sludge depth = 3.42 ftSludge flow into tank = 72.5 gpma. 2.26 psib. 2.73 psic. 3.80 psid. 4.54 psi

66. Given the following parameters, calculate how long a primary sludge pump should operate in minutes per hour.Plant flow = 675 gpmSludge pump = 42.0 gpmInfluent suspended solids (SS) = 348 mg/LEffluent SS = 129 mg/LSludge = 4.76% solidsa. 2.61 min/hrb. 3.78 min/hrc. 4.08 min/hrd. 4.44 min/hr

67. Determine the waste activated sludge (WAS) flow in gpm, given the following data.Influent flow = 1.82 mgdClarifier radius = 40.0 ftClarifier depth = 13.6 ftAerator = 0.425 mil galMLSS = 2,440 mg/LRAS SS = 6,945 mg/LSecondary effluent SS = 28.5 mg/LTarget solids retention time (SRT) = 10 days exactlya. 15.7 gpm WASb. 16.9 gpm WASc. 17.6 gpm WASd. 18.8 gpm WAS

68. What should the chemical feeder be set on in mL/min, given the following data?Polymer dosage = 8.80 mg/LPlant flow = 1,340 gpmSp gr of polymer = 13.03 lb/gala. 25.8 mL/min polymerb. 27.1 mL/min polymerc. 28.6 mL/min polymerd. 29.4 mL/min polymer


MATH QUESTIONS 193

69. What is the organic loading rate in lb BOD5/d/1,000 ft3 for a trickling filter that is 151 ft in diameter and 5.10 ft deep, if the primary effluent flow is 1,220 gpm and the BOD5 is 207 mg/L?a. 30.4 lb BOD5/d/1,000 ft3

b. 33.2 lb BOD5/d/1,000 ft3

c. 33.7 lb BOD5/d/1,000 ft3

d. 34.3 lb BOD5/d/1,000 ft3

70. What is the air-to-solids ratio for a dissolved air flotation (DAF) unit that has an air flow rate equal to 9.10 ft3/min, a solids concentration of 0.76%, and a flow of 0.185 mgd? Know: Air = 0.0807 lb/ft3 at standard temperature, pressure, and average composition.a. 0.090 air-to-solids ratiob. 0.105 air-to-solids ratioc. 0.117 air-to-solids ratiod. 0.75 air-to-solids ratio

71. How many lb/hr of air does a DAF unit receive, if the unit receives air at an average rate of 0.259 m3/min? Know: Air = 0.0807 lb/ft3 at standard temperature, pressure, and average composition and there are 35.3 ft3/m3.a. 0.529 lb/hr of airb. 1.2 lb/hr of airc. 44.3 lb/hr of aird. 49.8 lb/hr of air

72. Given the following data, determine the feed time for a centrifuge in minutes.Capacity of basket centrifuge = 24.9 ft3

Influent sludge flow = 61 gpmInfluent sludge solids concentration = 7,070 mg/LAverage solids concentration in the basket = 7.92%a. 4.6 minb. 14 minc. 27 mind. 34 min

73. Given the following data for blending compost (BC) with wood chips, calculate the percent of the blended compost.Bulk density of sludge = 1,690 lb/yd3

Sludge volume = 10.5 yd3

Sludge solids content = 17.7%Density of wood chips = 585 lb/yd3

Wood chip solids = 52.8%Mix ratio (MR) of wood chips to sludge = 3 to 1a. 35.6% solids in blended compostb. 42.3% solids in blended compostc. 48.6% solids in blended compostd. 53.9% solids in blended compost



74. Determine the number of acres used in the following land application of biosolids, given the following data.Hydraulic loading rate = 0.46 in./dFlow = 375,000 gpda. 30 acresb. 32 acresc. 36 acresd. 40 acres

75. A 1.00% stock polymer solution (10,000 ppm or 10,000 mg/L) is desired for performing a jar test. If the polymer has a specific gravity of 1.31 and is 35.1% polymer, how many milliliters of polymer are required to make exactly 1,000 mL stock solution?a. 20.3 mL of polymerb. 21.7 mL of polymerc. 22.4 mL of polymerd. 23.6 mL of polymer


195

Answers to Math Questions

1. a. 7.2°C2. d. 3103. a. 3.4% lime solution4. c. 166,000 ft3

5. b. 1.15 g/cm3

6. c. 29 days7. b. 0.45 ft/s8. d. 360 gpd/ft2

9. b. 17.3 gpm10. a. 33.0 gpd11. d. 1,280 lb/d sodium hypochlorite12. c. 106 lb BOD5/d/acre13. b. 0.831 mgd14. c. 94 mg/L soluble BOD5

15. d. 17.4 days16. b. 0.285 lb VSA/d/ft3

17. b. 0.273 VS ratio18. d. 5,050 ft3/day19. c. 8.87 days20. c. 4,084 lb of lime21. c. 8,800 lb MLSS22. a. 1.06 F/M ratio23. d. 35,800 gal of seed sludge24. a. 1.22 lb/d/ft2

25. a. 0.534 lb/hr/ft2

26. a. 1,060 lb/day of air27. d. 90.9%28. b. 3.7 lb/hr/ft2

29. a. 1.11 BDI30. c. 16.5 mg/L unseeded BOD5

31. b. 50.0% Mg(OH)2 solution32. b. 42.5% VS reduction33. d. 55.8% moisture reduction34. b. 638 lb/d VS35. b. 50 gallons of the 3.5% solution and 100 gallons of the 12.5% solution36. a. 9,420 ft3



37. a. 1.17 gpm38. d. 5.90 lb MLSS/d/ft2

39. a. 115 gpm40. d. 212 mhp41. c. 6.86 mg/L42. b. 137 gpd sodium hypochlorite43. c. 7.07 mL/min polymer44. c. 1,970 lb/d45. b. 9.6 days46. d. 164 lb BOD5/d/1,000 ft3

47. d. 18.6 days48. a. 0.290 VS ratio49. c. 190,00 people50. b. 3,220 lb MLSS51. c. 14.2 mA52. a. 2,220 gpd53. a. 1,060 gpd/ft54. c. 766 lb/d VS55. d. 10.9% final solution56. d. 4,160 lb MLSS57. b. 0.832 ft/s58. d. 33.5 days59. c. $98.09/d60. a. 158 gpd sodium hypochlorite61. a. 44.8 gpd of liquid alum solution62. d. 130 lb BOD5/d/acre63. b. 108 lb VS/d/1,000 ft3

64. c. 54 gpm65. a. 2.26 psi66. d. 4.44 min/hr67. c. 17.6 gpm WAS68. c. 28.6 mL/min polymer69. b. 33.2 lb BOD5/d/1,000 ft3

70. a. 0.090 air-to-solids ratio71. c. 44.3 lb/hr of air72. d. 34 min73. a. 35.6% solids in blended compost74. a. 30 acres75. b. 21.7 mL of polymer


197

Math Questions with Solutions

1. Convert 45 degrees Fahrenheit to degrees Celsius.Equation: °C = 5°C/9°F(°F – 32°F)Substitution: °C = 5°C/9°F(45 – 32) = 5/9(13) = 7.22°C, round to 7.2°C

2. If 157 is 37.4%, what is 74%?Write a ratio and solve for the unknown number, x.157/37.4% = x/74%x = (74%)(157)/37.4% = 310.64, round to 310

3. If 62 grams (g) of lime are dissolved in 1.75 liters (L) of water, what is the percent strength of the lime solution?Know: 1 L = 3,785 mL = 3,785 g

Equation: Percent strength = Number of g, Water g, Chemical(Number of g of chemical)(100%)

+Substitute and solve.Percent strength = (1.75 L, Water)(1,000 g/L) 62 g Lime

(62 g Lime)(100%)+

Percent strength = 1,812 g(62 g Lime)(100%) = 3.4% lime solution

4. If a stabilization pond is 318 ft long, 68.3 ft wide, and 7.65 ft deep, what is the number of cubic feet in a stabilization pond?The volume equation for a rectangular basin is: Volume = (Length)(Width)(Depth)Volume = (L)(W)(D) = (318 ft)(68.3 ft)(7.65 ft) = 166,153 ft3, round to 166,000 ft3

5. If a substance weighs 9.55 lb/gal, what is the density of a solution in g/cm3?Equation: Number of g/cm3 = (Number of lb/gal)(454 g/1 lb)(1 gal/3,785 cm3)Number of g/cm3 = (9.55 lb/gal)(454 g/1 lb)(1 gal/3,785 cm3)Number of g/cm3 = 1.1455 g/cm3, round to 1.15 g/cm3



6. An average of 61 gallons per day of screenings is removed from a screening pit that has a capacity of 8.75 yd3. How long will it take to fill the pit in days, if the screenings are not removed?First, convert yd3 to ft3.Know: 1 yd3 = 27 ft3

Number of ft3 = (8.75 yd3)(27 ft3/yd3) = 236.25 ft3

Number of days = 61 gal/d

(236.25 ft )(7.48 gal/ft )3 3 = 28.97 days, round to 29 days

7. What is the velocity in feet per second (ft/s) for water flowing through a channel that is 9.8 ft wide and 4.15 ft deep, and the flow is 18.5 ft3/s?Write the equation: Q (Flow) = (Area)(Velocity). Substitute known parameters:18.5 ft3/s = (9.8 ft)(4.15 ft)(Velocity), and solve for velocity by rearranging the equation.

Velocity = (9.8 ft)(4.15 ft)18.5 ft /s3

= 0.4549 ft/s, round to 0.45 ft/s

8. What is the hydraulic loading rate on a trickling filter in gallons per day per square foot (gpd/ft2), given the following data?Flow = 2.63 mgd Diameter of tricking filter = 100.1 ft Clarifier recirculation rate = 0.22 mgdFirst, determine the total flow in gallons per day (gpd) through the trickling filter.Total flow, gal = (2.63 mgd + 0.22 mgd)(1,000,000 gal/mil) = 2,850,000 gpdNext, determine the surface area in ft2 for the clarifier.Area = r2 where r = Diameter/2 = 100.1 ft/2 = 50.05 ftClarifier surface area, ft2 = (3.14)(50.05 ft)(50.05 ft) = 7,865.71 ft2

Lastly, calculate the hydraulic loading rate.

Hydraulic loading rate = Surface area, ftTotal flow, gpd

2

Hydraulic loading rate = 7,865.71 ft

2,850,000, gpd2

= 362.33 gpd/ft2, round to 360 gpd/ft2

9. What is the estimated biosolids pumping rate for the following wastewater system?Assume sludge = 8.34 lb/gal Plant flow = 1.27 mgd Removed biosolids = 1.08% Influent total suspended solids (TSS) = 317 mg/L Effluent TSS = 105 mg/L


MATH QUESTIONS WITH SOLUTIONS 199

Equation: Estimated pumping rate =

(Percent solids in sludge)(Sludge, lb/gal)(1,440 min/d)(Influent TSS, mg/L – Effluent TSS, mg/L)(Flow, mgd)(8.34 lb/gal)

Substitute known values and solve.

Estimated pumping rate = (1.08%/100%)(8.34 lb/gal)(1,440 min/d)

(317 TSS mg/L – 105 TSS, mg/L)(1.27 mgd)(8.34 lb/gal)

Estimated pumping rate = (1.08%/100%)(8.34 lb/gal)(1,440 min/d)(212 TSS mg/L)(1.27 mgd)(8.34 lb/gal)

Estimated pumping rate = 17.3 gpm

10. How many gpd of a 10.4% sodium hypochlorite solution are needed to disinfect a flow of 375,000 gallons, if the dosage required is 9.15 mg/L. Assume the solution weighs 8.34 lb/gal.First, convert gpd to mgd.

Number of mgd = 1,000,000/mil375,000 gpd = 0.375 mgd

Next, using the “pounds equation,” calculate the lb/d of chlorine needed.Equation: Chlorine, lb/d = (Dosage, mg/L)(mgd)(8.34 lb/gal)Chlorine, lb/d = (9.15 mg/L)(0.375 mgd)(8.34 lb/gal) = 28.617 lb/dSince the solution is not 100%, divide the percent hypochlorite into the lb/d of chlorine needed.Hypochlorite, lb/d =

10.4%/100%28.617 lb/d = 275.16 lb/d hypochlorite

Lastly, determine the gpd of hypochlorite solution needed.

Hypochlorite, gpd = 8.34 lb/gal275.16 lb/d = 32.99 gpd, round to 33.0 gpd sodium hypochlorite

11. A wastewater plant is treating 1.81 mgd at a chlorine dosage of 9.75 mg/L. If the sodium hypochlorite being used is 11.5% available chlorine, what is the chlorine usage in lb/d?Equation: Chlorine, lb/d =

Percent available chlorine/100%(Dosage, mg/L)(mgd)(8.34 lb/gal)


Chlorine, lb/d = 11.5%/100% Available chlorine

(9.75 mg/L)(1.81 mgd)(8.34 lb/gal)

Chlorine, lb/d = 1,279.8 lb/d, round to 1,280 lb/d sodium hypochlorite



12. A wastewater treatment pond receives a flow of 243,000 gpd. What is the organic loading rate in pounds of biochemical oxygen demand per day per acre (lb BOD5/d/acre), if the pond has a surface area of 4.12 acre-ft and the influent BOD5 concentration is 216 mg/L?First, convert gallons per day to mgd.

Number of mgd = 1,000,000/mil243,000 gpd

= 0.243 mgd

Next, determine the pounds of BOD5/d/acre using a modified version of the “pounds” equation.

Organic loading rate, lb BOD5/d/acre = ft-Surface area of pond, acre

(BOD , mg/L)(Flow, mgd)(8.34 lb/gal)5

Organic loading rate, lb BOD5/d/acre = 4.12 acre ft

(216 mg/L BOD )(0.243 mgd)(8.34 lb/gal)5

-

Organic loading rate, lb BOD5/d/acre = 106.25 lb BOD5/d/acre, round to 106 lb BOD5/d/acre

13. Calculate the influent flow to a trickling filter in mgd, if the BOD5 loading is 1,470 lb/d and the BOD5 is 212 mg/L.Equation: Number of lb/d BOD5 = (BOD5, mg/L)(Number of mgd)(8.34 lb/gal)Rearrange the equation to solve for mg/L.

Number of mgd = (Number of mg/L BOD )(8.34 lb/gal)Number of lb/d BOD

5

5

Substitute values and solve.

Number of mgd flow = (212 mg/L)(8.34 lb/gal)

1,470 lb/d = 0.831 mgd

14. What is the soluble BOD5 if the total BOD5 is 209 mg/L, the K factor is 0.58, and the suspended solids (SS) are 198 mg/L?Equation: Soluble BOD5 = Total BOD5 – (K factor)(Total SS)Substitute known values and solve.Soluble BOD5 = 209 mg/L BOD5 – (0.58 K factor)(198 mg/L SS)Soluble BOD5 = 209 mg/L – 114.84 mg/LSoluble BOD5 = 94.16 mg/L, round to 94 mg/L soluble BOD5



15. What is the hydraulic digestion time for a 49.8-ft diameter digester with a level of 10.85 ft and sludge flow of 9,105 gallons per day (gpd)?First, determine the volume of the digester in ft3.Volume, ft3 = (0.785)(Diameter)2(Depth, ft)Volume, ft3 = (0.785)(49.8 ft)(49.8 ft)(10.85 ft) = 21,123 ft3

Next, determine the number of gallons in the digester.Know: 1 ft3 = 7.48 gal/ft3

Number of gal = (21,123 ft3)(7.48 gal/ft3) = 158,000 galLastly, calculate the digestion time in days.

Equation: Digestion time, days = Influent sludge flow, gal/d

Number of gallons Substitute and solve.

Digestion time, days = 9,105 gal/d158,000 gal

= 17.35 days, round to 17.4 days

16. Determine the loading rate on a digester in pounds of volatile solids added per day per cubic foot (lb VSA/d/ft3), if the volume of sludge in the digester is 200,100 gal and the digester has an influent of 7,630 lb/d of volatile solids (VS).First, convert the volume of the digester from gallons to cubic feet.Number of ft3, digester = 200,100 gal/7.48 gal/ft3 = 26,751 ft3

Next, using the following formula, determine the loading rate on the digester.

Equation: Digester loading rate, lb VSA/d/ft3 = Volume of digester, ft

lb VSA3

Digester loading rate, lb VSA/d/ft3 = 26,751, ft

7,630 lb VSA3

= 0.285 lb VSA/d/ft3

17. Given the following data, calculate the volatile solids (VS) loading ratio on a digester.Sludge weight in digester = 164,740 lb VS loading = 1,305 lb/d Total solids (TS) percentage = 4.33% VS percentage = 67.1%Use expanded equation with percentages:

Digester VS ratio = (lb VS in digester)(TS %/100%)(VS %/100%)

VS added lb/d

Digester VS ratio = (164,740 lb VS)(4.33%/100% TS)(67.1%/100% VS)

1,305 lb/d, VS

Digester VS ratio = (164,740 lb VS)(0.0433 TS)(0.671 VS)

1,305 lb/d, VS

Digester VS ratio = 0.273 VS ratio



18. What must have been the gas production by a digester in ft3/d, given the following data?Volatile solids destroyed = 428 lb/d Gas produced in ft3/lb VS destroyed = 11.8 ft3/lb

Equation: Gas produced, ft3/lb VS destroyed = VS destroyed, lb/d

Gas production, ft /d3

Rearrange to solve for gas production in ft3/d.Gas production, ft3/d = (Gas produced, ft3/lb VS destroyed)(VS destroyed, lb/d)Substitute and solve.Gas production, ft3/d = (11.8 ft3/lb)(428 lb/d) = 5,050.4 ft3/d, round to 5,050 ft3/d

19. Given the following data, calculate the mean cell residence time (MCRT) for this activated sludge system.Aeration tank and final clarifier volume = 0.677 mil gal Mixed liquor suspended solids (MLSS) = 3,580 mg/L Suspended solids (SS) wasted = 1,910 lb/d Secondary effluent SS = 368 lb/d

Equation: MCRT, days = SS wasted, lb/d SS lb/d

(MLSS, mg/L)(mil gal)(8.34 lb/d)+


MCRT, days = 1,910 lb/d SS 368 lb/d SS

(3,580 mg/L)(0.677 mil gal)(8.34 lb/d)+

MCRT, days = 2,278 lb/d SS

(3,580 mg/L)(0.677 mil gal)(8.34 lb/d) = 8.87 days

20. If a sour digester has a volume of 188,000 gallons and a volatile acid concentration of 2,605 mg/L, how many pounds of lime will it take to neutralize the volatile acids?Know 1 mg/L of lime will neutralize 1 mg/L of volatile acidsFirst, convert the digester’s volume from gallons to mil gal.

Number of mil gal = 1,000,000/mil

188,000 gal = 0.188 mil gal

Equation: Lime, lb = (Volatile acids, mg/L)(mil gal)(8.34 lb/gal)Substitute and solve.Lime, lb = (2,605 mg/L)(0.188 mil gal)(8.34 lb/gal) = 4,084 lb of lime



21. The capacity of an aeration tank is 510,000 gallons. How many pounds of MLSS are being aerated, if the aeration tank is 78% full and the concentration of MLSS is 2,640 mg/L?First, determine how many gallons are in the aeration tank.Number of gallons = (510,000 gal capacity)(78%/100% full) = 397,800 gallonsNext, convert gallons to mil gal.


397,800 gal = 0.3978 mil gal

Next, determine the pounds of MLSS under aeration using the “pounds” equation.Equation: Number of lb = (MLSS, mg/L)(Number of mil gal)(8.34 lb/gal)Substitute values and solve.Number of lb MLSS = (2,640 mg/L MLSS)(0.3978 mgd)(8.34 lb/gal)Number of lb MLSS = 8,758.6 lb, round to 8,800 lb MLSS

22. What is the food-to-microorganism (F/M) ratio for an aeration tank with a volume of 419,500 gallons, if the primary effluent flow is 5.77 mgd, the mixed liquor volatile suspended solids (MLVSS) is 3,095 mg/L, and the BOD5 is 238 mg/L?First, convert the volume of wastewater in the tank to mil gal.Number of mil gal = 419,500 gal/1,000,000/mil = 0.4195 mil galNext, write the equation:

F/M = (mg/L MLVSS)(Volume of tank, mil gal)

(BOD , mg/L)(Flow, mgd)5


F/M = (3,095 mg/L MLVSS)(0.4195 mil gal)

(238 mg/L BOD )(5.77 mgd)5 = 1.06 F/M ratio

23. Given the following data, determine the seed sludge required in gallons.Digester has a radius of 24.8 ft Liquid level in digester is 16.5 ft Requires 15.0% seed sludgeFirst, determine the number of gallons in the digester.Volume, gal = r2(Depth, ft)(7.48 gal/ft3)Volume, gal = (3.14)(24.8 ft)(24.8 ft)(16.5 ft)(7.48 gal/ft3) = 238,352 gal



Next, use the following equation.

Seed sludge, gal = 100%

(Capacity of digester)(Percent seed sludge required)


(238,352 gallons)(15%)

Seed sludge, gal = 35,752.8 gal, round to 35,800 gal of seed sludge

24. A gravity thickener with a radius of 29.7 ft receives a flow of 0.0752 mgd. If the percent solids are 4.50%, what are the solids loading on the gravity thickener in lb/d/ft2?First, convert the mgd to gpd.Number of gpd = (0.0752 mgd)(1,000,000/mil) = 75,200 gpdKnow: Area of gravity thickener = πr2 where π = 3.14

Equation: Solids loading, lb/d/ft2 = (Gravity thickener area)(100%)

(Flow, gpd)(Percent solids)

Substitute and solve.

Solids loading, lb/d/ft2 = (3.14)(29.7 ft)(29.7 ft)(100%)

(75,200 gpd)(4.5%)

Solids loading, lb/d/ft2 = 1.22 lb/d/ft2

25. What is the solids loading for a dissolved air flotation (DAF) unit in lb/hr/ft2 that is 59.4 ft by 18.0 ft, with a sludge flow of 0.201 mgd and a waste-activated sludge (WAS) concentration of 7,830 mg/L, and the sludge weighs 8.71 lb/gal?First, determine the area of the DAF unit in ft2.DAF area, ft2 = (59.4 ft)(18.0 ft) = 1,069.2 ft2

Next, calculate the solids loading using the following equation.

Equation: Solids loading, lb/hr/ft2 = (DAF area, ft )(24 hr/d)

(WAS, mg/L)(mgd)(lb/gal, Sludge)2

Solids loading, lb/hr/ft2 = (1,069.2 ft DAF)(24 hr/d)

(7,830 mg/L, WAS)(0.201 mgd)(8.71 lb/gal, Sludge)2

Solids loading, lb/hr/ft2 = 0.534 lb/hr/ft2



26. If a DAF unit receives air at an average rate of 9.11 ft3/min, how many lb/d of air does it receive? Know: Air = 0.0807 lb/ft3 at standard temperature, pressure, and average composition.Equation: Air, lb/d = (Air flow, ft3/min)(1,440 min/d)(0.0807 lb/ft3, Air)Substitute and solve.Air, lb/d = (9.11 ft3/min)(1,440 min/d)(0.0807 lb/ft3)Air, lb/d = 1,058.65 lb/d, round to 1,060 lb/d of air

27. Given the following data, determine the removal efficiency of sludge solids on a disk centrifuge.Influent sludge solids = 5,917 mg/L Effluent sludge solids = 536 mg/LFirst, determine the amount of sludge solids removed.Sludge solids removed = 5,917 mg/L – 536 mg/L = 5,381 mg/LNow, determine the removal efficiency.

Percent removal efficiency = Influent solids, mg/L

(Solids removed, mg/L)(100%)

Percent removal efficiency = 5,917 mg/L

(5,381 mg/L)(100%) = 90.94%, round to 90.9%

28. Calculate the filter loading in lb/hr/ft2 on a vacuum filter that has a surface area of 275 ft2, a digested biosolids flow rate of 47 gpm, and a solids concentration of 4.33%.First, convert gpm to gal/hr.Number of gal/hr = (47 gpm)(60 min/hr) = 2,820 gal/hrNext, solve for filter loading.Equation: Vacuum filter loading, lb/hr/ft2 =

Vacuum filter area, ft(Biosolids, gal/hr)(8.34 lb/gal)(Percent solids)

2

Vacuum filter loading, lb/hr/ft2 = 275 ft

(2,820 gal/hr)(8.34 lb/gal)(4.33%/100%)2

Vacuum filter loading, lb/hr/ft2 = 3.703 lb/hr/ft2, round to 3.7 lb/hr/ft2



29. A settleability test result shows that 219 mL of activated biosolids settled in a 1-L graduated cylinder. If the concentration of mixed liquor suspended solids (MLSS) is 2,440 mg/L, what is the biosolids density index (BDI)?

Equation: BDI = (Settled biosolids, mL/L)(1,000 mg/g)

(MLSS, mg/L)(100)


BDI = (219 mL/L)(1,000 mg/g)

(2,440 mg/L)(100) = 1.11 g/mL, which reduces to 1.11 BDI

30. Calculate the unseeded BOD5 in mg/L, given the following data.Start of test bottle dissolved oxygen (DO) = 9.3 mg/L Bottle was incubated for 5 days in the dark at 20°C After 5 days DO = 3.8 mg/L Sample size = 100 mL Total volume = 300 mLGive answer to three significant figures.Equation: BOD5 unseeded, mg/L =

Sample volume, mL(Initial DO, mg/L – Final DO, mg/L)(Total volume, mL)

BOD5 unseeded, mg/L = 100 mL

(9.3 mg/L – 3.8 mg/L)(300 mL) = 16.5 mg/L

31. If 125 lb of magnesium hydroxide are dissolved 15.0 gallons of water, what is the percent strength of the solution?Equation: Percent strength =

(Number of gal)(8.34 lb/gal) Number of lb, Chemical(Number of lb, chemical)(100%)

+


Percent strength = (15.0 gal)(8.34 lb/gal) 125 lb

(125 lb)(100%)+

= 125.1 lb 125 lb

12,500 lb %+

Percent strength = 250.1 lb

12,500 lb % = 49.98%, round to 50.0% Mg(OH)2 solution



32. If the sludge entering a digester has a volatile solids (VS) content of 53.8% and the digester effluent sludge has a VS content of 40.1%, calculate the percent VS reduction.First, convert percentage to decimal form by dividing by 100%53.8%/100% = 0.538 and 40.1%/100% = 0.401Equation:Percent VS reduction =

[Percent influent VS – (Percent influent VS)(Percent effluent VS)](Percent influent VS – Percent effluent VS)(100%)

Percent VS reduction = 0.538 – (0.538)(0.401)(0.538 – 0.401)(100%)

= 0.538 – 0.215738

0.137(100%)

Percent VS reduction = 0.322262

13.7% = 42.51%, round to 42.5% VS reduction

33. What is the percent moisture reduction for a digester, if the raw biosolids is 7.18% solids and the digested biosolids solids is 14.9%?Equation: Percent moisture reduction =

[Percent influent moisture – (Percent influent moisture)(Percent moisture, after digestion)]

(Percent influent moisture – Percent moisture, after digestion)(100%)

First, convert the percentages for solids to moisture percent then to decimal form for easier substitution.Raw biosolids = 100% – 7.18% = 92.82%/100% = 0.9282Digested biosolids = 100% – 14.9% = 85.1%/100% = 0.851Substitute known values and solve.

Percent moisture reduction = {0.9282 – (0.9282)(0.851)}

(0.9282 – 0.851)(100%)

Simplify:Percent moisture reduction =

{0.9282 – 0.7898982}(0.0772)(100%)

Percent moisture reduction = 0.1383018

(0.0772)(100%) = 55.8% moisture reduction

34. Given the following data, how many lb/d of volatile solids (VS) are pumped to a digester?Pumping rate = 2,180 gpd Solids content = 5.51% Volatile solids = 60.1% Specific gravity of sludge = 1.06



First, determine the lb/gal for the sludge.Sludge, lb/gal = (8.34 lb/gal)(1.06) = 8.84 lb/galEquation:

VS, lb/d = 100% 100%

(Number of gpd to digester)(Percent solids)(Percent VS)(Sludge, lb/gal)

VS, lb/d = 100% 100%

(2,180 gpd Solids)(5.51%)(60.1% VS)(8.84 lb/gal)

VS, lb/d = 638.17 lb/d, round to 638 lb/d VS

35. A solution containing exactly 150 gallons of 9.5% hypochlorite is required. How many gallons of a 12.5% solution must be mixed with a 3.5% solution to make the required solution?Solve the problem using the dilution triangle.12.5% 6.0 6.0 parts of the 12.5% solution are required for every 9.0 parts. 9.5%3.5% 3.0 3.0 parts of the 3.5% solution are required for every 9.0 parts. 9.0 total parts

6.0 parts (150 gal) = 100 gallons of the 12.5% solution 9.0 parts

3.0 parts (150 gal) = 50 gallons of the 3.5% solution 9.0 parts 150 gallons

To make the 150 gallons of the 9.5% solution, mix 50 gallons of the 3.5% solution with 100 gallons of the 12.5% solution.

36. A soda ash tank is conical at the bottom and cylindrical at the top. If the diameter of the cylinder is 17.5 ft, with a depth of 34.5 ft, and the cone depth is 14.0 ft, what is the volume of the tank in ft3?First, find the volume of the cone in ft3.Volume, ft3 = 1/3(0.785)(Diameter)2(Depth)Volume, ft3 = 1/3(0.785)(17.5 ft)(17.5 ft)(14.0 ft) = 1,122 ft3

Next, find the volume of the cylindrical part of the tank.Volume, ft3 = (0.785)(Diameter)2(Depth) = (0.785)(17.5 ft)(17.5 ft)(34.5 ft) = 8,294 ft3

Lastly, add the two volumes for the answer.Total volume, ft3 = 1,122 ft3 + 8,294 ft3 = 9,416 ft3, round to 9,420 ft3



37. If a 10.0-ft diameter chemical tank drops 4.31 in. in exactly 3 hr, what is the pumping rate for the chemical in gpm?First, determine the amount in feet the tank level dropped.Drop, ft = (4.31 in.)(1 ft/12 in.) = 0.359 ftThen, determine the volume in gallons for the drop in level of the tank.Equation: Volume, gal = (0.785)(Diameter)2(Drop, ft)(7.48 gal/ft3)Substitute and solve.Volume, gal = (0.785)(10.0 ft)(10.0 ft)(0.359 ft)(7.48 gal/ft3) = 211 galNext, convert 3 hr to min.Number of minutes = (3 hr)(60 min/hr) = 180 minNow, calculate the pumping rate in gpm.Equation: Pumping rate = Flow, gal/Time, minPumping rate = 211 gal/180 min = 1.17 gpm

38. What is the solids loading rate for a secondary clarifier, given the following information?Radius of clarifier = 65.4 ft Primary effluent flow = 3,779,000 gpd Return sludge flow = 725,000 gpd Mixed liquor suspended solids (MLSS) = 2,110 mg/LFirst, determine the surface area of the clarifier.Know: Area = r2 where r equals the radius.Clarifier surface area, ft2 = (3.14)(65.4 ft)(65.4 ft) = 13,430.28 ft2

Next, determine the total flow in mgd.

Total clarifier flow = 1,000,000/mil

3,779,000 gpd 725,000 gpd+ = 4.504 mgd

Lastly, determine the solids loading rate of MLSS.

Solids loading rate, lb MLSS/d/ft2 = Surface area, ft

(MLSS, mg/L)(mgd)(8.34 lb/gal)2

Solids loading rate, lb MLSS/d/ft2 = 13,430.28 ft

(2,110 mg/L, MLSS)(4.504 mgd)(8.34 lb/gal)2

Solids loading rate, lb MLSS/d/ft2 = 5.90 lb MLSS/d/ft2



39. Determine the waste activated sludge (WAS) pumping rate in gpm, given the following data.Amount of WAS to be wasted = 4,380 lb/d WAS suspended solids concentration = 3,170 mg/LFirst, use the “pounds” equation to solve for the number of mgd.Equation: WAS, lb/d = (WAS, mg/L)(Number of mgd)(8.34 lb/gal)Rearrange the equation to solve for mg/L.

Number of mgd = (Number of mg/L WAS )(8.34 lb/gal)

(WAS, lb/d) Substitute values and solve.

Number of mgd = (3,170 mg/L WAS)(8.34 lb/gal)

(4,380 lb/d) = 0.16567 mgd

Now, convert mgd to gpm.

Number of gpm = (1,440 min/d)

(0.16567 mgd)(1,000,000/mil) = 115 gpm

40. Find the motor horsepower (mhp) for a pump with the following parameters.Motor efficiency (ME): 88.7% Pump efficiency (PE): 76.5% Total head (TH): 215 ft Flow: 3.81 mgdFirst, convert mgd to gpm.Gpm = (3.81 mgd)(1,000,000/mil)(1 d/1,440 min) = 2,645.8 gpmUse the following equation for this problem.

Equation: mhp = (3,960)(ME)(PE)

(Flow, gpm)(TH, ft) Substitute and solve.

mhp = (3,960)(88.7%/100% ME)(76.5%/100% PE)

(2,645.8 gpm)(215 ft)

mhp = 212 mhp

41. What is the chlorine dosage in mg/L for a wastewater plant, if the chlorinator setting is 135 lb/d and it treats 2,360,000 gpd?First, convert gpd to mgd.Number of mgd = (2,360,000 gpd) / (1,000,000/mil) = 2.36 mgdEquation: Number of lb/d = (Dose, mg/L)(Number of mgd)(8.34 lb/gal)Rearrange the equation to solve for the chlorine dosage in mg/L.

Chlorine dosage, mg/L = (Number of mgd)(8.34 lb/gal)

Number of lb/d

Chlorine dosage, mg/L = (2.36 mgd)(8.34 lb/gal)

135 lb/d = 6.86 mg/L



42. How many gpd of a 11.8% sodium hypochlorite solution are needed to disinfect a flow of 1,744,000 gallons, if the dosage required is 9.65 mg/L and the specific gravity (sp gr) of the hypochlorite is 1.04?First, convert gpd to mgd.

Number of mgd = 1,000,000/mil1,744,000 gpd

= 1.744 mgd

Next, determine the lb/gal for the hypochlorite solution.Hypochlorite, lb/gal = (8.34 lb/gal)(1.04 sp gr) = 8.6736Next, using the “pounds equation,” calculate the lb/d of chlorine needed.Equation: Chlorine, lb/d = (Dosage, mg/L)(mgd)(8.34 lb/gal)Chlorine, lb/d = (9.65 mg/L)(1.744 mgd)(8.34 lb/gal) = 140.36 lb/dSince the solution is not 100%, divide the percent hypochlorite into the lb/d of chlorine needed.

Hypochlorite, lb/d = 11.8%/100%140.36 lb/d = 1,189.49 lb/d hypochlorite


Hypochlorite, gpd = 8.6736 lb/gal1,189.49 lb/d = 137.14 gpd, round to 137 gpd sodium hypochlorite

43. Given the following data, calculate the feed rate of a polymer solution in mL/min.Influent flow = 875 gpm Polymer dose = 6.80 mg/L Polymer solution specific gravity = 1.29 Polymer percent purity = 40.5%First, convert gpm to mgd.

Number of mgd = 1,000,000/mil

(875 gpm)(1,440 min/d) = 1.26 mgd

Next, determine the lb/gal for the polymer.Polymer, lb/gal = (8.34 lb/gal)(1.29 sp gr) = 10.7586 lb/galNow, calculate the dosage using the following equation:

Polymer dosage, mg/L = (3,785 mL/gal)(mgd)(8.34 lb/gal)(Percent Polymer)

(mL/min)(1,440 min/d)(Polymer, lb/gal)

Rearrange the formula to solve for mL/min.Polymer feed, mL/min =

(1,440 min/d)(Polymer, lb/gal)(Polymer dosage, mg/L)(3,785 mL/gal)(mgd)(8.34 lb/gal)(Percent Polymer)


Polymer feed mL/min = (1,440 min/d)(10.7586 lb/gal)

(6.80 mg/L)(3,785 mL/gal)(1.26 mgd)(8.34 lb/gal)(40.5%/100%)

Polymer feed mL/min = 7.07 mL/min polymer



44. Given the following data, determine the amount of dry solids produced in lb/d.Flow = 1,248,000 gpd Influent suspended solids = 295 mg/L Primary effluent suspended solids = 106 mg/LFirst, determine the number of mg/L of suspended solids (SS) removed.SS removed, mg/L = 295 mg/L, influent – 106 mg/L effluent = 189 mg/L SS removedNext, convert gpd to mgd.


= 1.248 mgd

SS removed, lb/d = (SS removed, mg/L)(Number of mgd)(8.34 lb/gal)SS removed, lb/d = (189 mg/L SS)(1.248 mgd)(8.34 lb/gal)SS removed, lb/d = 1,967.17 lb/d, round to 1,970 lb/d

45. Determine the age of the sludge in an aeration tank, given the following data.Primary effluent flow = 1.27 mgd Volume of aeration tank = 435,000 gallons MLSS = 2,040 mg/L Suspended solids = 73 mg/LFirst, convert the volume of the aeration tank to mil gal.


435,000 gal = 0.435 mil gal

Equation: Sludge age, days = (SS, mg/L)(Flow, mgd)(8.34 lb/gal)

(MLSS, mg/L)(Volume of aeration tank)(8.34 lb/gal)

Sludge age, days = (73 mg/L)(1.27 mgd)(8.34 lb/gal)

(2,040 mg/L)(0.435 mil gal)(8.34 lb/gal)


46. What is the organic loading rate in lb BOD5/d/1,000 ft3 for a trickling filter that is 68.5 ft in diameter and 5.05 ft deep, if the primary effluent flow is 2.46 mgd and the BOD5 is 149 mg/L?First, determine the volume of the tricking filter in ft3.Volume, ft3 = (0.785)(Diameter)2(Depth, ft)Volume, ft3 = (0.785)(68.5 ft)(68.5 ft)(5.05 ft) = 18,601.25 ft3

Next, factor out 1,000 ft3 from the volume = (18.60125)(1,000 ft3)



Next, determine the pounds of BOD5/d/1,000 ft3 using a modified version of the “pounds” equation.

Organic loading rate, lb BOD5/d/1,000 ft3 = Volume of trickling filter, ft /1,000 ft(BOD , mg/L)(Flow, mgd)(8.34 lb/gal)

3 35

Organic loading rate, lb BOD5/d/1,000 ft3 = (18.60125)(1,000 ft )

(149 mg/L BOD )(2.46 mgd)(8.34 lb/gal)3

5

Organic loading rate, lb BOD5/d/1,000 ft3 = 164.34, round to 164 lb BOD5/d/1,000 ft3

47. What is the hydraulic digestion time for a digester that is 25.0 ft in radius, has a level of 13.7 ft, and has a sludge flow of 10,840 gallons per day (gpd)?First, determine the volume of the digester in gallons.Volume, gal = πr2(Depth, ft)(7.48 gal/ft3)Volume, gal = (3.14)(25.0 ft)(25.0 ft)(13.7 ft)(7.48 gal/ft3) = 201,109 galNext, calculate the digestion time in days.

Equation: Digestion time, days = Influent sludge flow, gal/d

Number of gallons Substitute and solve.

Digestion time, days = 10,840 gal/d201,109 gal


48. What is the ratio of volatile solids loading on a digester, if the digester has 59,980 kg of volatile solids (VS), 1,330 lb/d are pumped into it, percentage total solids (TS) are 4.89%, and percentage volatile solids (VS) are 70.8%?First, convert the number of kg to lb.Know: 1 kg = 2.205 lbNumber of lb = (59,980 kg)(2.205 lb/kg) = 132,255.9 lbUse the following expanded equation with percentages to solve for digester volatile solid ratio.

Equation: Digester VS ratio = (lb VS in digester)(TS %/100%)(VS %/100%)

VS added lb/d


Digester VS ratio = (132,255.9 lb)(4.89%/100% TS)(70.8%/100% VS)

1,330 lb/d

Digester VS ratio = (132,255.9 lb)(0.0489 TS)(0.708 VS)

1,330 lb/d = 0.290 VS ratio



49. Given the following data, calculate the population equivalent.Average wastewater flow for the day = 3.06 ft3/s BOD5/person = 0.22 lb/d BOD5 concentration in the wastewater = 2,525 mg/LFirst, convert ft3/s to mgd.Know: 1 ft3/s = 0.6463 mgdNumber of mgd = (3.06 ft3/s)(0.6463 mgd/ft3/s) = 1.9777 mgdUse the following equation to solve this problem.

Number of people = lb/d of BOD /person

(BOD , mg/L)(mgd)(8.34 lb/gal)5

5

Number of people = 0.22 lb/d

(2,525 mg/L BOD )(1.9777 mgd)(8.34 lb/gal)5

Number of people = 189,306 people, round to 190,000 people

50. How many pounds of MLSS are being aerated, if the aeration tank is 49.8 ft in diameter, with a sludge height of 13.5 ft and a concentration of MLSS of 1,905 mg/L, and the specific gravity of the MLSS is 1.03?First, determine how many gallons are in the aeration tank.Number of gallons = (0.785)(Diameter)2(Height, ft)(7.48 gal/ft3)Number of gallons = (0.785)(49.8 ft)(49.8 ft)(13.5 ft)(7.48 gal/ft3) = 196,591 galNext, convert gallons to mil gal.


196,591 gal = 0.196591 mil gal

Next, determine the lb/gal for the MLSS.Number of lb/gal, MLSS = (8.34 lb/gal)(1.03 sp gr) = 8.59 lb/galNext, determine the pounds of MLSS under aeration using a modified version of the “pounds” equation because the MLSS weighs more than water (8.34 lb/gal).Equation: Number of lb MLSS = (MLSS, mg/L)(Number of mil gal)(MLSS lb/gal)Substitute values and solve.Number of lb MLSS = (1,905 mg/L MLSS)(0.196591 mil gal)(8.59 lb/gal)Number of lb MLSS = 3,217 lb, round to 3,220 lb MLSS

51. A digester tank has a level capacity of 28.5 feet. Currently there is 18.2 feet of sludge water in the tank. What would the SCADA reading be on the board in mA for a 4 mA to 20 mA signal?Equation:Current process reading =

16 mA span(Live signal, mA – 4 mA offset)(Maximum capacity)



Substitute known values.

18.2 ft = 16 mA span

(Live signal, mA – 4 mA offset)(28.5 ft)

Rearrange formula to solve for the current number of milliamps.

Live signal, mA – 4 mA offset = 28.5 ft

(18.2 ft)(16 mA)

Live signal, mA = 28.5 ft

(18.2 ft)(16 mA) + 4 mA

Live signal, mA = 10.218 mA + 4 mA = 14.218 mA, round to 14.2 mA

52. A total of 2,860 gpd of sludge (primary sludge) with 5.64% solids content and weighing 8.43 lb/gal is pumped to a wastewater thickener tank. Determine the amount of sludge that should flow from the thickener tank in gpd, if the sludge (secondary sludge) is further concentrated to 7.18% solids and weighs 8.52 lb/gal. Note: For convenience, primary sludge is abbreviated to 1° sludge and secondary sludge to 2° sludge.Equation:(x gpd)(2° sludge lb/gal)(% 2° sludge) = (1° sludge, gpd)(1° sludge lb/gal)(% 1° sludge)

Substitute known values and solve.(x gpd)(8.52 lb/gal)(7.18%/100%) = (2,860 gpd)(8.43 lb/gal)(5.64%/100%)

x gpd = (8.52 lb/gal)(7.18%/100%)

(2,860 gpd)(8.43 lb/gal)(5.64%/100%) = 2,223 gpd, round to 2,220 gpd

53. What is the weir overflow rate in gpd/ft, if the flow is 0.266 mgd and the radius of the clarifier is 39.9 ft?First, calculate the length of the weir.Weir length, ft = 2(radius, ft)Weir length, ft = 2(3.14)(39.9 ft) = 250.572 ftNext, convert mgd to gpd.Number of gpd = (0.266 mgd)(1,000,000 gal/mil) = 266,000 gpdNext, solve for the weir overflow rate.

Equation: Weir overflow rate, gpd/ft = Weir length, ft

Flow, gpd

Weir overflow rate, gpd/ft = 250.572 ft

266,000 gpd = 1,061.57 gpd/ft, round to 1,060 gpd/ft



54. Given the following data, how many lb/d of volatile solids (VS) are pumped to a digester?Influent pumping rate = 2,850 gpd Solids content = 5.15% Volatile solids (VS) = 59.6% Specific gravity of sludge = 1.05First, determine the lb/gal for the sludge.Sludge, lb/gal = (8.34 lb/gal)(1.05 sp gr) = 8.757 lb/galEquation:

VS, lb/d = 100% 100%

(Number of gpd to digester)(Percent solids)(Percent VS)(Sludge, lb/gal)

VS, lb/d = 100% 100%

(2,850 gpd Solids)(5.15%)(59.6% VS)(8.757 lb/gal)

VS, lb/d = 766.04 lb/d, round to 766 lb/d VS

55. What percent hypochlorite solution would result, if 825 gallons of a 12.5% solution were mixed with exactly 175 gallons of a 3.5% solution? Assume both solutions have the same density and give answer to three significant numbers.First, find the total volume that would result from mixing these two solutions:Total volume = 825 gal + 175 gal = 1,000 galThen, write the equation:

(Concentration1)(Volume1) + (Concentration2)(Volume2) = (Concentration3)(Volume3)

Condensed as C1V1 + C2V2 = C3V3, where C1 and C2 = % Concentration of the two solutions before being mixed, V1 and V2 = Volume of the two solutions before being mixed, and C3 and V3 = the resulting % Concentration and Volume, respectively.Substitute known values and solve.

100%(12.5%)(825 gal)

100%(3.5%)(175 gal)

+ = 100%

C (1,000 gal)3

103.125 gal + 6.125 gal = 100%

C (1,000 gal)3 Solving for C3:

C3 = 1,000 gal

(103.125 gal 6.125 gal)(100%)+ = 1,000 gal

(109.25 gal)(100%)

C3 = 10.925 round to 10.9% final solution



56. Find the number of pounds of MLSS being aerated, if the aeration tank is 50.0 ft in diameter, with a sludge height of 14.2 ft and a concentration of MLSS of 2,280 mg/L, and the specific gravity of the MLSS is 1.05.First, determine how many gallons are in the aeration tank.Number of gallons = (0.785)(Diameter)2(Height, ft)(7.48 gal/ft3)Number of gallons = (0.785)(50.0 ft)(50.0 ft)(14.2 ft)(7.48 gal/ft3) = 208,449 galNext, convert gallons to mil gal.


208,449 gal = 0.208449 mgd

Next, determine the lb/gal for the MLSS.Number of lb/gal, MLSS = (8.34 lb/gal)(1.05 sp gr) = 8.757 lb/galNext, determine the pounds of MLSS under aeration using a modified version of the “pounds” equation because the MLSS weighs more than water.Equation: Number of lb MLSS = (MLSS, mg/L)(Number of mil gal)(MLSS lb/gal)Substitute values and solve.Number of lb MLSS = (2,280 mg/L MLSS)(0.208449 mil gal)(8.757 lb/gal)Number of lb MLSS = 4,161.88 lb, round to 4,160 lb MLSS

57. What is the flow velocity in feet per second (ft/s) for a trapezoidal channel, given the following data?Bottom width, w1 = 5.11 ft Water surface width, w2 = 8.73 ft Depth = 3.91 ft Flow = 22.5 ft3/sEquation: Flow (Q), ft3/s =

2(w w )1 2+ (Depth, ft)(Velocity, ft/s)

Rearrange the formula to solve for velocity in ft/s.

Velocity, ft/s = (w w )(Depth, ft)

2(Q, ft /sec)1 2

3

+ Substitute and solve.

Velocity, ft/s = (5.11 ft 8.73 ft)(3.91 ft)

2(22.5 ft /s)3

+ =

(13.84 ft)(3.91 ft)45 ft /s3

= 0.832 ft/s

58. What is the biosolids retention time (BRT) in days for a digester that is 70.1 ft in diameter, has a working level of 17.55 ft, and receives an average flow of 10.5 gpm?First, convert gpm to gpd.Digester influent, gpm = (10.5 gpm)(1,440 min/d) = 15,120 gpd



Next, determine the working volume in gallons for the digester.Equation: Digester volume, gal = (0.785)(Diameter, ft)2(Height, ft)(7.48 gal/ft3)Digester volume, gal = (0.785)(70.1 ft)(70.1 ft)(17.55 ft)(7.48 gal/ft3) = 506,389 galLastly, calculate the BRT.

Equation: BRT, days = Influent flow, gpd

Digester working volume, gal


BRT, days = 15,120 gpd506,389 gal

= 33.5 days

59. Given the following data, calculate the cost of running a pump in dollars and cents per day.Flow = 1.58 mgd TDH = 167 ft Motor efficiency = 89.9% Pump efficiency = 71.4% Cost in kilowatt hours = $0.076First, convert mgd to gpm.

Number of mgd = (1,440 min/d)

(1.58 mgd)(1,000,000/mil) = 1,097.2 gpm

Next, determine the horsepower (hp).

Equation: Motor hp = (3,960)(Motor efficiency)(Pump efficiency)

(Flow, gpm)(TDH, ft)

Motor hp = (3,960)(89.9%/100%)(71.4%/100%)

(1,097.2 gpm)(167 ft) = 72.09 hp

Now, calculate the cost of running the pump in dollars and cents.Equation: Cost, $/d = (Motor hp)(24 hr/d)(0.746 kW/hp)(Cost/kW-hr)Cost, $/d = (72.09 hp)(24 hr/d)(0.746 kW/hp)($0.076/kW-hr) = $98.09/d

60. A wastewater treatment plant is treating 2,080,000 gpd with a 12.5% sodium hypochlorite solution. If the dosage required is 9.75 mg/L and the specific gravity of the hypochlorite is 1.03, how many gpd of sodium hypochlorite are required?First, convert gpd to mgd.


= 2.08 mgd

Next, determine the lb/gal for the hypochlorite solution.Hypochlorite, lb/gal = (8.34 lb/gal)(1.03 sp gr) = 8.5902



Next, using the “pounds equation,” calculate the lb/d of chlorine needed.Equation: Chlorine, lb/d = (Dosage, mg/L)(mgd)(8.34 lb/gal)Chlorine, lb/d = (9.75 mg/L)(2.08 mgd)(8.34 lb/gal) = 169.1352 lb/dSince the solution is not 100%, divide the percent hypochlorite into the lb/d of chlorine needed.

Hypochlorite, lb/d = 12.5%/100%

169.1352 lb/d = 1,353.08 lb/d hypochlorite


Hypochlorite, gpd = 8.5902 lb/gal1,353.08 lb/d = 157.51 gpd,

round to 158 gpd sodium hypochlorite

61. What should the chemical feed pump be set on in gpd, given the following data?Plant’s treatment flow = 2,610,000 gpd Alum dose = 11.05 mg/L Alum = 5.37 lb/gal (assuming 48.5% purity of aluminum sulfate)First, convert gpd to mgd.


= 2.61 mgd

Next, calculate the number of lb/d of alum required.Equation: lb/d = (mg/L)(mgd)(8.34 lb/gal)Alum, lb/d = (11.05 mg/L)(2.61 mgd)(8.34 lb/gal) = 240.53 lb/d of dry alumNow, calculate the amount of liquid alum by dividing the amount of dry alum by 5.37 lb/gal.

Alum, gpd = 5.37 lb/gal240.53 lb/d = 44.8 gpd of liquid alum solution

62. Given the following data on a wastewater treatment pond, calculate the organic loading rate in lb BOD5/d/acre.Influent flow = 278,000 gpd Surface area of pond = 4.11 acre-ft Influent BOD5 concentration = 231 mg/LFirst, convert gallons per day to mgd.

Number of mgd = 1,000,000/mil278,000 gpd

= 0.278 mgd

Next, determine the pounds of BOD5/d/acre using a modified version of the “pounds” equation.

Organic loading rate, lb BOD5/d/acre = ftSurface area of pond,


-acre



Organic loading rate, lb BOD5/d/acre = ft4.11

(231 mg/L BOD )(0.278 mgd)(8.34 lb/gal)5

-acreOrganic loading rate, lb BOD5/d/acre = 130.31 lb BOD5/d/acre, round to 130 lb BOD5/d/acre

63. Given the following data, calculate the digester loading in lb volatile solids (VS)/d/1,000 ft3.Digester diameter = 49.9 ft Sludge level = 14.8 ft Influent sludge flow = 10,500 gpd Percent sludge solids = 4.88% Percent volatile solids = 70.1% Specific gravity of sludge = 1.04Equation: Digester loading, lb VS/d/1,000 ft3 =

(0.785)(Diameter) (Sludge level)(Flow, gpd)(8.34 lb/gal)(sp gr)(Percent sludge)(Percent volatile solids)

2

Substitute and solve.Digester loading, lb VS/d/1,000 ft3 =

(0.785)(49.9 ft)(49.9 ft)(14.8 ft)(10,500 gpd)(8.34 lb/gal)(1.04 sp gr)(4.88%/100%)(70.1%/100%)

Digester loading, lb VS/d/1,000 ft3 = 28,928.94 ft

3,115.49 lb VS/d3

Factor out 1,000 ft3 from the denominator and do not divide by 1,000 ft3, as it will become part of the units and not part of the calculation.

Digester loading, lb VS/d/1,000 ft3 = (28.92894)(1,000 ft )

3,115.49 lb VS/d3

Digester loading, lb VS/d/1,000 ft3 = 1,000 ft

107.69 lb VS/d3

, round to 108 lb VS/d/1,000 ft3

64. Calculate the required waste rate from an aeration tank in mgd and gpm, given the following data.Aeration tank volume = 0.585 mil gal Desired COD lb/MLVSS lb = 0.18 Primary effluent flow = 2.39 mgd Primary effluent COD = 134 mg/L Mixed liquor volatile suspended solids (MLVSS) = 3,588 mg/L Waste volatile solids (WVS) concentration = 4,075 mg/LFirst, find the existing MLVSS in pounds.Equation: Existing MLVSS, lb = (MLVSS, mg/L)(mil gal)(8.34 lb/gal)Existing MLVSS, lb = (3,588 mg/L)(0.585 mil gal)(8.34 lb/gal) = 17,505 lb MLVSS



Next, determine the desired MLVSS in pounds.

Equation: Desired MLVSS, lb = Desired COD lb/MLVSS lb

(Primary effluent COD, mg/L)(mgd)(8.34 lb/gal)

Desired MLVSS, lb = 0.18 COD lb/MLVSS lb

(134 mg/L)(2.39 mgd)(8.34 lb/gal) = 14,839 lb MLVSS

Next, subtract the existing MLVSS from the desired MLVSS to find the waste in pounds.Waste, lb = 17,505 lb – 14,839 lb = 2,666 lbNext, calculate the waste rate in mgd.

Equation: Waste rate, mgd = (WVS concentration, mg/L)(8.34 lb/gal)

Waste, lb

Waste rate, mgd = (4,075 mg/L)(8.34 lb/gal)

2,666 lb = 0.078445 mgd = 0.078 mgd

Lastly, calculate the waste rate in gpm.

Waste rate, gpm = 1,440 min/d

(0.078445 mgd)(1,000,000 gpd/mgd) = 54.48 gpm, round to 54 gpm

65. What is the pressure 5.25 ft from the bottom of a tank at exactly 48 hr after the pumping started, given the following data?Tank radius = 35.5 ft Initial sludge depth = 3.42 ft Sludge flow into tank = 72.5 gpmFirst, calculate the amount the pump added to the tank in 48 hr.Volume, gal = (72.5 gpm)(60 min/hr)(48 hr) = 208,800 galNext, determine the depth of sludge this would add to the tank.Equation: Volume, gal = (radius)2(Depth, ft)(7.48 gal/ft3)Rearrange the equation to solve for depth.

Depth, ft = (radius) (7.48 gal/ft )

Volume, gal2 3r

Depth, ft = 3.14(35.5 ft)(35.5 ft)(7.48 gal/ft )

208,800 gal3

= 7.054 ft

Next, add the level in feet at the beginning to the level in feet that was added.Total level, ft = 3.42 ft + 7.054 ft = 10.474 ft



Next, subtract the final level in feet from the level in feet that is in question.Level, ft = 10.474 ft – 5.25 ft = 5.224 ftFinally, determine the pressure 5.25 ft above the tank bottom.Pressure = (5.224 ft)(0.4335 psi/ft) = 2.26 psi

66. Given the following parameters, calculate how long a primary sludge pump should operate in minutes per hour.Plant flow = 675 gpm Sludge pump = 42.0 gpm Influent suspended solids (SS) = 348 mg/L Effluent SS = 129 mg/L Sludge = 4.76% solidsFirst, convert gpm to mgd.


(675 gpm)(1,440 min/d) = 0.972 mgd

Equation:Operating time, min/hr =

(Sludge pump, gpm)(Percent Solids)(24 hr/d)(Flow, mgd)(Influent SS, mg/L – Effluent SS, mg/L)(100%)


Operating time, min/hr = (42.0 gpm)(4.76%)(24 hr/d)

(0.972 mgd)(348 mg/L SS – 129 mg/L SS)(100%)

Operating time, min/hr = (42.0 gpm)(4.76%)(24 hr/d)

(0.972 mgd)(219 mg/L SS)(100%)

Operating time, min/hr = 4.44 min/hr

67. Determine the waste activated sludge (WAS) flow in gpm, given the following data.Influent flow = 1.82 mgd Clarifier radius = 40.0 ft Clarifier depth = 13.6 ft Aerator = 0.425 mil gal MLSS = 2,440 mg/L RAS SS = 6,945 mg/L Secondary effluent SS = 28.5 mg/L Target solids retention time (SRT) = 10 days exactlyFirst, determine the volume in mil gal for the clarifier and add to the aerator volume.

Equation: Clarifier, gal = 1,000,000/mil

(radius) (Depth, ft)(7.48 gal/ft )2 3r



Clarifier, gal = 1,000,000/mil

3.14(40.0 ft)(40.0 ft)(13.6 ft)(7.48 gal/ft )3 = 0.511 gal

Total volume = 0.511 + 0.425 = 0.936 mil galEquation:Target SRT =

(RAS SS mg/L)( mgd)(8.34 lb/gal) (Effluent SS, mg/L)(Flow, mgd)(8.34 lb/gal)(MLSS mg/L)(Clarifier, Aerator Volume, mil gal)(8.34 lb/gal)

x +


10 days SRT = x(6,945 mg/L)( mgd)(8.34 lb/gal) (28.5 mg/L)(1.82 mgd)(8.34 lb/gal)(2,440 mg/L)(0.936 mil gal)(8.34 lb/gal)

+

10 days SRT = x(6,945 mg/L)( mgd)(8.34 lb/gal) 432.60 lb/d19,047.23 lb MLSS

+

Rearrange the equation so that x mgd is in the numerator.

(6,945 mg/L)(x mgd)(8.34 lb/gal) + 432.60 lb/d = 10 days SRT

19,047.23 lb MLSS

(6,945 mg/L)(x mgd)(8.34 lb/gal) + 432.60 lb/d = 1,904.723 lb/dSubtract 432.60 lb/d from both sides of the equation.(6,945 mg/L)(x mgd)(8.34 lb/gal) = 1,472.123 lb/d

x mgd = (6,945 mg/L)(8.34 lb/gal)

1,472.123 lb/d = 0.025416 mgd WAS

Lastly, convert mgd to gpm.

WAS flow, gpm = 1,440 min/d

(0.025416 mgd)(1,000,000/mil) = 17.65 gpm, round to 17.6 gpm WAS

68. What should the chemical feeder be set on in mL/min, given the following data?Polymer dosage = 8.80 mg/L Plant flow = 1,340 gpm Sp gr of polymer = 13.03 lb/galFirst, convert gpm flow to mgd.


(1,340 gpm)(1,440 min/d) = 1.9296 mgd

Next, determine the lb/d of polymer using the pounds formula.Polymer, lb/d = (Dosage, mg/L)(mgd)(8.34 lb/gal)Polymer, lb/d = (8.80 mg/L)(1.9296 mgd)(8.34 lb/gal) = 141.62 lb/d



Next, calculate the number of gallons polymer used.

Polymer, gal = 13.03 lb/gal141.62 lb/d = 10.87 gal

Now, using the following equation, calculate the mL/min of polymer being used.

Equation: Number of mL/min = 1,440 min/d

(Number of gal used)(3,785 mL/gal)


Polymer, mL/min = 1,440 min/d

(10.87 gal)(3,785 mL/gal) = 28.6 mL/min polymer

69. What is the organic loading rate in lb BOD5/d/1,000 ft3 for a trickling filter that is 151 ft in diameter and 5.10 ft deep, if the primary effluent flow is 1,220 gpm and the BOD5 is 207 mg/L?First, determine the volume of the tricking filter in ft3.Equation: Volume, ft3 = (0.785)(Diameter)2(Depth, ft)Volume, ft3 = (0.785)(151 ft)(151 ft)(5.10 ft) = 91,283.8 ft3

Next, factor out 1,000 ft3 from the volume = (91.2838)(1,000 ft3)Next, convert gpm to mgd.


(1,220 gpm)(1,440 min/d) = 1.7568 mgd

Next, determine the pounds of BOD5/d/1,000 ft3 using a modified version of the “pounds” equation.Organic loading rate, lb BOD5/d/1,000 ft3 =

Volume of trickling filter, ft /1,000 ft(BOD , mg/L)(Flow, mgd)(8.34 lb/gal)

3 35

Organic loading rate, lb BOD5/d/1,000 ft3 = (91.2838)(1,000 ft )

(207 mg/L BOD )(1.7568 mgd)(8.34 lb/gal)3

5

Organic loading rate, lb BOD5/d/1,000 ft3 = 33.225, round to 33.2 lb BOD5/d/1,000 ft3

70. What is the air-to-solids ratio for a dissolved air flotation (DAF) unit that has an air flow rate equal to 9.10 ft3/min, a solids concentration of 0.76%, and a flow of 0.185 mgd?Know: Air = 0.0807 lb/ft3 at standard temperature, pressure, and average composition.

Equation: Air-to-solids ratio = (gpm)(Percent solids/100%)(8.34 lb/gal)

(Air flow, ft /min)(Air, lb/ft )3 3

Since equation requires gpm, first convert mgd to gpm.

Number of gpm = 1,440 min/d

(0.185 mgd)(1,000,000/mil) = 128.47 gpm



Now, using the air-to-solids equation, substitute and solve.

Air-to-solids ratio = (128.47 gpm)(0.76%/100%)(8.34 lb/gal)

(9.10 ft /min)(0.0807 lb/ft )3 3

= 0.090 air-to-solids ratio

71. How many lb/hr of air does a DAF unit receive, if the unit receives air at an average rate of 0.259 m3/min? Know: Air = 0.0807 lb/ft3 at standard temperature, pressure, and average composition and there are 35.3 ft3/m3.First, convert m3/min to ft3/min.Air flow, ft3/min = (0.259 m3/min)(35.3 ft3/m3) = 9.1427 ft3/minEquation: Air, lb/d = (Air flow, ft3/min)(60 min/hr)(0.0807 lb/ft3, Air)Substitute and solve.Air, lb/d = (9.1427 ft3/min)(60 min/hr)(0.0807 lb/ft3)Air, lb/d = 44.3 lb/hr of air

72. Given the following data, determine the feed time for a centrifuge in minutes.Capacity of basket centrifuge = 24.9 ft3 Influent sludge flow = 61 gpm Influent sludge solids concentration = 7,070 mg/L Average solids concentration in the basket = 7.92%First, convert the influent sludge solids concentration to percent.Know: 1% = 10,000 ppm or mg/L.

Percent sludge solids concentration = 10,000 mg/L / 1%

7,070 mg/L = 0.707%

Equation:Feed time, min =

(Sludge flow, gpm)(8.34 lb/gal)(Sludge solids conc.)(Centrifuge capacity)(7.48 gal/ft )(8.34 lb/gal)(Percent solids conc.)3

Feed time, min = (61 gpm)(8.34 lb/gal)(0.707%/100%)

(24.9 ft )(7.48 gal/ft )(8.34 lb/gal)(7.92%/100%)3 3

Feed time, min = 34.20 min, round to 34 min



73. Given the following data for blending compost (BC) with wood chips, calculate the percent of the blended compost.Bulk density of sludge = 1,690 lb/yd3

Sludge volume = 10.5 yd3

Sludge solids content = 17.7% Density of wood chips = 585 lb/yd3

Wood chip solids = 52.8% Mix ratio (MR) of wood chips to sludge = 3 to 1Equation: Percent solids BC =

(Sludge, yd )(lb/yd ) (Sludge, yd )(Mix ratio)(Wood chips, lb/yd )[(Sludge, yd )(lb/yd )(% solids) (Sludge, yd )(MR)(lb/yd )(% solids, chips)](100%)

3 3 3 3

3 3 3 3

++

Substitute known values and solve.Percent solids BC =

(10.5 yd )(1,690 lb/yd ) (10.5 yd )(3)(585 lb/yd )[(10.5 yd )(1,690 lb/yd )(17.7%/100%) (10.5 yd )(3)(585 lb/yd )(52.8%/100%)](100%)

3 3 3 3

3 3 3 3

+

+

Percent solids BC = 17,745 18,427.5

(3,140.865 9729.72)(100%)+

+ = 36,172.5

(12,870.585)(100%)

Percent solids BC = 35.58%, round to 35.6% solids in blended compost

74. Determine the number of acres used in the following land application of biosolids, given the following data.Hydraulic loading rate = 0.46 in./d Flow = 375,000 gpdFirst, determine the number of gallons per acre-inch.Number of gal/acre-in. = (43,560 ft2/acre)(1 ft/12 in.)(7.48 gal/ft3) = 27,152 gal/acre-in.

Equation: Hydraulic loading rate, in./d = in.) (Area, acres)

Flow, gpd-(27,152 gal/acre

Rearrange the equation and solve for area, in acres.

Area, acres = in.)(Hydraulic loading rate, in./d)


Area, acres = in.)(0.46 in./d)

375,000 gpd-(27,152 gal/acre

Area, acres = 30.02 acres, round to 30 acres



75. A 1.00% stock polymer solution (10,000 ppm or 10,000 mg/L) is desired for performing a jar test. If the polymer has a specific gravity of 1.31 and is 35.1% polymer, how many milliliters of polymer are required to make exactly 1,000 mL of the 1.00% stock solution?First, find the number of lb/gal of polymer.Polymer, lb/gal = (sp gr)(8.34 lb/gal) = (1.31 sp gr)(8.34 lb/gal) = 10.9254 lb/galNext, determine the number of g/mL.

Number of g/mL, Polymer = (3,785 mL/gal)(100%)

(10.9254 lb/gal)(35.1% Polymer)(454 g/lb)

Number of g/mL, Polymer = 0.460 g/mLConvert g/mL to mg/mL.Number of mg/mL = (0.460 g/mL)(1,000 mg/g) = 460 mg/mLNext, convert mL to liters by multiplying by 1,000.Number mg/L = (460 mg/mL)(1,000 mL/L) = 460,000 mg/LNext, determine the number of mL required.Equation: C1V1 = C2V2

(460,000 mg/L)(x, mL) = (10,000 mg/L)(1,000 mL)

x, mL = 460,000 mg/L

(10,000 mg/L)(1,000 mL) = 21.74 mL, round to 21.7 mL of polymer

Now, using a pipette, add 21.7 mL of the 35.1% polymer solution to a clean, dry 1,000-mL flask. Dilute the polymer to the 1,000-mL mark with deionized water. Add a magnetic stir bar and place the flask on a magnetic stirrer. Turn the magnetic stirrer on and mix this solution with the bar as vigorously as possible for at least 10 minutes.Thus, every 1 mL of polymer solution that is added to 1,000-mL sample of raw water will add a 10 mg/L dose (because of second dilution with raw water; 10,000 mg/L/ 1,000 mg/L raw water sample = 10 mg/L). If 3 mL of this stock solution were added to the 1,000-mL raw water sample, it would be a dose of 30 mg/L. If you are using the 2-L square jars, simply double the mL added for each 10 mg/L dosage increase desired. Another way is to feed the polymer neat by using a micropipette; pipette 0.0217 mL of polymer into a 1,000-mL raw water sample.


229

Appendix A

Common Conversion Factors

Length1 inch (in.) = 2.54 centimeters (cm)100 cm = 1 meter (m)1 m = 39.37 in.1 m = 3.281 feet (ft)1 yard = 0.9144 m1,000 m = 1 kilometer (km)1 km = 1.609 miles (mi)

Area1 acre (ac) = 43,560 square feet (ft2)1 acre-foot (acre-ft) = 43,560 cubic feet (ft3)2.4711 ac = 1 hectare (ha)1 ha = 0.4047 ac1 ha = 10,000 square meters (m2)1 square mile (mi2) = 640 ac

Volume2 pints (pt) = 1 quart (qt)8 pt = 1 gallon (gal)4 qt = 1 gal1 qt = 32 fl oz1 gal = 128 fl oz1 gal = 3.785 liters (L)1 L = 1.0567 qt1 L = 1,000 milliliters (mL)3,785 mL = 1 gal1,000 L = 1 cubic meter (m3)1 mL = 1 cubic centimeter (cm3)1 = 35.3 cubic feet (ft3)1 ft3 = 7.48 gal



1 million gallons (mil gal) = 3.07 acre-ft1 acre-ft = 325,829 gal

Weight1 gram (g) = 1,000 milligrams (mg)1,000 g = 1 kilogram (kg)1 pound (lb) = 454 gm1 lb = 7,000 grains (gr)1 kg = 2.205 pounds (lb)2,000 lb = 1 ton1 mg/L = 1 part per million (ppm)1 grain per gal (gpg) = 17.1 ppm

DensityWater has a density of 1 g per mL (1 g/mL) or 8.34 lb/gal or 62.4 lb/ft3

Pressure1 pound per square inch (psi) = 2.307 ft of water1 ft of water = 0.4335 psi1 atmosphere (atm) = 14.7 psi1 atm = 29.92 in. of mercury1 atm = 33.90 ft of water1 atm = 760 millimeters (mm) of mercury

Power1 horsepower (hp) = 0.746 kilowatts (kW)1 kW = 1.341 hp

Flow1 miner’s inch = 1.5 ft3/min1 ft3 = 448.8 gal/min (gpm)1 ft3/s = 0.6463 million gallons per day (mgd)1 mgd = 1.547 ft3/s

Concentration1% solution = 1 part in 100 parts1% solution = 10,000 parts per million (ppm)


APPENDIX A 231

1 ppm = 1 mg/L1 grain per gal (gpg) = 17.12 ppm

TemperatureDegrees Fahrenheit (°F) = (9°F/5°C)(°C) + 32°FDegrees Celsius (°C) = (°F – 32°F)(5°C/9°F)


233

Appendix B

Summary of Wastewater Treatment Equations (All Grades)

Area Equations

Area of a rectangle = (Length)(Width)

Area of a circle (tank) = (0.785)(Diameter)2 or πr2

Area of a parallelogram = (Base)(Height)

Area of a trapezoid = (Altitude)2

(Base Base )1 2+

Basic Electricity Formulas

Voltage = (amps)(Resistance, ohms)

Resistance, ohms = Voltage/amps

Current process reading = 16 milliamp span

(Live signal, mA – 4 mA offset)(Maximum capacity)

Biochemical Oxygen Demand Loading Equations

BOD5, lb/d = (BOD5, mg/L)(Number of mgd)(8.34 lb/gal)

Biochemical Oxygen Demand Unseeded and Seeded Equations

BOD5 unseeded, mg/L = Sample volume, mL

(Initial DO, mg/L – Final DO, mg/L)(Total volume, mL)

Seed correction, mg/L = Total volume, mL

(BOD of seed stock, mg/L)(Seed stock, mg/L)5

BOD5 seeded, mg/L = Sample volume, mL

(Initial DO, mg/L – Final DO, mg/L – Seed correction, mg/L)(Total volume, mL)



Biosolids Concentration Factor

CF = Percent influent biosolids

Percent thickened biosolids

Biosolids Pumping and Production Formulas

Estimated pumping rate = (Percent solids in sludge)(Sludge, lb/gal)(1,440 min/d)

(Influent TSS, mg/L – Effluent TSS, mg/L)(Flow, mgd)(8.34 lb/gal)

Biosolids, lb/mil gal = (Flow, mgd)(Number of days)

(Biosolids, gal)(8.34 lb/gal)

Biosolids, lb/mil gal = Flow, mil gal

(Biosolids, gal/d)(8.34 lb/gal)

Biosolids, wet tons/yr = 2,000 lb/ton

(Biosolids, lb/mil gal)(mgd)(365 d/yr)

Estimated pumping rate = (Percent solids in sludge)(Sludge, lb/gal)(1,440 min/d)

(Influent TSS, mg/L – Effluent TSS, mg/L)(Flow, mgd)(8.34 lb/gal)

Biosolids Retention Time Equation

BRT, days = Influent flow, gpd

Digester working volume, gal

Biosolids Volume Index and Biosolids Density Index Equations

BVI = MLSS, mg/L

(Settled biosolids, mL/L)(1,000 mg/g)

BDI = (Settled biosolids, mL/L)(1,000 mg/g)

(MLSS, mg/L)(100)

Centrifuge Thickening Equation

Hydraulic loading, gpd = (Sludge flow, gpm)(1,440 min/d)

Feed time, min = (Flow, gpm)(Solids concentration, %/100%)(8.34 lb/gal)(Capacity, ft )(Solids, %/100%)(7.48 gal/ft )(8.34 lb/gal)3 3


APPENDIX b 235

or simplified:

Feed time, min = (Flow, gpm)(Solids concentration, %)(Capacity, ft )(Solids, %)(7.48 gal/ft )3 3

Chemical Feed Solution Settings

Feed rate, mL/min = 1,440 min/d

(gpd)(3,785 mL/gal) or

Number of mL/min = 1,440 min/d

(Number of gallons used)(3,785 mL/gal)

Chemical Oxygen Demand Loading Formula

COD, lb/d = (COD, mg/L)(Number of mgd)(8.34 lb/gal)

Chemistry and Laboratory Equations

Moles = Gram formula weight

Grams of chemical

Percent of element in compound = Molecular Wt of compound

(Molecular Wt of the element)(100%)

BOD5 unseeded, mg/L = Sample volume, mL

(Initial DO, mg/L – Final DO, mg/L)(Total volume, mL)

Seed correction, mg/L = Total volume, mL

(BOD of seed stock, mg/L)(Seed stock, mg/L)5

BOD5 seeded, mg/L =

Sample volume, mL(Initial DO, mg/L – Final DO, mg/L – Seed correction, mg/L)(Total volume, mL)

Percent of an element in a compound = Molecular Wt of the compound

(Molecular Wt of the element)(100%)

Molarity = Liters solutionMoles solute

Normality (N) = equivalents of Solute

Number of liters of Solution-Number of gram



Dosage, mg/L = Sample size, mL

(Stock, mL)(1,000 mg/gram)(Concentration in g/L)

Percent VS = Weight of total solids, g(Solids lost, g)(100%)

Alum, mg/L = Sample size, mL

(Stock, mL)(1,000 mg/g)(Concentration in g/L)

Solids (ash), g = Sample and dish dried, g – Burnt sample, g

Circumference Formulas

Circumference = π(Diameter)

Circumference = 2π(radius) or 2πr

Common Conversion Factors

Gallons to pounds: Number of pounds (lb) = (Number of gal)(8.34 lb/gal)

Gallons to cubic feet: Number of ft3 = 7.48 gal

(Number of gal) (1 ft )3

Acre-feet to cubic feet: Number of ft3 = (Number of acre-ft)(43,560 ft3/acre-ft)

mgd to ft3/s: Number ft3/s = (Number of mgd) (1,000,000 gal) (1 ft3) (1 day) (1 min)

1 mil gal (7.48 gal) (1,440 min) (60 sec)

ft3/s to mgd: Number of mgd = (Number of ft3) (60 sec) (1,440 min) (7.48 gal) (1 mil gal)

sec min day ft3 1,000,000 gal

ft3/s to gpd = Number of gpd = (Number of ft3/s)(86,400 s/d)(7.48 gal/ft3)

gpm to ft3/s: Number of ft3/s = (60 s/min)(7.48 gal/ft )

Number of gpm3

ppm to Percent: Percent solution = 10,000 ppm

(Known ppm) (1%)

Gallons to liters: Number of liters = (Number of gal)(3.785 L/1 gal)


APPENDIX b 237

Composting Equations

Mixture’s % moisture = Sludge, lb Compost, lb

[(Sludge, lb)(% moisture) (Compost, lb)(% moisture)]100%+

+

Percent solids BC =

(Sludge, yd )(lb/yd ) (Sludge, yd )(Mix ratio)(lb/yd )(Sludge, yd )(lb/yd )(% solids, sludge) (Sludge, yd )(MR)(lb/yd )(% solids, chips)(100%)

3 3 3 3

3 3 3 3

+

+

Compost cycle, days = x Wet compost, lb/d

(Site capacity, yd )(Density of compost, lb/yd )3 3

Cycle time, days =

xPercent solidsDry solids, lb/d

+ x

Percent solids (Bulk density of wet sludge lb/yd )( Dry solids, lb/d)(MR)(Bulk density of wood chips, lb/yd )

(Capacity, yd )(Bulk density of compost lb/yd )

3

3

3 3

Percent moisture in mixture =

DB, lb/d Compost, lb/d[(DB, lb/d)(Percent moisture DB) (Compost lb/d)(Percent moisture compost)]100%

+

+

Density Equations

Density = Mass/Volume

Number of g/cm3 = (Number of lb/gal)(454 g/1 lb)(1 gal/3,785 cm3)

Detention Time Equation

Detention time, hr = Flow rate, gph

Volume, gal

Detention time, hr = Flow, gpd

(Volume, gal)(24 hr/d)

Dewatering Formulas

Total nonfilterable residue, mg/L = Total residue, mg/L – Total filterable residue, mg/L

Filter yield, lb/hr/ft2 = Area, ft

(Wet cake flow, lb/hr)(Percent solids/100%)2



Vacuum filter loading, lb/d/ft2 = Vacuum filter area, ft

(Biosolids, gpd)(8.34 lb/gal)(Percent solids)2

Digester Gas Production Formula

Gas produced, ft3/lb VS destroyed = VS destroyed, lb/d

Gas production, ft /d3

Gas produced, m3/lb VS destroyed = (VS destroyed, lb/d)(35.3 m /ft )

(Gas production, ft /d)3 3

3

Digester Loading Rate Equation

Digester loading rate, lb VSA/d/ft3 = Volume of digester, ft

lb VSA3

Where VSA = Volatile Solids Added

Digester loading, lb VS/d/ft3 =


2

Digester loading, lb VS/d/1,000 ft3 =


2

Digester Volatile Solids Ratio Formula

Digester VS ratio = lb VS in digester

VS added lb/d

Digester VS ratio = (lb VS in digester)(TS %/100%)(VS %/100%)

VS added lb/d

Dilution Triangle

Concentration (Conc.)1 Conc.2 – Conc. Desired = Number of parts Conc. Conc. Desired Conc.2 Conc.1 – Conc. Desired =

Total number of partsNumber of parts


APPENDIX b 239

Total number of parts(Number of gallons)(Number of parts of Conc. )1 = Number of gallons of Conc.1 required

Total number of parts(Number of gallons)(Number of parts of Conc. )2 = Number of gallons of Conc.2 required

Dissolved Air Flotation: Air Rate Flow Equations

Air, lb/d = (Air flow, ft3/min)(1,440 min/d)(0.0807 lb/ft3, Air)

Air, lb/d = (Air flow, ft3/min)(60 min/hr)(0.0807 lb/ft3, Air)

Dissolved Air Flotation: Air-to-Solids Ratio Equation

Air-to-solids ratio = (gpm)(Percent solids/100%)(8.34 lb/gal)

(Air flow, ft /min)(Air, lb/ft )3 3

Dissolved Air Flotation: Thickener Solids Loading Equations

Solids loading, lb/d/ft2 = DAF area, ft

(WAS, mg/L)(mgd)(Sludge, lb/gal)2

Solids loading, lb/hr/ft2 = (DAF area, ft )(24 hr/d)

(WAS, mg/L)(mgd)(Sludge, lb/gal)2

Dosage Formulas

Chlorine dose = Chlorine demand + Chlorine residual

Chemical feed, lb/d = (Flow, mgd)(Dosage, mg/L)(8.34 lb/gal) or rearranging to solve for dosage:

Dosage, mg/L = (mgd)(8.34 lb/gal)

lb/d

lb/d = (Percent purity/100%)

(mgd)(Dosage, mg/L)(8.34 lb/gal)

Above formula used when the purity of a substance or solution is less than 100%.

(mgd)(x, mg/L)(8.34 lb/gal) = (mgd)(Dosage, mg/L)(8.34 lb/gal)

Where x = unknown mgd



Chlorine, lb/d = Percent available chlorine/100%(Dosage, mg/L)(mgd)(8.34 lb/gal)

Chemical dosage, mg/L = (3,785 mL/gal)(mgd)(8.34 lb/gal)

(mL/min)(1,440 min/d)(Chemical, lb/gal)

Chemical dosage, mg/L = (3,785 mL/gal)(mgd)(8.34 lb/gal)(Percent Polymer)

(mL/min)(1,440 min/d)(Chemical, lb/gal)

Dry Chemical Feed Settings

Chemical, lb/d = 454 g/lb

(Number of g/min)(1,440 min/d)

Extrapolation: Used for Pipes Not Flowing Full

Division Factor = Ratio d/D

(High d/D – Low d/D)

Food/Microorganism Ratio Formula

F/M ratio = (mg/L MLVSS)(Volume in tank, mil gal)(8.34 lb/gal)


or as follows since 8.34 lb/gal in the numerator and denominator cancel each other out.

F/M ratio = (mg/L MLVSS)(Volume of tank, mil gal)

(BOD , mg/L)(Flow, mgd)5

Flow Rate Equations

Flow = Volume/Time

Q (Flow) = (Area)(Velocity) Example: Q, flow in ft3/s = (Area, ft2)(Velocity in feet per sec)

Flow in a pipe that changes size: (Area1, ft2)(Velocity 1, ft/s) = (Area2, ft2)(Velocity 2, ft/s)

Flow (Q), ft3/s = 2

(w w )1 2+ (Depth, ft)(Velocity, ft/s)


APPENDIX b 241

Gravity Thickener Solids Loading Formula

Solids loading, lb/d/ft2 = (Gravity thickener area)(100%)

(Flow, gpm)(1,440 min/d)(Percent solids)

Solids loading, lb/d/ft2 = (Gravity thickener area)(100%)

(Flow, gpd)(Percent solids)

Grit Removal Formula

Grit removal, ft3/mil gal = (7.48 gal/ft )(mil gal treated)Number of gallons removed

3

Hydraulic Digestion Time Equation

Digestion time, days = Influent sludge flow, gal/d

Number of gallons

Digestion time, days = Influent sludge flow, gal/day

(0.785)(Diameter) (Depth, ft)(7.48 gal/ft )2 3

Hydraulic Loading Rate Equation

Hydraulic loading rate = Surface area, ftTotal flow, gpd

2

Hydraulic loading rate, in./d = in.)(Area, acres)


Kilowatt Formulas

kW = (Number of hp)(0.746 kW/hp)

Kilowatt-hr/d = (hp)(0.746 kW/hp)(Operating time, hr/d)

kW = (Number of hp)(0.746 kW/hp)(Startup energy)

Lime Neutralization Formula

Lime, lb = (Volatile acids, mg/L)(mil gal)(8.34 lb/gal)



Mass Balance (Percent) Equation

Percent mass balance = Solids produced, lb/d

(Solids produced, lb/d Solids removed, lb/day)(100%)+

Mean Cell Residence Time Equations

MCRT, days = SS wasted, lb/d SS lb/d

(MLSS, mg/L)(mil gal)(8.34 lb/d)+

MCRT, days =

(WAS, mg/L)(Waste rate, mgd)(8.34 lb/gal) (TSS, mg/L)(Flow, mgd)(8.34 lb/gal)(MLSS, mg/L)(Aeration tank mil gal Clarifier tank mil gal)(8.34 lb/gal)

+

+

Mixture Formula

Percent mixture strength =

gal of solution gal of solutionSolution gal (Available %/100%) Solution gal (Available %/100%)(100%)

1 2

1 2

+

+

Nitrogen Loading Rate Equation

Nitrogen loading rate, lb/day = (Total Nitrogen, mg/L)(mgd)(8.34 lb/gal)

Nitrogen (Total) Equation

Total nitrogen (N) = Nitrate, mg/L + Nitrite, mg/L + TKN, mg/L

Operating Time Formulas

Operating time = Flow rate

Treated water

Operating time, min/hr = (Sludge pump, gpm)(Percent solids)(24 hr/d)

(Flow, mgd)(Influent SS, mg/L – Effluent SS, mg/L)(100%)

Organic Loading Rate Equations

Organic loading rate, lb BOD5/d/acre = ft


-Surface area of pond, acre


APPENDIX b 243

Organic loading rate, lb BOD5/d/1,000 ft3 = Volume of trickling filter, ft /1,000 ft(BOD , mg/L)(Flow, mgd)(8.34 lb/gal)

3 3

5

Particulate and Soluble Biochemical Oxygen Demand Formulas

Particulate BOD5, mg/L = (SS, mg/L)(K-value)

Soluble BOD5 = Total BOD5 – (K-value)(Total SS)

Total BOD5 = (Particulate BOD5)(K-value) + Soluble BOD5

Perimeter Formulas

Circumference = π(Diameter) = (3.14)(Diameter)

Rectangle = 2(Length) + 2(Width)

Plant Available Nitrogen Formulas

PAN, mg/L = [MR(TKN – NH3)] + [0.50(NH3)] + (NO3 + NO2)

PAN, dry tons/acre = PAN, lb/dry ton

Plant nitrogen required, lb/acre

PAN, lb/dry ton = [(Organic N, mg/kg)(MR) + (Ammonia N, mg/kg)(VR)](0.002 lb/dry ton)

Population Equivalent Equation

Number of people = lb/d of BOD /person

(BOD , mg/L)(mgd)(8.34 lb/gal)5

5

Population Loading Equation

Population loading, people/acre = Area of pond(s), acres

Number of people served

Pressure Formulas

psi = 2.31 ft/psiDepth, ft

Height, ft = (psi)(2.31 ft/psi)



psi = (Depth, ft)(0.433 psi/ft)

Pressure = Area, ftForce, lb

2

Pressure, lb/ft2 = (Height or Depth, ft)(Density, 62.4 lb/ft3)

wPressureA +

g2Velocity 2

A = w

PressureB + g2

VelocityB2

Percent Mixture Formulas

Percent mixture = strength Soln. , lb Soln. , lb

(Soln. , lb)(Avail %/100%) (Soln. , lb)(Avail %/100%)(100%)1 2

1 2

+

+

Percent mixture strength =

(Soln. , gal)(lb/gal) (Soln. , gal)(lb/gal)[(Soln. , gal)(lb/gal)(Avail %/100%) (Soln. , gal)(lb/gal)(Avail %/100%)]100%

1 2

1 2

+

+

Percent Recovery Equation

Percent recovery = Feed sludge TSS, %(Cake TS, % – Return flow TSS, %)

Cake TS, %(Feed sludge TSS, % – Return flow TSS, %)(100%)

Percent Reduction Equations

Percent VS reduction = Effluent – (Effluent)(Influent)

(Influent – Effluent)(100%)

Percent VM reduction = [Percent influent VM – (Percent influent VM)(Percent effluent VM)]

(Percent influent VM – Percent effluent VM)(100%)

Percent moisture reduction =

[Percent influent moisture – (Percent influent moisture)(Percent moisture, after digestion)](Percent influent moisture – Percent moisture, after digestion)(100%)

Percent Removal Formula

Percent ntu removal = Influent ntu

(Influent ntu – Effluent ntu)(100%) or

In(In – Out)(100%)

Percent BOD5 removal = Influent BOD

(Influent BOD – Effluent BOD )(100%)5

5 5 or In

(In – Out)(100%)


APPENDIX b 245

Percent removal efficiency = Influent solids, mg/L

(Solids removed, mg/L)(100%)

Percent Strength of Solutions and Solids

Percent strength = Number of lb, Water lb chemical(Number of lb of chemical)(100%)

+

Percent strength = Number of grams, water grams chemical

(Number of grams of chemical)(100%)+

Percent total solids = Sludge sample in grams

(Dry sample in grams)(100%)

Percent Settled Sludge and Solids Formulas

Percent settled sludge = Total sample vol., mL

(Settled sludge, mL)(100%)

or similarly:

Percent settleable solids = Sample size, mL

(Settled sludge, mL)(100%)

Percent inorganic solids = Sludge sample in grams

(Dry sample in grams)(100%)

(2° gpd)(2° sludge lb/gal)(x% 2° sludge) = (1° sludge, gpd)(1° sludge lb/gal)(% 1° sludge)

Where x = unknown %

Phosphate Loading Rate Equation

Phosphorus (P) loading rate, lb/d = (P, mg/L)(mgd)(8.34 lb/gal)

Pit Volume and Days-to-Fill Formulas

Screenings, ft3/mil gal = Number of mgdNumber of ft /d3

Number of days to fill = Screenings removed, ft /d

Pit volume, ft3

3



Pumping and Pumping Cost Formulas

mhp = (Motor efficiency)(Pump efficiency)

(whp)

Where mhp = motor horsepower and whp = water horsepower

mhp = (3,960)(Motor efficiency)(Pump efficiency)

(Flow, gpm)(TH, ft) Where TH = Total head

bhp = (1,714)(Pump efficiency)

(Flow, gpm)(Differential pressure, psi)

Brake hp = (hp)(Motor efficiency) Where hp = horsepower

Water hp = (mhp)(Motor efficiency)(Pump efficiency)

Water hp = (bhp)(Pump efficiency)

Motor hp = bhp/Motor efficiency

Brake hp = whp/Pump efficiency

Cost, $/d = (Motor hp)(24 hr/d)(0.746 kW/hp)(Cost/kW-hr)

Pumping Rate Equations

Pumping rate = Flow, gal/Time, min

Pump’s discharge rate, gpm = Time, min

Discharge, gal

Discharge rate, gpm = Influent flow, gpm + Level drop, gpm

Number of gal per stroke = (0.785)(Bore diameter, ft)2(Stroke, ft)(7.48 gal/ft3)

Ratios

Ratio = Plant influent flowRecirculated flow

Polymer dosage , mLSpeed setting , Percent

1

1 = Polymer dosage , mL

Speed setting , Percent2

2

Flow , mgdChlorine dosage , mg/L

1

1 = Flow , mgd

Chlorine dosage , mg/L2

2


APPENDIX b 247

Flow , gpmDigester solids , lb/d

1

1 = Flow , gpm

Digester solids , lb/d2

2

Speed setting , %Alum dosage , mL

1

1 = Speed setting , %Alum dosage , mL

2

2

Screenings Formula

Screenings, ft3/mil gal = Number of mgdNumber of ft /d3

Seed Sludge Equation


(Capacity of digester)(Percent seed sludge required)

Sludge Age (Gould) Equation

Sludge age, days = Solids added, lb/d

Solids under aeration, lb

Sludge age, days = (SS, mg/L)(Flow, mgd)(8.34 lb/gal)

(MLSS, mg/L)(Volume of aeration tank)(8.34 lb/gal)

Sludge Pumping Formula

Sludge, lb/d = (Pumping, min/day)(24 hr/d)(Pump rate, gpm)(8.34 lb/gal)(sp gr of sludge)

(Primary sludge, gal)(Primary sludge, lb/gal)(Percent PSS) =

(x Thickened sludge, gal)(Thickened sludge, lb/gal)(Percent TSS)

Sludge Removed Equation

SS removed, lb/d = (SS removed, mg/L)(Number of mgd)(8.34 lb/gal)

Sludge Volume Index and Sludge Volume Density Equations

SVI = MLSS, g/L(SS, mL)



SDI = SS, mL

(MLSS, g)(100%)

Sodium Absorption Ratio Formula

Sodium absorption ratio = [(0.5)(Ca Mg )]

Na/2 2 1 2++ +

+

Solids Balance (Digester) Equations

Total solids, lb/d = (Raw sludge, lb/day)(Percent solids)

Fixed solids, lb/d = Total solids, lb/day – VS, lb/d

Water in sludge, lb/d = Sludge, lb/day – Total solids, lb/d

Percent VSR = In – (In)(Out)

(In – Out)(100%)

Gas produced, lb/d = (Effluent VS, lb/day)(Percent VSR)

VS in digested sludge, lb/d = Influent VS, lb/d – Destroyed VS, lb/d

Total digested solids, lb/d = Percent digested VS

VS digested, lb/d

Fixed solids, lb/d = Total digested solids, lb/d – VS digested, lb/d

Digested sludge, lb/d = Digested sludge percent solids

Total digested solids, lb/d

Water in digested sludge, lb/d = Sludge, lb/d – Total solids, lb/d

Solids Loading Rate Equations

Solids loading rate = Area, ft

(MLSS, mg/L)(mgd)(8.34 lb/gal)2

Solids loading rate, lb/d/ft2 = (Surface area, ft )

(Percent solids)(Biosolids added, gpd)(8.34 lb/gal)2

Solids Produced Formula

Solids produced, lb/d = (BOD5 removed, lb/d)(0.85 lb solids/lb BOD5)


APPENDIX b 249

Solids Pumping Equations

Solids, lb/d = 100%

(Sludge, lb/d)(Percent solids)

Solids, lb/d = (Pumping, min/d)(24 hr/d)(Pump rate, gpm)(8.34 lb/gal)(sp gr of sludge)

Solids, lb/d = (Time, min/cycle)(cycles/d)(Pump rate, gpm)(8.34 lb/gal)(Percent solids)

Solids Retention Time Equation

Target SRT =

x(RAS SS mg/L)( mgd)(8.34 lb/gal) (Effluent SS, mg/L)(Flow, mgd)(8.34 lb/gal)(MLSS mg/L)(Clarifier, Aerator Volume, mil gal)(8.34 lb/gal)

+

Where x = unknown mgd

Solids Under Aeration

Number of lb, solids = (SS, mg/L)(Number of mil gal)(8.34 lb/gal)

Solution Formulas

(Concentration1)(Volume1) + (Concentration2)(Volume2) = (Concentration3)(Volume3) orabbreviating the above equation: C1V1 + C2V2 = C3V3

Percent HTH solution = (Number of gal)(8.34 lb/gal)

(lb HTH)(100%)

(Solution1, percent)(x gal, Solution1)(Solution1, lb/gal) =

(Solution2, percent)(Solution2, gal)(Solution2, lb/gal)

Specific Gravity Equations

Specific gravity (sp gr) = Density of substance/Density of water

sp gr = 8.34 lb/gal

Solute, lb/gal

sp gr = 62.4 lb/ft

Number of lb/ft3

3



Statistic Formulas

Average = Number of measurementsSum of all measurements

or

Arithmetic mean = Number of measurementsSum of all measurements

Median is the middle value.

Range = Largest value – Smallest value

Mode is the value that occurs most frequently. Note: There also can be two or more modes within a set of numbers.

Geometric mean = [(X1)(X2)(X3)(X4)…..(Xn)]1/n

Standard deviation = [Σf(X – X)2/n – 1]1/2

Surface Loading Formula

Surface loading rate = Number of ft

gallons per day (gpd)2

Suspended Solids Loading Equation

Suspended solids, lb/d = (SS, mg/L)(Number of mgd)(8.34 lb/gal)

Temperature Formulas

°C = 5°C /9°F (°F–32)

°F = 9°F /5°C (°C) + 32°F

Total Force and Hydraulic Press Equations

Total force, pounds = (Area, in.2)(psig) or Total force = (Pressure)(Area)

Pressure = Area, ft

Total force, lb2


APPENDIX b 251

Total Head Formulas

Total head, ft = Total static head, ft + Head losses, ft

TDH = Specific gravity

(Differential pressure)(2.31 ft/psi)

Vacuum Filter Operating Time Equation

Filter yield, lb/hr/ft2 =

Filter area ft(Filter operation, lb/d)(100%)

(Solids, lb/d)(Percent recovery)

2

Velocity Equations

Velocity, ft/s = (Width, ft)(Depth, ft)(60 sec/min)(7.48 gal/ft )

Flow, gpm3

or

Velocity, ft/s = (Area, ft )(60 sec/min)(7.48 gal/ft )

Flow, gpm2 3

Volatile-Acids-to-Alkalinity Ratio Formulas

Ratio = Volatile acids/Alkalinity

Volatile acids = (Alkalinity)(Ratio)

Digester VS ratio = lb VS in digester

VS added lb/d

Volatile Solids Destroyed Equation

VS destroyed, lb/d/ft3 = Digester capacity,1,000 ft

(Flow, gpd)(Sludge, lb/gal)(SSC, %)(VSC, %)(VSR, %)3

Volatile Solids: lb/d (Percent)

VS, lb/d = (Number of lb/d, sent to digester)(Percent VS/100%)

VS, lb/d = 100% 100%

(Number of gpd to digester)(Percent solids)(Percent VS)(8.34 lb/gal)



Volatile Solids Pumping Formula

VS, lb/d =

(Time, min/cycle)(cycles/day)(Pump rate, gpm)(8.34 lb/gal)(Percent, solids)(Percent VM)

Volume Equations

Volume of a basin in ft3 or m3 = (Length)(Width)(Depth)

Volume of a basin in gallons = (Length)(Width)(Depth)(7.48 gal/ft3)

Volume of a cone in ft3 or m3 = 1/3πr2(Height or Depth)

Volume of a circular tank in ft3 or m3 = πr2(Height) or Volume of a pipe = πr2(Length) or

Volume of a cylindrical tank in ft3 or m3 = (0.785)(Diameter)2(Height)

Volume of a trough in gallons = 2

(b b )1 2+ (Depth of water)(Length)(7.48 gal/ft3)

Where b1 = bottom width of trough or ditch and

b2 = surface width of trough or ditch

Volume, gal = 2 2

(Length Length )(Width Width )1 2 1 2+ + (Depth, ft)(7.48 gal/ft3)

Volume of sphere, ft3 = 3

4 r3r

Digester capacity, ft3 = π(radius)2(Height, ft)

Equation below is for a partially filled pipe division factor for determining flow:

Division factor = Ratio d/D

(High d/D – Low d/D)

Waste Activated Sludge Loading Rate Equation

WAS, lb/d = (WAS, mg/L)(Number of mgd)(8.34 lb/gal)

Waste Activated Sludge Pumping Rate Formula

Number of mgd = (Number of mg/L WAS)(8.34 lb/gal)

WAS, lb/d


APPENDIX b 253

Waste Rate Equation

Waste rate, mgd = (WVS concentration, mg/L)(8.34 lb/gal)

Waste, lb

The following two equations are used to find the waste rate (above equation).

Desired MLVSS, lb = Desired COD lb/MLVSS lb

(Primary effluent COD, mg/L)(mgd)(8.34 lb/gal)

Existing MLVSS, lb = (MLVSS, mg/L)(Aeration tank, mil gal)(8.34 lb/gal)

Waste rate, gpd = (Waste TSS, mg/L)(8.34 lb/gal)

(Solids produced, lb/d)(1,000,000/M)

Waste rate, lb/d =

Desired MCRTMLSS, mg/L[AT, mil gal CT, mil gal](8.34 lb/gal)+ – (TSS, mg/L)(Flow, mgd)(8.34 lb/gal)

Weir Length Formula

Weir length, ft = π(Diameter, ft)

Weir length, ft = 2π(radius, ft)

Weir and Surface Overflow Rate Equations

Weir overflow rate = Weir Length, ft

Flow, gpd

Surface overflow rate = Area, ft

Flow, gpd2


255

Appendix C

Wastewater Flowchart Diagrams

The following 12 diagrams show some of the most common types of wastewater treatment plants with the last diagram showing sludge processing. Wastewater plants shown in Figures 9 and 10 are newer technologies using membranes, nanofiltration, and electrodialysis.

Please note that not all the processes in any diagram are necessarily used. Also, some steps are not depicted. Of course, many more wastewater treatment plant types and process arrangements exist that are not shown. The number of different wastewater treatment plant arrangements is beyond the scope of this book. For further study of other wastewater treatment plants, please see the references.

Wastewater Treatment and Sludge Processing FlowchartsFigure 1 Flowchart of Typical Wastewater Treatment ProcessesFigure 2 Flowchart of Conventional Activated Sludge ProcessFigure 3 Flowchart of Contact Stabilization ProcessFigure 4 Flowchart of Activated Sequencing Batch ReactorFigure 5 Flowchart of Typical Wastewater Treatment Using Ponds after Secondary TreatmentFigure 6 Flowchart of Typical Wastewater Treatment Using Polishing Ponds in Series TreatmentFigure 7 Flowchart of Wastewater Treatment Using Rotating Biological Contactor ProcessFigure 8 Flowchart of Typical Wastewater Treatment Using an Oxidation DitchFigure 9 Flowchart of Typical Wastewater Treatment Using Electrodialysis or NanofiltrationFigure 10 Flowchart of Typical Wastewater Treatment Using a Membrane BioreactorFigure 11 Flowchart of Wastewater Treatment Nitrification ProcessFigure 12 Flowchart of Wastewater Treatment Plant Processing Sludge



Influent Flow Screening

Comminutor

FlowmeterFlow Equalization

Grit Chamber

Preaeration

Return ActivatedSludge

ActivatedSludgeProcess

ReturnSecondary

Sludge

PrimarySedimentation

Disinfection Discharge

TricklingFilter

Disinfection

Disinfection

Discharge

Discharge

SecondaryClarifier

Solids RemovalNitrogen Control

Phosphorus Removal

TertiarySludge

Management

SecondarySludge

Management

Figure 1 Flowchart of Typical Wastewater Treatment Processes


APPENDIX C 257


Comminutor


Grit Chamber

Preaeration

PrimarySettling


SecondarySludge

Management

TertiarySludge

Management

Disinfection

Discharge

ReturnActivated

Sludge

AerationTank

PrimaryClarifier

SecondaryClarifier

DigesterWaste

ActivatedSludge

Figure 2 Flowchart of Conventional Activated Sludge Process




Comminutor


Grit Chamber

Preaeration

PrimarySettling


PrimaryClarifier

Disinfection

Discharge

Discharge

Disinfection

ContactTank

SecondaryClarifier

SecondarySludge

Management

SludgeStabilization

Tank

Figure 3 Flowchart of Contact Stabilization Process


APPENDIX C 259


Comminutor


Grit Chamber

Digester

SequencingBatch

Reactor 1

SequencingBatch

Reactor 2

Thickening

SolidsHandling

Equalization

Filtration

Disinfection

Discharge

Figure 4 Flowchart of Activated Sequencing Batch Reactor




Comminutor


Grit Chamber

Digester

Thickening

SolidsHandling

Disinfection

Discharge

PrimaryTreatment

(Sedimentation)

SecondaryTreatment(BiologicalProcess)

SolidsReturn

Pond 1

Pond 2

Figure 5 Flowchart of Typical Wastewater Treatment Using Ponds after Secondary Treatment


APPENDIX C 261


Grit Chamber

Flowmeter

Digester

Thickening

SolidsHandling

Disinfection

Discharge

PrimaryClarificationTreatment

Pond 1

Pond 2

TricklingFilter

Figure 6 Flowchart of Typical Wastewater Treatment Using Polishing Ponds in Series Treatment




Grit Chamber

Flowmeter

OrganicRemoval

Disinfection

Discharge

PossibleRecycling

AerobicNitrification

AnoxicDenitrification

ClarifierUnit

Figure 7 Flowchart of Wastewater Treatment Using Rotating Biological Contactor Process


APPENDIX C 263


Grit Chamber

Flowmeter

Disinfection

Discharge

SecondaryClarifier

OxidationDitch

Figure 8 Flowchart of Typical Wastewater Treatment Using an Oxidation Ditch




Grit Chamber

Flowmeter

Disinfection

Discharge

SequencingBatch

Reactor

Waste SludgeHandling

ClothFilter

EnhancedFiltration

Electrodialysisor

Nanofiltration

Figure 9 Flowchart of Typical Wastewater Treatment Using Electrodialysis or Nanofiltration


APPENDIX C 265


Grit Chamber

Flowmeter

Disinfection

Discharge

Bioreactor

ConcentrateFrom

Membranes MembraneModules

Waste SludgeHandling

Figure 10 Flowchart of Typical Wastewater Treatment Using a Membrane Bioreactor




Comminutor


Grit Chamber

PrimaryTreatment Disinfection Discharge

Disinfection

Discharge

Clarifier

FermentationTank

TricklingFilter

AnaerobicContactor

Aerobic

Aerobic

Anoxic

Aerobic

SludgeManagement

Figure 11 Flowchart of Wastewater Treatment Nitrification Process


APPENDIX C 267

Figure 12 Flowchart of Wastewater Treatment Plant Processing Sludge

PreliminaryTreatment

PrimaryTreatment

SecondaryTreatment

Sludge

SludgeThickening

SludgeStabilization

SludgeConditioning

VolumeReduction

Dewatering: e.g.,Filter Presses or

Vacuum Filtration

SanitaryLandfill

LandApplication


269

Appendix D

Abbreviations

acre-ft acre-feetAIDS acquired immune deficiency syndromeamps amperesAT aeration tankavail. availableavg. averageAWWA American Water Works Associationb1 bottom width of trough or ditchb2 surface width of trough or ditchBC blending compostbhp brake horsepowerBOD or BOD5 biochemical oxygen demand (subscript 5 means it was a 5-day test)C concentration°C degrees CelsiusCERCLA Comprehensive Environmental Response, Compensation, and Liability ActCF concentration factorCFR Code of Federal RegulationsCOD chemical oxygen demandcm centimeter(s)CT clarifier tankD diameter or depth (note context)d dayDAF dissolved air flotationDB dewatered biosolidsDLD dedicated land disposalDO dissolved oxygeneffic. efficiencyemf electromotive force°F degrees FahrenheitF/M ratio food-to-microorganism ratioft foot or feetft/s feet per secondft2 square feet



ft3/s cubic feet per secondft3 cubic feetft3/min cubic feet per minuteft3/s cubic feet per secondg gram(s) or gravity (note context)gal gallon(s)gpcpd gallons per capita per daygpd gallons per daygph gallons per hourgpm gallons per minutegr-eq gram equivalent weightHHS Department of Health and Human Serviceshp horsepowerhr hoursHTH high test hypochloritein. inch(es)K-value amount of organic suspended solids in the total amount of suspended solids in a wastekg kilogramkW kilowatt(s)kW-hr kilowatt-hourL liter or length (note context)lb poundsm meterM million, or mole(s) (note context)m3 cubic metersmA milliamp(s)MCRT mean cell residence timeMCL maximum contaminant levelME motor efficiencymeq milliequivalentmg milligramsmil millionmil gal million gallonsmgd million gallons per daymg/L milligrams per litermhp motor horsepowermin minutemL milliliterMLSS mixed liquor suspended solidsMLVSS mixed liquor volatile suspended solidsMPN most probable number


APPENDIX D 271

MR mix ratioMSDS material safety data sheetN Normality, the number of gram-equivalent weights of solute in 1 liter of solutionNPDES National Pollutant Discharge Elimination Systemnm nanometer(s)ntu nephelometric turbidity unitoz ounce(s)Δp pressure dropPAN plant available nitrogenpH hydrogen ion concentrationPE pump efficiency% percentppm parts per millionpsi pressure per square inch absolutepsig pressure per square inch gaugePTFE polytetrafluoroethyleneQ flowr radiusRAS return activated sludgeRBC rotating biological contactorrpm revolutions per minutes second(s)SBR sequence batch reactorsec second(s)sed sedimentationsp gr specific gravitysoln. solutionSRT solids retention timeSS suspended solidsSSC sludge solids concentrationSDI sludge density indexSVI sludge volume indextemp. temperatureTDS total dissolved solidsTKN total Kjedahl nitrogenTS total solidsTSS total suspended solidsTH total headUSEPA US Environmental Protection AgencyUV ultravioletV volume or velocity or volt(s) (note context)



vol. volumeVM volatile matterVR volatilization rateVS volatile solidsVSC volatile solids concentrationVSR volatile solids reductionVSS volatile suspended solidsW widthw wattWAS waste activated sludgewhp water horsepowerWt weightWVS waste volatile solidsyr yearyd3 cubic yards


273

References

American Water Works Association. 2003. Basic Science Concepts and Applications. Third ed. Denver, Colo.: AWWA.

Arasmith, E.E. 1992. Pumps and Pumping. Albany, Ore.: ACR.California State University, Sacramento, California Water Pollution Control Association, and

United States. 2006. Advanced Waste Treatment. Fifth ed. Operator Training Manuals. Sacramento: California State University.

Giorgi, J. 2009. Math for Wastewater Treatment Operators Grades 1 and 2: Practice Prob-lems to Prepare for Wastewater Treatment Operator Certification Exams. Denver, Colo.: AWWA.

Giorgi, J. 2009. Math for Wastewater Treatment Operators Grades 3 and 4: Practice Prob-lems to Prepare for Wastewater Treatment Operator Certification Exams. Denver, Colo.: AWWA.

Giorgi, J. 2007. Math for Water Treatment Operators: Practice Problems to Prepare for Water Treatment Operator Certification Exams. Denver, Colo.: AWWA.

Giorgi, J. 2007. Math for Water Distribution System Operators: Practice Problems to Prepare for Distribution System Operator Certification Exams. Denver, Colo.: AWWA.

Giorgi, J. 2003. Operator Certification Study Guide: A Guide to Preparing for Water Treat-ment and Distribution Operator Certification Exams. Denver, Colo.: AWWA.

Kerri, K.D. 2004. Operation of Wastewater Treatment Plants: A Field Study Training Pro-gram. Volume I. Sixth ed. Operator Training Manuals. Sacramento: California State University.

Kerri, K.D. 2007. Operation of Wastewater Treatment Plants: A Field Study Training Pro-gram. Volume II. Seventh ed. Operator Training Manuals. Sacramento: California State University.

Kerri, K.D. 1993. Small Water System Operation and Maintenance: A Field Study Train-ing Program. Operator Training Manuals. Sacramento: Hornet Foundation, California State University.

Price, J.K. 1991. Applied Math for Wastewater Plant Operators. Mathematics for Water and Wastewater Treatment Plant Operators Series. Lancaster, Pa.: Technomic.

Price, J.K. 1991. Basic Math Concepts for Water and Wastewater Plant Operators. Math-ematics for Water And Wastewater Treatment Plant Operators Series. Lancaster, Pa.: Technomic.

Sarai, D.S. 2005. Basic Chemistry for Water and Wastewater Operators. Denver, Colo.: AWWA.Skoog, D.A. 2004. Fundamentals of Analytical Chemistry. Eight ed. Belmont, Calif.: Thomson-

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Spellman, F.R. 2004. Mathematics Manual for Water and Wastewater Treatment Plant Oper-ators. Boca Raton, Fla.: CRC.

Spellman, F.R., and J. Drinan. 2001. Pumping. Fundamentals for the Water and Wastewater Maintenance Operator Series. Lancaster, Pa.: Technomic.

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