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Applied Math for Operators

January 16, 2013

WEFCOM Discussion

WEF Members: Please join us on the Webcasts and Online Courses Community for a discussion after the webcast: Wefcom.wef.org/webcasts

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Wefcom.wef.org/webcasts

Speaker

Jeanette Brown, Senior Scientist, University of Connecticut Department of Chemical, Materials, and Biomolecular Engineering

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Content

• Introduction

• Area/Volume

• Flow Rate

• Force, Pressure, Head, hp

• Preliminary Treatment

• Primary Treatment

• Secondary Treatment

– Biofilm

– Activated Sludge

• Solids Processing

– Production

– Thickening

– Dewatering

– Stabilization• Chemical dosage

• Laboratory

• Spreadsheets

Introduction

• Math is critical to everything we do as operators

• Math is not difficult once you understand the basic concepts

• Math is not difficult if you use dimensional analysis

– Dimensional analysis is a mathematical system using conversion factors to move from one unit of measurement to a different unit of measurement. For example, if you need to calculate how many minutes in a day, you can use dimensional analysis to set all of your needed conversions into one problem.

– Also called cancelling of units

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Area/Volume

Area

• Area of a square or rectangle– A L*W l length;w width

• Area of a circle– (r=radius)

– 0.785

• Answers always expressed as square something-inches, feet, meters, etc.

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Area

Example 1: A rectangular settling tank is 22 ft in length and 11 ft in width. What is the area of that tank?

• A = L*W = 22 ft1 * 11ft1 = 242 ft2

Example 1: A circular settling tank is 20 ftdiameter. What is the area of that tank?

• A = 3.14*10 ft*10ft =314 ft2 or

• 0.785 = 0.785*20 ft*20ft =314 ft2

Volume

• Volume of a square or rectangular tank– V A*d d depth

• Volume of a circular tank– V *dalsoV A*dorV 0.785D2 *d

• Answers always expressed as cubic something-inches, feet, meters, etc

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Volume

Example 1: A rectangular settling tank is 22 feet in length,11 feet in width and 8 feet deep. What is the volume of that tank?

• V = L*W*d = 22 ft * 11 ft* 8 ft = 1936 ft3

Example 1: A circular settling tank is 20 feet diameter and 8 feet deep. What is the volume of that tank?

• *d V = 3.14*10 ft*10 ft*8 ft =2512 ft3 or

• 0.785 = 0.785*20 ft*20 ft*8 ft =2512 ft3

Volume

The volume of the circulate settling tank is 2512 ft3. How many gallons does it hold?

1 ft3 of water = 7.48 gallons

Vgallons = 2512 ft3 * 7.48 gal = 18,790 gal

1 ft3

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Weight and Volume Relationship

• 1 ft3 of water = 7.48 gallons

• 1 gallon of water = 8.34 lbs

• 1 ft3 of water = 62.4 lbs

1 ft3 x 7.48 gallons x 8.34 lbs

1 ft3 gallon

= 62.4 lbs

• Every cubic foot of water weights 62.4 lbs

Flow Rate

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Flow Rates

• One of the most common parameters used– ft3/s (CFS); gallons/minutes (gpm), million

gallons per day (MGD), m3/day

– Volume over time

• Q (flow) = A (area) * V (velocity)

Flow Rates

• Flow in a channel

Example:

What is the flow in this channel in ft3/s (CFS)?

1. Convert the width in inches to feet

36 in * 1 ft = 3 ft

12 in

2. Calculate the wetted area

A = L*W = L*d = 3 ft *1.6 ft = 4.8 ft2

3. Calculate the flow

Q= Av = 4.8 ft2 * 2.8 ft/s = 13.4 ft3/s

v=2.8 ft/s

dwater= 1.6 ft

w=36 inches

Hchannel= 2.5 ft

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Flow Rates

• Flow in a pipeline, flowing full

What is the flow in this pipe in ft3/s (CFS)?

1. Convert the width in inches to feet

8 in * 1 ft = 0.67 ft

12 in

2. Calculate the wetted area

= 3.14 *0.67 ft* 0.34 ft = 1.4 ft2

3. Calculate the flow

Q= A.v = 1.4 ft2 * 2.1 ft/s = 2.9 ft3/s

(More complicated for a pipe flowing partially full.

D=8 in

v= 2.1 ft/s

Force, Pressure, Head, Horsepower

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Force, Pressure

• Force-the push exerted by water on any surface used to confine it. Units of force are typically pounds, tons, kilograms, grams. The force of water on the bottom of a tank is the weight of water pressing down on it.

• Pressure-the force per unit area, typically pounds per square inch (psi).

Example

• If you had 1 ft3 of water, what is the force acting on the bottom of the container? – 62.4 lbs

• What is the pressure in psi?– 62.4 lbs/144 in2= 0.433 psi

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Head

• Static head-the vertical distance in height between the point of entry to the system and the highest point of discharge.

• Friction head-represents the head required by the system to overcome the frictional resistance of flow in pipes, valves, fittings, etc.

• Total dynamic head-this is the total energy required to lift wastewater from one level to another. It is the sum of static head, velocity head and friction loss.

Pressure/Work/Horsepower

• Pressure– 1 psi = 2.31 ft (at 1 ft of depth each square

inch has 0.433 pounds so at depth of 2.31 ftthere is one pound of weight on each square inch, 2.31 ft/psi)

• Work: lifting of a weight a vertical distance, ft-lbs

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Pressure/Work/Horsepower

• Power: rate of doing work and expressed in ft-lbs/min• Horsepower: unit of power

– 1 hp = 33,000 ft-lbs/min = 746 wattsExample: How many kWh does a 20 hp motor use if it running 24 hours per day?20 hp * 746 watts = 14,920 watts * 1 kW = 14.9 kW

hp 1000 wattsSince the motor runs 24 hrs/day14.9 kW * 24 hrs/d = 357.6 kWh (daily)If you are paying $0.08/kWh, how much does it cost per day to run the motor:

357.6 kWh/d * $ 0.08/kWh = $28.61/d

Horsepower

• Work is done by lifting water

HP = Q, gal/min x lift, ft x 8.34 lbs/gal x HP/33,000 ft-lbs/min

• This is the horsepower necessary to lift water. This is called water horsepower.

• Can also be written:

Water hp (whp) = (Q,gpm)(H,ft)/3960

Water horsepower is the is the actual horsepower available to pump water.

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Pumping Calculations

• Pumps are not 100% efficient.– Brake horsepower-power supplied to the motor is greater than

the motor transmits.

Brake hp= (Q, GPM)(H, ft)/(3960)(Ep)

= whp/(Ep)

• Motors are not 100 % efficient. – Motor horsepower-horsepower supplied to the pump. It is

greater than the water horsepower.

Motor hp (mhp)= (Q, GPM)(H, ft)/(3960)(Ep)(Em)

=bhp/Em

Horsepower

Example: If you want to pump 1200 gpm at a head of 130ft, what is the whp, bhp and mhp if the pump is 83% efficient and the motor is 85% efficient?

• Water, hp=whp = (Q, gpm*H, ft)/3960

• Brake (pump), HP = (Q, GPM)(H, ft)/(3960)(Ep)=whp/Ep

• Motor, HP = (Q, GPM)(H, ft)/(3960)(Ep)(Em)=bhp/(Em)

Whp = (1200 gal/min * 130 ft)/3960 = 39.4 hp

bhp = 39.4 hp/0.83 = 47.5 hp

mhp = 47.5/0.85 = 55.8 hp

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Preliminary Treatment

Screenings

• Typically about 0.5 to 12 ft3 of screenings are removed per million gallons depending on the types of screens and wastewater. Need to know how much your plant is producing.

Example: Your plants on average removes 8.0 ft3 every day. Your daily flow is 4.5 MGD. How many ft3/MGD does your plant remove?

Screenings Remove/MG = 8.0 ft3/4.5 MGD = 1.8 ft3/MG (daily average)

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Screenings• Storage

If you have storage for screenings, you need to calculate how much storage you have. If you storage tank hold 250 gallons, how many days of screening would it hold?

250 gal* ft3 = 33.4 ft3

7.48 gal

Plant produces 1.8 ft3/MGD; Q = 3.0 MGD

Total Screenings/day = 1.8 ft3/MGD* 3.0 MGD = 5.4 ft3

Storage in days= 33.4 ft3/5.4 ft3/d = 6.2 days

Grit Removal

• Two important factors; velocity and volume removed.

• Velocity:

Example:

What is the velocity in the channel if the flow rate is 3.2 MGD?

Convert MGD to ft3/s = 3,200,000 gal * ft3 * day = 4.9 ft3/s

day 7.48 gal 86,400 s

Q = A*v

v = Q/A A = d * W = 1.6 ft * 3 ft = 4.8 ft2

v = 4.9 ft3/s = 1.02 ft/s

4.8 ft2

dwater= 1.6 ft

w=36 inches

Hchannel= 2.5 ft

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Grit Removal

• How many cubic yards of grit are removed per year if the removal rate is 1.8 ft3/MG and the average annual daily flow is 4.2 MGD?

• Calculate how many ft3 are removed per day-– 1.8 ft3/MG * 4.2 MGD = 7.6 ft3/d

• Calculate annual production– 7.6 ft3/d * 365 d/yr = 2759 ft3/yr

• Convert to yd3

– 1 yd3 * 2759 ft3/yr = 102 yd3/yr

27 ft3

Primary Treatment

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Primary Sedimentation

• Hydraulic detention time

• Weir Overflow rate

• Organic loading rate

• Sludge Pumping

• Process Efficiency

• Surface overflow rate

Detention Time

• dt = V/Q

– V= volume of tank in gallons

– Q=plant flow in gallons per day

• Typical unit for detention time is hours.

Detention time describes a period sufficient enough to remove almost all of the settleable solids.

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Detention Time

Example: The flow to primary sedimentation is 4.9 MGD. There are two settling tanks. Each tank 80 feet long, 40 feet wide and 10 feet deep. What is the detention time in hours?

• dt = V/Q

• Calculate the volume per tank and convert to gallons.

– V = L*W*D = 80 ft *40 ft *10 ft = 3200 ft3

– 3200 ft3 * 7.48 gal/ft3 = 239,360 gallons

• Since there are two tanks; half the flow or double the volume– dt = V/Q = 239,360 gal/2,450,000 gallons/day

– = 0.98 day *24 hr/d = 2.3 hrs

Detention Time

• Example-Circular Tank

– Diameter of settling tank = 100 ft

– Depth =13 ft

– 2 units-each receiving 9.5 MGD

• dt = V/Q

• V *d perunit– V= (50 ft)2*3.14*13 ft =102,050 ft3 *7.48 gal/ ft3

– V=763,334 gal

• dt=763,334 gal/9,500,000 gal/day=0.08 day * 24 hr/day

• dt=1.9 hrs

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Weir Overflow Rate

Velocity of water over the weirs

WOR = Q, gpd

Weir length, ft

Example: Primary clarifier has a diameter of 80 ft. The plant flow is 3.5 MGD. What is the WOR?

• Calculate the circumference of the clarifier, that is the weir length. = 3.14*80 ft = 251 ft

• WOR = 3,500,000 gpd / 251 ft = 13,944 gpd/ft

Process Loading

• Plant Loadings, lbs/day

lbs/day = Q, MGD * 8.34 lbs/million gallons-mg/L* C, mg/L

lbs/day = Q, MGD * 8.34 lbs/MG-mg/L* C, mg/L

Example: If a plant has an influent BOD5 of 250 mg/l and a flow of 9 MGD, how many pounds per day are entering the plant

lbs/day BOD5 = 9 MG /day x 8.34 lbs/MG-mg/L x 250 mg/L = 18,765 lbs/day

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Process Loading

lbs/day = Q, gpd * 8.34 lbs/gallon* C, decimal fraction

Example: If 50,000 gallons per day of solids are being pumped from a tank at a concentration of 4 % solids, how many pounds of solids were pumped that day?

– Decimal fraction of 4% = 0.04

• lbs of solids/day = 50,000 gallons/day x 8.34 lbs/gallon x 0.04 = 16,680 lbs/day

Process Efficiency

• How well is the process working– Per unit

– Entire treatment train

– Measured as percent removal

– Removal, % = (in-out)/in X 100

Example-What is the % removal of TSS in a primary settling tank.

– Influent TSS = 250 mg/L

– Primary Effluent TSS = 110 mg/L

• % Removal = (250 mg/L-110 mg/L)/250 mg/L*100

= 56%

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Surface Overflow Rates (SOR)

• Used to design and process control

• SOR, gal/ft2.day =Q,gpd/A,ft2

• A = 7,000 ft2 (Per unit)

• Q=5,500,000 gal/d (Per unit)

• SOR = (5,500,000 gal/d)/ 7,000 ft2

= 785 gal/ft2.day

Secondary Treatment

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Secondary Treatment

• Biofilm– Trickling Filter

• Hydraulic and organic loading

• Recirculation Ratio

– RBC• Hydraulic and Organic

Loading

• Activated Sludge– SRT

– F/M

– SVI

– SLR

– Wasting

Trickling Filter Calculations

• Hydraulic loading– GPD/ft2

Hydraulic loading = Q, gpd/A, ft2

Example:Q = 3.24 MGDTF diameter = 100 ft, Radius = 50 ftHydraulic loading = 3,240,000 gpd/

3.14(50ft)(50ft)= 413 gpd/ft2

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Trickling Filter Calculations

• Organic loading

Organic Loading = mass/volume

= lbs BOD/d/1000 ft3

= Q, MGD*8.34 lbs/MG-mg/L*BOD, mg/l

Volume, 1000 ft3

Trickling Filter Calculations

• Example: Primary effluent BOD = 197 mg/LQ = 1.5 MGDDepth = 5 ft Diameter = 100 ft= (BOD, mg/l)(8.34 lbs/million gallons-mg/l)(Q, MGD)

Volume, 1000 ft3

Volume = 3.14 (r2)(h)/1000 = 3.14(50ft)(50ft)(5ft)/1000= 39(1000 ft3)

OLR= 1.5 MG/d*8.34 lbs/MG-mg/L*197 mg/L/ 39 (1000 ft3) = 63 lbs/d (1000 ft3)

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Trickling Filter Calculations

• Recirculation Ratio

– Typcial values

• 0.5:1, 1:1, 2:1

Recirculation ratio = QRecycle/QPriEff

Examples:

QPriEff = 2.5 MGD Qrecycle = 4.25 MGD

RR = 4.25/2.5 = 1.7

RBC Calculations

• Hydraulic loading-gpd/ft2

– Typically 1.5 to 6 gpd/ft2

• Organic loading, lbs BOD/d/1000 ft2

– SBOD 2.5 to 4 lbs/d/1000 ft2

– TBOD 6 to 8 lbs/d/1000 ft2

SBOD = TBOD - PBOD

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RBC Calculations

• Hydraulic Loading Example– QPriEff = 1.85 MGD

– Media Area = 600,000 ft2

Hydraulic Loading = 1,850,000 gpd/600,000 ft2

Hydraulic Loading = 3.1 gpd/ft2

RBC Calculations

• Organic Loading Example– QPriEff = 1.0 MGD

– Media Area = 500,000 ft2 500 (1000 ft2)

– TBOD = 340 mg/l; PBOD= 120 mg/l; SBOD=340-120 = 220 mg/L

Organic Loading = SBOD, lbs/d / Media area, ft2

Organic Loading = (1.0 MG/d)(8.34 lbs/MG-mg/L)(220 mg/L)/(500 (1000 ft2))

= 3.7 lbs SBOD/d/1000 ft2

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Solids Retention Time

• SRT-How long the solids remain in the system.– Dependent of growth rate of the organisms

(bacteria).

– Used as a process monitoring, control and design parameter.

Solids Retention Time

• SRT (MCRT) = Solids in biological system divided by the solids leaving the system.– For BOD removal, typically 3 to 5 days

– For nitrogen removal, greater than 8 days.

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Solids Retention Time

Solids (bio)= Vbio, MG x 8.34 lbs/MG-mg/L x MLSS, mg/L

Solids leaving the system = Qwas MG/d x 8.34 lbs/MG-mg/L x Cwas, mg/L + Q, MG/d x 8.34 lbs/MG-mg/L x TSSeff,

mg/L

Therefore;

SRT =(Vbio (MG) x 8.34 lbs/MG-mg/L x MLSS, mg/l)/(Qwas MGD x 8.34 lbs/MG-mg/L x Cwas, mg/L) + ( Q, MGD x 8.34 lbs/MG-mg/L x TSSeff)

SRT Example

• Given:– Vbio = 1.4 MG

– Q = 3.1 MGD

– MLSS = 2650 mg/L

– Qwas = 70,000 gpd

– TSSeff = 20 mg/L

– WAS =RAS = 5960 mg/L

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Example

• Solidsbio= (1.4 MG x 8.34 lbs/MG-mg/L x 2650 mg/L) = 30,941 lbs

• Solids leaving system = (0.07 MG x 8.34 lbs/MG-mg/L x 5960 mg/L) + (3.1MG/d x 8.34 lbs/MG-mg/L x 20 mg/L) = 3997 lbs/day

• SRT = 30,941 lbs/3997 lbs/d = 7.7 days

F/M Ratio

• F/M Ratio– Food to microorganism ratio

– Mass of food entering biological reactors, lbs BOD

– Mass of microorganisms in the biological reactors, lbs MLVSS

– MLVSS typically is 80% of the MLSS

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F/M Ratio

• F/M Ratio

• Food = Q, MGD x 8.34 lbs/MG-mg/L x BOD, mg/L

• Microorganisms = Vbio, MG x 8.34 lbs/MG-mg/L x MLVSS, mg/L

F/M = (Q, MGD x 8.34 lbs/MG-mg/L x BOD, mg/L)/(Vbio x 8.34 lbs/MG-mg/L x MLVSS, mg/L)

F/M Ratio

Example: MLSS=3000 mg/L; 82% volatile; Q = 16.5 MGD; V = 7.5 MG; PE BOD = 135 mg/L

F/M = (Q, million gallons/d x 8.34 lbs/million gallons-mg/L x BOD, mg/L)/(Vaer x 8.34 lbs/million gallons-mg/L x MLVSS, mg/L)

F/M = (16.5 MGD) * 8.34 lbs/ MG-mg/L * 135 mg/L

(7.5 MG) * 8.34 lbs/MG-mg/L * (3000*0.82) mg/L

F/M = 2228 = 0.12

18450

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WAS and RAS Rates

• RAS is the source of microorganisms• Microorganisms use the food (BOD) to make

new cells. For every pound of BOD used, there are about 0.5 to 0.6 pounds of new cells produced. This is known as Cell Yield.

• WAS is used to remove excess cells (solids) from the process.

WAS and RAS Rates

• How do I know how much to waste or return?– Based on mixed liquor concentration– Based on F/M ratio– Based on SRT (MCRT)

• Depends on what you are trying to achieve.

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WAS and RAS Rates

• Typically, RAS is 30-50% influent Q, but can be higher. Designers generally give capability of returning 100%, but high rates can negatively impact clarifier performance.

• Calculate WAS by rearranging the SRT formula solving for WAS.

WAS Rate

SRT = lbs of solids biological systemWAS, lbs/d + TSSEff, lbs/d

WAS = Solids biological system- TSSEff

SRT

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WAS Rate

WAS, lbs/d = Solids biological- TSSEff

SRT

WAS, MGD = (WAS, lbs/d)/(8.34 MG-mg/L x WAS,mg/L)

Waste time, minutes = WAS, gpd/ Pump rate, gpm

WAS Rate Example

• SRT = 10 days

• MLSS = 3100 mg/L, Vbio =3.1 MG

• TSSEff = 8 mg/l, WAS = 8600 mg/l

• Q = 18.6 MGD

Sbio = 3100 mg/L x 8.34 lbs/million gallons/d-mg/L x 3.1 million gallons= 80147 lbs/d

TSSeff, lbs/d = (8 mg/L x 8.34 lbs/million gallons/d-mg/L x 18.6 million gallons/d) =1241 lbs/d

WAS, lbs/day = 80147 lbs/d–1241 lbs/d

10

WAS = 6774 lbs/d

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WAS Rate Example

WAS = 6774 lbs/d

WAS, MGD = 6774 lbs/d / (8.34 lbs/MG-mg/Lx 8600 mg/L)

WAS = 0.119 MGD =119,000 gpd

Pump rate = 250 gpm

Pump time = 119,000 gal/250 gpm = 476 min* hr/60 min = 7.9 hrs.

Surface Overflow Rates (SOR)

• Used to design and process monitoring and control

• SOR, gal/ft2.day =Q,gpd/A,ft2

• A = 7,000 ft2 (Per unit)

• Q=5,500,000 gal/d (Per unit)

• SOR = (5,500,000 gal/d)/ 7,000 ft2

= 785 gal/ft2.day

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Solids Loading Rate

SLR = (Q,MGD + R, MGD) x MLSS, mg/L x 8.34 lbs/MG-mg/L/ A, ft2

SLR = Solids loading rate (lbs/d-ft2 )Q = Plant flow, MGDR = RAS flow rate, MGDMLSS = Mixed liquor suspended solids, mg/LA = Clarifier surface area, ft2

SLR Example

• Q = 18.2 MGD, R = 7.7 MGD• MLSS = 3350 mg/L, • A = 39,978 ft2 (total area)

SLR = (18.2 + 7.7) MG/d x 8.34 lbs/MG-mg/L x 3350, mg/L / 39,978 ft2

SLR = 18 lbs/d-ft2

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Solids Processing

Solids Production

• From primary settling:

Example: The influent SS is 250 mg/L, the flow is 3.1 MGD and a removal rate of 65%, how many pounds of primary sludge is generated? At 3% solids, how many gallons/day?

lbs/d ssin = 3.1 MG * 8.34 lbs * 250 mg/L =6464 lbs/d

day MG-mg/L

PS = 0.65 * 6464 lbs/d = 4201 lbs/d

gpd = 4201 lbs/d = 16,792 gpd

8.34 lbs/gal * 0.03

PSTSSin SSout

SSremoved

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Solids Production

• For secondary settling: Must take into account bacterial growth, Y

• Assume Y = 0.47 This means that for every pound of BOD consumed by the bacteria that 0.47 pounds of solids are produced

Example: Primary Effluent BOD=124 mg/L; Final Effluent BOD = 8 mg/L, Q = 2.1 MGD

lbs/d BOD = 2.1 MG * 8.34 lbs * (124-8) mg/L = 2032 lbs/d

day MG-mg/L

= 2032 lbs/d * 0.47 lbs/lbs = 955 lbs/d

Sludge Thickening Problem

How many gallons per day of thickened sludge given the following:

Primary Sludge = 3.1% ; Qps = 4,800 gal/d;

Thickened sludge = 5.3 %Qps * 8.34 lbs/gal * %/100 = Qts * 8.34 lbs/gal * %/100

4800 gal/d * 8.34 lbs/gal * 0.031 = Qts * 8.34 lbs/gal * 0.053

1241 lbs/d =Qts * 8.34 lbs/gal * 0.053

Qts = 1241 lbs/d Qts = 2808 gal/d8.34 lbs/gal * 0.053

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GBT/BFP

• Hydraulic capacity is fixed based on belt size.• Variables include sludge characteristics, belt

porosity, polymer, belt speed.• Typical hydraulic loading per meter belt width

– 1 m 100 to 250 gpm– 1.5 m 150-375 gpm– 2.0 m 200-500 gpm– 3.0 m 300 to 750 gpm

• Typically 4-9% solids with a WAS feed concentration of 0.5 to 1 %

GBT Problem

• What is the loading rate in gpm on a 2-m GBT if you are thickening 4,500 lbs/d of WAS and the GBT’s are operating 8 hrs/day with a feed solids concentration of 3000 mg/L?

Q = 4500 lbs/d8.34 lbs/gal-mg/l * 3000 mg/L

Q= 0.18 mgd = 180,000 gal(Day = 8 hours = 480min)

Q = 180,000 gal = 375 gpm per 2-m GBT480 min

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Anaerobic Digestion

Calculate Volatile Solids to the Digester

VS, lbs/d = TS, lbs/d * (% VS/100)

Example: A total of 4500 gpd of solids is pumped to the digester. It is 3.5 % solids and is 72% volatile.

VS, lbs/d = 4500 gpd * 8.34 lbs/gal * 0.035 *0.72 = 946

VA:ALK Ratio

• Volatile Acid to Alkalinity Ratio. The VA:ALK ratio is an indicator of the progress of digestion and the balance between the acid fermentation and methane fermentation microorganisms.

Example: The VA concentration in the sludge is 195 mg/l and the alkalinity is 2150 mg/L, calculate VA:ALK

VA:ALK = 195 mg/L = 0.09

2150 mg/L

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Volatile Solids Reduction

• %VS Reduction = in – out x 100

in – (in*out)

Example: Sludge entering digester is 70% volatile and leaving the digester is 52% volatile. What is the VS reduction?

• %VS Reduction = 0.70– 0.52 x 100

0.70– (0.70*0.52)

= 0.18 x 100 = 54% VS

0.34

Chemical Feed

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Hypochlorite Dose

Chlorine dose = chlorine demand + chlorine residual

Example: The plant effluent had a chlorine demand of 8 mg/L. If the desired residual is 0.3 mg/L, what is the chlorine dose.

• Chlorine dose = 8 mg/L + 0.3 mg/L = 8.3 mg/L

The hypochlorite has 65 % available chlorine, how many lbs/d of hypochlorite is required for this chlorine demand if the plant fowis 1.8 MGD?

lbs Cl2 /d = 1.8 MG * 8.34 lbs * 8.3 mg/L = 125 lbs Cl2 /day

day MG-mg/L

Hypochlorite, lbs/d = lbs/day Cl2 = 125 lbs Cl2 /day = 192 lbs/d

% Available Cl2/100 0.65 avail Cl2

Polymer Feed

• How many pounds of dry polymer must be added to 38 gallons of water to make a 0.5 % solution?% Strength = lbs chemical * 100

(lbs water + lbs of chemical)0.5 = x lbs of chemical * 100

(38 gal * 8.34 lbs/gal) + x lbs of chemical) 0.5 = 100 x

(317 + x)0.5 (317 + x) =100 x ; 158 + 0.5 x = 100x158 = 100x – 0.5 x; 158 = 99.5 x x = 158 x = 1.6 lbs of chemicals

99.5

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Laboratory

Settleable Solids

% Settleable Solids = mL settle solids x100

2000-mL Sample

Example: The MLSS settled to 420 mL, what is the percent settleable solids?

% Settleable Solids = 420 mL x 100 = 21 %

2000-mL Sample

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Sludge Volume Index

• SVI= a measurement of the ability of the floc to settle

• Typically 80 to 150

• (2200 mg = 2.2 g)

• SVI = ml/MLSS,g

• SVI = 210 ml/2.2 g = 95

210 ml

Biochemical Oxygen Demand

BOD5= Initial DO, mg/L – DO after 5 days, mg/L

Sample Volume, mL/BOD Bottle Volume, mL

Example: Sample volume = 5 mL; BOD Bottle = 300 mL;

Initial DO = 7.2 mg/L; DO after 5 days = 3.5mg/L

BOD5= 7.2 mg/L – 3.5 mg/L = 3.7 mg/L = 222 mg/L

5 mL/300 mL 0.017

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Total Solids

Total/Volatile Solids

Example:

Wt of Sample & Dish: 75 g sample; 31g after drying; 27.5 g ash

Wt of Dish: 22.08 g

% TS = 31 g -22.08 g x 100 = 8.92 g/ 52.9g * 100 = 16.9 %

75 g- 22.08 g

% VS = 27.5 g -22.08 g x 100 = 5.42 g/ 8.92g * 100 = 60.8 %

31-22.08) g- 22.08 g

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QUESTIONS?

WEFCOM Discussion

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