Filter Loading and Backwash Rates Well Yields and Chlorine Dosage in

Waterworks Operation

Math for Water TechnologyMTH 082

Lecture Chapter 4 & 8- Applied Math for Water Plant Operators

Loading Rate Calculations (pg 62-66); Well yield (163-177)Mathematics Chapter 19-20 &23- Basic science Concepts and

Applications (pg 183-190; 197-203)

Objectives

1. Filter overviews2. Filter loading and backwash rates3. Well yield 4. Chlorine dosage for a new well

Reading assignment: Chapter 4 & 8- Applied Math for Water Plant Operators Loading Rate Calculations (pg 62-66); Well yield (163-177)Mathematics Chapter 19-20 &23- Basic science Concepts and Applications (pg 183-190; 197-203)

Conventional Treatment • Conventional Treatment – common treatment steps used to remove turbidity from the initial source water.

2. Flocculation 3. Sedimentation 4. Filtration1. Coagulation

Pretreatment

Raw Water

Rapid or flash Mixing

Alum, polymer

SlowFast

SludgeWashwater

5. Clear Well Disinfection

Distri

bution

ClearwellBackwash

pumps

ChlorinationOzone

UV

High Rate FilterDual or Multi Media Filter

-4 times faster then rapid sand -granular activated carbon-anthracite coal-garnet sand and gravel-backwash ready (70 hr longevity)-excellent water quality

High Rate Filters

Anthracite Coal

Fine sand

Garnet sand

High Rate Filter

Dualmedia Filter

Anthracite Coal

Fine sand

Garnet sand

Monomedia Coarse sand

Monomedia Filter

Dual-media and multimedia filters

Requir

e an ex

t...

Can opera

te at.

..

Canno

t red

uce ...

Do not req

uire...

5% 5%0%

91%1. Require an extremely deep bed2. Can operate at three or four times the

rate of sand filters3. Cannot reduce turbidity4. Do not require backwashing

When mixed media filters composed of garnet, sand, and

crushed anthracite coal are used, which of the following

describes their placement in the filter bed?

Anthrac

ite co

a...

Garn

et on to

p,...

Sand on t

op, a...

Anthracite

coa..

.

8%

92%

0%0%

1. Anthracite coal on top, garnet in the middle, and sand on the bottom

2. Garnet on top, anthracite coal in the middle, and sand on the bottom

3. Sand on top, anthracite coal in the middle, and garnet on the bottom

4. Anthracite coal on top, sand in the middle, and garnet on the bottom

In a filter using gravel, anthracite, and sand, the

anthracite should be?

The top

laye

r ...

Benea

th the g

r...

Betwee

n the s

a...

Mixe

d with

the..

.

88%

0%12%

0%

1. The top layer of media2. Beneath the gravel3. Between the sand and the gravel4. Mixed with the sand

Pilot Filters

Anthracite Coal

Fine sand

Garnet sand

The main action of a mixed media filter is:

Straining

Disinfec

ting

Coagulat

ing

None of th

e ab..

.

87%

9%4%0%

1. Straining2. Disinfecting3. Coagulating4. None of the above

Backwash• Suspended particles entrapped by filter media. • Accumulation occurs:

– head loss within the filter to reach excessively high levels (6 to 8 feet of hydraulic head).

- Particles pass through the filter, water turbidities reach unacceptable levels

- Rule Backwash at 0.1 NTU- SWTR Allows 0.3 NTU.

Head Loss• Clean filter =0 psi= one foot of head loss on a

new filter• As filter clogs more negative pressure• Pressure builds in a linear fashion

-2.5 to -4.0 psi = 6 to 10 ft of head loss• More clogged greater the head loss• Remember 1 ft of water column = 0.433 psi• 2.31 ft of water for 1 psi change

Filter Backwash• Some plants use head loss, some use

time• Some plants use operator knowledge

and turbidity• Each operator has their own scheme!!• Parents and diapers--

The most critical criterion for determining when a mixed media filter should be backwashed is:

Filter e

ffluen

...

Flow ra

te

Head lo

ss

Visual

inspec

t...

52%

4%

39%

4%

1. Filter effluent quality2. Flow rate3. Head loss4. Visual inspection of the filter surface

Filtration Rate CalculationUnit Filter Run= (Total gal filtered gal) Filter Area (sq ft)

Filtration Rate= (flow gpm) Area (sq ft) Units will be gpm ft2!

Backwash Rate = (flow gpm) Area (sq ft)

Units will be gpm ft2!

Downward

Upward

The total water filtered during a filter run (between backwashes) is 2,950,000 gal. If the filter is 15 ft by 20 ft, What is the

unit filter run volume (UFRV)?

9.83

3 g/ft2

101.7

g/ft2

9833

g/ft2

0.98

33 g/ft2

25%

0%

75%

0%

L= 15 ft, W=20 ft; Rate 2,950,000 galA=L X WUFVR= (Total gallons filtered g) Area (sq ft)

A= 20 ft X 15 ft = 300 ft2

UFVR= (2,950,000 gal) 300 (sq ft) UFVR = 9833 g/ft2

GivenFormula

Solve:

1. 9.833 g/ft2

2. 101.7 g/ft2

3. 9833 g/ft2

4. 0.9833 g/ft2

The total water filtered during a filter run (between backwashes) is 4.8 MG. If the

filter is 20 ft by 30 ft, What is the unit filter run volume (UFRV)?

960

00 g/

ft2

8000

g/ft2

800 g

/ft2

0.00

8 g/ft2

0% 0%0%

100%

L= 20 ft, W=30 ft; Rate 4.8 MGA=L X WUFVR= (Total gallons filtered g) Area (sq ft)

A= 20 ft X 30 ft = 600 ft2

UFVR= (4,800,000 gal) 600 (sq ft) UFVR = 8000 g/ft2

GivenFormula

Solve:

1. 96000 g/ft2

2. 8000 g/ft2

3. 800 g/ft2

4. 0.008 g/ft2

A filter 20 ft by 25 ft receives a flow of 1940 gpm. What is the filtration rate in

gpm/ft2?

3.9

gpd/ft3

3.9 g

pm/ft2

0.25

gpm/ft2

0.25

gpd/ft3

11%0%6%

83%

L= 20 ft, W=25 ft; Rate 1940 gpmA=L X WFiltration Rate= (flow gpm) Area (sq ft)

A= 20 ft X 25 ft = 500 ft2

Filtration Rate= (1940 gpm) 500 (sq ft) Filtration Rate = 3.9 gpm/ft21. 3.9 gpd/ft3

2. 3.9 gpm/ft2

3. 0.25 gpm/ft2

4. 0.25 gpd/ft3

GivenFormula

Solve:

A filter 20 ft by 35 ft receives a flow of 1530 gpm. What is the filtration rate in

gpm/ft2?

2.2

gpm/ft2

2.2 g

pm/ft3

0.45

gpm/ft2

0.45

gpd/ft3

89%

0%0%11%

L= 20 ft, W=35 ft; Rate 1530 gpmA=L X WFiltration Rate= (flow gpm) Area (sq ft)

A= 20 ft X 35 ft = 700 ft2

Filtration Rate= (1530 gpm) 700 (sq ft) Filtration Rate = 2.2 gpm/ft21. 2.2 gpm/ft2

2. 2.2 gpm/ft3

3. 0.45 gpm/ft2

4. 0.45 gpd/ft3

GivenFormula

Solve:

A filter 25 ft by 30 ft receives a flow of 3.3 MGD. What is the filtration rate in

gpm/ft2?

3.1 g

pm/ft3

0.32

gpm/ft2

3.1 g

pm/ft2

0.32

gpd/ft3

0% 0%

100%

0%

L= 25 ft, W=30 ft;Rate 3.3 MG 1,000,000 gal 1 Day D 1MG 1440 minA=L X WFiltration Rate= (flow gpm) Area (sq ft)

A= 25 ft X 30 ft = 750 ft2

Filtration Rate= (2292 gpm) 750 (sq ft) Filtration Rate = 3.1 gpm/ft2

1. 3.1 gpm/ft3

2. 0.32 gpm/ft2

3. 3.1 gpm/ft2

4. 0.32 gpd/ft3

GivenFormula

Solve:

GivenFormula

Solve:

A filter 25 ft by 10 ft has a backwash rate of 3400 gpm. What is the filter backwash

rate in gpm/ft2?

13.1

gpm/ft3

0.07

4 gpm

/ft2

13.6

gpm/ft2

136 g

pd/ft3

0% 0%

100%

0%

L= 25 ft, W=10 ft; Rate 3400 gpmA=L X WBackwash Rate = (flow gpm) Area (sq ft)

A= 25 ft X 10 ft = 250 ft2

Filter Backwash Rate= (3400 gpm) 250 (sq ft) Filter Backwash Rate = 13.6 gpm/ft2

1. 13.1 gpm/ft3

2. 0.074 gpm/ft2

3. 13.6 gpm/ft2

4. 136 gpd/ft3

GivenFormula

Solve:

A filter 20 ft by 15 ft has a backwash rate of 4.5 MGD. What is the filter backwash

rate in gpm/ft2?

10.4

gpm/ft3

0.01

gpm/ft2

10.4

gpm/ft2

0.01

gpd/ft3

6% 0%

94%

0%

L= 20 ft, W=15 ft; ft;Rate 4.5 MG 1,000,000 gal 1 Day D 1MG 1440min A=L X WBackwash Rate = (flow gpm) Area (sq ft)

A= 20 ft X 15 ft = 300 ft2

Filter Backwash Rate= (3125 gpm) 300 (sq ft) Filter Backwash Rate = 10.4 gpm/ft21. 10.4 gpm/ft3

2. 0.01 gpm/ft2

3. 10.4 gpm/ft2

4. 0.01 gpd/ft3

Well Problems• Drawdown ft = pumping water level – static water level ft

• Well yield = Flow gallons duration of Test, min

• Specific yield, gpm/ft = (Well yield gpm) (Drawdown ft)

• Well casing disinfectionlbs= (dose mg/L Cl2)(water in well casing MG)(8.34 lb/gal)Chlorine lbs = chlorine lbs % available chlorine 100

GivenFormula

Solve:

Before the pump is started the water level is measured at 140 ft. The pump is then started. If the pumping water level is determined to be 167 ft, what is the

drawdown in ft?

307 f

t -2

7 ft

27 ft 0

ft

0% 0%

100%

0%

Static WL= 140 ft, Pumped WL=167 ft

Drawdown ft = pumping water level – static water level ft

Drawdown = 167 ft- 140 ftDrawdown = 27 ft

1. 307 ft2. -27 ft3. 27 ft4. 0 ft

GivenFormula

Solve:

During a five minute test for well yield, a total of 740 gallons are removed from the

well. What is the well yield in gpm?

67 gp

m

148 g

pm

3700

gpm 0

gpm

0% 4%0%

96%

total = 740 gal, time = 5 minutes

Well yield = Flow gallons Duration of Test, min

Well yield = 740 gallons = 148 gpm 5 min 1. 67 gpm2. 148 gpm3. 3700 gpm4. 0 gpm

GivenFormula

Solve:

How many lbs of calcium hypochlorite (65% available chlorine) is required to disinfect a well if the casing is 18 inches in diameter and 220 ft long, with water level at 100 ft from the top of the well? The desired dose is

50 mg/L?

2 lbs

1 lbs

.02 l

bs

0.65

lbs

24%

6%0%

71%

Cl= 65/100 D=18 in=1.5 ft Well 220-100 =120 ft220 ft - 100 ft = 120 ft water in well(0.785)(D2)(H) = ft3

(0.785)(1.5 ft)(1.5 ft) (120 ft)(7.48 gal/ft3)= 1585 gal(50 mg/L)(.001585 MG)(8.34 lb/gal) = 1.01lbs 65/100

1. 2 lbs2. 1 lbs3. .02 lbs4. 0.65 lbs

Today’s objective: Filter Loading Rates, Filter Backwash Rates, Well yield and

chlorine dosage of new wells been met?

Strongly

Agree

Agree

Neutra

l

Disagree

Strongly

Disagree

0% 0% 0%0%0%

1. Strongly Agree2. Agree3. Neutral4. Disagree5. Strongly Disagree