Lin S.D. Water and Wastewater Calculations Manual

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Step 4. Estimate blended ﬂows

Using C

⫽ 0.367 kg/m

, estimate TDS concentration of the product water.

The TDS in the blended water is limited to 0.45 kg/m

Mass balance equation would be

⫹ Q

) (0.45 kg/m

) ⫽ Q

(2.6 kg/m

) ⫹ Q

(0.367 kg/m

)

2.15Q

⫺ 0.083Q

⫽ 0 (1)

and

⫹ Q

⫽ 0.25 m

⫽ 0.25 ⫺ Q

(2)

Substituting (2) into (1)

2.15Q

⫺ 0.083 (0.25 ⫺ Q

) ⫽ 0

2.067Q

⫽ 0.02075

⫽ 0.010 (m

/s)

then Q

⫽ (0.25 ⫺ 0.01) m

/s ⫽ 0.24 m

The water supply will use blending 0.24 m

/s of RO product water and 0.01 m

of feedwater (4%).

Step 5. Compute the required membrane area

Example 5: Determine the sizes of a 4 MGD (15,140 m

/d) RO system to pro-

duce a ﬁnal blended water TDS of <300 mg/L. The design average TDS in feed-

water is 1100 mg/L. The other design criteria (Crittenden et al., 2005) and

conditions are given below:

5 32,880 m

0.24 m

/s 3 1000 kg/m

7.3 3 10

kg/m

A 5

Public Water Supply 455

Stage 1 Stage 2

2400 m

480 m

12,000 m

9600 m

/d 1920 m

Bypass 3100 m

14,620 m

Plant average ﬂow rate, Q 15,100 m

Feedwater pressure 1035 kN/m

(150 psi, 10.3 bar)

Feedwater temperature 20ºC

System recovery factor, R 80% for both stages 1and 2

Salt rejection, Rej 96%

System design criteria

Number of RO skids 3

Capacity per skid 5040 m

Membrane area per element 32.5 m

Element per pressure vessel 7 maximum

Number of stages per skid 2

Average permeate ﬂux for stage 1 22 L/m

⋅ h

Average permeate ﬂux for stage 2 17 L/m

⋅ h

solution:

Step1. Calculate expected permeate TDS concentration in stage 1, C

, using

Eq. (5.194b)

Step 2. For stage 1 per skid: Calculate permeate ﬂow rate (Q

), concentrate

ﬂow rate Q

, and TDS in concentrate C

(by mass balance)

⫽ QR ⫽ 5040 m

/d ⫻ 0.80 ⫽ 4030 m

⫽ 5040 m

/d ⫺ 4030 m

/d ⫽ 1010 m

then 1010 m

/d ⫻ C

⫹ 4030 m

/d ⫻ 55 mg/L ⫽ 5040 m

/d ⫻ 1100 mg/L

⫽ (5040 ⫻ 1100⫺4030 ⫻ 55)/1010

⫽ 5270 (mg/L)

Step 3. For stage 2 per skid: Same as Steps 1 and 2, but using feedwater

ﬂow rate of 1010 m

/d and TDS of 5270 mg/L (which is actual feed to second

stage), we calculate TDS concentration and ﬂow rate, for both permeate (C

and Q

) and concentrate (C

and Q

) for the stage 2

⫽ 5270 mg/L ⫻ 0.04/0.8 ⫽ 264 mg/L

⫽ 1010 m

/d ⫻ 0.8 ⫽ 808 m

/d (say 810 m

/d )

⫽ 1010 m

/d ⫺ 810 m

/d ⫽ 200 m

and 200 m

/d ⫻ C

⫹ 810 m

/d ⫻ 264 mg/L ⫽ 1010 m

/d ⫻ 5270 mg/L

⫽ 25,540 mg/L

Step 4. Compute combined permeate ﬂow rate Q

, and its TDS concentra-

tion C

⫽ Q

⫹ Q

⫽ 4030 m

/d ⫹ 810 m

/d ⫽ 4840 m

5 55 mg/L

5 C

s1–Rejd/R 5 1100 mg/L s1–0.96d/0.80

456 Chapter 5

and Q

⫽ Q

⫹ Q

4840 m

/d ⫻ C

⫽ 4030 m

/d ⫻ 55 mg/L ⫹ 810 m

/d ⫻ 264 mg/L

⫽ 90 mg /L

Step 5. Determine the fraction (F) of the plant ﬂow rate for bypass ﬂow Q

from feedwater to blend with the ﬁnal permeate and to meet ﬁnished water

TDS <300 mg/L requirement (same Q

ﬁnal

)

F ⫻ 1100 mg /L ⫹ (1⫺ F) ⫻ 90 mg/L ⫽ 1 ⫻ 300 mg/L

F ⫽ 0.208

⫽ 0.208 ⫻ 5040 m

/d ⫽ 1050 m

Step 6. Compute feedwater ﬂow rate to be treated (Q

) by RO system to

meet the TDS <300 mg/L requirement

⫽ 5040 m

/d ⫺ 1050 m

/d ⫽ 3990 m

/d (say 4000 m

/d )

Step 7. Determine RO system per skid

Feedwater ﬂow rate ⫽ 4000 m

Permeate ﬂow rate ⫽ 3200 m

Concentrate ﬂow rate ⫽ 800 m

System recovery ⫽ 80%

Number of RO skids ⫽ 3

Permeate capacity per skid ⫽ 3200 m

Step 8. Sizing RO for stage 1 per skid

Capacity per skid ⫽ 4000 m

/d ⫽ 4000 m

/d ⫻ 1000 L/m

÷ 24 h/d

⫽ 167,000 L/h

Permeate ﬂow for each element ⫽ 32.5 m

⫻ 22 L/m

⋅ h ⫽ 715 L/h

Number of elements ⫽ (167,000 L/h) / (715 L/h) ⫽ 237

Provide 240 elements. Using 6 elements per pressure vessel, 40 pressure ves-

sels are needed.

Step 9. Sizing RO for stage 2 per skid

Capacity per array ⫽ 4000 m

/d ⫻ 0.2 ⫽ 800 m

/d ⫻ 1000 L/m

÷ 24 h/d

⫽ 33,300 L/h

Flux rate for each element ⫽ 32.5 m

⫻ 17 L/m

⋅ h ⫽ 552.5 L/h

Number of elements ⫽ (33,300 L/h)/(552.5 L/h) ⫽ 60

Public Water Supply 457

Provide 60 elements. There will be 10 pressure vessels that use 6 elements

per pressure vessel.

16.4 Silt density index

Silt density index (SDI) test involves water ﬁltering in the dead end

mode with no cross-ﬂow at a constant pressure of 207 kPa (30 psig). It

uses disc membranes rated at 0.45 ␮m for a 15-min test with a 500-mL

water sample. The SDI is calculated as (Chellam et al., 1997):

(6.150)

where SDI ⫽ silt density index

⫽ time initially needed to ﬁlter 500 mL of sample

⫽ time needed to ﬁlter 500 mL at the end of the 15-min

test period

The SDI has been widely used as a rough estimation of the potential

for colloidal fouling of NF and RO systems. Generally, the feedwaters

for NF and RO should have SDI of 4 or less. (Most manufacturers have

a 4 as warranty limit.)

Example: The time initially required to ﬁlter 500 mL of dual-media ﬁlter

efﬂuent is 14.5 s. The time required to ﬁlter 500 mL of the same water sample

at the end of the 15-min test period is 35 s. Calculate the SDI.

solution:

17 Residual from Water Plant

A water treatment plant not only produces drinking water, but is also

a solids generator. The residual (solids or wastes) comes principally

from clariﬁer basins and ﬁlter backwashes. These residuals contain

solids which are derived from suspended and dissolved solids in the

raw water with the addition of chemicals and chemical reactions.

Depending on the treatment process employed, waste from water

treatment plants can be classiﬁed as alum, iron, or polymer sludge from

coagulation and sedimentation; lime sludge or brine waste from water

5 3.9

SDI 5 a1 2

100

5 a1 2

14.5 s

35 s

b 3

100

SDI 5 a1 2

100

458 Chapter 5

softening; backwash wastewater and spent granular activated carbon

from ﬁltration; and wastes from iron and manganese removal process,

ion exchange process, diatomaceous earth ﬁlters, microstrainers, and

membranes.

The residual characteristics and management of water plant sludge

and environmental impacts are discussed in detail elsewhere (Lin and

Green, 1987).

Prior to the 1970s, residuals from a water treatment plant were dis-

posed of in a convenient place, mostly in surface water. Under the 1972

Water Pollution Control Act, the discharge of a water plant residuals

requires a National Pollution Discharge Elimination System (NPDES)

permit. Residuals directly discharged into surface water is prohibited

by law in most states. Proper residual (solids) handling, disposal, and/or

recovery is necessary for each waterwork nowadays. Engineers must

make their best estimate of the quantity and quality of residuals gen-

erated from the treatment units.

Example 1: A conventional water treatment plant treats an average ﬂow

of 0.22 m

/s (5.0 MGD). The total suspended solids (TSS) concentration in

raw (river) water averages 88 mg/L. The TSS removal through sedimenta-

tion and ﬁltration processes is 97%. Alum is used for coagulation/sedimen-

tation purpose. The average dosage of alum is 26 mg/L. Assume the

aluminum ion is completely converted to aluminum hydroxide, (a) Compute

the average production of alum sludge. (MW: Al ⫽ 27, S ⫽ 32, O ⫽ 16, and

H ⫽ 1). (b) Estimate the volume of residual produced daily, assuming the

speciﬁc gravity of 4.0% solids is 1.02.

solution:

Step 1. Determine quantity (q

) settled for TSS

TSS ⫽ 88 mg/L ⫽ 88 g/m

⫽ ﬂow ⫻ TSS ⫻ % removal

⫽ 0.22 m

/s ⫻ 88 g/m

⫻ 0.97

⫽ 18.8 g/s

Step 2. Determine A1(OH)

generated, q

The reaction formula for alum and natural alkalinity in water using Eg. (5.51a):

(SO

)

⋅ 18H

O ⫹ 3Ca(HCO

)

→ 2A1(OH)

⫹ 3CaSO

⫹ 6CO

⫹ 18H

27 ⫻ 2 ⫹ 3(32 ⫹ 16 ⫻ 4) ⫹ 18(18) 2(27 ⫹ 17 ⫻ 3)

⫽ 666 ⫽ 156

The above reaction formula suggests that each mole of dissolved Al

3⫹

dosed

will generate 2 moles of A1(OH)

Public Water Supply 459

Step 3. Compute total residual generated q for (a)

Step 4. Compute the volume generated daily for (b)

Volume ⫽ 1740 kg/d ÷ (1000 kg/m

⫻ 1.02 ⫻ 0.04)

⫽ 42.6 m

or Volume ⫽ 3818 lb/d ÷ (62.4 lb/ft

⫻ 1.02 ⫻ 0.04)

⫽ 1500 ft

Example 2: A river raw water is coagulated with a dosage of 24 mg/L of fer-

rous sulfate and an equivalent dosage of 24 mg/L of ferrous sulfate and an equiv-

alent dosage of lime. Determine the sludge generated per m

of water treated.

solution: From Eqs. (5.51 f) and (5.51 g), 1 mole of ferrous sulfate reacts with

1 mole of hydrated lime and this produces 1 mole of Fe(OH)

precipitate.

MW of FeSO

⋅ 7H

O ⫽ 55.85 ⫹ 32 ⫹ 4 ⫻ 16 ⫹ 7 ⫻ 18

⫽ 277.85

MW of Fe(OH)

⫽ 55.85 ⫹ 3 ⫻ 17

⫽ 106.85

By proportion

sludge

F.S. dosage

106.85

277.85

5 0.3846

5 3817 lb/d

5 1740

2.194 lb

1 kg

5 1740 kg/d

5 20.14

1 kg

1000 g

86,400 s

1 day

5 20.14 g/s

q 5 q

1 q

5 18.8 g/s 1 1.34 g/s

5 1.34 g/s

5 26 g/m

3 0.22 m

/s 3

156

666

5 alum dosage 3 flow 3 c

156 g AlsOHd

666 g alum

460 Chapter 5

Note: Mgal ⫽ million gallons.

Example 3: A 2-MGD (0.088 m

/s) water system generates 220 lb/Mgal

(26.4 mg/L) solids in dry weight basis. The residual solids concentration by

weight is 0.48%. Estimate the rate of residual solids production and pump-

ing rate if the settled residual is withdrawn every 5 min/h.

solution:

Step 1. Calculate residual production

Step 2. Calculate pump capacity required

7.63 gpm ⫻ 60/5 ⫽ 92 gpm

17.1 Residual production and density

The quantity of residuals (called sludge) generated from clariﬁers are

related to coagulants used, raw water quality, and process design. The

residual production for whole water plant as the weight per unit volume

treated (g/m

or lb/Mgal) and its solids concentration should be deter-

mined. These values with the average ﬂow rate will be used for the esti-

mation of daily residual production. Then the residual production rate

can be employed for the design of residual (sludge) treatment units,

commonly intermitted sand drying bed. The density of residuals (pounds

per gallon) can be estimated using the following equation:

(5.151a)w 5

8.34

1 c/r

sUS customary unitsd

s220 lb/Mgalds2 MGDd

s8.34 lb/galds0.0048d

5 10,990 gpd 5 7.63 gpm

5 77.0 lb/Mgal

5 9.23

0.0022 lb

1 g

1 m

2.642 3 10

Mgal

5 9.23 g/m

5 9.23

1000 mg

1000 L

5 9.23 mg/L

5 24 mg/L 3 0.3846

Sludge 5 F.S. dosage 3 0.3846

Public Water Supply 461

(5.151b)

where w ⫽ density of residuals, lb/gal or kg/L

⫽ weight fractional percent of solids

c ⫽ weight fractional percent of water

␳

⫽ speciﬁc gravity of solid

␳ ⫽ speciﬁc gravity of water (approximately 1.0)

Example 1: Calculate the sludge density (lb/gal) for a 6% sludge composed

of solids with a speciﬁc gravity of 2.48.

solution:

Step 1. Find speciﬁc gravity of 6% residual (sludge) ␳

Weight fraction of sludge, solids, and water are 1.0, 0.06, and 0.94, respectively.

Using mass balance, we obtain

Step 2. Determine the density of the residue

Note: This solution is a kind of proof of Eq. (5.151).

Example 2: A water plant treats 4.0 MGD (0.175m

/s) and generates 240 lb

of residuals per million gallons (Mgal) treated (28.8 g/m

). The solid concen-

tration of the residuals is 0.5%. Estimate: (a) the daily production of residu-

als, (b) the pump capacity, if the pump operates for 12 min each hour, and (c)

the density of the residuals, assuming the speciﬁc density of the residual is

2.0 with 4% solids.

solution:

Step 1. Determine residual production rate q

(a)

5 16.0 gpm

5 23,020 gpd

q 5

s4 MGDds240 lb/Mgald

s0.005ds8.34 lb/gald

5 1037 g/L

5 8.65 lb/gal

Density 5 r

3 8.34 lb/gal 5 1.037 3 8.34 lb/gal

5 1.037

0.06

0.94

0.06

2.48

0.94

1.0

w 5

1 c/r

sSI unitsd

462 Chapter 5

Step 2. Find pump capacity q

(b)

Step 3. Estimate residual density w, using Eq. (5.151)

(c)

Example 3: In a 1 MGD (0.0438 m

/s) waterworks, 1500 lb of dry solids

are generated per million gallons of water treated. The sludge is concen-

trated to 3% and then applied to an intermittent sand drying bed at a 18-in

depth, with 24 beds used per year. Determine the surface area needed for

the sand drying bed.

solution:

Step 1. Calculate yearly 3% sludge volume, v

v ⫽ 1500 lb/Mgal ⫻ 1 Mgal/d/0.03

⫽ 50,000 lb/d

⫽ 50,000 lb/d ⫻ 365 d/(62.41b/ft

)

⫽ 292,500 ft

⫽ 8,284 m

Step 2. Calculate area A (D ⫽ depth, n ⫽ number of applications/year)

v ⫽ ADn

or A ⫽ v/Dn ⫽ 292,500 ft

/(1.5 ft ⫻ 24)

⫽ 8125 ft

⫽ 755 m

18 Disinfection

Disinfection is a process to destroy disease-causing organisms, or

pathogens. Disinfection of water can be done by boiling the water,

ultraviolet radiation, and chemical inactivation of the pathogen. In the

water treatment processes, pathogens and other organisms can be

partly physically eliminated through coagulation, ﬂocculation, sedi-

mentation, and ﬁltration, in addition to the natural die-off. After ﬁl-

tration, to ensure pathogen-free water, the chemical addition of

5 8.51 lb/gal

w 5

8.34

0.04/2.0 1 0.96/1.0

5 80 gpm

5 16 gpm 3

60 min

12 min

Public Water Supply 463

chlorine (so called chlorination), rightly or wrongly, is most widely used

for disinfection of drinking water. This less expensive and powerful dis-

infection of drinking water provides more beneﬁts than its short coming

due to disinfection by-products (DBPs). DBPs have to be controlled. The

use of ozone and ultraviolet for disinfection of water and wastewater

is increasing in the United States.

Chlorination serves not only for disinfection, but as an oxidant for

other substances (iron, manganese, cyanide, etc.) and for taste and odor

control in water and wastewater. Other chemical disinfectants include

chlorine dioxide, ozone, bromine, and iodine. The last two chemicals are

generally used for personal application, not for the public water supply.

18.1 Chemistry of chlorination

Free available chlorine. Effective chlorine disinfection depends upon its

chemical form in water. The inﬂuencing factors are temperature, pH, and

organic content in the water. When chlorine gas is dissolved in water,

it rapidly hydrolyzes to hydrochloric acid (HCl) and hypochlorous

acid (HOCl)

⫹ H

O ↔ H

⫹

⫹ Cl

⫺

⫹ HOCl (5.152)

The equilibrium constant is

(5.153)

The dissolution of gaseous chlorine, Cl

2(g)

, to form dissolved molecu-

lar chlorine, Cl

2(aq)

follows Henry’s law and can be expressed as (Downs

and Adams, 1973)

(5.154)

where [Cl

2(aq)

] ⫽ molar concentration of Cl

⫽ partial pressure of chlorine in atmosphere

The distribution of free chlorine between HOCl and OCl

⫺

is presented

in Fig. 5.10. The disinfection capabilities of HOCl is generally higher

than that of OCl

⫺

(Water, 1978).

(5.155) 5 4.805 3 10

expa

2818.48

H 5 Henry’s law constant, mol/L atm

2sgd

2saqd

Hsmol/L atmd

[Cl

2saqd

]

5 4.48 3 10

at 258C sWhite, 1972d

][Cl

][HOCl]

[Cl

2saqd

]

464 Chapter 5