Lin S.D. Water and Wastewater Calculations Manual

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Estimation of the expected softener hardness. Since 40 mg/L NaOH

reduces 100 mg/L of the total hardness, the expected total hardness of

at the softener, TH

, can be estimated as:

⫽ TH

– (NaOH

cal

– f ) ⫻ 100/40 (5.94e)

where TH

⫽ raw water total hardness, mg/L as CaCO

NaOH

cal

⫽ calculated average NaOH dose, mg/L

f ⫽ correction factor, 20 mg/L NaOH

40 ⫽ molecular weight of NaOH, g/mol

100 ⫽ molecular weight of CaCO

, g/mol

Example: Given: TH

⫽ 250 mg/L as CaCO

; NaOH

cal

⫽ 98 mg/L. Determine

the total hardness at the softener, TH

solution: Using Eq. (5.94e)

⫽ TH

– (NaOH

cal

– f ) ⫻ 100/40

⫽ 250 mg/L – (98 – 20) mg/L ⫻ 2.5

⫽ 55 mg/L

Estimation of the expected treated water hardness. The expected treated

water hardness, TH

, can be estimated by the following formula:

⫽ TH

– (NaOH

avg

– f ) ⫻100/40 (5.94f)

where NaOH

avg

⫽ calculated average NaOH dose for entire water

ﬂow including bypass and ﬁltration ﬂow rate, mg/L

Others ⫽ as previously mentioned

Example: Given: TH

⫽ 250 mg/L as CaCO

; NaOH

avg

⫽ 88 mg/L. Determine

the expected total hardness in the treated water, TH

solution: Using Eq. (5.94f)

⫽ TH

– (NaOH

avg

– f ) ⫻100/40

⫽ 250 mg/L – (88 – 20) mg/L ⫻ 2.5

⫽ 80 mg/L

NaOH dosing pumps.

The actual delivery rate of a NaOH dosing pump

is estimated as follows:

Pump rate ⫽ maximum pump capacity ⫻ efﬁciency

⫻ % stroke adjustment (5.94g)

Public Water Supply 415

Example: Given: maximum pump capacity ⫽ 720 L/h, speed effic-

iency ⫽ 0.85, and stroke adjustment ⫽ 50%. Estimate the actual delivery

rate of the NaOH dosing pump.

solution: Using Eq. (5.94g)

Actual pumping rate ⫽ 720 L/h ⫻ 0.85 ⫻ 0.50

⫽ 306 L/h

Sulfuric acid dose rate.

Sulfuric acid is dosed into the treated water to

neutralize the excess of NaOH. The dose rate is calculated as:

dose rate ⫽ acid ﬂow rate ⫻ concentration /

treated water ﬂow rate (5.94h)

Example: Given: required H

ﬂow rate ⫽ 60 L/h; H

(50%) concen-

tration ⫽ 690 g/L; and the treated water ﬂow rate ⫽ 1900 m

/h. Compute the

dose rate in mg/L of H

solution: Using Eq. (5.94h)

1900 m

/h ⫽ 1,900,000 L/h

dose rate ⫽ 60 L/h ⫻ 690 g/L / 1,900,000 L/h

⫽ 0.0218 g/L

⫽ 21.8 mg/L

Seeding sand demand.

A lack of sand supply for crystallization has a

negative impact on the efﬁciency of the softening (then more NaOH is

needed to get the desired total hardness). This excess dosage of NaOH

to be neutralized triggers an additional sulfuric acid demand.

The seeding sand is added according to the demanded ratio between

effective sand size and pellet size. If the effective sand size or pellet size

is altered, the sand dosing rate also must be altered.

It is essential that very ﬁne, ﬁne, and smaller sand particles should

be removed from sand prior feeding to the reactors. If not, the ﬁne par-

ticles would be ﬂushed out from the reactor due to up to 100 m/h oper-

ation velocity there and enter the rapid ﬁltration units, causing

additional permanent head losses. Those ﬁne particles will not be

removed during the rapid ﬁlter backwash sequence from the ﬁlter, as

the ﬁlter backwash velocity is only 40 to 50 m/h. The separation of the

very ﬁne, ﬁne, and smaller sand particles from large particles is effec-

tive by washing the sand with a velocity of 120 m/h for a set time.

Sand can be fed (must be dry from a silo with free ﬂow) to a reactor

batchwise at a time. The expected sand dosing rates range from 3 to 7 mg

416 Chapter 5

of sand per liter of water treated. Using the pellet-to-sand weight ratio,

the demanded dosage of seeding sand can be calculated as:

Required sand dosage, mg/L ⫽ (TH

– TH

)/pellet-

to-sand ratio (5.94i)

where TH

and TH

are stated previously.

Example: Determine the required sand dosage and sand butch per volume

under the following conditions: The total hardness of the raw water and

treated water are respective 250 and 88 mg/L as CaCO

; the pellet-to-sand

ratio is 25 to 1. The surface area of the sand washer is A ⫽ 0.2 m

. Speciﬁc

density of sand ␥ ⫽ 1.55 g/L.

solution: Using Eq. (5.94i)

Step 1. Determine the required sand dosage, using Eq. (5.94i)

Required sand dosage ⫽ (TH

⫺ TH

)/pellet-to-sand ratio

⫽ (250 mg/L ⫺ 88 mg/L)/25

⫽ 6.48 mg/L

Step 2. Determine sand butch per volume

Sand dosing is effected batchwise with ﬁxed amount of sand per batch.

The height of sand after washing in the sand washer should be measured

frequently.

Washer (nozzle bottom) depth measured from the top ⫽ 235 cm

Sand level after washing measured from the top ⫽173 cm

Difference, D ⫽ 62 cm ⫽ 0.62 m

Volume of washed sand, V ⫽ A ⫻ D ⫽ 0.2 m

⫻ 0.62 m ⫽ 0.124 m

⫽ 124 L

Weight of sand, W ⫽ V ⫻␥⫽124 L ⫻ 1.55 kg/L ⫽192.2 kg

⫽ 192.2 ⫻ 10

Step 3. Calculate the water can be treated with one batch of sand

Effected volume ⫽ W/sand dosage ⫽ 192.2 ⫻ 10

mg/6.48 mg/L

⫽ 29.66 ⫻ 10

⫽ 29,660 m

13 Ion Exchange

Ion exchange is a reversible process. Ions of a given species are dis-

placed from an insoluble solid substance (exchange medium) by ions of

another species dissolved in water. In practice, water is passed through

Public Water Supply 417

the exchange medium until the exchange capacity is exhausted and

then it is regenerated. The process can be used to remove color, hard-

ness (calcium and magnesium), iron and manganese, nitrate and other

inorganics, heavy metals, and organics.

Exchange media which exchange cations are called cationic or acid

exchangers, while materials which exchange anions are called anionic

or base exchangers.

Common cation exchangers used in water softening are zeolite, green-

sand, and polystyrene resins. However, most ion exchange media are

currently in use as synthetic materials.

Synthetic ion exchange resins include four general types used in water

treatment. They are strong- and weak-acid cation exchangers and

strong- and weak-base anion exchangers. Examples of exchange reac-

tions as shown below (Schroeder, 1977):

Strong acidic

2R—SO

H ⫹ Ca

2⫹

↔ (R—SO

)

Ca ⫹ 2H

⫹

(5.95)

2R—SO

Na ⫹ Ca

2⫹

↔ (R—SO

)

Ca ⫹ 2Na

⫹

(5.96)

Weak acidic

2R—COOH ⫹ Ca

2⫹

↔ (R—COO)

Ca ⫹ 2H

⫹

(5.97)

2R—COONa ⫹ Ca

2⫹

↔ (R—COO)

Ca ⫹ 2Na

⫹

(5.98)

Strong-basic

2R—X

NOH ⫹ ↔ (R—X

⫹ 2OH

–

(5.99)

2R—X

NCl ⫹ ↔ (R—X

⫹ 2Cl

–

(5.100)

Weak-basic

2R—NH

OH ⫹ ↔ (R—NH

)

⫹ 2OH

–

(5.101)

2R—NH

Cl ⫹ ↔ (R—NH

)

⫹ 2Cl

–

(5.102)

where in each reaction, R is a hydrocarbon polymer and X is a speciﬁc

group, such as CH

The exchange reaction for natural zeolites (Z ) can be written as

(5.103)Na

Z 1 •

4 Mg

¶Z 1 2Na

418 Chapter 5

In the cation-exchange water softening process, the hardness-causing

elements of calcium and magnesium are removed and replaced with

sodium by a strong-acid cation resion. Ion-exchange reactions for soft-

ening may be expressed as

(5.104)

where R represents the exchange resin. They indicate that when a hard

water containing calcium and magnesium is passed through an ion

exchanger, these metals are taken up by the resin, which simultaneously

gives up sodium in exchange. Sodium is dissolved in water. The normal

rate is 6 to 8 gpm/ft

(350 to 470 m/d) of medium.

After a period of operation, the exchanging capacity would be

exhausted. The unit is stopped from operation and regenerated by back-

washing with sodium chloride solution and rinsed. The void volume for

backwash is usually 35% to 45% of the total bed volume.

The exchange capacity of typical resins are in the range of 2 to 10 eq/kg.

Zeolite cation exchangers have the exchange capacity of 0.05 to 0.1 eq/kg

(Tchobanoglous and Schroeder, 1985). During regeneration, the reaction

can be expressed as:

(5.105)

The spherical diameter in commercially available ion exchange resins

is of the order of 0.04 to 1.0 mm. The most common size ranges used in

large treatment plant are 20 to 50 mesh (0.85 to 0.3 mm) and 50 to 100

mesh (0.3 to 0.15 mm) (James M. Montgomery Consulting Engineering,

1985). Details on the particle size and size range, effective size, and

uniform coefﬁcient are generally provided by the manufacturers.

The afﬁnity of exchanges is related to charge and size. The higher the

valence, the greater afﬁnity and the smaller the effective size the greater

the afﬁnity. For a given sense of similar ions, there is a general order of

afﬁnity for the exchanger. For synthetic resin exchangers, relative afﬁni-

ties of common ions increase as shown in Table 5.6.

The design of ion exchange units is based upon ion exchange equilibria.

The generalized reaction equation for the exchange of ions A and on a

cation exchange resin can be expressed as

⫺

⫹

⫹ B

n⫹

↔ Rn

⫺

n⫹

⫹

(5.106)

where R

⫺

⫽ an anionic group attached to exchange resin

⫹

, B

n⫹

⫽ ions in solution

fR 1 2NaCl

R 1

fCl

R 1

sHCO

fR 1 •

2 NaHCO

2NaCl

Public Water Supply 419

The equilibrium expression for this reaction is

(5.107)

where K

A → B

⫽⫽selectivity coefﬁcient, a function of

ionic strength and is not a true constant

–

⫹

], [R

–

n⫹

] ⫽ mole fraction of A

⫹

and B

⫹

exchange resin,

respectively, overbars represent the resin

phase, or expressed as and

⫹

], [B

n⫹

] ⫽ concentration of A

⫹

and B

⫹

in solution,

respectively, mol/L

, q

⫽ concentration of A and B on resin site,

respectively, eg/L

, C

⫽ concentration of A and B in solution,

respectively, mg/L

[B][A],

][A

]

420 Chapter 5

TABLE 5.6 Selectivity Scale for Cations on Eight Percent Cross-

Linked Strong-Acid Resin and for the Anions on Strong-Base Resins

Cation Selectivity Anion Selectivity

⫹

1.0 HPO

2⫺

0.01

⫹

1.3 CO

2⫺

0.03

⫹

2.0 OH

⫺

(type I) 0.06

2⫹

2.5 F

⫺

0.1

2.6 0.15

⫹

2.9 CH

COO

⫺

0.2

⫹

3.2 0.4

⫹

3.3 OH

⫺

(type II) 0.65

2⫹

3.3 1.0

2⫹

3.5 Cl

⫺

1.0

2⫹

3.7 CN

⫺

1.3

2⫹

3.8 1.3

3.9 1.6

2⫹

3.9 Br

⫺

2⫹

4.0 4

2⫹

4.1 I

⫺

2⫹

5.0 9.1

2⫹

5.2 17

2⫹

6.5 100

⫹

8.5

2⫹

9.9

2⫹

11.5

2⫹

13.0

SOURCES: James M. Montgomery Consulting Engineering (1985),

Clifford (1990)

CrO

SeO

HSO

BrO

HCO

The selectivity constant depends upon the valence, nature, and con-

centration of the ion in solution. It is generally determined in labora-

tory of speciﬁc conditions measured.

For monovalent/monovalent ion exchange process such as

The equilibrium expression is (by Eq. (5.107))

(5.108)

and for divalent/divalent ion exchange processes such as

then using Eq. (5.107)

(5.109)

Anderson (1975) rearranged Eq. (5.107) using a monovalent/monovalent

exchange reaction with concentration units to equivalent fraction as

follows:

1. In the solution phase: Let C ⫽ total ionic concentration of the solu-

tion, eq/L. The equivalent ionic fraction of ions A

⫹

and B

⫹

in solu-

tion will be

(5.110)

(5.111)

then

or (5.112)

2. In the resin phase: Let ⫽ total exchange capacity of the resin per

unit volume, eq/L. Then we get

(5.113)

⫽ equivalent fraction of the A

⫹

ion in the resin

and

(5.114)X

5 [R

]/C or [R

] 5 C X

5 [R

]/C or [R

] 5 C X

] 5 1 2 [B

]

] 1 [B

] 5 1

5 [B

]>C or [B

] 5 CX

5 [A

]>C or [A

] 5 CX

NaR

CaR

[Ca

R][Na

]

[Na

[Ca

]

2sNa

d 1 Ca

1 2Na

NaR

[Na

][H

]

][Na

]

1 Na

4 Na

1 H

Public Water Supply 421

also

(5.115)

Substitute Eqs. (5.110), (5.111), (5.113), and (5.114) into Eq. (5.107)

which yields

(5.116)

Substituting for Eqs. (5.112) and (5.115) in Eq. (5.116) gives

(5.117)

If the valence is n, Eq. (5.117) will become

(5.118)

Example 1: Determine the meq/L of Ca

2⫹

if Ca

2⫹

concentration is 88 mg/L

in water.

solution:

Step 1. Determine equivalent weight (EW)

EW ⫽ molecular weight/electrical charge

⫽ 40/2

⫽ 20 g/equivalent weight (or mg/meq)

Step 2. Compute meq/L

meq/L ⫽ (mg/L)/EW

⫽ (88 mg/L)/(20 mg/meq)

⫽ 4.4

Example 2: A strong-base anion exchange resin is used to remove nitrate ions

from well water which contain high chloride concentration. Normally bicar-

bonate and sulfate is presented in water (assume they are negligible). The total

s1 2 X

5 K

n21

s1 2 X

1 2 X

5 K

1 2 X

5 K

sCX

dsCX

sCX

dsCX

[CX

][CX

]

[CX

][CX

]

5 1 2 X

1 X

5 1

422 Chapter 5

resin capacity is 1.5 eq/L. Find the maximum volume of water that can be

treated per liter of resin. The water has the following composition in meq/L:

2⫹

⫽ 1.4 CI

⫺

⫽ 3.0

2⫹

⫽ 0.8 ⫽ 0.0

⫹

⫽ 2.6 ⫽ 1.8

Total cations ⫽ 4.8 Total anions ⫽ 4.8

solution:

Step 1. Determine the equivalent fraction of nitrate in solution

Step 2. Determine selective coefﬁcient for sodium over chloride from Table 5.5

Step 3. Compute the theoretical resin available for nitrate ion by Eq. (5.117)

It means that 71% of resin sites will be used

Step 4. Compute the maximum useful capacity Y

Step 5. Compute the volume of water (V ) that can be treated per cycle

Example 3: A strong-acid cation exchanger is employed to remove calcium

hardness from water. Its wet-volume capacity is 2.0 eq/L in the sodium form.

If calcium concentrations in the inﬂuent and efﬂuent are 44 mg/L (2.2 meq/L)

and 0.44 MG/l, respectively, ﬁnd the equivalent weight (meq/L) of the com-

ponent in the water if given the following:

5 4429 gal of water/ft

of resin

5 592

1 gal

3.785 L

28.32 L

1 ft

5 592 L of water/L of resin

V 5

1065 meq/L of resin

1.8 meq/L of water

Y 5 1.5 eq/L 3 0.71 5 1.065 eq/L 5 1065 meq/L

5 0.71

1 2 X

5 4 3

0.38

1 2 0.38

5 2.45

5 4/1 5 4

5 1.8/4.8 5 0.38

Public Water Supply 423

Cations Anions

2⫹

2.2 2.9

2⫹

1.0 Cl

–

3.1

⫹

3.0 0.2

Total cations 6.2 Total anions 6.2

solution:

Step 1. Determine the equivalent fraction of Ca

2⫹

Step 2. Find the selectivity coefﬁcient K for calcium over sodium from

Table 5.5

Step 3. Compute the theoretical resin composition with respect to the cal-

cium ion

Using Eq. (5.118), n ⫽ 2

This means that a maximum of 51% of the resin sites can be used with cal-

cium ions from the given water. At this point, the water and resin are at

equilibrium with each other.

Step 4. Compute the limiting useful capacity of the resin (Y )

Step 5. Compute the maximum volume (V ) of water that can be treated

per cycle

5 3472 gal of water/ft

of resin

5 464 L of water/L of resin

V 5

1020 meq/L of resin sfor Cad

2.2 meq/L of calcium in water

5 1.02 eq/L sor 1020 meq/Ld

Y 5 2.0 eq/L 3 0.51

5 0.51

s1 2 X

5 2.6

0.35

s1 2 0.35d

5 2.15

/C 5 6.2/6.2 5 1.0

5 5.2/2.0 5 2.6

5 2.2/6.2 5 0.35

HCO

424 Chapter 5