Lin S.D. Water and Wastewater Calculations Manual

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13.1 Leakage

Leakage is deﬁned as the appearance of a low concentration of the unde-

sirable ions in the column efﬂuent during the beginning of the exhaus-

tion. It comes generally from the residual ions in the bottom resins due

to incomplete regeneration. Leakage will occur for softening process. A

water softening column is usually not fully regenerated due to inefﬁcient

use of salt to completely regenerate the resin on the sodium form. The

leakage depends upon the ionic composition of the bed bottom and the

composition of the inﬂuent water.

A detailed example of step-by-step design method for a ﬁx bed ion

exchange column for water softening is given by Beneﬁeld et al. (1982).

Example: The bottom of the water softener is 77% in the calcium form after

regeneration. The strong-acid cation resin has a total capacity of 2.0 eq/L and

selective coefﬁcient for calcium over sodium is 2.6. Determine the initial cal-

cium leakage the water composition has as follows (in meq/L):

2⫹

⫽ 0.4 CI

–

⫽ 0.4

2⫹

⫽ 0.2 ⫽ 0.4

⫹

⫽ 1.2 ⫽ 1.0

Total cations ⫽ 1.8 Total anions ⫽ 1.8

solution:

Step 1. Determine the equivalent fraction of calcium ion from the given:

Using Eq. (5.118)

Step 2. Calculate the initial calcium leakage y

5 0.010 meq/L

y 5 0.0056 c 5 0.0056 3 1.8 meq/L

0.77

s1 2 0.77d

5 2.6a

0.002

s1 2 X

5 0.0056

5 2.6

C 5 2 meq/L 5 0.002 eq/L

5 2.0 eq/L

5 0.77

HCO

Public Water Supply 425

13.2 Nitrate removal

Nitrate, is a nitrogen–oxygen ion that occurs frequently in nature

as the result of interaction between nitrogen in the atmosphere and

living things on earth. It is a portion of the nitrogen cycle. When plant

and animal proteins are broken down, ammonia and nitrogen gas are

released. Ammonia is subsequently oxidized to nitrite and nitrate

by bacterial action (Fig. 1.1).

High concentration of nitrate in drinking water may cause methe-

moglobinemia, especially for infants. Nitrates interfere with the body’s

ability to take oxygen from the air and distribute it to body cells. The

United States EPA’s standard for nitrate in drinking water is 10 mg/L.

In Illinois, the water purveyor must furnish special bottle water (low

nitrate) to infants if the nitrate level of the ﬁnished water exceeds the

standard.

The sources of nitrates to water are due to human activities.

Agricultural activities, such as fertilizer application and animal feed-

lots, are major contributors of nitrate and ammonia to be washed off

(runoff ) and percolate to soil with precipitation. The nitrates that are

polluted can subsequently ﬂow into surface waters (streams and lakes)

and into groundwater. Municipal sewage efﬂuents improperly treat

domestic wastewaters, and draining of septic tanks may cause nitrate

contamination in the raw water source of a waterworks. Identiﬁcation

of pollution sources and protective alternatives should be taken.

If the nitrate concentration in the water supply is excessive, there are

two approaches for solving the problem, i.e. treatment for nitrate

removal and nontreatment alternatives. Nontreatment alternatives

may include (1) use new water source, (2) blend with low nitrate waters,

(3) connect to other supplier(s), and (4) organize in a regional system.

Nitrate is not removed by the conventional water treatment processes,

such as coagulation–ﬂocculation, sedimentation, ﬁltration, activated

carbon adsorption, and disinfection (chlorination and ozonation). Some

treatment technologies, such as ion exchange, reverse osmosis, electro-

dialysis, microbial denitriﬁcation, and chemical reduction, can remove

nitrates from drinking water. An ion exchange process design is given

as the following examples.

sNO

5 0.50 mg/L as CaCO

5 0.20 mg/L as Ca

100 as CaCO

40 as Ca

5 0.20 mg/L as Ca

5 0.010

meq

20 mg as Ca

meq

426 Chapter 5

Two types (ﬁxed bed and continuous) of ion exchangers are used. The

ﬁxed bed exchange is used mostly for home and industry uses and is con-

trolled by a ﬂow totalizer with an automatic regeneration cycle at

approximately 75% to 80% of the theoretical bed capacity. Continuous

ion exchangers are employed by larger installations, such as water-

works, that provide continuous product water and require minimum bed

volume. A portion of the resin bed is withdrawn and regenerated out-

side of the main exchange vessel.

In the design of an ion exchange system for nitrate removal, raw water

quality analyses and pilot testing are generally required. The type of

resin and resin capacity, bed dimensions, and regenerant requirement

must be determined. Basic data, such as the design ﬂow rate through the

exchanger, inﬂuent water quality, total anions, and suggested operating

conditions for the resin selected (from the manufacturer) must be known.

Analysis of water quality may include nitrate, sulfate, chloride, bicar-

bonate, calcium carbonate, iron, total suspended solids, and total organic

carbon. For Duolite A-104, for example, the suggested design parameters

are listed below (Diamond Shamrock Co., 1978):

Parameters Recommended values

Minimum bed depth 30 in

Backwash ﬂow rate 2–3 gpm/ft

Regenerant dosage 15–18 lb sodium chloride (NaCl) per ft

resin

Regenerant concentration 10–12% NaCl by weight

Regenerant temperature Up to 120°F or 49°C

Regenerant ﬂow rate 0.5 gpm/ft

Rinse ﬂow rate 2 gpm/ft

Rinse volume 50–70 gal/ft

Service ﬂow rate Up to 5 gpm/ft

Operating temperature Salt form, up to 180°F or 85°C

pH limitation None

In design processes, resin capacity, bed dimensions, and regenerant

requirements must be computed. Resin capacity determines the quan-

tity of resin needed in the ion exchanger and is computed from a pilot

study. Data is provided by the manufacturer. For example, the operat-

ing capacity of Duloite A-104 resin for nitrate removal is quite depend-

ent upon the nitrate, sulfate, and total anion concentrations to calculate

the corrected resin capacity. The raw or uncorrected resin capacity is

determined by using the manufacturer’s graph (Fig. 5.7). This capacity

must be adjusted downward to reﬂect the presence of sulfate in the

water (Fig. 5.8), because sulfate anions will be exchanged before nitrate

(US EPA, 1983).

Once the adjusted resin capacity is determined for the water to be

treated, the required bed volume of the ion exchange resin can be

Public Water Supply 427

428 Chapter 5

Figure 5.7 Relationship of NO

/TA and unadjusted resin capacity for A-104 resin (US

EPA, 1983).

Figure 5.8 Correction factor for A-104 resin (US EPA, 1983).

calculated. The bed volume is the amount of nitrate that must be

removed in each cycle divided by the adjusted resin capacity. Using this

bed volume and minimum depth requirement, the surface area of resin

can be calculated. A standard size containment vessel can be selected

with the closest surface area. Using this standard size, the depth of the

resin can be recalculated. The ﬁnal height of the vessel should be added

to a bed expansion factor during backwashing with a design tempera-

ture selected (Fig. 5.9).

Once the bed volume and dimensions are computed, the regenera-

tion system can be designed. The regeneration system must deter-

mine: (1) the salt required per generation cycle, (2) the volume of brine

produced per cycle, (3) the total volume of the brine storage tank, and

(4) the time needed for the regeneration process. The design of an ion

exchanger for nitrate removal is illustrated in the following three

examples.

Example 1: A small public water system has the maximum daily ﬂow rate

of 100,000 gal/d, maximum weekly ﬂow of 500,000 gal/week, and maximum

nitrate concentration of 16 mg/L. The plant treats 100,000 gal of water and

operates only 7 h/d and 5d/week (assume sufﬁcient storage capacity for week-

end demand). Finished water after ion exchange process with 0.6 mg/L of

nitrate–nitrogen (NO

-N) will be blended with untreated water 16 mg/L of

-N to produce a ﬁnished water of 8 mg/L or less of NO

-N. The NO

-N stan-

dard is 10 mg/L. Determine the ﬂow rate of ion exchanger in gpm and the

blending rate.

Public Water Supply 429

Figure 5.9 Bed expansion curve for A-104 resin (modiﬁed from US EPA, 1983).

solution:

Step 1. Compute the quantity of water that passes the ion exchanger, q

Let untreated ﬂow ⫽ q

and c, c

, and c

⫽ NO

-N concentrations in ﬁnished,

treated (ion exchanger) and untreated waters, respectively.

Then

c ⫽ 8 mg/L

⫽ 0.6 mg/L

⫽ 16 mg/L

by mass balance

(a)

since

(b)

Substituting (b) into (a)

Step 2. Compute ﬂow rate in gpm to the exchanger (only operates 7 h/d)

Step 3. Compute blending ratio, br

Example 2: The given conditions are the same as in Example 1. The anion

concentration of the inﬂuent of the ion exchanger are: NO

-N ⫽ 16 mg/L,

⫽ 48 mg/L, Cl ⫽ 28 mg/L, and HCO

⫽ 72 mg/L. Determine the total

quantity of nitrate to be removed daily by the ion exchanger and the ratios

of anions.

> 0.52

br 5

51,950 gpd

100,000 gpd

5 124 gpm

5 51,950

gal

24 h

7 h

1 day

1440 min

5 51,950 sgpdd

15.4q

5 800,000

06q

1 16s100,000 2 q

d 5 800,000

5 100,000 2 q

0.6q

1 16q

5 100,000 3 8

1 c

5 100,000c

430 Chapter 5

solution:

Step 1. Compute total anions and ratios

Concentration, Milliequivalent Concentration,

Anions mg/L weight, mg/meq meq/L Percentage

-N 16 14.0 1.14 27.7

48 48.0 1.00 24.4

Cl 28 35.5 0.79 19.2

HCO

72 61.0 1.18 28.7

Total 4.11

In the last column above for NO

-N

Step 2. Determine nitrate to be removed daily

The total amount of nitrates to be removed depends on the concentrations of

inﬂuent and efﬂuent, c

and c

and can be determined by

nitrates removed (meq/d) ⫽ [(c

⫺ c

) meq/L] ⫻ q

(L/d)

Since

If the ion exchanger is going to operate with only one cycle per day, the quan-

tity of nitrate to be removed per cycle is 215,700 meq.

Example 3: Design a 100,000 gpd ion exchanger for nitrate removal using

the data given in Examples 1 and 2. Find resin capacity, resin bed dimension,

and regenerant requirements (salt used, brine production, the total volume

of the brine storage tank, and regeneration cycle operating time) for the ion

exchanger.

solution:

Step 1. Determine the uncorrected (unadjusted) volume of resin needed

(a) In Example 2, the nitrate to total anion ratio is 27.7%. Using 27.7% in Fig.

5.7, we obtain the uncorrected resin capacity of 0.52 meq/(NO

mL) resin.

(b) In Example 2, the sulfate to total anion ratio is 24.4%. Using 25% in

Fig. 5.8, we obtain the correction factor of resin capacity due to sulfate is 0.70.

5 215,700 meq/d

Nitrates removed 5 s1.14 2 0.043d meq/L 3 51,950 gpd 3 3.785 L/gal

5 0.043 meq/L

5 0.6 mg/L 4 14 mg/meq

5 1.14 meq/L

% 5 1.14 3 100/4.11 5 27.7

Public Water Supply 431

0.52 meq/mL ⫻ 0.70 ⫽ 0.364 meq/mL

(d) Convert meq/mL to meq/ft

Step 2. Determine the bed volume, BV

In Example 2, nitrate to be removed in each cycle is 215,700 meq

Step 3. Check the service ﬂow rate, SFR

In Example 1, the ion exchange unit ﬂow rate ⫽124 gpm

This value exceeds the manufacturer’s maximum SFR of 5 gpm/ft

. In

order to reduce the SFR, we can modify it either by: (a) increasing the

exchanger operating time per cycle, or (b) increasing the bed volume.

(a) Calculate adjusted service time, AST

Calculate the adjusted ﬂow rate for the new cycle

In Example 1, q

⫽ 51,950 gpd or 51,950 gal/cycle

Alternatively

(b) Increase the bed volume and retain the initial unit ﬂow rate

say 5 25 ft

5 24.7 ft

Adjusted BV 5 20.93 ft

3 s5.9 gpm/ft

/5.0 gpm/ft

5 104 gpm

5 51,950 gal/cycle/s8.3 h/cycle 3 60 min/hd

then, flow rate 5 q

/AST

5 8.3 h/cycle

AST 5

7 h

cycle

5.9 gpm/ft

5.0 gpm/ft

5 5.9 gpm/ft

SFR 5 124 gpm/BV 5 124 gpm/20.93 ft

5 20.93 ft

BV 5 215,700 meq 4 10,305 meq>ft

5 10,305 meq/ft

0.364 meq/mL 3 3785 mL/gal 3 7.48 gal/ft

432 Chapter 5

In comparison between alternatives (a) and (b), it is more desirable to

increase BV using 7 h/d operation. Thus BV ⫽ 25 ft

is chosen.

Step 4. Determine bed dimensions.

(a) Select the vessel size

Based on the suggested design speciﬁcation, the minimum bed depth (z)

is 30 in or 2.5 feet. The area, A, would be

For a circular vessel, ﬁnd the diameter

The closest premanufactured size is 3.5 ft in diameter with a correspon-

ding cross-sectional area of 9.62 ft

. The resin height, h, would then be

recalculated as

(b) Adjusting for expansion during backwash

Since allowable backwash ﬂow rate is 2 to 3 gpm/ft

. If the backwash

rate is selected as 2.5 gpm/ft

with the backwash water temperature of

50°F, the percent bed expansion would be 58% as read from the curve in

Fig. 5.9. Thus, the design height (H ) of the vessel should be at least

Note: We use 3.5 ft diameter and 4.11 ft (at least) height vessel.

Step 5. Determine regenerant requirements

(a) Choice salt (NaCl) required per regeneration cycle since regenerant dosage

is 15 to 18 lb of salt per cubic foot of resin. Assume 16 lb/ft

is selected.

The amount of salt required is

We need 400 lb of salt per cycle.

5 400 lb

Salt 5 16 lb/ft

3 25 ft

5 4.11 ft

H 5 2.6 ft 3 s1 1 0.58d

5 2.6 ft

h 5 25 ft

/9.62 ft

5 3.57 ft

D 5 24A/p

5 24 3 10.0 ft

/3.14

A 5 pD

5 10.0 ft

A 5 BV/z 5 25 ft

/2.5 ft

Public Water Supply 433

(b) Calculate volume of brine used per cycle

On the basis of the speciﬁcation, the regenerant concentration is 10% to

12% NaCl by weight. Assume it is 10%, then

The speciﬁc weight of salt is 2.165

Note: The total volume of brine generated is 454 gal per cycle. The total

volume of the brine storage tank should contain sufﬁcient capacity for 3 to 4

generations. Assume 3 cycles of brine produced should be stored, then the total

brine tank volume (V ) would be

(d) Calculate time required for regeneration cycle

The regeneration ﬂow rate (Q) is 0.5 gpm/ft

resin speciﬁed by the manu-

facturer. Then

The regeneration time (t) is the volume of the brine per cycle divided by

the ﬂow rate Q, thus

t 5 454 gal/12.5 gal/min

5 36.3 min

Q 5 0.5 gpm/ft

3 25 ft

5 12.5 gpm

V 5 454 gal/cycle 3 3 cycle

5 1362 gal

5 454 gal

Total volume in gallons 5 60.65 ft

3 7.48 gal/ft

5 60.65 ft

Total volume of the brine 5 57.69 ft

1 2.96 ft

5 2.96 ft

Volume of salt 5 400 lb/s62.4 lb/ft

3 2.165d

5 57.69 ft

Volume of water 5

wt. of water

density of water

3600 lb

62.4 lb/ft

5 3600 lb

Weight of water 5 s4000 2 400d lb/cycle

5 4000 lb

total wt. of brine 5 400 lb 3 100/10

Salt concentration, % 5

wt. of salt 3 100

total wt. of brine

434 Chapter 5