Lin S.D. Water and Wastewater Calculations Manual

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Public Water Supply 345

Using Henry’s law

p ⫽ Hx

x ⫽ p/H ⫽ 0.024 atm/170atm

⫽ 1.41 ⫻ 10

⫺4

Step 2. Convert mole fraction to mass concentration

In 1 L of water, there are 1000/18 mol/L ⫽ 55.6 mol/L.

Thus

where n ⫽ number of mole for CHC1

⫽ number of mole for water

Then nx ⫹ n

x ⫽ n

Molecular weight of CHC1

⫽ 12 ⫹ 1 ⫹ 3 ⫻ 35.45 ⫽ 119.4

Concentration (C ) of CHC1

C ⫽ 119.4 g/mol ⫻ 7.84 ⫻ 10

⫺3

mol/L

⫽ 0.94 g/L

⫽ 940 mg/L

n 5

1 2 x

55.6 3 1.41 3 10

1 2 1.41 3 10

5 7.84 3 10

mol/L

x 5

n 1 n

TABLE 5.4 Henry’s Law Constant and Temperature Correction Factors

Henry’s constant at

Gas 20°C, atm ∆H, 10

kcal/kmol J

Ammonia 0.76 3.75 6.31

Benzene 240 3.68 8.68

Bromoform 35 — —

Carbon dioxide 1.51 ⫻ 10

2.07 6.73

Carbon tetrachloride 1.29 ⫻ 10

4.05 10.06

Chlorine 585 1.74 5.75

Chlorine dioxide 54 2.93 6.76

Chloroform 170 4.00 9.10

Hydrogen sulﬂde 515 1.85 5.88

Methane 3.8 ⫻ 10

1.54 7.22

Nitrogen 8.6 ⫻ 10

1.12 6.85

Oxygen 4.3 ⫻ 10

1.45 7.11

Ozone 5.0 ⫻ 10

2.52 8.05

Sulfur dioxide 38 2.40 5.68

Trichloroethylene 550 3.41 8.59

Vinyl chloride 1.21 ⫻ 10

——

346 Chapter 5

Note: In practice, the treated water is open to the atmosphere. Therefore

the partial pressure of chloroform in the air is negligible. Therefore the

concentration of chloroform in water is near zero or very low.

Example 2: The ﬁnished water in an enclosed clear well has dissolved

oxygen of 6.0 mg/L. Assume it is in equilibrium. Determine the concentration

of oxygen in the air space of the clear well at 17°C.

solution:

Step 1. Calculate Henry’s constant at 17°C (290 K)

From Table 5.4 and Eq. (5.18)

Step 2. Compute the mole fraction of oxygen in the gas phase

Oxygen (M.W. ⫽ 32) as gas i in water

⫽ 6 mg/L ⫻ 10

⫺3

g/mg ⫻ 10

L/m

⫼ 32 g/mol

⫽ 0.1875 mol/m

At 1 atm for water, the molar density of water is

⫽ 55.6 kmol/m

From Eq. (5.17)

Step 3. Calculate partial pressure P

⫽ y

⫽ 0.131 ⫻ 1

⫽ 0.131 atm

0.131

38,905 atm 3 3.37 3 10

1 atm

0.1875 mol/m

55.6 3 10

mol/m

5 3.37 3 10

5 1 atm

log H 5

2⌬H

1 J

21.45 3 10

kcal/kmol

s1.987 kcal/kmol Kds290 Kd

1 7.11

5 4.59

H 5 38,905 atm 3

103.3 kPa

1 atm

4.02 3 10

kPa

Step 4. Calculate oxygen concentration in gas phase C

⫽ n

Using the ideal gas formula

6.1 Gas transfer models

Three gas-liquid mass transfer models are generally used. They are two-

ﬁlm theory, penetration theory, and surface renewal theory. The ﬁrst

theory is discussed here. The other two can be found elsewhere

(Schroeder, 1977).

The two-ﬁlm theory is the oldest and simplest. The concept of the

two-ﬁlm model is illustrated in Fig. 5.1. Flux is a term used as the mass

transfer per time through a speciﬁed area. It is a function of the driv-

ing force for diffusion. The driving force in air is the difference between

the bulk concentration and the interface concentration. The ﬂux gas

0.131 atm

0.08206 atm L/mol K 3 s290 Kd

5 5.50 3 10

mol/L

5 5.50 3 10

mol/L 3 32 g/mol

5 0.176 g/L

5 176 mg/L

V 5 n

Public Water Supply 347

Gas film

Bulk gas

(a)

Liquid film

Bulk liquid

Liquid film

Bulk liquid

(b)

Air film

Bulk air

Figure 5.1 Schematic diagram of two-

ﬁlm theory for (a) gas absorption;

(b) gas stripping.

348 Chapter 5

through the gas ﬁlm must be the same as the ﬂux through the liquid ﬁlm.

The ﬂux relationship for each phase can be expressed as:

(5.19)

where F ⫽ ﬂux

W ⫽ mass transfer

A ⫽ a given area gas-liquid transferal

t ⫽ time

⫽ local interface transfer coefﬁcient for gas

⫽ concentration of gas in the bulk of the air phase

⫽ concentration of gas in the interface

⫽ interface transfer coefﬁcient for liquid

⫽ local interface concentration at equilibrium

⫽ liquid-phase concentration in bulk liquid

Since P

and C

are not measurable, volumetric mass transfer coefﬁcients

are used, which combine the mass transfer coefﬁcients with the inter-

facial area per unit volume of system. Applying Henry’s law and intro-

ducing P* and C* which correspond to the equilibrium concentration

that would be associated with the bulk-gas partial pressure P

and bulk-

liquid concentration C

, respectively, we can obtain

F ⫽ K

⫺P*) ⫽ K

(C* ⫺ C

) (5.20)

The concentration differences of gas in air and water is plotted in Fig. 5.2.

From this ﬁgure and Eq. (5.19), we obtain

From Eq. (5.20)

Thus

5 P

2 P*

2 P* 5 sP

2 P

d 1 sP

2 P*d

5 P

2 P

1 s

2 C

F 5

dtA

5 k

2 P

d 5 k

2 C

Public Water Supply 349

Similarly, we can obtain

In dilute conditions, Henry’s law holds, the equilibrium distribution

curve would be a straight line and the slope is the Henry’s constant:

⫽ s

⫽ H

Finally the overall mass transfer coefﬁcient in the air phase can be

determined with

(5.21)

For liquid phase mass transfer

(5.22)

where 1/k

and 1/k

are referred to as the liquid-ﬁlm resistance and

the gas-ﬁlm resistance, respectively. For oxygen transfer, l/k

is con-

siderably larger than l/k

, and the liquid ﬁlm usually controls the

oxygen transfer rate across the interface.

The ﬂux equation for the liquid phase using concentration of mg/L is

(5.23a)

mass transfer rate:

(5.23b)

or (5.23c)

5 K

asC* 2 C

5 K

2 C

F 5

5 K

2 C

Mole

fraction

of gas

in air

Concentration

Figure 5.2 Concentration differences of gas in air and water phases.

350 Chapter 5

where K

⫽ overall mass-transfer coefﬁcient, cm/h

A ⫽ interfacial area of transfer, cm

V ⫽ volume containing the interfacial area, cm

⫽ concentration in bulk liquid at time t, mg/L

C* ⫽ equilibrium concentration with gas at time t as

⫽ HC

mg/L

a ⫽ the speciﬁc interfacial area per unit system volume, cm

⫺1

a ⫽ overall volumetric mass transfer coefﬁcient in liquid h

⫺1

or [g ⋅ mole/(h ⋅ cm

· atm)]

the volumetric rate of mass transfer M is

(5.24)

(5.25)

where N ⫽ rate at which solute gas is transferred between phases,

g ⋅ mol/h

Example 1: An aeration tank has the volume of 200 m

(7060 ft

) with 4 m

(13 ft) depth. The experiment is conducted at 20°C for tap water. The oxygen

transfer rate is 18.7 kg/h (41.2 lb/h). Determine the overall volumetric mass

transfer coefﬁcient in the liquid phase.

solution:

Step 1. Determine the oxygen solubility at 20°C and 1 atm

Since oxygen in air is 21%, according to Dalton’s law of partial pressure, P is

P ⫽ 0.21 ⫻ 1 atm ⫽ 0.21 atm

From Table 5.3, we ﬁnd

H ⫽ 4.3 ⫻ 10

atm

The mole fraction x from Eq. (5.14) is

In 1L of solution, the number of moles of water n

Let n ⫽ moles of gas in solution

x 5

n 1 n

1000

5 55.6 g

mol/L

x 5

0.21

4.3 3 10

5 4.88 3 10

5 K

a sP 2 P *2d

M 5

N 5 K

asC * 2 Cd

Public Water Supply 351

Since n is so small

Step 2. Determine the mean hydrostatic pressure on the rising air bubbles

Pressure at the bottom is p

1 atm ⫽ 10.345 m of water head

Pressure at the surface of aeration tank is p

⫽ 1 atm

Mean pressure p is

Step 3. Compute C* ⫺ C

C* ⫽ 2.71 ⫻ 10

⫺4

g ⋅ mol/L ⋅ atm ⫻ 1.193 atm

⫽ 3.233 ⫻ 10

⫺4

g ⋅ mol/L

The percentage of oxygen in the air bubbles decrease, when the bubbles rise

up through the solution, assuming that the air that leaves the water contains

only 19% of oxygen.

Thus

Step 4. Compute K

Using Eq. (5.24)

Equation (5.23c) is essentially the same as Fick’s ﬁrst law:

(5.23d)

5 K

asC

2 Cd

5 9.98 h

584 g

mol/h

200 m

3 1000 L/m

3 2.925 3 10

mol/L

a 5

VsC* 2 Cd

N 5 18,700 g/h 4 32/mole 5 584 g

mol/h

5 2.925 3 10

mol/L

C * 2 C 5 3.223 3 10

mol/L 3

p 5

1 p

1 atm 1 1.386 atm

5 1.193 atm

5 1.386 atm

5 1 atm 1 4m 3

1 atm

10.345 m

5 s1 1 0.386d atm

5 2.71 3 10

mol/L

5 55.6 g

mol/L 3 4.88 3 10

n >n

352 Chapter 5

where ⫽ rate of change in concentration of the gas in solution,

mg/L ⋅ s

a ⫽ overall mass transfer coefﬁcient, s

⫺1

⫽ saturation concentration of gas in solution, mg/L

C ⫽ concentration of solute gas in solution, mg/L

The value of C

can be calculated with Henry’s law. The term C

⫺ C is

a concentration gradient. Rearrange Eq. (5.23d) and integrating the

differential form between time from 0 to t, and concentration of gas

from C

to C

in mg/L, we obtain

(5.24a)

or (5.24b)

When gases are removed or stripped from the solution, Eq. (5.24b)

becomes

Similarly, Eq. (5.25) can be integrated to yield

(5.25a)

a values are usually determined in full-scale facilities or scaled up

from pilot-scale facilities. Temperature and chemical constituents in

wastewater affect oxygen transfer. The temperature effects on overall

mass transfer coefﬁcient K

a are treated in the same manner as they

were treated in the BOD rate coefﬁcient (section 8.3 of Chapter 1.2). It

can be written as

(5.26)

where K

(T )

⫽ overall mass transfer coefﬁcient at temperature T ºC,

⫺1

(20)

⫽ overall mass transfer coefﬁcient at temperature 20ºC,

⫺1

Values of ␪ range from 1.015 to 1.040, with 1.024 commonly used.

sTd

5 K

s20d

T220

2 p *

5 K

aRTt

2 C

5 e

2sK

atd

2 C

5 e

2 C

52K

2 C

5 K

Mass transfer coefﬁcient are inﬂuenced by total dissolved solids in the

liquid. Therefore, a correction factor ␣ is applied for wastewater

(Tchobanoglous and Schroeder, 1985):

(5.27)

Value of ␣ range from 0.3 to 1.2. Typical values for diffused and mechan-

ical aeration equipment are in the range of 0.4 to 0.8 and 0.6 to 1.2.

The third correction factor (␤) for oxygen solubility is due to particu-

late, salt, and surface active substances in water (Doyle and Boyle, 1986):

(5.28)

Values of ␤ range from 0.7 to 0.98, with 0.95 commonly used for waste-

water.

Combining all three correction factors, we obtain (Tchobanoglous and

Schroeder 1985)

(5.29)

where AOTR ⫽ actual oxygen transfer rate under ﬁeld operating

conditions in a respiring system, kg ⋅ O

/kW ⋅ h

SOTR ⫽ standard oxygen transfer rate under test conditions at

20ºC and zero dissolved oxygen,

kg ⋅ O

/kW ⋅ h

␣, ␤, ␪ ⫽ deﬁned previously

⫽ oxygen saturation concentration for tap water at ﬁeld

operating conditions, g/m

⫽ operating oxygen concentration in wastewater, g/m

s20

⫽ oxygen saturation concentration for tap water at

20ºC, g/m

Example 2: Aeration tests are conducted with tap water and wastewater at

16ºC in the same container. The results of the tests are listed below. Assume

the saturation DO concentrations (C

) for tap water and wastewater are the

same. Determine the values of K

a for tap water and wastewater and a values

at 20ºC.

Assume ␪ ⫽ 1.024

solution:

Step 1. Find DO saturation concentration at 16ºC

From Table 1.2

⫽ 9.82 mg/L

AOTR 5 SOTRsadsbdsu

T220

2 C

s20

b 5

C *swastewaterd

C *stap waterd

a 5

a swastewaterd

a stap waterd

Public Water Supply 353

Step 2. Calculate ⫺ ln(C

⫺ C

)/(C

⫺ C

), C

⫽ 0

The values of ⫺ ln(C

⫺ C

)/(C

⫺ C

) calculated are listed with the raw test data

Step 3. Plot the calculated results in Step 2 in Fig. 5.3.

Step 4. Find the K

a at the test temperature of 16ºC, (T ), from the above table

For tap water

For wastewater

5 0.64 h

a 5

1.28

120 min

60 min

1 h

5 1.00 h

a 5

2.01

120 min

60 min

1 h

354 Chapter 5

DO concentration, mg/L

Contact time, min Tap water Wastewater Tap water Wastewater

0 0.0 0.0 0 0

20 3.0 2.1 0.36* 0.24

40 4.7 3.5 0.65 0.44

60 6.4 4.7 1.05 0.65

80 7.2 5.6 1.32 0.84

100 7.9 6.4 1.69 1.05

120 8.5 7.1 2.01 1.28

*ln (C

⫺ C

)/(C

⫺ C

) ⫽ ln(9.82 ⫺ 3.0)/(9.82 ⫺0) ⫽ 0.36

2 ln

2 C

0.5

1.5

2.5

0 20 40 60 80 100 120

TIME, min

−ln (C

− C

)/(C

− C

)

Tap water

Wastewater

Figure 5.3 Functional plot of the data from aeration test.