Lin S.D. Water and Wastewater Calculations Manual

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Step 5. Convert the values of K

a at 20ºC using ␪ ⫽ 1.024

Using Eq. (5.26)

For tap water

(T)

⫽ K

(20ºC)

␪

20⫺T

For wastewater

Step 6. Compute the ␣ value using Eq. (5.27)

6.2 Diffused aeration

Diffused aeration systems distribute the gas uniformly through the

water or wastewater, in such processes as ozonation, absorption, activated-

sludge process, THM removal, and river or lake reaeration, etc. It is more

costly using diffused aeration for VOC removal than the air stripping

column.

The two-resistance layer theory is also applied to diffused aeration.

The model proposed by Mattee-Müller et al. (1981) is based on mass

transfer ﬂux derived from the assumption of diffused bubbles rising

in a completely mixed container. The mass transfer rate for diffused

aeration is

(5.30)

where F ⫽ mass transfer rate

⫽ gas (air) ﬂow rate, m

/s or ft

⫽ ﬂow rate of liquid (water), m

/s or ft

F 5 Q

a1 2 exp

5 0.64

0.70 h

0.10 h

a 5

a for wastewater

a for tap water

5 0.70 h

s20d

5 0.64 h

s1.024d

20216

5 1.10 h

5 1.00 h

s1.024d

20216

s20d

5 K

sT d

202T

Public Water Supply 355

⫽ unitless Henry’s constant (see, Section of Design of Packed

Tower)

⫽ efﬂuent (exit) gas concentration, µg/L

a ⫽ overall mass transfer coefﬁcient, per time

V ⫽ reaction volume (water), m

or ft

Assuming that the liquid volume in the reactor is completely mixed

and the air rises as a steady state plug ﬂow, the mass balance equation

can be expressed as

(5.31a)

(5.31b)

where C

⫽ initial concentration, µg/L

If ␪ >> 1, the transfer of a compound is with very low Henry’s constant

such as ammonia. Air bubbles exiting from the top of the liquid surface

is saturated with ammonia in the stripping process. Ammonia removal

could be further enhanced by increasing the air ﬂow. Until ␪ < 4, the

exponent term becomes essentially zero. When the exponent term is

zero, the air and water have reached an equilibrium condition and the

driving force has decreased to zero at some point within the reactor

vessel. The vessel is not fully used. Thus the air-to-water ratio could be

increased to gain more removal.

On the other hand, if ␪ << 1, the mass transfer efﬁciency could be

improved by increasing overall mass transfer coefﬁcient by either

increasing the mixing intensity in the tank or by using a ﬁner diffuser.

In the case for oxygenation, ␪ < 0.1, the improvements are required.

Example: A groundwater treatment plant has a capacity of 0.0438 m

(1 MGD) and is aerated with diffused air to remove trichloroethylene with

90% design efﬁciency. The detention time of the tank is 30 min. Evaluate the

diffused aeration system with the following given information.

T ⫽ 20ºC

⫽131 µg/L , (expected C

⫽ 13.1 µg/L)

⫽ 0.412

a ⫽ 44 h

⫺1

u 5

1 1 H

[1 2 exp s2 ud]

1 1 H

[1 2 exp s2 K

aV/H

]

356 Chapter 5

Public Water Supply 357

solution:

Step 1. Compute volume of reactor V

Step 2. Compute Q

, let Q

⫽ 30 Q

⫽ 30V

Step 3. Compute ␪

Step 4. Compute efﬂuent concentration C

with Eq (5.31b)

This exceeds the expected 90%.

6.3 Packed towers

Recently, the water treatment industry used packed towers for stripping

highly volatile chemicals, such as hydrogen sulﬁde and VOCs from

water and wastewater. It consists of a cylindrical shell containing a

support plate for the packing material. Although many materials can

be used as the packing material, plastic products with various shapes

and design are most commonly used due to less weight and lower cost.

Packing material can be individually dumped randomly into the cylin-

der tower or ﬁxed packing. The packed tower or columns are used for

mass transfer from the liquid to gas phase.

Figure 5.4 illustrates a liquid-gas contacting system with a down-

ward water velocity L containing c

concentration of gas. The ﬂows are

counter current. The air velocity G passes upward through the packed

material containing inﬂuent p

and efﬂuent p

. There are a variety of

mixing patterns, each with a different rate of mass transfer. Removal

of an undesirable gas in the liquid phase needs a system height of z and

a selected gas ﬂow rate to reduce the mole fraction of dissolved gas from

to c

. If there is no chemical reaction that takes place in the packed

% removal 5

s131 2 10d 3 100

131

5 92.4

5 10.0 mg/L

131

mg/L

1 1 0.412 3 30[1 2 exps 2 3.56d]

1 1 H

[1 2 exps 2 ud]

u 5

44 3 79

0.412 3 2370

5 3.56

5 30 3 79 m

5 2370 m

5 79 m

V 5 0.0438 m

/s 3 60 s/min 3 30 min

358 Chapter 5

tower, the gas lost by water should be equal to the gas gained by air. If

the gas concentration is very dilute, then

(5.32)

where L ⫽ liquid velocity, m/s, m

/(m

⋅ s), or mol/(m

⋅ s)

G ⫽ gas velocity, same as above

⫽ change of gas concentration in water

⫽ change in gas fraction in air

From Eq. (5.23d), for stripping (c and c

will be reversed):

(5.23e)

then

(5.33)

Multiplying each side by L, we obtain

(5.34)

where dz is the differential of height.

The term (c ⫺ c

) is the driving force (DF) of the reaction and is constantly

changing with the depth of the column due to the change of c with time.

Integrating the above equation yields

(5.35)

z 5

dsDFd

dz 5

dsc 2 c

Ldt 5 dz 5

Ldc

asc 2 c

dt 5

asc 2 c

5 K

asc 2 c

⌬p

⌬c

L⌬c 5 G⌬p

Partial

Pressure

Concentration

Figure 5.4 Schematic diagram of packed column.

Public Water Supply 359

The integral of DF is the same as the log mean (DF

) of the inﬂuent

(DF

) and efﬂuent driving forces (DF

). Therefore

(5.36)

and

(5.37)

where z ⫽ height of column, m

L ⫽ liquid velocity, m

⋅ h

, c

⫽ gas concentration in water at inﬂuent and efﬂuent,

respectively, mg/L

a ⫽ overall mass transfer coefﬁcient for liquid, h

⫺1

, DF

⫽ driving force at inﬂuent and efﬂuent, respectively, mg/L

⫽ log mean of DF

and DF

, mg/L

Example 1: Given

T ⫽ 20ºC ⫽ 293 K

L ⫽ 80 m

water (m

⋅ h)

G ⫽ 2400 m

air/(m

⋅ h)

⫽ 131 ␮g/L trichloroethylene concentration in water at

entrance

⫽ 13.1 ␮g/L CCHCl

concentration at exit

a ⫽ 44 h

⫺1

Determine the packed tower height to remove 90% of trichloroethylene

by an air stripping tower.

solution:

Step 1. Compute the molar fraction of CCHCl

in air p

Assuming no CCHCl

present in the air the entrance, i.e.

⫽ 0

MW of CCHCl

⫽ 131, 1 mole ⫽131g of CCHCl

per liter.

5 13.1 mg/L 5 0.1 3 10

mol/m

swith 90% removald

5 1 3 10

mol/m

5 131 mg/L 5 131

1 mol/g

131 3 10

mg/g

1 m

2 DF

ln sDF

>DF

z 5

Lsc

2 c

aDF

360 Chapter 5

Applying Eq. (5.32)

80 m

air/(m

⋅ h)(1 ⫺ 0.1) ⫻ 10

⫺3

mol gas/m

air ⫽ 2400 m

/(m

⋅ h)( p

⫺ 0)

⫽ 3.0 ⫻ 10

⫺5

mol gas/m

air

Step 2. Convert p

in terms of mol gas/mol air

Let V ⫽ volume of air per mole of air

From Step 1

Step 3. Compute DF for gas entrance and exit

At gas inﬂuent (bottom)

⫽ 0

⫽ 13.1 ␮g/L ⫽ 0.0131 mg/L

⫽ 0

⫽ c

⫺ c

⫽ 0.0131 mg/L

At gas efﬂuent (top)

From Table 5.3, Henry’s constant H at 20ºC

H ⫽ 550 atm

Convert atm to atm L/mg, H

5 7.55 3 10

atm

L/mg

55,600 3 MW

550 atm

55,600 3 131 mg/L

5 7.2 3 10

mol gas/mol air

5 131 mg>L 5 0.131 mg/L

5 to be determined

5 3.0 3 10

mol gas/m

air

5 3.0 3 10

mol gas

air

0.024 m

air

1 mol air

5 7.2 3 10

mol gas/mol air

V 5

nRT

s1 moleds0.08206 L atm/mol Kd 3 s293 Kd

1 atm

5 24.0 L

5 0.024 m

L⌬c 5 G⌬p

Public Water Supply 361

Step 4. Compute DF

From Eq. (5.37)

Step 5. Compute the height of the tower z

From Eq. (5.36)

Example 2 In a groundwater remediation study, volatile organic carbon

removal through the vapor phase, the total hydrocarbons analyzer measured

1,1,1,-trichloroethane (Jones et al., 2000). The extraction pump and air strip-

per combined air ﬂow averaged 0.40m

/min (14ft

/min). The concentration of

1,1,1,-trichloroethane averaged 25 parts per million by volume (ppmv) of air.

Determine the amount of 1,1,1,-trichloroethane removed daily. Temperature

is 20ºC.

solution

Step 1. Calculate 1,1,1,-trichloroethane concentration using the ideal gas law

T ⫽ 20°C ⫽ (20 ⫹ 273)K ⫽ 293 K

⫽ 41.6 mol/m

This means 41.6 mole of total gases in lm

of the air.

The MW of 1,1,1,-trichloroethane (CH

CC1

) ⫽ 133

One ppmv of 1,1,1,-trichloroethane ⫽ 41.6 ⫻ 10

⫺6

mol/m

⫻ 133g/mol

⫻ 1000 mg/g

⫽ 5.53 mg/m

25 ppmv of 1,1,1,-trichloroethane ⫽ 5.53 mg/m

⫻ 25

⫽ 138 mg/m

n 5

s1 atmd s1000 L/mg

0.082sL

atm/mol-Kd 3 293 K

z 5

Lsc

2 c

aDF

80 m/h 3 s0.131 2 0.0131d mg/L

44 h

3 0.0487 mg/L

5 4.4 m

2 DF

ln sDF

/DF

0.1215 mg/L 2 0.0131 mg/L

ln s0.1215/0.0131d

5 0.0487 mg/L

5 0.1215 mg/L

5 c

2 c

5 0.131 mg/L 2 0.0095 mg/L

5 0.0095 mg/L

7.2 3 10

mol gas/mol air 3 1 atm

7.55 3 10

atm

L/mg

Step 2. Calculate the daily removal

Daily removal ⫽ 138 mg/m

⫻ 0.40 m

/min ⫻ l440 min/d

⫽ 79500 mg/d

⫽ 79.5 g/d

Design of packed tower. Process

design of packed towers or columns is

based on two quantities: the height of the packed column, z, to achieve

the designed removal of solute is the product of the height of a transfer

unit (HTU) and the number of transfer unit (NTU). It can be expressed

as (Treybal, 1968):

z ⫽ (HTU)(NTU) (5.38)

The HTU refers the rate of mass transfer for the particular packing

materials used. The NTU is a measure of the mass transfer driving

force and is determined by the difference between actual and equilib-

rium phase concentrations. The height of a transfer unit is the constant

portion of Eq. (5.35):

(5.39)

The number of transfer units is the integral portion of Eq. (5.35). For

diluted solutions, Henry’s law holds. Substituting the integral expres-

sion for NTU with p

⫽ 0, the NTU is

(5.40)

where

(5.41)

⫽ stripping factor, unitless when H

is unitless

, c

⫽ mole fraction for gas entrance and exit, respectively

Convert Henry’s constant from terms of atm to unitless:

(5.42)

⫽ H/4.56T

1 L of water

55.6 mole

5 cH

atmsmol gas/mol aird

mol gas/mol water

d a

1 mol air

0.082T atm L of air

R 5

NTU 5

R 2 1

dsR 2 1d 1 1

HTU 5

362 Chapter 5

Public Water Supply 363

When T ⫽ 20ºC ⫽ 293 K

⫽ H/4.56 ⫻ 293 ⫽ 7.49 ⫻ 10

⫺4

H, unitless (5.42a)

G ⫽ superﬁcial molar air ﬂow rate (k mol/s ⋅ m

)

L ⫽ superﬁcial molar water ﬂow rate (k mol/s ⋅ m

)

The NTU depends upon the designed gas removal efficiency, the air-

water velocity ratio, and Henry’s constant. Treybal (1968) plotted the

integral part (NTU) of Eq. (5.35) in Fig. 5.5. By knowing the desired

removal efficiency, the stripping factor, and Henry’s constant, the

NTU in a packed column can be determined for any given stripping

factor of the air to water ﬂow rate ratios. It can be seen from Fig. 5.5

that when the stripping factor, R, is greater than 3, little improvement

for the NTU occurs.

Example: Using the graph of Fig. 5.5 to solve Example 1 above. Given:

T ⫽ 20°C ⫽ 293 K

L ⫽ 80 m

water/(m

column cross section h)

G ⫽ 2400 m

air/(m

column cross section h)

⫽ 131 ␮g/L of CCHCl

⫽ 13.1 ␮g/L of CCHCl

a ⫽ 44 h

⫺1

solution:

Step 1. Compute HTU with Eq. (5.39)

Step 2. Compute H

, the unitless Henry’s constant at 20°C

From Table 5.3

H ⫽ 550 atm

Using Eq. (5.42a)

⫽ 7.49 ⫻ 10

⫺4

H ⫽ 7.49 ⫻ 10

⫺4

⫻ 550

⫽ 0.412

Step 3. Compute the stripping factor R using Eq. (5.41)

⫽ 12.36

R 5

0.412 3 2400 m/h

80 m/h

HTU 5

80 m/h

44 h

5 1.82 m

364 Chapter 5

Figure 5.5 Number of transfer units for absorbers or strippers with constant

absorption or stripping factors (D. A. Cornwell, Air Stripping and aeration. In:

reprinted with permission of McGraw-Hill).

Step 4. Find NTU from Fig. 5.5

Since

Using R ⫽ 12.36 and c

Ⲑc

⫽ 0.1, from the graph in Fig. 5.5, we obtain

NTU ⫽ 2.42

13.1 mg/L

131 mg/L

5 0.1