Chapter 9 Part 1 HWM 2nd Ed Solutions

7/30/2019 Chapter 9 Part 1 HWM 2nd Ed Solutions

1/12

Hazardous Waste management, 2nded. Instructors Manual Chapter 9:Physicochemical Processes

2001McGraw-Hill, Inc.

Page 1 of 12

CHAPTER

9

PHYSICO:CHEMICAL PROCESSES

Supplemental Questions:

In the first line of the opening quote, what is "it"? i.e. what were they proposing to sweep for seven years?

The sands at the seashore.

9 -1. Derive Eq. 9-20 starting with Eq. 9-16.

Equation 9-16

But Equation 9-19

Let U = C(R-1); dU = (R-1) dc

Divide numerator and denominator by Cout

Equation 9-20

=

Cin

Cout eqCC

dCNTU R

CCC t

ou

eq

=

)(

1

)(R

CCC

dC

RR

R

CCC

dC

NTUoutout ==

+=+=touout

CRC

dCR

CCRC

dCR

)1(

= UUdU

ln

CinC

CoutCout

tou

CRCR

R

CRC

dCR

R

R =

=+

=

+

= ))1(ln(1)1(

)1(

1

}{ ))1(ln())1(ln(1

outoutoutin CRCCRCR

R++

=

))1(

)1(ln(

1 outout

outin

CRC

CRC

R

R

+

+

=

+

+

=

out

out

out

out

out

out

out

in

C

CR

C

C

C

CR

C

C

R

R

)1(

)1(

ln1


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Page 2 of 12

9-2. Provide a preliminary design of an air stripping column to remove toluene

from ground water. Levels of toluene range from 0.1 to 2.1 mg/L and this

must be reduced to 50 g/L. A hydrogeologic study of the area indicates thata flow rate of 110 gal/min is required to ensure that contamination not spread.

Laboratory investigations have determined the overall transfer constant, KLa =

0.020 sec--1

. Use a column diameter of 2.0 feet and an air to water ratio of

15. Specifically determine: Liquid loading rate, stripping factor, height of

the tower and provide a sketch of the unit indicating all required

appurtenances.

Assumptions: Temp = 20C = 293K;

Influent: Cin = 2.1 mg/L

Effluent: Cout = 0.05 mg/LColumn diameter = 2.0 ft = 0.61 m;

1 mole of water = 18 g.

Henry's constant: From App. B:

A = 5.13 B = 3.02 x 103 = 3020

= 0.235 (dimensionless)

1) Liquid Loading Rate:Cross-sectional Area of column

Mass Rate

F = 45 (Table 9-2)

2) Stripping Factor:dimensionless

3) Height of Transfer Unit:

)(

exp TB

A

H

=

molmatmeH /1064.5 33)

293

302013.5(

==

KKmolmatmRT

HH

293/10205.8

1064.535

3

==

22

292.04

)61.0(m=

2sec1806.0

(sec)60

(min)1785.3

min

1100.1

ft

lb

gal

Lgal

L

Kg

=

2

3

2 sec

1320)

18

1)(

10)(

sec

8.23(

m

mol

g

mol

kg

g

m

kgL

=

=

525.515235.0 ==

=

W

A

Q

QHR

mKM

LHTU

aLw

187.1020.055600

1320=

==


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Page 3 of 12

4) Number of Transfer Units:

transfer units

5) Height of packing in column:

9 - 3 . Using the data in problem 9-2, determine the pressure drop through the tower.

1) Select 2" Rasching Rings as packing.2) Calculate Pressure Drop. The parameters for Figure 9-5:

A = Air Density = (@20C)

W = Water Density = 62.3 lb/Ft3 (@20C)

F = 45 (Table 9-2)

Ordinate =

Abscissa =

This point intersects the curves of Fig 9.5 at dp = 0.25 inches of water per foot of packing depth (2050 Pa/m) . Thispressure drop is the lower limit of the recommended range of 0.25 to 0.5 inches of water per foot of packing depth,

indicating that the air flow rate could be increased significantly without causing flooding.Pressure Drop = (Ht. of packing in column Z) x (dp)

= (23) x 0.25 = 5.75 in = 0.021 psi =1430 Pascals

9-4. Using the data in problem 9-2, determine the impact on effluent quality by varying the air towater ratio and the packing height.

Molecular weight: 92.2 g/mol

Boiling point : 111 C

Molal Volume at boiling point: 0.1182 L/molHenry's Constant: 0.19000

765.4525.3

)1525.3(05.0

1.2

ln1525.3

525.31)1)((

ln)1

( =

=

+

=

R

RCout

Cin

R

RNTU

ftHTUNTUZ 6.18656.5187.1765.4 ====

22

sec

874.42048.0

sec

8.23

Ft

lb

m

KgL

=

=

sec

2129.0

sec

85.212)15(19.14)(

3mL

W

AQWQA ====

33

075.0205.1

Ft

lb

m

Kg=

223

3

3

sec

1806.0

sec

881.0

291.0

)205.1

)(sec

2129.0(

Ft

lb

m

Kg

m

m

Kgm

G

=

==

00976.0)17.32)(3.62)(075.0(

)45()1806.0(

)(

22

=

=g

FG

WA

936.03.62

075.0

1806.0

874.45.05.0

=

=

W

A

G

L


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Page 4 of 12

Temperature Constant: 3517 K

Solution:(1) Choose a column packing = Ceramic 2.0 in Ranching Ring(2) Choose Design temperature and pressure

T = 20C = 68FP = 101.3 kpa = 1.0 ATM

(3) Choose a Liquid's loading rateL = 23.8 kg/M2 s

(4) Choose a range of A/W ratios:

(5) Choose a range of packing depth

(6) Indicate contaminantToluene

The problem was solved by putting the above data into the program AIRSTRIP. This program may beordered from: AIRSTRIP, 3209 Garner Street, Ames, IA 50010.The computer will then calculate the effluent concentration of toluene over the indicated range of A/W

ratios and depth. The next page shows the screen display for the above reference problem. An effluent

concentration of 30.3 g/L is chosen corresponding to A/W = 20 and Z = 7.1m. The F9 key producesthe second display.

Toluene Concentration In 2.1 mg/LC-Raschig Rings 50.8 mm Atmospheric Pressure 101.3 k

Design Temperature 20.0 C Liquid Loading Rate 23.8 kg/m2 sMinimum Packing Depth 4.6 meter Minimum A/W Ratio 10.0

Maximum Packing Depth : 9.1 meter Maximum A/W Ratio 30.0

Concentration Remaining (mg/L)

Packing Depth A/W = 10 A/W = 15 A/W = 20 A/W = 25 A/W = 30(meter)

4.6 0.2 0.2 0.1 0.1 0.15.7 0.2 0.1 0.1 0.1 0.06.9 0.1 0.1 0.0 0.0 0.08.0 0.1 0.0 0.0 0.0 0.09.1 0.0 0.0 0.0 0.0 0.0

R 1.9 2.8 3.8 4.7 5.7dP(Pa/m) 79 149 246 381 563

F10 Toggle to English units F 1 Help F7 Quit ProgramF9 Continue with design procedure F3 Main menu Esc to go back

3010 W

A

mZm 0.96.4


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Page 5 of 12

AIRSTRIP Release 1.2 Summary of Selected Design Copyright 1988

Contaminant TolueneConcentration In 2.1 mg/L

Concentration Out 30.3 g/LPercentage Removed 98.6 %Packing C-Rasching Rings 50.8 mm

Water Temperature 20.0 C.Atmospheric Pressure 101.3 kPaPacking Depth 7.1 meter

Liquid Loading Rate 23.8 kg/m2 sAir/Water Ratio 20Stripping Factor 3.8Air Pressure Gradient 246 Pa/m

F 10 Toggle to English units F1 Help F7 Quit ProgramF6 Save design "P" print report F3 Main menu Esc to go back

9-5. Determine the overall mass transfer coefficient, KLa, given the following datafrom an air stripping column: contaminant = tetrachloroethene(C2CL4); liquid

mass loading rate = 10.2 kg/m2 s; packing =1" polyethylene Tri-paks

; air

mass loading rate = 1.5 kg/m2 s; temperature = 6C; diffusion coefficient in

water = 1.1 x 10-6

cm2/s.

Given: Contaminant = tetrachloroethene (C2Cl4)Liquid mass loading rate = 10.2 kg/m2 s

Air mass loading rate = 1 kg/m2 s

Packing = 1 polyethylene Tri-packs

Temperature = 6CDiffusion coefficient in water = 1.1 x 10-6 cm2/s

Assume: Column Diameter = 3mSolution:According to Onda Correlations (eqn. 9-5):

Let at total packing area = 279 m2/m3 (table 9-2)

dp nominal packing diameter =

L liquid mass loading rate = 10.2 kg/m2 s

L liquid density = 997 kg/m2 @ 6 C

L viscosity of water = 1.4728 x 10-3 kg/m2s @ 6C

g acceleration due to gravity = 9.81m/sec2

( ) 4.03

23

1

0051.0 ptLL

L

LwL

LL da

Da

L

gK

=

min

minin 051.0

37.3922 ==

31

33

2

81.9104728.1

997

wLaK


6/12



Page 6 of 12

DL is obtained by The Wilke-Chang method (Sec. 3.2)

The Molar Volume: (See Table 3-4, p. 97) of C2Cl4

C = 2 x 14.8 = 29.6 Cl = 4 x 24.6 =128.098.4

Total V = 29.6 + 98.4 = 128.0 cm3/mo1

Using (eqn. 3-13)

From equation 9-6:

= 89.18 m2/m

3

KL =19.86 x 10-4 ms-1

From equation 9-7:

KLa = 0.065 (sec-1) A = 124 B = 4.92 x 103

H = exp [12.4 (4.92 x 103) / 279] = 0.00533

( ) 4.05.0

332

3051.0279

997

104728.1

104728.1

2.100051.0

=

Lw

Da

sec10215.5

10sec)128(14728.1

)6273(1006.5 2924

22

6.0

7 m

cm

mcm

=

+=

=

2.02

05.0

2

21.075.0

45.11tLLLt

ctw

a

L

g

L

a

Leaa

=

2.02

05.0

2

21.0

3

75.0

279075.0997

2.10

81.9977

2.10

104728.1279

2.10

075.0

033.045.11279 e

( )

4.0

5.0

9

332

3 051.027910215.5997

104728.1

104728.118.89

2.10

0051.007.41

=

LK

( ) 23

17.0

23.5

= pt

aa

a

atat

a daDCa

a

Da

K

23

1

6

57.0

56)051.0279(

101.1247.1

1081.1

1081.1279

5.123.5

101.1279

=

aK

1364 1094.410427.136.282.5323.510069.3 == msKa

233.027910205.8

00533.05

=

==

RT

HH


7/12



Page 7 of 12

KLa = 0.065 (sec-1)

9-6. Recalculate KLa from Example 9-1 incorporating the following changes: H is given as0.0704 (dimensionless) at a temperature of 6C, at = 138 m

2/m

3.

This surface tension of water is a function of temperature. From "Tables of Physical and Chemical Constants

and Some Mathematical Functions", Kaye, G. W. C. and Labe., T. H., London, Longman Publishers, 1986:

At 0C, = 0.076

At 10C, = 0.0742

At 6C,

diameter = 3 ft = 0.91m =

Unit weight of water = 8.34 lb/gal

At = 138 m

2

/m

3

39.1518.891084.19

1

18.891094.4233.0

1143

=

+

=aKL

075.007492.06010

0018.0076.0 ==

+=

44.0075.0

033.0

==

c

22

650.04

)91.0(m=

sec49.16

650.0

204.2

134.8

sec60

min1

min170

22 =

=m

Kg

m

lb

kg

gal

lbgal

L

sec10404.910

sec09404.0

26

2

22m

cm

mcmDG

==

sec379.0

sec60

min1

48.7min170

33 ft

gal

ftgalQW ==

sec99.1

650.0

205.102832.0sec

9.37

22

33

33

=

=m

kg

m

m

kg

ft

mft

G

sec106

10sec106

210

24

226 m

cm

mcmDL

==

min

mindp 051.0

37.392 ==


8/12



Page 8 of 12

Reynolds No.:

(dimensionless)

Froude No.:

(dimensionless)

Weber No.:

(dimensionless)

From equation 9-6:

From equation9-5:

KL = 1.546 x 10-4 m/sec

Viscosity of air@0C =1.71 x 10-5

NS/m2

Viscosity of air @ 10 C =1.76 x 10 -5 NS/m2Viscosity of air @ 6 C =171 + (0.5) x 6 = 174

@6C .G =1.74 x 10-5 NS/m2

From equation 9-7:

KG= 1.717 x 10-2 ms-1

Approximately 1/3 smaller than the value @ 20C calculated in example 9-1.

13.81

sec104728.1138

sec49.16

3

2

=

=

m

kgm

kg

a

L

Lt

3

2

2

2

2

10848.381.9)997(

138)49.16( =

=

g

atL

L

0264.0138075.0)997(

)49.16(2

22

=

=tL a

L

( ){ } 322.005.031.075.0 5.7449.74)0264.0(10848.3)13.81()44.0(45.11138m

mea

w===

4.0

5.0

10

332

3

31

3)051.0138(

106997

1074.1

10473.15.74

49.160051.0

81.910473.1

997

=

LK

)10404.9)(138( 6GK

23

1

60

57.0

5)0051.0)(138(

)10404.9()265.1(

1074.1

1074.1138

99.123.5

=

)5.74)(10564.1(

1

)5.74)(10717.1)(0704.0(

111144

+

=+

=aKaKHaK LGL

sec93.961

=aKL

)(sec0103.0 1=aKL


9/12



Page 9 of 12

9-7. Design an air stripping column to remove TCE from water. The initial concentration of TCEis 1.3 mg/L and this must be reduced to 75 g/L. Use the following criteria: water flow rate: 350gal/min; water temperature: 16C; air temperature: 23C; packing: use 2" Intalox saddles.

Liquid Loading Rate:

Select column diameter = 2 Ft

Area of column = 0.292 m2

Mass Rate = 1.0 Kg/L x 350 gal/min

= 1.0 Kg/L x (3503.785/60 L/sec) = 22.08 Kg/secMass Loading = 22.08/0.292 = 75.62 Kg/secm

2

L = (75.62 Kg/secm2 )(103 g/Kg)(1/18 mol/g) = 4201.11 mol/secm2

Stripping Factor:

For TCE KL = 20.4 cm/hr

H = exp(9. 7 - 4308/T) = exp(9.7 - 4308/289. 3) = 5. 566 10 9-3

H' = H/RT = 5.566 9

-3

/(8. 206 9

-5

x 289. 3) = 0.234 (dimensionless)R = H'(QA/QW ) = 0. 234(70) = 16. 38 (dimensionless)

Height of Transfer Unit:

For 2" IntoloxSaddles

a = 118 m2/m3KLa = (20. 4cm/hr)(118 m

2/m

3) = 20. 4 x 118 x 10

-2/(3600) sec

-1= 0. 0067 sec

-1

Number of Transfer Units:

NTU = (R/ R-1) ln{ ( (Cin/Cout) (R-1)+1)/R}

= (1.07) ln {( (1300/75) (16.3-1) +1)/16.3} = 2.99 Transfer Units

Height of Packing in Column:

Z = NTU x HTU = (2.99)(11.28) = 33.72m

Note: In an actual design, a safety factor of 2% would be added the height of packing raised to the next wholenumber!

Z = 33.72 (1.2) = 40.464 = 41 m

But the tower height should be between l and 15 meters

Provide three towers of 14 m (45 feet).

maKM

LHTULW

28.11)0067.0)(55600(

4201 ===


10/12



Page 10 of 12

9 - 8 . Demonstrate that the stripping factor used in this book is numerically similar to that incommon use by chemical engineers:

R = H'G/L

H'= Henry's constant (dimensionless)

G = gas loading rate (kmol/hr)

L = liquid loading rate (kmol/hr)

First note that the chemical engineering equation defines the stripping factor in terms of molar flow rates,while equation 9-20 defines R in terms of volumetric flow rates:

(i)

where Ha = Henry's constant in Atmosphere

G, L = Loading Rates (Kmol/sm2)

(ii)

(From eqn. 9-20)

where H' = Henry's Constant (Dimensionless)QA = Air Flow Rate (m

3/sec)

Qw = Water Flow Rate (m3/sec)

Note: at 20C

Mw = molar vol of water = 55,600 mol/m3

R= 8.25x10-5 atmm3/molk

From eqn (i):

From eqn (ii):

( From the Chemical engineering literature (reference 6):

L

GHR a=

=

W

A

Q

QHR

1340

A

W

A H

RTM

HH ==

AH

R

L

G=

AN

A

H

R

Q

Q

H

R )1340(==

)105.7( 4=W

A

A Q

Q

H

R

)105.7( 4=W

A

Q

Q

L

G

m

m

L

GmS =


11/12



Page 11 of 12

where

Gm = molar air flow rate (Kmol/sm2)

Lm = molar liquid air flow rate (Kmol/s m2)

M = slope (i.e. dimensionless Henry's constant, H')

Gm = QAA / mol wt of airLm = Qww /mol wt of water

A=1. 205 Kg/m3 air @ 20 C

w = 998.2 Kg/m3 water @ 20 C

MWair= 28.97 Kg/mol

MWwater=18 Kg/mol

@20C, the value of the dimensionless stripping factor is the same for both equation.

( )4105.797.28

18

2.998

205.1 ===W

A

W

A

air

water

W

A

W

A

m

m

Q

Q

Q

Q

MW

MW

Q

Q

L

G


12/12



Page 12 of 12

9-9. 100 mL of a solution with a TOC (total organic carbon) concentration of 0.5% is

placed

in each of flue containers with activated carbon and shaken for 48 hours. The samples

are filtered and the concentration of TOC measured, yielding the following analyses:Container: 1 2 3 4 5Carbon (grams): 10 8 6 4 2TOC (mg/L): 42 53 85 129 267

Determine the Freundlich constants, K and n, and plot the isotherm.

Sample Volume, V = Liters 0.1Initial Concentration, Ci= mg/L 5000

M Cf X = (Ci-Cf)V Cf X/M Log(X/M) Log Cfgrams mg/L mg mg/L

10 42 495.8 42 49.6 1.70 1.62

8 53 494.7 99 61.8 1.79 1.726 85 491.5 212 81.9 1.91 1.934 129 487.1 310 121.8 2.09 2.11

2 267 473.3 510 236.7 2.37 2.43

Slope = 1/n = 0.834442 0.3340686 Constant= Log KStd Error 0.036627 0.0726537 Std Error of constantR-Squared 0.994253 0.0234578 Std Error o y

F 519.0196 3 d. f.SS reg 0.285601 0.0016508 SS residual

K = 2.16 mg/gram

1/n = 0.834

Documents

Chapter 9 Part 1 HWM 2nd Ed Solutions