Chapter 9 Part 1 HWM 2nd Ed Solutions

(1)

2001McGraw-Hill, Inc.

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

C

eq

dC

NTU

_R

C

_eq

=

−

out

)

(

1 )

(

R

C

dC

R

C

dC

NTU

out out

−

=

−

=

_∫

∫

=

₋

₊

+

−

=

t ou out

C

R

C

dC

R

C

RC

dC

R

)

1 (

∫

= U U dU ln Cin C Cout C out t ou

C

R

C

R

C

R

C

dC

R

= =

+

−

=

+

−

=

∫

ln(

(

1 )

)

1 )

1 (

)

1 (

1 }

{

ln( ( 1) ) ln( ( 1) )

1 Cin R Cout Cout R Cout

R R ₋ ₊ ₋ ₋ ₊ − = ) ) 1 ( ) 1 ( ln( 1 _out _out out in C R C C R C R R + − + − − =             + − + − − = out out out out out out out in C C R C C C C R C C R R ) 1 ( ) 1 ( ln 1

(2)

2001McGraw-Hill, Inc. 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 µg/L. A hydrogeologic study of the area indicates that

a flow rate of 110 gal/min is required to ensure that contamination not spread.

Laboratory investigations have determined the overall transfer constant, K

L

a =

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 = 20°C = 293K;

Influent:

C

in

= 2.1 mg/L

Effluent:

C

out

= 0.05 mg/L

Column 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 10

3

= 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

T B A

H

=

−

mol

m

atm

e

H

293 )

5 .

64

10

3 3

/

3020 13 . 5 (

⋅

×

=

∴

− −

K

mol

m

atm

RT

H

293 /

10

205 .

8

10

64 .

5

3 5 3

×

⋅

×

=

′

₋ − 2 2 292 . 0 4 ) 61 . 0 ( m = π 2 sec 1806 . 0 (sec) 60 (min) 1 785 . 3 min 110 0 . 1 ft lb gal L gal L Kg ⋅ = ⋅ × × 2 3 2 sec 1320 ) 18 1 )( 10 )( sec 8 . 23 ( m mol g mol kg g m kg L ⋅ = ⋅ = 525 . 5 15 235 . 0 × = =       ′ = W A Q Q H R m K M L HTU a L w 187 . 1 020 . 0 55600 1320 ₌ × = =

(3)

2001McGraw-Hill, Inc. 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:

ρA = Air Density = (@20°C) ρW = Water Density = 62.3 lb/Ft 3_(@20°C) 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) . This pressure 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 to water ratio and the packing height.

Molecular weight: 92.2 g/mol Boiling point : 111 °C

Molal Volume at boiling point: 0.1182 L/mol Henry's Constant: 0.19000 765 . 4 525 . 3 ) 1 525 . 3 ( 05 . 0 1 . 2 ln 1 525 . 3 525 . 3 1 ) 1 )( ( ln ) 1 ( =           ₋       − =           ₋ ₊ − = R R Cout Cin R R NTU ft HTU NTU Z = ⋅ = 4.765×1.187=5.656=18.6 2 2 sec 874 . 4 2048 . 0 sec 8 . 23 Ft lb m Kg L ⋅ = × ⋅ = sec 2129 . 0 sec 85 . 212 ) 15 ( 19 . 14 ) ( 3 m L W A QW QA= = = = 3 3 075 . 0 205 . 1 Ft lb m Kg = 2 2 3 3 3 sec 1806 . 0 sec 881 . 0 291 . 0 ) 205 . 1 )( sec 2129 . 0 ( Ft lb m Kg m m Kg m G ⋅ = ⋅ = = 00976 . 0 ) 17 . 32 )( 3 . 62 )( 075 . 0 ( ) 45 ( ) 1806 . 0 ( ) ( 2 2 = × = g F G W A ρ ρ 936 . 0 3 . 62 075 . 0 1806 . 0 874 . 4 0.5 5 . 0 =             =             W A G L

ρ

(4)

Temperature Constant: 3517 K Solution:

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

T = 20°C = 68°F

P = 101.3 kpa = 1.0 ATM (3) Choose a Liquid's loading rate

L = 23.8 kg/M2• s

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

(5) Choose a range of packing depth

(6) Indicate contaminant Toluene

The problem was solved by putting the above data into the program AIRSTRIP. This program may be ordered 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 µg/L is chosen corresponding to A/W = 20 and Z = 7.1m. The F9 key produces the second display.

Toluene Concentration In 2.1 mg/L

C-Raschig Rings 50.8 mm Atmospheric Pressure 101.3 k

Design Temperature 20.0° C Liquid Loading Rate 23.8 kg/m2• s

Minimum 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 0.1 5.7 0.2 0.1 0.1 0.1 0.0 6.9 0.1 0.1 0.0 0.0 0.0 8.0 0.1 0.0 0.0 0.0 0.0 9.1 0.0 0.0 0.0 0.0 0.0 R 1.9 2.8 3.8 4.7 5.7 dP(Pa/m) 79 149 246 381 563

F10 Toggle to English units F 1 Help F7 Quit Program

F9 Continue with design procedure F3 Main menu Esc to go back 30 10 ≤ ≤ W A

m

Z

m

9 .

0

6 .

4 ≤

≤

(5)

Contaminant Toluene

Concentration In 2.1 mg/L

Concentration Out 30.3 µg/L Percentage Removed 98.6 %

Packing C-Rasching Rings 50.8 mm

Water Temperature 20.0 °C. Atmospheric Pressure 101.3 kPa

Packing Depth 7.1 meter

Liquid Loading Rate 23.8 kg/m2• s

Air/Water Ratio 20

Stripping Factor 3.8

Air Pressure Gradient 246 Pa/m

F 10 Toggle to English units F1 Help F7 Quit Program F6 Save design "P" print report F3 Main menu Esc to go back

9-5. Determine the overall mass transfer coefficient, KLa, given the following data

from 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 = 6°C; 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 = 6°C

Diffusion coefficient in water = 1.1 x 10-6 cm2/s Assume: Column Diameter = 3m

Solution:

According to Onda Correlations (eqn. 9-5):

Let at total packing area = 279 m

2

/m3 (table 9-2)

dp nominal packing diameter =

L liquid mass loading rate = 10.2 kg/m2• s

ρL liquid density = 997 kg/m 2

@ 6° C

µL viscosity of water = 1.4728 x 10-3 kg/m2•s @ 6°C

g acceleration due to gravity = 9.81m/sec2

(

)

0.4 3 2 3 1 0051 . 0 _t _p L L L L w L L L a d D a L g K _            =      

ρ

µ

ρ

m in m in in 0.051 37 . 39 2 2 = × = 3 1 3 3 2 81 . 9 10 4728 . 1 997       × × − w La K

(6)

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 = 128.0 98.4 ∴ Total V = 29.6 + 98.4 = 128.0 cm3 /mo1 Using (eqn. 3-13) From equation 9-6: = 89.18 m2/m3 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

(

)

0.4 5 . 0 3 3 2 3 279 0.051 997 10 4728 . 1 10 4728 . 1 2 . 10 0051 . 0 _ × ×      × ×       × × × = − − − L w D a sec 10 215 . 5 10 sec ) 128 ( 14728 . 1 ) 6 273 ( 10 06 . 5 9 2 2 4 2 2 6 . 0 7 m cm m cm − − × = × × + × × =                                           − − = −0.05 ₂ 0.2 2 2 1 . 0 75 . 0 45 . 1 1 t L L L t c t w a L g L a L e a a σ ρ ρ µ σ σ                           × ×       ×       × ×       − − = − − 2 . 0 2 05 . 0 2 2 1 . 0 3 75 . 0 279 075 . 0 997 2 . 10 81 . 9 977 2 . 10 10 4728 . 1 279 2 . 10 075 . 0 033 . 0 45 . 1 1 279 e

(

)

0.4 5 . 0 9 3 3 2 3 279 0.051 10 215 . 5 997 10 4728 . 1 10 4728 . 1 18 . 89 2 . 10 0051 . 0 07 . 41 _ ×      × × ×       × × × = × − − − − L K

(

)

2 3 1 7 . 0 23 . 5 _ −            = t p a a a a t a t a _a_d D C a a D a K µ µ 2 3 1 6 5 7 . 0 5 6 (279 0.051) 10 1 . 1 247 . 1 10 81 . 1 10 81 . 1 279 5 . 1 23 . 5 10 1 . 1 279 − − − − −  ×      × × ×       × × = × × a K 1 3 6 4

10

94 .

4

10

427 .

1

36 .

2

82 .

53

23 .

5

10

069 .

3 ×

−

×

−

=

×

− −

=

ms

K

_a 233 . 0 279 10 205 . 8 00533 . 0 5 _× = × = = ′ ₋ RT H H

(7)

2001McGraw-Hill, Inc. KLa = 0.065 (sec

-1 )

9-6. Recalculate KLa from Example 9-1 incorporating the following changes: H’ is given as

0.0704 (dimensionless) at a temperature of 6°C, at = 138 m2/m3.

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 0°C, σ = 0.076

At 10°C, σ = 0.0742 ∴At 6°C,

diameter = 3 ft = 0.91m =

Unit weight of water = 8.34 lb/gal

At = 138 m2/m3 39 . 15 18 . 89 10 84 . 19 1 18 . 89 10 94 . 4 233 . 0 1 1 4 3_× + _× _× = × × = ∴ ₋ ₋ a KL 075 . 0 07492 . 0 6 0 10 0018 . 0 076 . 0 × = =      − − + = σ 44 . 0 075 . 0 033 . 0 ₌ = σ σc 2 2 650 . 0 4 ) 91 . 0 ( m = π sec 49 . 16 650 . 0 204 . 2 1 34 . 8 sec 60 min 1 min 170 2 2 = ⋅ × × × = m Kg m lb kg gal lb gal L sec 10 404 . 9 10 sec 09404 . 0 2 6 2 2 2 m cm m cm DG − × = × = sec 379 . 0 sec 60 min 1 48 . 7 min 170 3 3 ft gal ft gal QW = × × = sec 99 . 1 650 . 0 205 . 1 02832 . 0 sec 9 . 37 2 2 3 3 3 3 ⋅ = × × = m kg m m kg ft m ft G sec 10 6 10 sec 10 6 2 10 2 4 2 2 6 m cm m cm D_L = × − × = × − m in m in dp 0.051 37 . 39 2 × = =

(8)

2001McGraw-Hill, Inc. 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@0°C =1.71 x 10-5 N•S/m2 Viscosity of air @ 10° C =1.76 x 10-5 N•S/m2 Viscosity of air @ 6° C =171 + (0.5) x 6 = 174 ∴@6°C µ.G =1.74 x 10-5 N•S/m2 From equation 9-7: KG= 1.717 x 10 -2 ms-1

Approximately 1/3 smaller than the value @ 20°C calculated in example 9-1. 13 . 81 sec 10 4728 . 1 138 sec 49 . 16 3 2 = ⋅ × × × ⋅ = − m kg m kg a L L t

µ

3 2 2 2 2 10 848 . 3 81 . 9 ) 997 ( 138 ) 49 . 16 ( ₌ _× − × × = g at L L

ρ

0264 . 0 138 075 . 0 ) 997 ( ) 49 . 16 ( 2 2 2 = × × = t L a L

σ

ρ

(

)

[

]

{

}

3 2 2 . 0 05 . 0 3 1 . 0 75 . 0

5 .

74

49 .

74 )

0264

.

0 (

10

848 .

3 )

13 .

81 (

)

44 .

0 (

45 .

1

138 m

m

e

a

_w

=

×

−

×

− −

×

=

4 . 0 5 . 0 10 3 3 2 3 3 1 3 ₉₉₇ ₆ ₁₀ (138 0.051) 10 74 . 1 10 473 . 1 5 . 74 49 . 16 0051 . 0 81 . 9 10 473 . 1 997 _×       × × ×       × × =       × × − − − − − L K ) 10 404 . 9 )( 138 ( × −6 G K 2 3 1 60 5 7 . 0 5 (138)(0.0051) ) 10 404 . 9 ( ) 265 . 1 ( 10 74 . 1 10 74 . 1 138 99 . 1 23 . 5 ₋ − − −       × × ×       × × = ) 5 . 74 )( 10 564 . 1 ( 1 ) 5 . 74 )( 10 717 . 1 )( 0704 . 0 ( 1 1 1 1 4 4 − − + _× × = + ′ = a K a K H a KL G L sec 93 . 96 1 ₌ a KL ) (sec 0103 . 0 −1 = a K_L

(9)

9-7. Design an air stripping column to remove TCE from water. The initial concentration of TCE is 1.3 mg/L and this must be reduced to 75 µg/L. Use the following criteria: water flow rate: 350 gal/min; water temperature: 16°C; air temperature: 23°C; 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 (350•3.785/60 L/sec) = 22.08 Kg/sec Mass Loading = 22.08/0.292 = 75.62 Kg/sec•m2

L = (75.62 Kg/sec•m2 )(103 g/Kg)(1/18 mol/g) = 4201.11 mol/sec•m2 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" Intolox Saddles a = 118 m2/m3 KLa = (20. 4cm/hr)(118 m 2 /m3 ) = 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 whole number!

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).

m a K M L HTU L W 28 . 11 ) 0067 . 0 )( 55600 ( 4201 ₌ = =

(10)

9 - 8 . Demonstrate that the stripping factor used in this book is numerically similar to that in common 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/s•m2) (ii)

(From eqn. 9-20)

where H' = Henry's Constant (Dimensionless)

QA = Air Flow Rate (m

3 /sec) Qw = Water Flow Rate (m

3 /sec) Note: at 20°C

Mw = molar vol of water = 55,600 mol/m 3 R= 8.25x10-5 atm•m3/mol•k

From eqn (i):

From eqn (ii):

( From the Chemical engineering literature (reference 6):

L G H R = a         ′ = W A Q Q H R 1340 A W A H RT M H H′= = A H R L G = A N A H R Q Q H R (1340) = = ′ ) 10 5 . 7 ( × −4 = ∴ W A A Q Q H R ) 10 5 . 7 ( × −4 = W A Q Q L G m m

L

G

m

S

=

⋅

(11)

where

Gm = molar air flow rate (Kmol/s•m 2

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

2 ) M = slope (i.e. dimensionless Henry's constant, H') Gm = QAρA / mol wt of air Lm = Qwρw /mol wt of water ρA=1. 205 Kg/m 3 air @ 20° C ρw = 998.2 Kg/m 3 water @ 20° C MWair = 28.97 Kg/mol MWwater =18 Kg/mol

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

(

4

)

10

5 .

7

97 .

28

18

2 .

998

205 .

1 _⋅

₌

_⋅

_×

−

⋅

=

⋅

=

W A W A air water W A W A m m

Q

MW

Q

L

G

ρ

(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 5

Carbon (grams): 10 8 6 4 2

TOC (mg/L): 42 53 85 129 267

Determine the Freundlich constants, K and n, and plot the isotherm. Sample Volume, V = Liters 0.1

Initial Concentration, Ci= mg/L 5000

M Cf X = (Ci-Cf)V Cf X/M Log(X/M) Log Cf

grams 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 1.72 6 85 491.5 212 81.9 1.91 1.93 4 129 487.1 310 121.8 2.09 2.11 2 267 473.3 510 236.7 2.37 2.43

Slope = 1/n = 0.834442 0.3340686 Constant= Log K Std Error 0.036627 0.0726537 Std Error of constant R-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