Chapter 9 Part 1 HWM 2nd Ed Solutions

Hazardous Waste management, 2nd ed. Instructors’ Manual

Chapter 9:Physicochemical Processes

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 Cin

NTU = NTU =

dC But ∫ C − Ceq Cout

∫

C eq =

C − C ou t R

Equation 9-19

1 dC dC = R∫ C − Cout R (C − C − Cout ) (C − ) R R

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

∫

dU = ln U U

dC dC = R∫ RC − C + Cout C ( R − 1) + Cou t

= R∫ =

=

R C ( R − 1) + Cout ln( in ) R −1 Cout ( R − 1) + Cout

C = Cin R R ( R − 1)dC C R C = − + ln( ( 1 ) ) out C = Cout R − 1 ∫ C ( R − 1) + Cou t R − 1

Cout  Cin ( R − 1) +  R C Cout ln  out = C C R −1 out  out ( R − 1) + C C  out  out

     

Divide numerator and denominator by Cout

=

R {ln(Cin ( R − 1) + Cout ) − ln(Cout ( R − 1) + Cout )} R −1

Equation 9-20

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9-2.

Chapter 9:Physicochemical Processes

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, 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 = 20°C = 293K; Influent: Cin = 2.1 mg/L Effluent: Cout = 0.05 mg/L Column diameter = 2.0 ft = 0.61 m; 1 mole of water = 18 g.

Henry's constant: From App. B:

H = exp

B ( A− ) T

∴H = e H′ =

B = 3.02 x 103 = 3020

A = 5.13 ( 5.13−

3020 ) 293

= 5.64 × 10 − 3 atm ⋅ m 3 / mol

H 5.64 × 10−3 = RT 8.205 × 10−5 atm ⋅ m 3 / mol ⋅ K × 293K

= 0.235 (dimensionless)

1)

Liquid Loading Rate:

Cross-sectional Area of column

Mass Rate L=(

2)

4

= 0.292m 2

1(min) 1.0 Kg 110 gal L lb × × 3.785 ⋅ = 0.1806 L min gal 60(sec) sec⋅ ft 2 F = 45 (Table 9-2)

1320mol 23.8kg 10 3 g 1mol )= )( )( 2 18 g kg sec⋅ m 2 sec⋅ m

Stripping Factor:

 QA R = H ′ Q  W 3)

π (0.61) 2

 dimensionless   = 0.235 × 15 = 5.525 

Height of Transfer Unit:

HTU =

L 1320 = = 1.187m M w KLa 55600 × 0.020 2001McGraw-Hill, Inc. Page 2 of 12

Hazardous Waste management, 2nd ed. Instructors’ Manual

4)

Chapter 9:Physicochemical Processes

Number of Transfer Units:  Cin   2.1   ( Cout )( R − 1) + 1   3.525   0.05 (3.525 − 1)  R NTU = ( ) ln   ln  =  = 4.765 R R −1  3.525   3.525 − 1       

5)

transfer units

Height of packing in column: Z = NTU ⋅ HTU = 4.765 × 1.187 = 5.656 = 18.6 ft

9-3.

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

1) 2)

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

L=

4.874lb 23.8 Kg × 0.2048 = 2 sec⋅ Ft 2 sec⋅ m

QA = QW (

A 212.85L 0.2129m 3 = ) = 14.19(15) = W sec sec

ρA = Air Density = 1.205Kg = 0.075lb (@20°C) m3 Ft 3 ρW = Water Density = 62.3 lb/Ft3 (@20°C)

0.2129m 3 1.205Kg )( ) 3 0.881Kg 0.1806lb sec m G= = = 3 0.291m sec⋅ m 2 sec⋅ Ft 2 (

F = 45 (Table 9-2) Ordinate =

Abscissa =

(0.1806) 2 × ( 45) G2F = = 0.00976 ρ A ( ρW ) g (0.075)(62.3)(32.17)

 L  ρ A     G  ρ W

   

0.5

 4.874  0.075  =    0.1806  62.3 

0.5

= 0.936

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 2001McGraw-Hill, Inc. Page 3 of 12

Hazardous Waste management, 2nd ed. Instructors’ Manual

Chapter 9:Physicochemical Processes

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: A 10 ≤ ≤ 30 W (5)

Choose a range of packing depth

4.6m ≤ Z ≤ 9.0m (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 C-Raschig Rings 50.8 mm Design Temperature Minimum Packing Depth Maximum Packing Depth :

20.0° C 4.6 meter 9.1 meter

Concentration Remaining (mg/L) Packing Depth A/W = 10 (meter) 4.6 0.2 5.7 0.2 6.9 0.1 8.0 0.1 9.1 0.0 R 1.9 dP(Pa/m) 79 F10 F9

Concentration In Atmospheric Pressure Liquid Loading Rate Minimum A/W Ratio Maximum A/W Ratio

Toggle to English units Continue with design procedure

A/W = 15

A/W = 20

A/W = 25

0.2 0.1 0.1 0.0 0.0 2.8 149

0.1 0.1 0.0 0.0 0.0 3.8 246

0.1 0.1 0.0 0.0 0.0 4.7 381

F1 F3

Help Main menu

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F7 Esc

2.1 mg/L 101.3 k 23.8 kg/m2• s 10.0 30.0

A/W = 30 0.1 0.0 0.0 0.0 0.0 5.7 563 Quit Program to go back

Hazardous Waste management, 2nd ed. Instructors’ Manual

AIRSTRIP Release 1.2

Summary of Selected Design

Contaminant Concentration In Concentration Out Percentage Removed Packing Water Temperature Atmospheric Pressure Packing Depth Liquid Loading Rate Air/Water Ratio Stripping Factor Air Pressure Gradient F 10 F6

Copyright 1988

Toluene 2.1 mg/L 30.3 µg/L 98.6 % C-Rasching Rings 50.8 mm 20.0 °C. 101.3 kPa 7.1 meter 23.8 kg/m2• s 20 3.8 246 Pa/m

Toggle to English units Save design "P" print report

9-5.

Chapter 9:Physicochemical Processes

F1 F3

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F7 Esc

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 Column Diameter = 3m

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

 ρL KL  µ g  L

   

1

3

 L = 0.0051 a µ  w L

2in = 2in × Let

Quit Program to go back

   

2

3

 µL  ρ D  L L

 0.4  (a t d p ) 

m = 0.051m 39.37in

at dp

total packing area = 279 m2/m3 (table 9-2) nominal packing diameter =

L ρL µL g

liquid mass loading rate = 10.2 kg/m2• s liquid density = 997 kg/m2 @ 6° C viscosity of water = 1.4728 x 10-3 kg/m2•s @ 6°C acceleration due to gravity = 9.81m/sec2

K Law

2

3

997     −3 × 9.81   1.4728 × 10

1

3

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Hazardous Waste management, 2nd ed. Instructors’ Manual

Chapter 9:Physicochemical Processes

DL is obtained by The Wilke-Chang method (Sec. 3.2) 2

  3  1.4728 × 10 −3  10.2    = 0.0051 ×  −3  1.4728 × 10 × a w   997 × DL  The Molar Volume: (See Table 3-4, p. 97) of C2Cl4

−0.5

× (279 × 0.051)

0.4

C = 2 x 14.8 = 29.6 Cl = 4 x 24.6 = 98.4 128.0 ∴ Total V = 29.6 + 98.4 = 128.0 cm3/mo1 Using

=

(eqn. 3-13)

2 m2 5.06 × 10 −7 × ( 273 + 6) cm 2 −9 m 5 . 215 10 × = × sec 10 4 cm 2 sec 1.14728 × (128) 0.6

From equation 9-6: aw

0.75    L  σ c    = a t 1 − e − 1.45  a µ σ     t L   

   

0.1

 L2   ρ 2g  L

   

−0.05

 L2   ρ σa  L t

   

0.2

     

− 0.05 0.75 0.1     10.2 10.2 2 10.2 2   0.033        = 279 1 − e  − 1.45    −3 2    997 × 0.075 × 279   0.075   279 × 1.4728 × 10   977 × 9.81     

= 89.18 m2/m3 10.2   K L × 41.07 = 0.0051 ×  −3   89.18 × 1.4728 × 10 

2

3

 1.4728 × 10 −3  −9  997 × 5.215 × 10

  

−0.5

(279 × 0.051)0.4

KL =19.86 x 10-4 ms-1 From equation 9-7:  a Ka = 5.23 a t Da  at µ a

  

0.7

 µa   C a Da

  

1

3

(a d )

−2

t

p

Ka 1.5   = 5.23 −6 −5  279 1 . 81 10 279 × 1.1 × 10 × ×  

0.7

 1.81 × 10 −5  −6  1.247 × 1.1 × 10

  

1

3

( 279 × 0.051) − 2

K a = 3.069 × 10 −4 × 5.23 × 53.82 × 2.36 × 1.427 × 10 −6 = 4.94 × 10 −3 ms −1 KLa = 0.065 (sec-1)

A = 124

B = 4.92 x 103

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

H′=

H 0.00533 = =0 .233 2001McGraw-Hill, Inc. RT 8.205 × 10 −5 × 279 Page 6 of 12

   

0.2

    

Hazardous Waste management, 2nd ed. Instructors’ Manual

Chapter 9:Physicochemical Processes

KLa = 0.065 (sec-1)

∴

1 1 1 = + = 15.39 K La 0.233 × 4.94 × 10 −3 × 89.18 19.84 × 10 − 4 × 89.18

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: σ = 0.076 σ = 0.0742

At 0°C, At 10°C,

− 0.0018   × 6 = 0.07492 = 0.075  10 − 0 

σ = 0.076 +  

∴At 6°C,

σ c 0.033 = = 0.44 σ 0.075 2 diameter = 3 ft = 0.91m = π (0.91) = 0.650m 2

4

Unit weight of water = 8.34 lb/gal 170 L=

gal lb 1kg 1 min × × 8.34 × Kg gal 2.204lb min 60 sec = 16.49 2 2 m ⋅ sec 0.650m

At = 138 m2/m3 QW = 170

gal ft 3 ft 3 1 min × × = 0.379 min 7.48 gal 60 sec sec

DG = 0.09404

37.9 G=

2 cm 2 m2 −6 m 9 . 404 10 × 10 = × sec sec cm 2

ft 3 m3 kg × 0.02832 3 × 1.205 3 kg sec ft m = 1.99 2 m ⋅ sec 0.650m 2

D L = 6 × 10 −6

d

p

= 2in ×

2 cm 2 m2 −10 m 6 10 × = × sec 10 4 cm 2 sec

m = 0.051m 39.37in

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Hazardous Waste management, 2nd ed. Instructors’ Manual

Chapter 9:Physicochemical Processes

Reynolds No.:

16.49

L

kg m ⋅ sec 2

(dimensionless)

= = 81.13 kg at µ L 138 × 1.4728 × 10 −3 × Froude No.: m ⋅ sec

(16.49) 2 × 138 L2 at = = 3.848 × 10 −3 (dimensionless) (997) 2 × 9.81 ρL2 g Weber No.:

L2

ρ Lσa t

=

(16.49) 2 = 0.0264 (dimensionless) (997) 2 × 0.075 × 138

From equation 9-6:

{ [

a w = 138 × 1 − e − 1.45(0.44) 0.75 × (81.13) 0.1 × (3.848 × 10 −3 )

−0.05

]}

× (0.0264) 0.2 = 74.49 = 74.5 m

From equation9-5: 997   KL   −3 1 . 473 × 10 × 9 . 81  

1

3

16.49   = 0.0051 −3  74 . 5 1 . 473 10 × ×  

2

3

 1.74 × 10 −3    −10   997 × 6 × 10 

−0.5

(138 × 0.051) 0.4

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 × 10 −6 )1 1 1 (1138)(9.404 = + = + K La H ′K G a K La (0.0704)(1.717 × 10 − 4 )( 74.5) (1.564 × 10 − 4 )( 74.5) 1 0.7  3 1.99 1.74 × 10 −5     (138)(0.0051) − 2  = 5.23 −5  − 60   138 × 1.74 × 10   (1.265) × (9.404 × 10 ) 

KG= 1.717 x 10-2 ms-1 Approximately 1/3 smaller than the value @ 20°C calculated in example 9-1. K L a = 0.0103(sec −1 )

1 = 96.93 sec K La

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2

m3

Hazardous Waste management, 2nd ed. Instructors’ Manual

9-7.

Chapter 9:Physicochemical Processes

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 m2/m3 ) = 20. 4 x 118 x 10-2/(3600) sec-1 = 0. 0067 sec-1 HTU =

4201 L = = 11.28m (55600)( 0.0067) MW K La

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). 2001McGraw-Hill, Inc. Page 9 of 12

Hazardous Waste management, 2nd ed. Instructors’ Manual

9-8.

Chapter 9:Physicochemical Processes

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)

R=

H aG L

where

Ha G, L

= =

Henry's constant in Atmosphere Loading Rates (Kmol/s•m2)

(ii)  QA R = H ′ Q  W

where

   (From eqn. 9-20) 

H'

= = =

QA

Qw

Henry's Constant (Dimensionless) Air Flow Rate (m3/sec) Water Flow Rate (m3/sec)

Note: at 20°C Mw = molar vol of water = 55,600 mol/m3 R= 8.25x10-5 atm•m3/mol•k

H′=

From eqn (i):

From eqn (ii):

HA HA = 1340 M W RT G R = L HA Q R R (1340) = A = ′ H QN HA

∴

QA R = (7.5 × 10 − 4 ) HA QW

QA G ( 7.5 × 10 − 4 ) = L QW

S = m⋅

Gm Lm

( From the Chemical engineering literature (reference 6): 2001McGraw-Hill, Inc. Page 10 of 12

Hazardous Waste management, 2nd ed. Instructors’ Manual

Chapter 9:Physicochemical Processes

where Gm = Lm = M =

molar air flow rate (Kmol/s•m2) molar liquid air flow rate (Kmol/s• m2) 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 air @ 20° C 3

ρw = 998.2 Kg/m3 water @ 20° C MWair = 28.97 Kg/mol MWwater =18 Kg/mol

Gm Q A ρ A MWwater Q A 1.205 18 Q = ⋅ ⋅ = ⋅ ⋅ = A ⋅ (7.5 × 10 −4 ) Lm QW ρW MWair QW 998.2 28.97 QW ∴@20°C, the value of the dimensionless stripping factor is the same for both equation.

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9-9.

Chapter 9:Physicochemical Processes

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 = Initial Concentration, Ci= M grams 10 8 6 4 2 Slope = 1/n = Std Error R-Squared F SS reg

K 1/n

= =

Cf mg/L 42 53 85 129 267 0.834442 0.036627 0.994253 519.0196 0.285601

2.16 0.834

Liters mg/L

0.1 5000

X = (Ci-Cf)V mg 495.8 494.7 491.5 487.1 473.3

Cf mg/L 42 99 212 310 510

X/M

0.3340686 0.0726537 0.0234578 3 0.0016508

Constant= Log K Std Error of constant Std Error o y d. f. SS residual

49.6 61.8 81.9 121.8 236.7

mg/gram

2001McGraw-Hill, Inc. Page 12 of 12

Log(X/M) Log Cf 1.70 1.79 1.91 2.09 2.37

1.62 1.72 1.93 2.11 2.43