tutorial solutions topic 5

Separation Process A

Mass Transfer

©HERIOT-WATT UNIVERSITY B48BD October 2014 v1

Handed out: Week

Handed in: Week

TUTORIAL ANSWERS - TOPIC 5

Question 1

A = Benzene vapour ; B = Air

No mention of stagnant film. Therefore, we can assume Fick’s law of diffusion

given by.

dx

dcD A

ABA J

Rearranging the terms and integrating:

2

1

2

1

..JA

A

C

C

AAB

x

x

A dcDdx

Steady state; Diffusion flux (JA) = constant. Diffusivity (DAB) is taken as constant:

2

1

2

1

JA

A

C

C

AAB

x

x

A dcDdx

2212 ..J AAABA ccDxx

12

12.J

xx

ccD AAABA

We do not have cA at points 1 and 2. Supposing Ideal Gas Mixture we can then calculate:

V

nc A

A TR

P

V

nTRnVP AA

A.

...

2

©HERIOT-WATT UNIVERSITY B49CA March 2015 v1

Combining both equations we can write:

TR

Pc A

A.

We can then rewrite the flux as:

12

12

12

12

..

.J

.J

xxTR

PPD

xx

ccD AAAB

AAAAB

A

Data provided:

DAB = 8.8 10-6 m2s-1

P = 1 bar = 1 bar x (105 Pa /1 bar) = 100103 Pa or N m-2

T = 25°C= 25+273 = 298 K

x2 – x1 = 4 mm = 4 x 10-3

yA1 = 0.10;

yA2 = 0.05

Dalton’s Law

i

i

y

PP ii yPP

Where:

iP partial pressure of component “i”

iy mole fraction of component “i”

PaPP AA 13

1 10.010100

PaPA3

1 1010

PaPP AA 23

2 05.010100

3


PaPA3

2 105

12

12

..

.J

xxTR

PPD AAAB

A

mKmolKmNmolKJ

mNPasmA 31111

233126

104298314.8

1010105.108.8J

1231044.4J smmolA

Question 2:

A = CO2 and B = Aqueous solution

Ideal gas; Stagnant film

1

2

12 A

AABA

PP

PPLn

xxTR

DPN

Data provided:

DAB = 7.17 10-6 m2s-1

P = 2 bar = 2 bar (105 Pa / 1 bar) = 200 103 Pa or N m-2

T = 300 K

x2 – x1 = 1 mm x (1 / 103 cm) = 10-3 cm

yA1 = 0.30;

yA2 = 0

4


Dalton’s Law

i

i

y

PP ii yPP

Where:

iP partial pressure of component “i”

iy mole fraction of component “i”

PaPP AA 13

1 30.010200

PaPA3

1 1060

PaPP AA 23

2 010200

PaPA 02

1

2

12 A

AABA

PP

PPLn

xxTR

DPN

Pa

PaLn

mKmolKJ

smPaN A 33

3

311

1255

106010200

010200

][10][300.][314.8

].[102.102

121 .1005.2 smmolNA

5


Question 3

Ideal gas; A = Water ; B = Air

21

12

AA

BLM

ABA PP

PxxTR

DPN

Data provided:

P = 1 atm = 101.3 kPa ≈ 1.01 x 105 Pa

T = 293K

DAB = 25 x 10-6 m2. s-1

x2 – x1 = 0.15 m

PA1 = 2.31 x 10-2 atm = 2.31 x 10-2 atm (atm 1

Pa 10 x 1.01 5

) = 2.33103 Pa

PA2 = 0

R = 8.314 11311 kmolKmPamolKJ

1

2

21

1

2

12

1

2

12

A

A

AA

A

A

AA

B

B

BBBLM

PP

PPLn

PP

PP

PPLn

PPPP

P

PLn

PPP

Pa

Pa

PaLn

PaPBLM

3

33

3

3

10100

1033.2103.101

0103.101

01033.2

6


01033.2101001015293314.8

1025103.101 3

2

63

AN

1241061.1 smmolNA

Unit m - +

Alternatively:

Ideal gas; A = Water ; B = Air

1

2

12 A

AABA

PP

PPLn

xxTR

DPN Stefan Diffusion

Data provided:

P = 1 atm = 101.3 kPa ≈ 1.01 x 105 Pa

T = 293K

DAB = 25 x 10-6 m2. s-1

x2 – x1 = 0.15 m

PA1 = 2.31 x 10-2 atm = 2.31 x 10-2 atm (atm 1

Pa 10 x 1.01 5

) = 2.33103 Pa

PA2 = 0

R = 8.314 11311 kmolKmPamolKJ

7


33

3

2

63

1033.2103.101

0103.101

1015293314.8

1025103.101LnN A

1241061.1 smmolNA

Question 4

Ideal gas; A = Benzene ; B = Air; Ideal Gas; Stagnant Film

1

2

12 A

AABA

PP

PPLn

xxTR

DPN

Data provided:

P = 101300 Pa

R = 8.314 11311 molKmNmolKJ

T = 295 K

DAB = 8 x 10-6 m2. s-1

x2 – x1 = 3 mm = 3 x 10-3 m

PA1 = 13.3 kPa = 13.3103 Pa

PA2 = 0

d = 5 m

hA = 5 mm = 5 x 10-3 m

ρA,295 K = 880 kg.m = 88 x 10 g.m

101300

NA

R = 8.314 JK-1mol-1 [= Nm K-1

mol-1]

x2-x1 = 3 mm (1 [m]/103 [mm]) = 310-3 m

DAB = 8 10-6 m2s-1

8


T = 295 K

PA1 = 13.3 kPa (103 [Pa]/1 [kPa]) = 13.3103 Pa

PA2 = 0 Pa

d = 5 m

hA= 5 mm (1 [m]/103 [mm]) = 510-3 m

ρA, 295 K = 880 kgm-3 (103 [g]/1 [kg]) = 88104 gm-3

A = Benzene vapour ; B = Air ; Ideal Gas ; Stagnant Film

Area for mass transfer = Cross sectional area of the tank

22

4

dA r

36 2 1

1 1 3 3 3

101 3 10 0101300 8 10

8 314 295 3 10 101 3 10 13 3 10A

Pa PaPa m sN Ln

N m K mol K m Pa Pa

. [ ] [ ][ ] [ ]

. [ ] [ ] [ ] . [ ] . [ ]

2 2 11 55 10 mol m s .

2

2 1 1

AABA

A

c cc DN Ln

x x c c

( )1 1

101300

8 314 295

Pan PcV R T N m K mol K

[ ]

. [ ] [ ]

341 3 mol m .

11 1 1

13300

8 314 295A

A

P Pac

R T N m K mol K

[ ]

. [ ] [ ]35 4 mol m . 2 0Ac

3 33 6 2 1

3 3 3

41 3 041 3 8 10

3 10 41 3 5 4A

mol m mol mmol m m sN Ln

m mol m mol m

. [ ] [ ]. [ ] [ ]

[ ] . [ ] . [ ]

2 2 11 55 10 mol m s .

2 2 11 55 10AN mol m s .

22

4A A A

dV h r h

219 6 m .2 25

4

mA

[ ]

9


Rate of mass transfer = Rate

Mass of benzene initially in the tank = mA

Time required for evaporation of benzene = tA

Question 5

A = NH3 ; B = Air ; Ideal Gas

Data provided:

P = 1 bar= 100000 Pa

T = 25 ° = 298 K

x2-x1 = 5 cm 1 [m]/102 [cm]) = 510-2 m

DAB = 23.6 10-6 m2s-1

cA1 = 5% v/v

cA2 = 2% v/v v/v = mol%

yA1 = 0.05 yA2 = 0.02

Ideal gas:

Dalton’s law: Pi=Pyi

PA1 = 5103 Pa PA2 = 2103 Pa

2 2 2 119 6 1 55 10ARate A N m mol m s . [ ] . [ ] 1 13 04 10 mol s . [ ]

2 224 3 3

295 295

588 10 5 10

4 4A A K A A K A

mdm V h g m m

, ,

[ ][ ] [ ]

1

86394

6 12 6 1A

A

A

m gn

M g mol

[ ]

[ ]1108 mol

86394 g

1 1

1107

3 04 10A

A

A

n molt

N mol s

[ ]

. [ ]3643 s

10


a) Equimolar counter-diffusion

b) Diffusion through stagnant air

4 2 15 72 10AN mol m s .

4 2 15 92 10AN mol m s .

1 2

2 1

ABA A A

DN P P

R T x x

( )

( )

6 2 13 3

1 1 2

23 6 105 10 2 10

8 314 298 5 10A

m sN Pa

N m K mol K m

. [ ]( ) [ ]

. [ ] [ ] [ ]

2

2 1 1

AB AA

A

P D P PN Ln

R T x x P P

( )

2

2 1 1

AB A

A

c D c cLn

x x c c

( )

3 6 2 1 3 3

1 1 2 3 3

100 10 23 6 10 100 10 2 10

8 314 298 5 10 100 10 5 10A

Pa m s PaN Ln

N m K mol K m Pa

[ ] . [ ] [ ]

. [ ] [ ] [ ] [ ]

11


Question 6

A = Chloroform ; B = Inert gas ; Ideal gas

2

2 1 1

AABA

A

P PD PN Ln

R T x x P P

( )

2

2 1 1

AAB

A

c cD cLn

x x c c

( )

2

1

AA ABA

A A

c cD cdhN LnM dt h c c

0 01

f fh t

A AB

h A A

M D c ch dh Ln dtc c

2 2

0

1

2 A ABf f

A A

M D c ch h Ln tc c

0 0 0

1

22 A AB

f f f

A A

M D c ch h h h h Ln tc c

( ) ( )

00

0

1 1

21

2 2

f

f

f A AB A AB

A A A A

t hh h

h h M D c M D cc cLn Lnc c c c

( )( )

y = a x + b

12


Data provided:

P = 1 bar= 100000 Pa

T = 20 ° = 293 K

MA = 119.39 gmol-1

ρA, 293 K = 1489 kgm-3 (103 [g]/1 [kg]) = 14.89105 gm-3

PA1 = 20.67 kPa (1000 Pa / 1 kPa) = 20.67103 Pa

5 2 11 01 10ABD m s .

2

1

1 648581342 A AB

A A

a s mM D c cLn

c c

[ ]

1 1

100000

8 314 293

Pan PcV R T N m K mol K

[ ]

. [ ] [ ]

341 05 mol m .

31

1 1 1

20 67 10

8 314 293A

A

P Pac

R T N m K mol K

. [ ]

. [ ] [ ]

38 49 mol m .

1 2 3 3

5 3 3

1

1 1

2 2 119 39 64858134 41 05 41 05

14 89 10 41 05 8 49

AB

A

A A

DM a c g mol s m mol m mol mc LnLn

g m mol mc c

. [ ] [ ] . [ ] . [ ]

. [ ] . . [ ]

13


Question 7:

a) Diffusion through stagnant air

A=CO2 ; B=Ethylene

Data provided:

P = 2 bar 100000 Pa / 1 bar) = 200 103 Pa

T = 40 °C = 40+273= 313 K

x2-x1 = 0.12 mm (1 [m]/103[mm]) = 1210-5 m

DAB = 7.04 10-6 m2s-1

Ideal gas:

yA01 = 0.20 yAi1 = 0.08


PA01= yA01 P = 0.20 200 103 [Pa] = 40 103 Pa

PAi1= yAi1 P = 0.08 200 103 [Pa] = 16 103 Pa

1 2 16 30 10AN mol m s .

1

2 1 01

AiABA

A

P PP DN Ln

R T x x P P

( )

3 6 2 1 3 3

1 1 5 3 3

200 10 7 04 10 200 10 16 10

8 314 313 12 10 200 10 40 10A

Pa m s PaN Ln

N m K mol K m Pa

[ ] . [ ] [ ]

. [ ] [ ] [ ] [ ]

14


Part b)

A=CO2 ; B=Ethylene

Film coefficient can be kG, hG or hD1, whereas overall mass transfer coefficients are always K.

A=CO2 ; B=Ethylene

CA=6.0810-7PA

Equilibrium relationships

PA01 CAe2

PAe1 CA02

PAi1 CAi2

Data provided:

CA02 = 4 10-3 kmolm-3

PA01 = 40 103 Nm-2

PAi1 = 16 103 Nm-2

5 1 12 63 10Gh mol N s .

01 1A G A AiN h P P ( )

01 1

AG

A Ai

Nh =

P P( )

1 2 1

3 3 2

6 3 10

40 10 16 10

mol m s=

Pa N m

. [ ]

( ) [ ]

15


CA02=6.0810-7PAe1

CAe2=6.0810-7PA01

5 1 11 89 10GK mol N s .

2 13 1 10LK m s .

021 76 08 10

AAe

CP

.

3 3

7

4 10

6 08 10

kmol m

[ ]

.

2

1 6579AeP N m

01 1A G A AeN K P P ( )

01 1

AG

A Ae

NK =

P P( )

1 2 1

3 3 2

6 3 10

40 10 6 58 10G

mol m sK =

Pa N m

. [ ]

( . ) [ ]

PAe1

CA02

PA01

CAe2

7 3 2

2 6 08 10 40 10AeC = N m . [ ]

3 3

2 24 32 10AeC kmol m .

2 02A L Ae AN K C C ( )

2 02

AL

Ae A

NK =

C C( )

1 2 1

3

6 3 10

24 32 4L

mol m sK =

mol m

. [ ]

( . ) [ ]

16


CAi2=6.0810-7PAi1

Question 8

A = Ammonia

yA01 = 0.021


PA01= yA01 P = 0.021 100 103 [Pa] = 2.1 103 Pa

CA02 = 1.2 %mass

a)

kG = hG = 2 10-8 kmolN-1s-1

1 11 1 10Lh m s .

7 3 2

2 6 08 10 16 10AiC = N m . [ ]

3 3

2 9 73 10AiC kmol m .

2 02A L Ai AN h C C ( )

2 02

AL

Ai A

Nh =

C C( )

1 2 1

3

6 3 10

9 73 4L

mol m sh =

mol m

. [ ]

( . ) [ ]

02

100A

A

A

mass

CM

% 3

1

1 2 997100

17

kg m

kg kmol

. [ ]

[ ]

30 704 kmol m .

17


We are told that 80% or the overall resistance (1/KG) lies in the gas film:

Part b)

CA [kmolm-3]=0.0006PA [Pa]

PAe1 CA02

CA02=610-4PAe1

80G

G

1

h100

1

K

5 1 11 6 10GK mol N s .

80

100G G

1 1

h K

80

100G GK h 5 1 180 2 10

100mol N s ( ) [ ]

021 46 10

AAe

CP

3

4

0 704

6 10

kmol m

. [ ]

18


Part c)

CA [kmolm-3]=0.0006PA [Pa]

PA01 CAe2 CAe2=610-4PA01

If 80% of resistance if in gas phase, then 20% must be in liquid phase:

2 2 11 44 10AN mol m s .

5 12 6 10LK m s .

2

1 1173AeP N m

01 1A G A AeN K P P ( ) 5 1 1 3 31 6 10 2.1 10 1 2 10mol N s Pa . [ ] ( . ) [ ]

4 3 2

2 6 10 2 1 10AeC = N m . [ ]

3

2 1 26AeC kmol m .

2 02A L Ae AN K C C ( )

2 02

AL

Ae A

NK =

C C( )2 2 1

3 3

1 44 10

1 26 10 704L

mol m sK =

mol m

. [ ]

( . ) [ ]

1 1 1

L G L

=K H h h

20L

L

1

h100

1

K

20

100L L

1 1

h K

100

20L Lh K 5 1100 2 6 10

20m s ( . ) [ ]

2 02A L Ai AN h C C ( )

2 02A

Ai A

L

NC C

h

19


Part d)

3

1 1 38 10AiP Pa .

01 1A G A AiN h P P ( )

01 1A

A Ai

G

NP P

h

1 01A

Ai A

G

NP P

h

2 2 13 2

1 5 1 1

1 44 102 1 10

2 10Ai

mol m sP Pa N m

mol N s

. [ ]. [ ]

[ ]

3

2 815AiC mol m

2 02A

Ai A

L

NC C

h

2 2 13

2 4 1

1 44 10704

1 3 10Ai

mol m sC mol m

m s

. [ ][ ]

. [ ]

20


Question 9

Diffusion through stagnant air

A=Oxygen ; G-L

Data provided:

P = 2 bar (100000 Pa / 1 bar) = 200 103 Pa

Diameter O2 bubble (d) = 2 mm 1 [m]/103[mm]) = 210-3 m

Ideal gas:

yA01 = 0.20


PA01= yA01 P = 0.20 200 103 [Pa] = 40 kPa (kNm-2)

KL = 810-3 ms-1

MA = 32 kgkmol-1

90% diffusional resistance in the L phase

PG=2.34CL

PG=kNm-2

CL=mgl-1

Part a) NA

PA01 CAe2

2 02A L Ae AN K C C ( )

2 34G LP C .01 22 34A AeP C . 01

22 34

AAe

PC

.

-240[kN m

2 34

]

.1

2 17 1AeC mg l .

3 3

2 3 3 3

10 1 117 1

1 10 32Ae

dm g molmgC

l dm m mg g

. 30 53 mol m .

3 3

02 3 3 3

10 1 18

1 10 32A

dm g molmgC

l dm m mg g

30 25 mol m .

3 1 38 10 0 53 0 25AN m s mol m [ ] ( . . )[ ]3 2 12 28 10AN mol m s .

21


Part b) KG

PAe1 CA02

Part c) kG and kL

Total resistance = Gas phase resistance + Liquid phase resistance

7 1 11 04 10GK mol N s .

6 1 11 04 10Gh mol N s .

3 18 88 10Lh m s .

01 1A G A AeN K P P ( )01 1

AG

A Ae

NK =

P P

( )

2 34G LP C . 1 022 34Ae AP C . -12 34 8[mg l . ] 2

1 18 72AeP kN m . 3 218 72 10 N m .

3 2 1

3 2 1 2

2 28 10

40 18 72 10G

mol m sK =

Pa N m kg m s

. [ ]

( . ) [ ]

10G

G

1

h100

1

K

10

100G

G

K

h

100

10G Gh K

7 1 1100 1 04 1010

Gh mol N s . [ ]

100 90L

L

1

h

1

K

90

100L

L

K

h

100

90L Lh K

3 1100 8 1090

Lh m s [ ]

22


Part d) PAi1 and CAi2

Overall transfer rate oxygen bubble→liquid (R)

3 2

1 37 81 10AiP N m .

3

2 0 79AiC mol m .

8 12 87 10R mol s .

01 1A G A AiN h P P ( )01 1

AA Ai

G

NP P =

h

1 01A

Ai A

G

NP P

h

3 2 13 2

1 6 1 1

2 28 1040 10

1 04 10Ai

mol m sP Pa N m

mol N s

. [ ][ ]

. [ ]

2 02A L Ai AN h C C ( )2 02

AAi A

L

NC C =

h

2 02A

Ai A

L

NC C

h

3 2 13

2 3 1

2 28 100 53

8 88 10Ai

mol m sC mol m

m s

. [ ]. [ ]

. [ ]

AR= N A

A Area for mass transfer buble sphere ( )

232 104

2

m

[ ]

2

42

d 24 r

5 21 26 10 m .

3 2 1 5 22 28 10 1 26 10R mol m s m . [ ] . [ ]

5 21 26 10 m .

23


Question 10

A=Propanol ; (Aqueous (1) to Organic (n-decanol) (2)) L-L

Data provided:

T = 20 °C = 20+273= 293 K

Internal Diameter Cell (d) = 12 cm (1 [m]/102[cm]) = 1210-2 m

Q(I) = Q(II) = 3 mlmin-1 (cm3min-1) (1 [m3]/106[cm3]) (1 [min]/60[s]) = 510-8 m3

s-1

Q(III) = Q(IV) = 3 mlmin-1 (cm3min-1) (1 [m3]/106[cm3]) (1 [min]/60[s]) = 510-8 m3

s-1

CA1= 1.5CA2

CA1= mass % concentration

CA2= mass % concentration

a) Overall propanol transfer rate

Mass Flows

Propanol (A) Mass Balance

3998III IV kg m ( ) ( )

3830I II kg m ( ) ( )

160 09AM kg kmol .

1 20oI II I Cm m Q ( ) ( ) ( ) ,

8 3 1 35 10 830m s kg m [ ] [ ] 5 14 2 10 kg s .

5 15 10 kg s 8 3 1 35 10 998m s kg m [ ] [ ]2 20oIII IV III Cm m Q ( ) ( ) ( ) ,

mass In mass Out

A I A III A II A IVm m m m ( ) ( ) ( ) ( )A I I A III III A II A IV IVw m w m m w m ( ) ( ) ( ) ( ) ( ) ( ) ( )

5 1 5 12 5 2 10 5 10 5 10100 100

A IIkg s m kg s ( ). .[ ] [ ]

7 12 10A IIm kg s ( )

24


CA01 and CA02 are always the leaving streams

Overall mass transfer = Propanol lost in aqueous phase Qaqueous

Part b) CA(II)

From previous section (see Table)

c) K1 and K2

CAe1 CA02

CA1= 1.5CA2

CAe1= 1.5CA02

CAe1= 1.50.48 = 0.72 (mass %)

30 07A IIC kmol m ( ) .

A III A IV IIIR C C Q ( ) ( ) ( )( ) 3 8 3 10 42 0 35 5 10kmol m m s ( . . )[ ] [ ]

6 13 5 10R mol s .

AR= N AA

RN =A

A Area for mass transfer cylinder circle ( ) 2

2

d 2212 10

2

m

[ ] 2 21 13 10 m .

4 2 13 09 10= mol m s .6 1

2 2

3 5 10

1 13 10A

mol sN =

m

. [ ]

. [ ]

2r

1

01 1

A

A Ae

NK =

C C

1

1

Ae III

Ae

A

wC =

M

( ) 3

1

0 72 998100

60 09

kg m

=kg kmol

. [ ]

. [ ]

30.12= kmol m

4 2 1

1 3 3 3

3 09 10

0 35 10 0 12 10

mol m sK =

mol m

. [ ]

( . . ) [ ]

1 01 1A A AeN K C C ( )

25


CA01 CAe2

CA1= 1.5CA2

CA01= 1.5CAe2

CAe2=2.1/ 1.5 = 1.4 (mass %)

6 1

1 1 34 10K m s .

6 1

2 2 58 10K m s .

2

2 02

A

Ae A

NK =

C C

2 2 02A Ae AN K C C ( )

2

2

Ae I

Ae

A

wC =

M

( ) 3

1

1 4 830100

60 09

kg m

=kg kmol

. [ ]

. [ ]

30.19= kmol m

4 2 1

2 3 3 3

3 09 10

0 19 10 0 07 10

mol m sK =

mol m

. [ ]

( . . ) [ ]

tutorial solutions topic 5

Documents

atm atm

pa pa2

v1 question

papp aa

v1 papa32

air ideal gas stagnant

ideal gas mixture

aaabapppplnxxtrdpn data