304_pdfsam_(3rd Edition) Phillip C. Wankat-Instructor's Solution Manual - Separation Process Engineering_ Includes Mass Transfer Analysis-Prentice Hall (2011)

SPE 3rd Edition Solution Manual Chapter 13 New Problems and new solutions are listed as new immediately after the solution number. These new problems are: 13.A12, 13.A13, 13.D3, 13.D5, 13D6, 13D10, 13D22, 13D30-13D34, 13D36-13D42, 13.E2, 13.E3, 13.G1, 13.G2 . Chapters 13 and 14 from the 2nd edition were rearranged to place all the extraction material into chapter 13 and the material for other separations in Chapter 14. Thus, the numbers of many problems have changed. 13.A3.

The amount of solvent should be increased. This will decrease F/S and move the mixing point M towards S. As a result the saturated extract product E N will be moved down (less solute). The difference point ∆ will be moved towards the triangular diagram. The combined effect will be that fewer stages are required. By adjusting F/S a condition requiring exactly two stages can be found.

13.A5.

The vertical axis will be the extract phase and the hypotenuse will be the raffinate phase. These will be connected by tie lines. Usual procedure can be used.

13.A7.

Situation where E = R and ∆ point is at infinity. All operating lines are parallel. However, this does not correspond to minimum number of stages in extraction.

13.A.11 c. 13.A12. a. C will be spread out and go into both raffinate and extract streams. b. C will concentrate around the feed edge. If C is very dilute in the feed, can concentrate C. Then by stopping the feed but continuing to flow solvents, solutes A and B can be removed. Solute C can now be collected by withdrawing a stream near the feed stage. 13.B1.

Specify:

T, p, z A , z B , F, x Ao , x B , y A N , y BN plus:

y B1 , R, E, N F x B N , R, E, N F x A N , R, E, N F R, E, y B1 , x A N N, N F , y A1 , E N, N F , x B N , R N, N F , x A N , R, etc. Could also not be given one of standard variables (such as solvent concentration). 13.B2.

a). One can build stages which are cross-flow (e.g. see Figure 12-12) within a countercurrent cascade. This effectively increases stage efficiency. Not that upward flowing less dense liquid will be mixed. b.) Build chambered stages within a counter-current cascade to prevent mixing of the dense liquid and give better cross flow on each stage. c.) Put in baffles to prevent MIXING of both less and more dense liquids. This will be more effective if counter-current is arranged so that flow across stages is always in same direction (see sketch)

319

13.C.7. Start by defining ∆ and the coordinates of ∆ as: E o R1 , x A E o y Ao R1x A1 , x D E o y Do nd rd Removing ∆ from 2 and 3 equations we obtain

Assume that E o

Rj

xA

E o y Ao

R 1 x A1

Eo

R1

(13-43a)

xD

E o y Do

R 1 x D1

Eo

R1

(13-43b)

R 1 . Next write the three independent mass balances around stages 1 to j.

Ej ,

1

R1x D1

xA

R j 1x A j 1

E j y Aj ,

xD

R j 1x D j 1

E j y Dj

These equations are now in a form similar to the form of the mixing equations developed previously. To develop the three point form of a straight line use the first equation to remove ∆ from the other two equations, solve for R j 1 E j in each of these equations, and finally set the results equal to each other. The development proceeds as follows: Use

Ej

R j 1 to remove ∆ from the other mass balances.

Ej

Rj

1

xA

Solve for R j

1

R j 1x A j 1 Ej ,

Rj Ej

1

E j yA j , E j yAj

xA

xAj 1

xA

,

Rj Rj Ej

1

1

xD

R j 1x Dj 1

yDj

xD

xDj 1

xD

Finally, set these equations equal to each other. yA j x A yDj x D yA j x A rearrange to: xAj 1 xA xDj 1 xD xDj xD

yAj 1

xA

x Dj 1

xD

E j y Dj

This last equation says that the slope of the line between the points

xD , xA

y D j , y A j and

is equal to the slope of the line between the points x D j 1 , x A j 1 and x D , x A

and

thus the lines are colinear. Furthermore, the lever-arm rule is valid for this system. 13.D1.

a. If we have a single column with only pure solvent then

320

R

y

x

E 44, R

is on op. line. E

R

1

2.273 0.0037 . Thus, cannot get to x o

Op line intersect equilibrium at x b. Now E

and point x N , y N

x N yN 1 E 100 , Slope R E

0.001, 0.0 0.012 .

74 and R E 1.35 .

44 30

44 0

0.001, but y N

30 0.004

0.00162 74 At x 0.001, equilibrium value of y 1.613 x 0.001613 . Alternative works, but have pinch point and need very large number of stages.

Still want x N

1

c. This alternative (Say use 25 kg/min of 0.004 butanol) 44 0 25 .004 yN 1 0.00145 which is below equilibrium point. 69 Now R E 100 69 1.449 m equil 1.613 Thus, this will work. Obtain

0.00145 69

0.012 100

0.001 100

0.0174 69 Op line closer to equilibrium – require lot more stages. 20 0.004 If use 20 kg/min of 0.004 butanol: y N 1 0.00125 64 0.00125 64 0.011 100 R E 100 64 1.5625 , y1 0.01844 64 Will also work. Becoming close to pinch at top equil y1* 0.019356 y1

If 15 kg/min, y N

15 .004 1

0.0010, R E

59 1.0010 59

y1

0.011 100

100 59 1.6949

0.1964

59

y1*

m eq

0.19356

Won’t work. Thus, there is a small range where option c will work, but with many stages. 13.D2.

R

20, E

20, x IN

Kremser equation

Eq (13-11)

x F , y IN

0, m

R

20

mE

8.333 20

y1 yN

y1 8.3333 x F

y 1

1

* 1

y1*

1

8.3333 x F

8.3333, b

0.12 , y1*

R mE N 1 R mE 1-0.12 1- 0.12

3

0, N

mx 0

b

2 8.3333 x F , y N+1

7.3460 x F , → y1

0

0.9873 x F

321

Rx F

Mass balance,

Recovery = 1 x N x F

Rx N

Ey1 , x N

0.01269 x F 20 0.9873 , which is higher than 0.963 obtained in cross-flow.

13.D3. New Problem in 3rd edition. From M.B.

R .013 Where R = 100 and the unknowns are E and yout.

y out

and Equilibrium:

S

1.613 x out E

E .001

1.613 .007

R .013 .007 y out

20 x F 19.746 x F

R .007

0.01129

100 0.006

0.001

E y out

0.01129 0.001

58.309 kg h

Alternative Solution: 1 Equilibrium Stage

y 1.613x

1.613 y

y out

0.01129 R

from graph

E

E

.001

x out

.007

x

R

0.001 0.01129

E

0.013 0.007 R 1.715

1.715

58.309

.013

Another alternative solution:

322

E, y1

x

y

x0

y2

y1

1

Or

Op. Line:

Eq.

R, x 0

R E

Slope

y1

Points x1 , y 2 ,

R E

x0

x 0 , y1

But x1 and y1 unknown

x where y 2

E

x1

On Op. Line

R, x1

E, y 2

R

y2

y1,in

y1,IN

x1

T & E in this configuration

R Slope known, N = 1 If Eq. line is straight, can Use Kremser with N=1. E Both representations are correct. Treating similar to a flash is easier. 13.D4.

Since concentrations are low, use wt. fractions and total flow rates. Equilibrium: y 0.828 x or m 0.828

R

550 lb h, E

mE R

700 lb h, x 0

1.0538 and R mE * 1

y

b n

0.828 .0097 mE 1 R

yN

R 1

E

R

x0

E

xN n

N

.0003

550

0

1

0.0003,

.00046

0.0075

0.00803

y N 1 y1* y1 y1* n

y1

0.00046, y N

.94893

mx IN

Kremser Eq. (13-11b), N

0.0097, x N

R mE

0.0097

700 .0077316 .0538 .0004716

mE R

550 700

1.5038 33.6

n .94893

13.D5. New problem in 3rd edition. Part a) Can do with Kremser eq or graphically. y m x b, m 0.828, b 0 R 400 R 400, E 560, mE 0.828 560

x0

0.005, x N

0.0003, y N+1

0.0001,

0.862664 mE

1

R

.862664

1.159

323

R

Since

1.0 , can use equation such as 13-11b mE y1* m x 0 b 0.828 0.005 0.00414

n

y N 1 y1* y1 y1*

mE R

1

N

n

n 1 1.1592 N where

y1

yN

R

x0

1

E 1.52637

N

mE R

R mE

0.0001 0.00414 0.0034571 0.00414 xN

1.1592

n0.862664 400 0.0001 0.005 0.0003 560

0.003457

10.332 0.14773 Alternate solution: Eq. (12-28) becomes L

R, V

yN

x *N

1

E , N

b

0.0001

m

b)

n

1

x *N x *N

x0 xN

R mE

0.000120773 , N

0.828

yN

R E

0.0001

1

xN

y1 Part b.

R

x0

E 100

yN

R 1

0.005

E

140 Kremser Eq.

slope x0

0.0003

13.D6. New problem in 3rd edition. Part a. Ey N 1 Rx 0 Ey1 Rx N

y1

min

E

0.005

10.332

.14773

0.828 x 0 y1*

R

Slope = 0.828

n mE R

1.52637

y1*

Equil. y

R mE

min

yN

x0

E MIN

xN

1

0.828 0.005

0.00414

0.00414 0.0001 0.005 0.0003

R 0.85957

400 0.85957

0.85957

465.3 kg h

Ext. M.B.

xN

0.0002

100 140

0.0005

0.003414

324

Convert

x

L

R

x

L

R

y

y

yN

E, x *N

V

1

m

100

0.0002 1.208

0.00016556

0.5913 mV mE 1.208 140 Lots of different forms can be used. n N

For example

n

1

N

0.4087

N Part c.

x *N x *N

L mV

n mV L n

Becomes

x0 x0

L 1 mV

x0 xN

R mE

x *N x *N

R mE

n mE R

0.005 0.00016556 0.0005 0.00016556 1 n 0.5913

0.5913

1.8717 0.5254 Eq.

y EQ

3.6

y 1.208x

1.208x

0.00604

y

R E MIN

y N 1 0.0002 y 1.208x xN R

0.0005

x

slope

x0

0.005

0.00604 0.0002

1.29777 0.005 0.0005 R 100 E MIN 77.05 1.29777 1.29777 Maximum extract out y EQ x 0 0.00604. Part d. The roles of extract and diluents are switched in the two problems, which changes the definitions of y and x. E MIN

13.D7. Equilibrium:

N

30, R

500, y N

y

0.828x, m

1

0.0002, x 0

0.828, x *N

yN

1

0.0111, x N .828

0.00037

0.00024155

325

Since rather dilute and linear equilibrium use one of the Kremser equations.

n N

Where

E 500

x *N x *N

R ME

(12-28 (modified))

mE n R

x0

x *N

xN

x *N

R/mE 1.21

83.756 . Solution is trial-and-error.

Calculated N Negative-Not possible Need

0.8626 .929 .9435 .945015

700 650 640 639

xo xN

R mE

1

R mE

1

17.01 E too high 26.175 E too high 29.84 E too high 30.30 E too low

By linear interpolation need E ~ 639.6 kg/h. Can use other forms of the Kremser equation. Was 13.D10 in 2nd edition. x is raffinate R L Convert Kremser y y extract, V E 13.D8.

a)

Use 12-31

xN

x *N

Other forms OK x 0

* N

x *N

xN

1

x0 1

b)

x

yn

KE R KE R

m

1 mV L mV L

1

K

1

K, b

1 KE R

N 1

1

R E

x0

3

30.488 25 100 30.488 25

1

R x0 xN

0.00001376

3

100

Can use External balance or Kremser to find y out

y1

KE R

1

1

Ey N

0

0

0.00092

3

Ey1

13.D9.

mx

100 25

y1

xN 0.00092 0.00001379

0.003625

Assume very dilute, R = 1500 kg/h, E = 750 kg/h Equil. Y K d X becomes y K d x From Table 13-3. K d,oleic 99% recovery oleic:

m oleic

4.14, K d,linoleic

.99 .0025 1500

md,linoleic

2.17

y1,oleic 750 → y 1,oleic

.00495

326

Use Kremser, Eq. (13-11b).

y1*

m oleic E

4.14 750

R

1500 4.14 .0025

m oleic x 0,oleic n

N

For linoleic acid:

yN

1

0, N

y N 1 y1* y1 y1*

mE R

1

1500

m lin E

750 2.17

Can use Eq. (13-11a):

y1 .00651 Recovery of linoleic:

m L x 0,L

yN

y1*

1

mE R

5.44

.9216 ,

2.17 003 R mE N R 1 mE .07834

.00651

1

y1*

y1

0.01035

R n mE

R

5.44, y1*

2.07

.00651

.40866 Re c .003 1500

1

.00124796 → y1linoleic

0.00526

.00526 750 → Rec = 0.877

th

13.D10. New problem in 4 edition. Analytical or graphical solution OK. Stage 1 F1x F1 E i Equilibrium

x 2,out

1.02 x1

Fx F1

1.02 E R1 x1

Mix with Feed 2

1.02 E

R1

1.02 50

R 1 x1

E2

y 2,in

E 2 y 2,out

y 2,out

1.02 x 2,out

0.0099338 100 R2

R 3,in

R 1 x1

100 0.015

F xF

R 1 x1 1.02 E 2

Ey1,out

y1,out

x1

Stage 2

y1,in

151

100

0.0099338

R 2 x 2,out , R1

0.006579, R 2

R2

R1

F1

F1

100

100

100 70 170

327

x 3,in x 3,in

Stage 3

x 3,out

x 4,out y 4,out

0.006579 100

0.005 70

170

0.0059286

E3

y3,in

E 3 y3,out

R 3 x 3,out

1.02 x 3,out

R 3 x 3,in 1.02 E 3

R 4 x 4,out y 4,out

x F2 F2

R 3,in

R 3 x 3,in y3,out

Stage 4

x 2,out R 2

E4

170 0.0059286 R3

y 4,in

1.02 50

E 4 y 4,out

170

0.00456

R 4 x 4,out , R 4

R3

170

1.02 x 4,out

R 4 x 4,out 1.02 E 4 1.02 x 4out

170 0.00456

R4

51 170

0.003508

0.003578

328

329

13.D11.

R

R

F

2501, E

Equilibrium: K D

E 1000

1.57 . For dilute this becomes m

xN R

Abietic Acid Recovery:

xN

.0475

.0475

R

2501

Top op. Eq.:

y

Goes through pt x 0 Bottom Op. Eq.: y

0.0000190 , y1

R E

x

R

x

yN

R 1

y1

E

x3

1

0.17594 Rx in

.0475

.05 F x F

.05 1.0 .05

E

1000

R

2500

E

1000

0.0000025

y 4 0, y*1

1

1.2399

0.00742

Ey in

2.5

x N through point x N , y N

1 1.2399

0 0.00742

y1 Overall bal.

0.00742

.95 1.0 0.5

x0

E

m 1.613, R mE 1.2399 , y N Eq. (13-11a)

13.D13. a.

R

y1

0, y1 . Slope

E Need 8 ½ stages (see Figure).

13.D12.

.95 F x F

K D in wt. frac. units.

Ey out

mxin

1

0 , R E

0.00742

0.17594

4

0.00742 10 0.0046

R

0.006114 5 0.006114 10

0.1 1000 0.003

xy . 90% recovery, 10% left

x

,out

2.501

0.00154

0.3 kg out

0.0003

330

O xy

For ortho, y max

95% recovery, 5% left

0.15 0.005

0.25 , x O,out

0.00025

0.00075 R E

For para

0.05 1000 0.005

0.00075 0 max,ortho

y max

0.08 0.003

0.00024

R

0.00024 0

E

0.003 0.0003

max,para

b. The p-xylene recovery controls.

E 1.5 11250 16875 ,

0.1579

0.005 0.00025

1000

E min

0.08888

R

0.08888 11, 250

0.0592592 E Can use Kremser eq. (13-11b) for ρ-xy to find N

n N

mE 1 R

y N 1 y1* y1 y1*

mE R

R mE m 0.080, R E 0.0592592, y N 1 0 , y1* mx o,p 0.080 0.003 0.00024 Mass balance: 90% entering ρ-xy leaves w. solvent. 0.9 1000 0.003 y1 0.00016 wt frac 16,875 R 0.0592592 R mE .080 1.35 0.74074, n 0.300106 , R 0.0592592 mE 0.080 mE n

331

n

0 0.00024 0.00016 0.00024

.35

N

1.35

n 0.30

4.012 0.300106 0.30016 Note: Can use other forms of Kremser eq if desired. c. For o-xy check if recovery > 95% R 1 * y1 unknown, y N 1 0 y1 y1 mE Eq. (13-11a) N 1 y1* mx 0 0.15 0.005 0.00075 y N 1 y1* R 1 mE R 0.0592592 0.39506, N 4.012 mE 0.15 y1

yN

1

1

y1* 1

R mE N R mE

External M.B.

y1*

1

Ey1

R xN

Rx 0

xN

0.39506

5.012

0.00075

0.0029194

R x0

5

Ey1 R

a)

1

Ey N 1

% Recovery 13.D14. (was 14.D4. in 2nd ed.)

1 0.39506

0.00075

Ey1 Rx 0

16875 .00029194 1000 100

S

10.0

2

MF

F

15.0

3

SM

7.3584 E 5

98.53%

Once have M, use trial-and-error to find tie through M. (final result is shown). This gives E and R. y A .115, yw 0.04, xA .23, xw .73. b) Plot raffinate, R x A

.1 . Find tie line through this point (not trial-and-error). This gives E. Draw Line ER. Intersection with line SF gives M. S S MF . Find S 85.7 kg/h. F 15.0 SM

332

13.D15.

Since dilute, use Kremser equations. Assume units are weight fractions. a) Column 1 at 40ºC. x N 0.0008, N 11,, x 0 0.01, E 1000,, R 100 Equilibrium: m

0.1022, thus y1*

mx 0

0.001022. Kremser (Eq. 13-11a):

1 1.022 0.93664 12 y N 1 0.001022 1 1 1.022 This simplifies to: y1 .093664y N 1 .00092628 y1

1

0.001022

External MB: y N 1E Rx o which simplifies to:

yN

y1E Rx N , y N 1

1

1000 Solve 2 eqs and 2 unknowns: y1,coll b) Column 2 at 25ºC: y N

y1,col2

yN

1,col1

1,col2

y1,col1

.6929 10 5 , x 0

1

1000

1 1000 y1

.08

1000 y1 .92

0.00092693, y N+1,coll

0.6929 10

5

0.00092693 ,

0, N

9, m

0.0328, E 1000, y1*

mx 0

0

Use Kremser to solve for R´. This is trial and error. For example, Using Eq. (13-11a), R R 1 1 * 0.0328 1000 y1 y1 mE 0.007475 N 1 10 y N 1 y1* R R 1 1 mE 32.8

R

50

60

50.5

50.35

RHS 0.007855 0.001981 0.007307 0.007467 Within error R´ = 50.35 y N 1E R x 0 y1E .92693 0 .006929 xN R 50.35

0.0183

333

c) Could be practical if m’s were larger, and have bigger shift in m. A similar scheme is used commercially for citric acid. Not practical here since have to pump around too much solvent. In addition, benzene is carcinogenic and would probably not be used as solvent.

R E 10 8 1.25, R mE

13.D16. a.)

* 1

y y1

m x A0

1.613 0.01

0.01613

0.0002 0.01613 xA

1.25 1.613 0.77495 0.01613. Use Eq. (13-11a),

1 0.77495 1 x A0

0.77495 E

yN

E 1

R R b.) Graphical check works fine (not shown)

yj

13.D17.

x6 Note:

x6

R Ej

xj

y IN

0.27044 → y1

7

R E0

y1

x j 1,

7.02498 E

R

10

Ej

2

0.01182 4

5

0.0018 (See graph)

x N,countercurrent

0.000702 even though use more total solvent.

334

13.D18. (was 14.D2. in 2nd ed.) Lever arm rule:

Plot S, F, R and E. Draw lines SF and RE. Intersection is point M.

S

MF

20.3

F

SM

4.5

Or Mass Bal. S + R = M and S y A Solve simultaneously

4.511 → S

F xA

100 4.511

M x A ( S .15

451.1 kg/h

.5 F .21 M )

S = 483.3

335

Difference is due to accuracy in reading numbers. Lever-Arm Rule more accurate!

13.D19.

Equil.

Kd

Acetone

y0 xN

FD

1000 .9

yA x A 0 1

0, x 1

0.10 wt frac

900 kg/h water, FS FD FS

Equil.

Y0

0.287 0.158 1.816

900 1364.1

0.005 X N+1

X1 0.10

.9 1371 .995

0.005 0.995

0.00503

0.1111

1364.1 kg/h chloroform.

0.6598

336

XA

xA 0

yA = 1.86 xA

0

0

YA 0

0.01 0.03

0.0101 0.0309

0.01816 0.05448

0.01850 0.0576

0.05

0.0526

0.0908

0.09987

0.07

0.0753

0.1271

0.1456

0.09

0.9890

0.1634

0.1954

0.1

0.1111

0.1816

0.2219

External M.B.

FD FS

XN

1

Y6

FD FS

X1

YN or YN

0.6598 0.1111

0.06999, y N Results pretty close to 13.D43. 2

1 2

vs 2

2 3

0.6598 .00503

YN 1 YN

0.0655

w i accuracy of graphs.

Note: The graph below should read acetone, not acetic acid as the solute.

337

13.D20. a) Batch Operation – Mix together & settle. Find fraction recovered: R R Operating Eq.: y x x 0 , R 5, S 4, x 0 x F S S Which is, y 1.25 x 1.25 x F Equilibrium

y

8.333 x, m

Eq. (13-21) written for batch

8.333

x

Rˆ Sˆ x 0 m

Frac. Rec 1 0.1304 0.8696 b) Continuous solvent addition: Sˆ 1 n x t ,final x t ,feed Eq. (13-28) Rˆ t m

x t,final x F Recovery = 99.87%.

exp

0.8 8.33

y iN

1.25 x F

Rˆ Sˆ

0.8

0

9.583

1 8.333

n

0.1304 x F

x t ,final xF

0.00127

338

13.D21 (Was 14.D1 in 2nd ed.) a. Let A = methylcyclohexane and D = n-heptane. Mass Balances: F1 F2 S M or M 350

F1 x AF

1

Then

F2 x AF

S y AS

2

x AM

1

F1 x A F

F2 x A F

F1 x D F

M F2 x D F

1

x DM

M x AM , F1 x DF

1

2

2

F2 x DF

S y DS

2

M x DM

S y AS

100 .6

50 .2

0

S y DS

350 100 4 50 .8

0

M

350

0.2 0.229

Plot M. Find tie line through M. (See figure.) This gives location of points E and R. Find x DR 0.48, x AR 0.42, y AE 0.06, y DE 0.05 .

b.

Mass balances: M

E R and Mx AM

Ey AE

Rx AR

Solving simultaneously: E = 214 and R = 136 kg/h 13.D.22. New problem in 3rd edition. 1 Af D s2 4 0.411 and Pperf 2

With interface at center, heavy phase flow area is 1 D s D 5 2.630 2

r

θ

Chord

.1

.4115

Ds 2

0.5115

Center

Interface

α

(length = C) arc

θ

r

.1 α C/2

r

C

2

2

.1

C 2

1.00326 m 339

2

Draw right triangle for interface below center to calculate new perimeter. 0.1 .1 sin .1955 11.274 r .5115 Then angle of arc, 180 2 157.452 3.14159 0.5115 157.452 r Length of arc 1.4056 180 180 Perf C arc length 2.4089m Mensuration formulas are from CRC Standard Mathematical Table. Re settler

4Q

c

Perf

4 0.006 998

c

2.4089 0.95 10

3

10, 466

Interference somewhat more likely than in Example 13-5. 13.D23 (was 14.D7. in 2nd ed.) Pyrdine F x AF Plot M on line FS . y p 0.223,

a) F S 500 300 M S y AO 500 .3 0 M x AM → x AM

150 800

0.1875

By T & E find tie line through M (Use Conjugate line) x p 0.84 ; y w 0.02, x w 0.84 ;

Mass balances: R1

E1

800 , 0.84R

M

0.02E

0.43M

Solve simultaneously, E1 ~ 400, R1 ~ 400 (Note: More accurate than pyrdine values.)

R 1 S2

b)

R1x A1

S2 y A0 60

x pyr M 2

700

Find tie line by T & E: y pyr2 MB:

R 2xw2

E2 yw 2 R2

Solve simultaneously: E 2

400 300

700

400 0.15

0.053 ; y w 2

M x m2w → 0.945 R 2 M

60

M 2 x AM 2

0.086

0.120; x pyr2

E2

M2

0.005, x w 2

0.005 E 2

0.945

700 0.48

700

346 and R 2

354

340

341

13.D24 (was 14.D10. in 2nd ed.) a) Feed 40% MCH 55% n-heptane, F = 200. Solvent 95% aniline & 5% n-heptane, Stotal 600 . S F M 800

S

Lever arm rule:

3

F

FM MS

. Find M (Easy way is divide line FS into 4 parts)

Use tie line through M to find points E & R (T & E) Extract: y MCH ~ 0.045, Raffinate:x MCH ~ 0.36 wt fracs Mass balance E + R = 800 = M and lever arm rule Solve simultaneously: b)

gives points R1 and E1.

Find:

F S R1

MR

R

ME

. Measure distances on figure.

R = 124.61 kg/h, E = 800 – R = 675.39

2 stage cross flow. Stage 1: F = 200, ρ = 300,

Mass balance 500

E

M

R1

S

3

FM

F

2

MS

. Find point M. Tie line through M

E1 and lever arm rule

207.04 kg h , E1

R1

E 1M 1

E1

M 1R 1

292.95

Note: Isotherms are very sensitive. Thus, calculation is not extremely accurate. Stage 2: Mass balance R 1 S2

M2

507.04

R2

E 2 and lever arm

S2

M 2R1

E1

M 2S 2

Find M 2 and from tie line through M 2 find R 2 . Then can find R2 and E2 from mass balance (given above) and new application of lever arm rule, Solving simultaneously, R 2

R2

E 2M

E2

R 2M

196.16 kg h. E 2

310.88

342

13.D25 (was 14.D9. in 2nd ed.) a. Draw lines from S to F and from R 1 to E N . Intersection gives point M (see Figure). Then from lever-arm,

b.

S

FM

F

SM

1.25 → S

∆ is at intersection of lines E N R N

1

1.25 2000

2500

and E 0 R1 . Then step off stages as shown. Need 2 stages.

343

13.D26. (was 14.D6. in 2nd ed.) Guess a value for M and step off stages. Repeat until need 3 stages. After three trials found M shown in Figure. This required 3 1/10 stages which is close enough. Extract Composition: Acetic Acid = 10.5%, Water = 3.5%. Raffinate Composition: Acetic Acid = 5%, Water = 93% Solvent Flow Rate: F S F Raffinate Flow Rate:

R1 E 0 EN

Extract Flow Rate:

SM SF 15 57

F S R1

13.D27 (was 14.D12. in 2nd ed.)

y AE

E0

MF

F

E0M

Lever arm rule:

1.112

y wE0

0

R1

M

R1

R 1 5600, R 1

2000 5600 772

6830

0 (Pure solvent)

. Step off stages

211.2 kg/h & lever arm:

Solve simultaneously, R1

770 kg/h.

. Find M. Line RM intersects sat’d extract at E N , y A N

Lines F E N & R1E 0 intersect at M.B. E N

E0

2000 S 2000 → S = 5600 kg/h

64.25, E N

0.18

3 more than enough. Need ~ 2 ¼

EN

MR 1

R1

ENM

2.287 (from graph).

146.95 kg/h

344

345

13.D28 (was 14.D14. in 2nd ed.)

To find ∆: 1) Plot E N and R N

2)

Ej

Rj

EN

1

E N x AN

xA

RN

F

1

1500

1

R N 1x A N 1

0.06666

3) ∆ is on line through points E N and R N 1 . Plot ∆. Or, use lever-rule.

RN

1

R N 1E N EN

1.5

Step off three stages starting at point E N . This gives points R 1 x A1

Mass Balance: E 0

0.275, x D1

RN

E 0 0.13

and

R1

1

E N → E0

1000 0.4

Solving simultaneously, R 1 13.D29 (was 14.D16. in 2nd ed.)

0.675 and E 0 y A1

R1

EN

R 1 0.275

655 kg/h, E 0

.13, and y D 0

RN

1

0.0 .

R 1 1500

2500 0.2

2155 kg/h

a) Plot Points F, S, E N and R 1

Find ∆ point at intersection of lines FE N and R 1S 2 stages is more than enough. (see graph) b)

Draw lines FS and E N NOT calc. value E 2 R1 . Intersection is mixing point M

F

dist. S to M

S F 0.786 1000 0.786 1272 kg/h.

dist F to M

Mass balance Give S

F + S = M and Lever arm

0.786

346

Alternate: Overall MB, F S M and Diluent mass balance, 650 F x F,D S yS,D M x M,D 0.28 M

M

Solve simultaneously:

2321 and S

1321 kg/h. But this is less accurate.

13.D.30. New problem in 3rd edition. Equation (13-59) becomes Qc /Ai < ut /(1 + safety factor). Using the equals sign and solving for the safety factor Sf we have, Sf = ut Ai / Qc -1 = 0.00172 (1.0)(4.0)/.006 – 1 = 0.1467 where Ai = Ds Ls. Thus safety factor is 14.67% instead of 20%. This may still be acceptable. 50 13.D.31. New problem in 3rd edition. Soln. A. Kremser Soln. R mE 0.30998 1.0 161.3 R 50, E 100, m 1.613, b 0, y 2 0.0, x 0 0.01 For example, Use 13-11.

yN

y1 x1 Soln. b.)

1

y1*

y1

* 1

y

1

1

R mE N R mE

y1*

1

mx 0 becomes

0.01613 0.01613 .7633696

y1 m

b

0.01613

y1 0.01613 0 0.01613

1 0.30998 1

0.30998

2

0.7633696

0.00381684

0.00381684

0.0023663 1.613 Do mass balances and equilibrium for single stage.

347

Sy IN Fx F Sy Fx 0 0.5 100y 50x also y x 1.613 . Solve simultaneously and obtain identical result. Soln. c. Do graphically as single stage system. Soln. d. Do graphically as counter-current system, N=1. Solution is valid, but awkward. 13.D.32. New problem in 3rd edition. Fixed Dispersed Phase. Q sol Q feed Q feed Q tol At feed conditions tol Q sol Q tol Q feed Q feed Q feed Q feed

0.6 .006

Q sol Q feed Q sol .006 .6 .006 1 Q feed Equation 13-48 operation in ambivalent range. a)

.6

tol

.3 L

1 0.3

From Example 13.5

L

L

H

0.375

H

L

0.625

1.6

Q sol Q feed 1 Q sol Q feed .375

865 0.95 10 998. 0.59 10

3

0.3

3

1.10235

0.375

The

1.10235 0.6614 0.625 Either phase can be dispersed. 1.0 b) 0.5 , also ambivalent range d 2.0 .5 Either phase dispersed 1.10235 1.10235 .5 2.0 c) .6667. According to 13-48 at border. d 3.0 .6667 water probably dispersed 1.10235 2.2 .3333 5.0 d) .8333 Equation (13-48), water (heavy) dispersed. d 6.0 0.8333 1.10235 5.5 water dispersed. 0.16667 13.D.33. New problem in 3rd edition. t re s Vliq Qd Qc 1.5 min 90s

Qd

Qc

0.0072 m3 s , Vliq

90 s 0.0072 m3 s

Note that there is a 1 inch air gap at top Vliq H t 0.0254 d 2tan k 4 0.648 , H t

Vliq

2d tan k

0.0254

Using Goal Seek d tan k

d 2tan k 4

0.648m3

2d tan k

0.648

0.7489 and H tan k

1.4978

348

13.D.34. New problem in 3rd edition. N = 500 rpm = 8.335 rps d i 0.20 d tan k 0.2 0.8279 0.16558 m

Use water values for Re L ,estimate

M

d i2 N

998 kg m3 and

w

2

0.16558

L

8.335 998

Curve b in Figure 13-32 again predicts a constant N p0 Then from Equation (13-52), P

P

4.0

M

2

8.335

N P0

.16558

2

M

5

40

d 5i g c where g c

1.0

0.95 10 3 kg m s

w

240, 064

3

0.95 10

L

M

0.034587

N

1.0 and

8.335. (A)

M

will be fairly close to c 998 since Q W 5QToluene (see Equation (13-53)). W The series of messy terms for Equation (13-56a) can be calculated. Since the tank dimensions and physical properties are the same as in Example 13-5, the only term on the RHS of Equation (13-56a) that is different is P. Thus the result in the same as Equation B in Example 13-5, d 0.0576 P 0.3 (B) In addition to Equations A and B, we need to solve Equation (13-53) (C) 1 d c 865 d 998 1 d M d d M

Solving equations A, B and C with Goal Seek we obtain Then solving Equation C,

M

d

1

d

d

0.146 and

d

0.146 865

c

d

d,feed

0.874.

0.854 998 978.6

Equation (B) P 0.3 2.876 P 33.84 W. d 0.05076 nd 13.D35 (was 14.D11. in 2 edition) From Eq. (12-46), E1 K 1 E2K 2 B1 1 , C1 , D1 R 0 x 0 R1 R2 (Eq. 6-6) For 1 < j < N A j

1, B j

(Eq. 12-48) For Stage N A N Example 13-4: R 0

1000, x A0

For Acetic Acid, K A j

1, B N

0.35, x D,0

E jK j

1

Rj

ENKN

1

E j 1K j

, Cj

RN

Rj

, DN

6 , EN

0.65, N

1

FN z N 1

, Dj

Fjz j

0

1

E N 1y N

1475, yA,N

1

1

0, yD,N+1

0

y Aj x A j : Use Fig. 14-4 to estimate K A, j .

K A1 K A4

0.03 0.1 0.12

0.5

0.3, K A 2

0.15 0.14

0.33, K A 3

0.09 0.21 0.16

0.43

0.5, K A 5 0.5, K A 6 0.5, 0.24 0.28 0.32 For first guess assume constant E 1475 and R 1000. Then C1 D1

B1

1

E 1K A1 R1

E 2K 2

1475

R2

1000

R D x A ,0

1

0.33

1000 0.35

1475 0.3 1000

1.4425

0.48675 350

349

and so forth with D6 1

DN

475 0

2

0 . Thus matrix for acetic acid is,

3

-0.48675

0

1475

1475

4

0

0

0

0

0

0

0

1.4425

2

-1

3

0

4

0

0

-1

5

0

0

0

-1

6

0

0

0

0

1000

.33

.43

1000 1

-1

1475 1000

1475

.43

0.5

1000

1

6

0

1

1

5

1475 1000

1475

0.5

1000 1475

1

1000

0

0.5

1475

0.5

1000

1

-1

1475 1000

0.5

0.5

13.D.36. Part a. New problem in 3rd edition. See figure

Forg

C Aq ,0

FAq

* org ,1

Min

Forg,Min

Forg

b.

C

0.736 FAq

1.4 147.2

Operating line goes through CAq,N

Corg,1 See Figure.

C Aq ,N C

* org ,N 1

0.10 0.008 0.133 0.008

0.736 200 L h

206.08 ,

147.2

Forg

206.08

FAq

200

0.008 and Corg,N

1

0.736 L h

1.0304

0.008 with slope 1.0304.

0.097

3 stages more than enough.

~2

3 4

stages needed.

350

Part c.

MW Zr NO3

4

91.22 4 14.0067 3 15.994

MW water

2 1.00797

15.994

18.00994

351

Basis 1 liter 0.10 mol Zr NO3

4

Have

33.917g Zr NO3

and

1000 g

33.917

4

966.083 g water

966.083 18.00994 53.64 mol water .1 Mole frac. Zr NO 3 4 0.00186 53.64 .1 33.9179 Mass frac. Zr NO 3 4 0.033917 1000 g System is dilute if consider mole fraction, less so if use mass fractions. If densities are constant, then constant flow rates is valid. Even with variable density, solving problem with mole fractions and constant molar flow rates would be accurate. This would require converting equilibrium data to mole fractions. Use of fractions with concentrations in mol/L is NOT correct. which is

13.D.37. New problem in 3rd edition. Part a. HETPlarge-scale = HETPpilot (Dlarge/Dpilot)0.38 = (0.24 m) (1.1 m/.05 m).038 = 0.78 m flarge-scale = fpilot (Dpilot/Dlarge)0.14 = (1.4 s-1)(.05 m/1.1 m)0.14 = 0.91 s-1 Part b. HETPlarge-scale = HETPpilot (Dlarge/Dpilot)0 = HETPpilot = 0.24 m flarge-scale = fpilot (Dpilot/Dlarge)0 = fpilot = 1.4 s-1 c. Use of the more conservative design developed for difficult systems (n 1 = 0.38, n2 = 0.14) results in a much higher HETP and thus a much taller column and more expensive column than use of the design procedure for easy systems (n1 = 0, n2 = 0). Considerably more data is needed for a large variety of systems to determine best design practice. If a variable speed motor is used in the large-scale system the difference in predicted optimum frequency is not as serious because the system can tuned to find the optimum frequency. New problem in 3rd edition.

13.D.38.

MWwater

18.02,

F 1.0 kmol hr ,

MWtoluene

S

92.14 , m

0.06 kmol hr.

C toluene

C raffinate

C water

0.00023 ,

x IN

20.8

y IN

0

→ x out Fx in / F Sm Note m m. m is equilibrium in mole fraction units. Assume extract has properties toluene and raffinate properties of water. F x IN

m

Fx out

Sy out and

C extract

kmole benzoic m 3 extract 20.8 kmole benzoic m 3 raffinate

Units on m are

y out

m x out

1 865 kg tol m 3

92.14 kg toluene kmol toluene

1 998 kgW m 3

18.02 kgW kmol W

122.71

kmol benzoic kmol extract kmol benzoic kmol raffinate

352

1.0 0.00023

x out

1.0

0.0000275 , y out

0.06 122.71

If use m

20.8 find x out

13.D.39. Feed is 0.1 1 equil. stage

1 .00023

122.71 0.0000275 1 .06 20.8

0.00337

0.000102, WRONG!

New problem in 3rd edition. CC 4 , 0.9 AA. F 10 kmol h . Solvent pure. S 10 kmol h.

Lever arm:

S

10

F

10

1

FM SM

x F,CC

, Alternatively

x M ,CC

4

x M ,CC

4

x S,CC

4

4

S F

1

Then x M,CC 4 0.05 Find Mixing Point M. [The figure is shown at the end of problem 13D39 as the single stage mixing line.] Phases split along the line –TE to find the line through M Rafinate: x CC 4 0.041, x AA 0.54 . Extract: yCC 4 0.095, y AA 0.07 Overall Balance: E+R+=F+S+=20 CCℓ4 Balance: .095E+0.041R = (0.0) S+0.1 for F=1.0 Solve simultaneously, R 16.6667, E 20 R 3.3333 NOTE: Since CCℓ4 mole fracs can be read more accurately, the CCℓ4 balance is probably more accurate than the acetic acid balance equations.

13.D.40.

S1

S 2

CCℓ4 1

2

E1

R1 = R2 single stage = 16.6667 Mix with S2 = 10 (pure)

E2

x M 2 ,acetic

16.6667

R1

SM 2

10

S

R 1M 2

x M2,AA

R2

x R 1 ,acetic

x M 2 ,AA

x Sacetic x M 2 acetic

0.54

0

x M 2 ,AA

.54 1.6667 2.6667 .3375

Find M2 and by trial and error find a tie line though M2. See figure on next page. Extract 2, yCC 4 0.046 y AA 0.065 Raffinate 2,

R2 CC

4

E2

x CC

4

x AA

R 1 S 16.6667 10

balance

Substitution

R2

0.018

19.40 and E 2

0.018R 2 0.018 R 2

0.57

26.6667

0.0046E 2

0.041 16.6667

0.046 26.6667 R 2

0.0 10

0.68333

7.16 kmol h .

353

354

13.D41. New problem in 3rd edition. R N 1 F 10, x CC 4,N 1 0.1, x AA,N

E0

S 14.5,

1

1.0 , y CC

y TEA,0

Mixing. Use lever arm rule.

1.45

x M ,CC

14.5

S

FM

10

F

SM

xN

1,CC

4

0.9 0.091

4,N

xN

1,CC

x M ,CC

S y S,CC F SF 1

4

4

x M ,CC

4

y S,CC

.1

4

4

4

1.45 0

0.041

2.45

Find M. Draw E N MR 1 line. See figure on next page. Raffinate:

x1,CC

4

0.008

x1,AA

.58

Passing Streams E N R N 1 & E O R 1 intersect at . Very close to parallel. Use parallel lines to step off stages. Estimate # Stages = 3. Flow rates 24.5 F S E 3 R 1

CC

4

balance. F .1 E3

S 0

1.0

E 3 .091

R1 0.008

1.0 24.5 0.008

9.69 kmol h , R 1 24.5 9.69 14.81 .091 .008 Can compare to 13.G.2 Part c. Extract 10.066 and Raffinate 14.433 Extract Mole fraction y TEA 0.841 x CC 4 0.0913 y AA 0.067 Raffinate Mole fraction

x TEA Two results are reasonably close.

.418

x CC

4

0.0056

x AA

.577

355

356

13.D42. a. First, plot points EN and R1 on the saturated extract and saturated raffinate curves, respectively. Second, Find point Δ at the intersection of lines FENΔ and R1SΔ. Third, step off equilibrium stages. Need about 3. See graph. Part b. Easiest: use the lever-arm rule. Find mixing point M at the intersection of lines FS and ENR1.Then S FM 0.81 F 1235kg / h F SM Can also write 3 mass balances (overall, pyridine, and water) and solve for the unknown flow rates F, EN and R1. Unfortunately, this will not be very accurate because it is difficult to read the water values accurately. 13D.43 (was 14.D5. in 2nd ed.)

Plot points for F, S

Use lever-arm rule to find point M.

E 0 , and R 1 (on saturated raffinate line)

E0M

F

1000

S 1371 FM Line R 1M intersects the saturated extract curve at E N . x acetone

0.067 .

Lines FE N and R 1E 0 intersect at ∆ (a second piece of paper was attached to find ∆ accurately). Step off stages. 3 is more than sufficient. Need about 2 & 2/3 stages. This is close to the 2 + ½ estimated in problem 13.D19 with a McCabe-Thiele analysis.

357

358

K Dm

13.E1. Since

E

K Do

y m,N Estimate: E

E

0.05, K Do R 1

20, E

200, F 1

1 ortho goes up column and since K D M

yo,N

1

0, x m,0

E .52F and R

200.52 and R

Recoveries:

0.15, R

20.48,

.92 .52 1 x ortho,N

x o,0

0

R .48F R 20

E 200.52 E y ortho,1 or y ortho,1

0.09974 and

E R

1 meta goes down.

R

20.48

E

200

0.1024

0.002386

0.00203

.94 .48 1 Rx meta,N or x meta,N .02179 Plot equilibrium curves and operating lines (see Figure) Feed cannot be 3rd stage since cannot get x m N desired. Cannot be 5 as will be past intersection of R E and meta op lines. Thus feed must be 4th stage. Do not get match of total number of stages. Need 8 1/3 for ortho and ~ 5 2/3 for meta. A very slight adjustment of recovery meta will change this. (Meta is approaching a pinch point at feed stage). 93% recovery was not enough. Therefore, need ~ 93.5% recovery with ~ 8 stages.

359

13.E.2. New problem in 3rd edition. Part a.

x N,p Part b.

xy

0.04 .004

Paraxylene:

96% recovery. 4% p-xy left in diluent

0.00016 wt. frac. y

Ka

0.080

m, E

20, 000, R

x Eq. (12-28) converted to extraction notation is convenient. L

n

x0 xN

R mE

1

N

n

x0 n N Part c.

x *N x *N

mE R

0.00016,

0.004 0.00016

ortho-xy m

.625

xN

xN

x *N

x0

* N

x

R

yN

x *N

V

R

1000

mE

0.080 20, 000

2.3025

m

E

1

0.

Thus

0.006, x *N

0,

0.625

4.899

0.470

0.150, x 0

1000

R mE

n 1 .625

Eq. (12-31) Converted:

Part d. m-xy

0.004, x N

.375

,

mE

0.150 20000

R

1000

3

1 mE R 1

mE R

N 1

1 3

0.006

1.842 E - 5 1 35.899 Alternative Solutions are presented below for meta-xylene. m 0.050, x 0 0.005, x*N 0, N 4.899 E 20,000,

mE R

R

1000, b

0

0.05 20, 000

1 1000 Must use special form. But the L mV 1 form in terms of x is not available. Thus, need to derive, or translate or find in another source. Looking at development of Eq. (12-12). N x x0 x N Solving for N,

N

x0

xN x

Where Δx is determined in same way Δy was determined for Eq. (12-12), L y1 x0 b y1 V x x j x j 1 const x0 L V L V Alternatively,

x

yN

1

L xN V L V

b

yN

1

L V

xN

360

Translating to this extraction problem,

x0

N And solving for xN, x N

xN

L V x0

x x0 0.005 N 1 5.899

R E , yN

1

0,

x

xN

xN xN 0.0008476

Alternative Solution: Redefine terms to match Eq. 12-12 [Relating y to solvent and x to raffinate is arbitrary. Switch these definitions.] y N 1 meta xylene in hexane 0.005

y1 m

1

1

Kd

0.05

mxy out is unknown

20, b yN

1

y1

0, L N y1

E

x 0 is now inlet solvent

20, 000; L V

x0

b

L

20, 000

mV

20 1000

4.899 y1

20 0

0

1 , V 1000 0

361

Solve for y1,

yN

y1

0.005

1

5.899

5.899

0.0008476

This is actually x N in normal notation. Part e. Shown for normal notation.

pxy equil slope = 0.080

y EQ

0.080 .004

0.00032

y

yN

1

0 x 0,pxy

x

x N ,pxy

0.004

0.00016 0.00032 0

Slope Operating line

R

Slope

E MIN

0.004 0.00016 R 0.08333 , E MIN 0.08333

0.08333 1000 0.08333

12, 000 kg h

13.E.3. New problem in 3rd edition. Part a. Plot the equilibrium data and points F and S. Straight line from power F to point S passes through mixing point M. Since amounts of F and S are equal, M is at the half-way point of the line. Find tie line through M by trial-and-error. This is difficult since tie line is very sensitive. Approximately, raffinate x AR and extract y AE Mass Balances:

0.326

x DR

0.575

0.046

y DE

0.058

E R S E 40 R 40 E Ey AE Rx A,S Sy A,s Fx AF 20 0

Solve simultaneously, E

18.0 kg , R

20 .4

8

22.0 kg.

Part b. First add solvent until reach saturated raffinate curve at intersection with FS line. Initial Raffinate x AR 0.36, x D 0.54

R INIT x AR R init

Fx AF SINIT x AS

8 x AR

8 0.36

20 .4

SINIT 0

8

22.22 kg

SINIT R INIT F 2.22 kg Second, use Eq. (13-27) for the continuous solvent addition batch extraction.

362

x t ,final ,A

S R

x t ,feed ,A

dx t ,A yA

x t,feed,A is the raffinate after solvent addition to form two phases x t,feed,A

0.36 , x t ,final,A

x A,initial raffinate

0.292

From equilibrium find values y A (extract), Approximate values are:

x A,t

yA 0.048 0.046 0.045

0.36 0.326 0.292 0.292

1y 20.8 21.7 22.22

dx A

0.36 0.292

yA

6

0.36

Sadded

1.47R t

20.8 4 21.7

1.47

Eq. A

In the derivation R t is assumed constant, R t

Sadded

22.22

R t,INIT

22.22 kg

32.66 kg

With this approximation E Sadded . 32.66 kg Solute mass balance R t x A,INIT Sadded y A,added R t x A,final

y A,added

y A ,Avg

0, x A,INIT

Ey A,Avg

0.36, x A,final

22.22 0.36 0.292 32.66

0.292

0.046

If we do not assume R is constant, then Eq. (13-27) is x t ,A

Sadded

d R x t ,A

dS added 0

yA

x t ,INITIAL ,raf

We would need to do a numerical integration with a calculation of R x t,A versus y A . this can be done, but is challenging. 13.G.1. New problem in 3rd edition. Extract 1: flow 3.90769, xTRA Raffinate 1: flow Extract 1: Raffinate 2:

0.84986, xcarbontet

16.03923, x TEA

flow flow

0.085102, xAcetic acid

0.41361, x carbontet

11.63396, x TEA

0.91426, x carbontet

14.40527, x TEA

0.065042

0.041332, x Acetic Acid 0.036586, x Acetic Acid

0.54506 0.049149

0.41633, x carbontet =0.016472, x Acetic Acid =0.56719 .

Entering carbon tet 0.10 10 1.0 kmoles hr Leaving in raffinate

0.016472 14.40527

In Out in Raffinate Extracted % extracted = 76.27%

0.23728

0.7627

363

13.G.2. New problem in 3rd edition. Part a, 3 stage cross-flow. All flow rates are kmol/h Total Flow rate TEA flow CCl4 flow Acetic flow Extract 1 3.961 3.366 .3371 .2576 Extract 2 11.634 10.637 .4256 .5718 Extract 3 11.052 10.419 .1554 .4777 Raffinate 3 13.353 5.580 .0819 7.693 Carbon tet remaining in raffinate 3 is 0.0819 kmol/h. Since carbon tet feed was 1.0 kmol/h, 0.9181 kmol/h was extracted. Fraction extracted = 0.9181/1.0 = 0.9181. Part b. 3 stage counter-current with S = 10 kmol/h. Extract 1 4.9142 3.723 .7242 Raffinate 3 15.086 6.277 .2758

.4672 8.533

Carbon tet remaining in raffinate 3 is 0.2758 kmol/h. Since carbon tet feed was 1.0 kmol/h, 0.7242 kmol/h was extracted. Fraction extracted = 0.7242/1.0 = 0.7242. Part c. 3 stage counter-current with S set to give same fraction extracted as in part a (0.9181) and outlet raffinate carbon tet flow rate of 0.0819 kmol/h. This is trial-and-error. First trial: S = 20 and CCl4 raf 3 flow rate = 0.0289 Second trial: S = 18 and CCl4 raf 3 flow rate = 0.04045 Third trial: S = 16 and CCl4 raf 3 flow rate = 0.0590 Fourth trial: S = 14 and CCl4 raf 3 flow rate = 0.0908 Fifth trial: S = 14.5 and CCl4 raf 3 flow rate = 0.08104 This is close enough. Final Results: Extract 1 10.066 8.469 0.9189 .6786 Raffinate 3 14.433 6.031 0.0810 8.321

364

SPE 3rd Edition Solution Manual Chapter 14 New Problems and new solutions are listed as new immediately after the solution number. These new problems are:14.A3, 14.A4, 14.C5, 14.D6, 14.D9, 14.D11, 14.D15-14D17, 14.E2, 14.E3. Chapters 13 and 14 from the 2nd edition were rearranged to place all the extraction material into chapter 13 and the material for other separations in Chapter 14. Thus, the numbers of many problems have changed. 14.C.5. New problem in 3rd edition. Part a. y y, x x, m 1, F U, S where F, U, S, O, R, E are kg Eq. (13-27b) becomes U U y x x F y IN O O U and (13-21) x x IN y IN 1 U O and y = x O Part b. Eq. (13-29b) becomes O 1 n x t ,final x t ,feed U K Where K y x at equilibrium = 1.0 in washing.

n 14.D1. (was 13D29 in 2nd ed.)

a) Translate eq. (12-28),

U mO

1

N

O, R

U, E

x *N x *N

x0 xN

O

U mO

n mO U

Note: x in wt frac. translates to x in kg m 3 if densities are constant. Densities cancel. For washing equilibrium is equal overflow & underflow concentrations. Thus, m = 1, b = 0 yN 1 b H 2SO 4 x *N y N 1 0, x 0 1.0, x N 0.09 m U 40 mO 1 0.8 and 1.25 mO 1.0 50 U 0.8

n

1.0 0 0.09 0

1 0.8

N

0.8 4.96

n 1 0.8

b) HCℓ Use Eq. (12-31) or (14-8)

xN

x0 N

1 1

xN

x *N

x0

* N

mO U mO U

4.96,

N 1

x

1 1

0.75

mO

1.0 50

U

40

mO U mO U

N 1

1 1.25 1 1.255.96 1.25, x *N HC

0.0674 kg m 3 yN

1HC

m

b

0

Alternative: 363

xN

Note:

x0

xN x0

HC

=0.09 H 2 SO 4

Thus, if one is clever and realizes change will be same for HCℓ & H 2SO 4

since

mO U

& N are identical , don’t need to use Kremser eqn for part b.

14.D2. (was 13.D22 in 2nd ed.) a. 1000 cc sand = 400 cc underflow liquid. This is about 400 g = 0.4 kg liquid. Equil: y = x. Use nomenclature of Table 13-4. U U Operating Eq. y j xj 1 y in x out O O U .4 Slope 0.8. Goes through point (y = 0, x = 0.002) O .5 Overall bal. O yin U x in U x out O y out

O

.4 0.035

.4 .002

.5 yout → y out

.0140 .0008

Need 6 2/3 Stages – See Graph (Can also use Kremser eq.)

b. Mass Balance: Op. Eq.:

U xj yj

O j y jin U Oj

xj

U xj y jin

0.0264

O jy j U

Oj

xj

1

U

2 slope , x out 0.002 (see graph) O Obtain approximately same separation, but use much more wash water. (was 13D23 in 2nd ed.) U

14.D3.

1

.5

0.4, O

0.2,

364

y

y

4

y

3

2

y 1

2

1

U=3

x

4

3

U=3

4

O y

Basis:

O 2 3 y 0 in 3

2

4

in 4

0

O y

2

in 2

2 0

O 2 1 y 0 in 1

1 kg CaCO 3 solids

Feed Mole frac. can be arbitrary. Pick x 0

U x iN

M.B.

O yiN U

y out

O y out

U

x iN y iN O O y out , x out at Equil (y = x) line

x in , yin

Point Slope Op line See graph. Find Recovery

x out

U x out

0.01 as basis

1

x in , 0 is on op line

U

3

O

2

x4

0.00127

x0

0.01

0.127

x4

1 0.127 0.873 x0 Recovery is significantly better with counter-current process.

365

14.D4.

(was 13D24 in 2nd ed.)

0.8 0.8 0.2

0.8,

1

4, O

4000 kg/h

366

U F1

1000

kg h

dry solids

U F2 UT In section 2: Slope

U F1

UT

yj

O

Intermediate feed at x

Slope

xj

L liquid L solid

kg

.8

h

.2

U F2 1

1.0

3200

1600

U F1 O

xF xj

1

kg liquid h

kg liquid

2.5 kg liquid 4800 h UT y0 x1 O

h

4800 4000 1.2 Goes through point y0 , x1

UT O

In Section 1: y j

2000

4

kg liquid L liquid kg solid 2.5 L solid

1.0

0, 0.006

0.02 U F1

yN

U F1 O 1600 4000

O

xN

1

y N , x N+1

0.4 . Goes through point

Also intersects Section 2 op. line at feed line. (Or calculate y N from mass balance). Equilibrium is y = x. Step off stages (see Figure). Need 5.4 equilibrium stages. Opt. Feed is 4 th.

14.D5.

(was 13D25 in 2nd ed.)

F1: 1000

kg dry

0.8, 1

0.2

1 1 kg dry L solid h 2.25 0.2 L L under flow kg dry 1 1 F2 : 2000 4000 h 2.5 0.2

2000 L h

L h

F1 5 wt %

F2 2 wt %

367

2000L

Liquid Volumes:

total 0.8 liq

h

3200 L liq

4000 0.8 liq underflow

U0

ON

1

4000 kg h , y N x0

144 4800 Ext. MB,

U

0,

1

f

4000

1.2,

O

4800

ON 1 y N y1

U0 x 0

1

h

1600

U0x0

U

O

O

U

xN

O

mx 0

n 1 0.8333 N

5 wt %

where

kg liq h

1.0 kg L

f

0.05

kg N a 0H kg liq

3200 0.02

0.006

UNx N

Convert to Kremser O V, U L, m 1, y1* Eq. (12-30)

kg liquid

U 0x 0

0.030 , x N specified

h

2 wt %

h 4800

FT : Total liqd h

L liq

1600

O1 y1 x0

x0

xN

4800 4000

0.030 0.006

0.030, mV L

0 0.030 0.0288 0.030 n 1.2

4000 4800

0.0288

0.8333

0.83333 8.83 or 9 stages

Use 2 feeds! 14.D.6. New problem in 3rd Edition. 2.5 kg wet is 1 kg dry solids-insoluble, and 1.5 kg underflow liquid. 1 kg dry solids Part a. 10 kg total 4 kg dry insoluble solids 2.5 kg total

1.5 kg liquid

6 kg liquid. , Ov 10 kg liquid. kg dry solids Before 1st mixing: 0.05 frac BaS 6 kg liquid 0.3 kg BaS 0.3 kg BaS 0.01875 mass frac in U & Ov. 1st Mix: 16 kg liquid total U

4 kg dry solids

Settle – (6 kg liquid in U) 2nd Mix Pure Water

0.01875

0.1125 kg BaS

0.1125 kg BaS

0.00703 mass frac in U & Ov. 16 kg liquid Settle – (6 kg liquid in U) 0.00703 0.0421875 kg BaS 0.0421875kg 0.00264 mass frac BaS in U and Ov. 3rd Mix Pure Water 16 kg liquid Part b. Result is same. Can also be done graphically. Part c. Countercurrent. Easiest solution approach is to use Kremser equation. x N x *N 1 m Ov U N 1 * x 0 x N 1 m Ov U

368

External M.B.

N

3, x 0

xN

x0

x0U

y N 1Ov

y1 14.D7.

0.05, m 1, Ov

1

30 6

1

30 6

U x0

(was 13D27 in 2nd ed.) Operating Eq.:

xNU

xN

U

Ov

Basis 1000 cc wet sand. .4

O

xj

1

m

0

0.0003205

6 0.05 0.0003205 30

xj

y j,in

U O

vol water

xj

y j wt. fractions 1

.4

O = 0.2 kg. Thus, each operating line has slope

y j,in , x j

x0

0.035, y N,in

yN

xN

0.002, yS,in

y N , y 2,in

1,in

0.009936

1000 cm 3 wet sand

vol sand wet

Each op line goes through pt.

yN

y1Ov

Equilibrium:

yj

U

0.05 0.00641

4

6, x *N

30, U

.2

1.0 g

kg

cc

1000 g

0.4 kg

2.

1

yN

2,in

0

y N 1 , y1,in

yN

2

Start at stage N where x N = 0.002. Find y N then work backwards to stage N-2. This gives inlets for first 3 stages so can then work forward (see Figure). Note: that stages 5 and N-2 are not connected. 8 stages gives more than enough separation, but 7 is not enough.

369

14.D8.

(was 13D28 in 2nd ed.) Use Kremser equation Fsolv Fsolid .95, y mx is equilibrium with m = 1.18, and N = 11. Recovery is 1 x N

x0 .

Eq. (12-31) becomes x *N

m Fsolv

yN * N * N

1

xN

x

xN

x0

x

x0

0 .

m

m Fsolv Fsolid

1 1

xN

1.18 .95 1.121. Then

Fsolid

N 1

m Fsolv Fsolid

1 1.121

x0

1

12

1.121

0.041

Thus Recovery = 0.959 14.D.9. New problem in 3rd Edition. Assume FSolid and Fsolvent are constant despite removal of sugar from solid.

FSolid

FSolid xF Fsolv

Eq. (13-21) becomes x a.

Fsolv

1.0, x F

3.0,

0.055, y solv,IN

y solv ,IN

mE

FSolid Fsolv

13

0.055 1 1.18 0.01211 , y 3 3 b. x = 0.004. Solve for Fsolv .

xm E

Fsolv Fsolv

FSolid

mE x

L Slurry stream

0.055 0.004

1.0

ySolv,IN

14.D10. (was 13D30 in 2nd ed.)

1.18 .004

G H 2 stream

120 lb h

x 1 x

Yin

G Yin X out

y

Y

lb CH 4 lb H 2

Yout 120

10.805 kg

CONSTANT

CONSTANT

,

100

0

100 lb h of H 2

Operating Line. Must work in weight ratios. Y

x

0.0143 wt frac.

x

x

mE x

1.18 0.01211

y solv,IN

xF xF

1.18,

FSolid Fsolv

x

FSolid

0, m E

1 y in

in

,

30 100

X in

L G

X x in

1 x in

0.30, Yout

L G

X in

Yout

0 lb CH 4 lb H 2

out out

.05 .95

.0527

L X out

0.30 0.0527

0.206 , x out

X

.206

1 X

1.206

0.171 370

Operating line becomes, L Y X Yout where G Equilibrium Curve:

L

120

G

100

y = 1.2 x becomes

x 0 .05 .10 .15 .20 .25

Y Y 1

1.2

X X 1

→ Y

1.2 X 1 .2 X

Plot Y vs X X Y 0 0 .0526 .0038 .1111 .1364 .1765 .2195 .2500 .3158 .3333 .4286

See Figure for Plot. Need 5 1/8 stages. 14.D.11. New problem in 3rd Edition. 10,000 kg h wet solids, 1 frac vol dry solids. Basis 1 m 3 wet solids : Weight liquid + weight solids

1.20 and goes through X in , Yout .

1.0

frac. vol. liquid ,

1000 kg m3 1.0 1

1500 kg m 3

0.4

400 kg Thus

400 1300

900 kg 1300 kg total m 3 wet solids.

of weight is underflow liquid, U

14.D12. (was 13D32 in 2nd ed.)

Fsolv Fsolid

400 1300

10, 000

kg h

3076.9 .

1.36

371

Op. Eq.:

Fsolid

y

Fsolv

x

Fsolid

y1

Fsolv

Where y and x are kg m 3 . y

xF

x0

m E x is equilibrium.

x 0 , x N 1 .975 x F .025 x 0 , y N 1 0, x *N 0, N 5, Fsolv Fsolid Can use any of Kremser equations such as Eq. (12-31). m Fsolv 1 * xN xN Fsolid 1 1.36 m 0.025 N 1 6 * x0 xN 1 1.36 m m Fsolv 1Fsolid

1.36

Which becomes: 0.1582 m 6 1.36 m 0.975 0 Find m = 1.313 1 1.313 1.36 .025 0.025005 which is OK. Check: 6 1 1.313 1.36 14.D13. (was 13D33 in 2nd ed.) a)

x

Use Eq. (13-21),

Rˆ Sˆ 10 12.5

Rˆ Sˆ x 0

0.8, m E

m

1.18

y in

Rˆ Sˆ

g L in liqd

, x

g L in solid Frac. Rec. 1 0.404040 0.5959596 x t ,final Sˆ 1 n b) Eq. (13-29b) mE x t ,feed Rˆ

x t,final

x F exp

1.25 1.18

, equil. y

m E x, y in

0.8 x F

0

1.18 0.8

0.8 1.98

0

xF

0.4040 x F

0.228779 x F , Frac Rec = 1 – 0.228779 = 0.7712

14.D14. (was 13D34 in 2nd ed.)

BaSO 4 coal BaS 2 CO 2 Equil: Soln conc in underflow = soln conc in overflow. Thus really washing Equil : y x, m 1, b 0

U

350

O

2075

kg h

in sol.

kg h

, y in

Eq. (12-29) 14.D.15.

1.5

kg solution kg insoluble solid

0.0, x *N N

0, y1 n

xN

U0 x0

525 kg soln., x in xN

O x *N

x0

n L mV

525 2075

x *0

0.20, x out

0.2 0.00001

0.00001 0.0506, x *0

n 0.00001 .2 .0506 n 525 1.0 2075

New problem in 3rd Edition. With 1000 kg/h dry solids U 1.5 1000

0.0506

6.99 or 7.0

1500 kg h

a) Can use Kremser eq. with large N to find Ov Min or a sketch

372

y1*

Equilibrium is y

0

y1*

U

y yN

x

U 0

0

Ov

b. Ov 1.2 Ov Min

U

1500

Ov

1782

Kremser: Eq. (12-28)

x0

Min

x0

xN

Min

1 .99 .15

0.15

xN

0.15 0.0015

1500

Ov Min

.15

0

10101

1.0101

1485

0.0015

1782

0.84175

y

y

V

Ov

x

x

L

U

xN

0.0015, x *N n

0, x 0

.15, m 1, U Ov x0 xN

U m Ov

1

N

n

x *N x *N

0.84175

U m Ov

m U Ov

0.15 0 .84175 .0015 0 2.81 N 16.33 1 .17227 n .84175 In theory, can use McCabe-Thiele, but it is difficult to accurately step off this large number of stages. U 1500 c. Ov 2000, .75 m 1 Ov 2000 n 1 .75 100 .75 N 11.29 1 n .75 n 1 .84175

N act

15

E overall

N eq

11.29

N sub actual

15

0.753

For m E use N = 15 and change mE with same equation

n N

1

.75 mE n

100

.75 mE

mE .75 373

Vary mE until N = 15. m E .911 On a McCabe-Thiele diagram this is trial and error. Kremser is much easier. 14.D.16. New problem in 3rd Edition. Part a. U 2 kg, O 2 kg, x IN 0.06, y IN 0 Solution (translation of Eq. (13-21)) is U x x IN y IN 1 U O 1 .06 0 2 .03 O Part b. Want x 0.005 O is unknown, x IN 0.06, y N 0, U 2 Solve for O

x

O

U

U

x

O

O

x IN

y IN ,

0.06 0.005

2

O

x

y IN

U

x IN

O

x

O

14.D.17. New problem in 3rd Edition. K = 1 Eq (13–28) becomes

O

x t ,final

x t ,feed exp

Part b.

U

2, U

x t,final

O 14.D18.

O U

2, x t,feed

O Part c.

2, x t,feed

0.06 e

x IN x

x y IN

22 kg water

0.005 0

Part a.

U

U

0.06 1

n x t ,final x t ,feed

0.02207

0.06, x t,final

0.005

U n x t ,final x t ,feed

2 n

0.005 0.06

4.97 kg

x in Part a.

O normal batch in Part b.

One equilibrium stage. F 1000, x A N+1

E 0 y A,0

.2, S

662, y AS

y DS

0

F x A,N+1

0.12 (same as Example 14-2) E0 F Plot M. By trial and error find tie line through M (Final result shown in Figure). y A1 .238, y D1 0; x A1 .078, x D1 .656 x A,M

Flow rates: Diluent balance:

R1x D1

R1

F x D,N+1 x D1

E1

M R1

F x D,N+1

1219.5

1662 1219.5

442.5

374

14.D19.

This problem is essentially a repeat of Example 14-2, except using exactly 3 stages. Clearly, x A1 0.04 since now have more stages. F, E 0 and M are unchanged. Problem is trial-and-error. Guess location of R 1 . Find E N and ∆. Step off 3 stages and see if have correct location of E N .

x A1

14.D20.

0.026 and y A3

The third and final trial is shown in the figure.

0.38.

Although this is leaching, this cross-flow problem is very similar to cross-flow extraction. We can derive R j 1 x A j 1 E j,in y A j,in x A Mj R j 1 E j,in

M Stage 1:

Rj

R0

1

E j,in where R j

1000, E1,in

M j x A Mj

421, x A0

xAj

yA j yA j

.2 y a1,in

0 , x AM1

200 1421 .1407

Find M on line SR 0 at x AM1 (see Figure). By trial-and-error find tie line through M. 375

This gives E1 and R 1. Find y A1

R1 Stage 2: x A M 2

y A2

.18, x A2 R2

Stage 3: x A M 3

R3

.35, x A1

1421 .1407 .35

0

0.085 1254.9 421 .058, from tie line , M 2

1675.9 0.085 .18 .058 .18 1305 .058

1421

1254.9

.113 .35

1254.9 .113

.113, M1

0

1726

1675.9

1305.0

0.044 , y A,3

1726 0.044 .09 .03 .09

14.D21. a. Basis 1 kg mix in underflow: x NaC values

.09, xA3

.03, M3

1726

1323.3 kg/h

0.8 1.0 crystals

0.2 yNaC

Since crystals are pure NaCℓ, NaOH is in liquid only. Since 20% of the underflow is liquid, x NaOH 0.2 y NaOH . Generate equilibrium table.

376

x NaOH

Soln (y) Mass frac NaOH

0 0.004 0.008 0.012 0.016 0.020 0.024 0.028 0.032 0.036

0 .02 .04 .06 .08 .10 .12 .14 .16 .18

y NaC

x NaC

.270 .253 .236 .219 .203 .187 .171 .156 .141 .126

.854 .8506 .8472 .8438 .8406 .8374 .8342 .8312 .8282 .8252

Feed is 45 wt% NaCℓ crystals. x values: NaCℓ (soln) = 0.5193, NaOH (soln) = 0.099, water 1-0.51930.099 = 0.3817. Since feed is 55% liquid, x F,NaOH 0.55 y NaOH 0.099 y NaOH 0.099 0.55 0.18 , y NaC 0.126 From the equilibrium data F = 100, S = 20, Plot F & S and find M. FM 20 , SM 100 Tie line through M gives E & R. E RM 1.119 R EM (measured on figure) E R 120 1.119 R R 120 R 56.63 kg/min, E 63.37 kg/min

R : Raffinate

0.833

E : Extract y NaC

x NaC , 0.026

0.16, y NaOH

x NaOH

0.135

The underflow is z wt frac crystals (Pure NaCℓ) + (1-z) wt frac solution y NaC 0.16 is soln in equil

z 1.0

Thus,

1 z 0.16 z

0.333 0.16

0.84 was 80% solids in problem statement. c.

Same M. Plot

R1

draw line

R1

0.833

M to

EN

EN

R1

80.1% OK

.

2 stages more than sufficient

120

1.137 R 1

EN

R 1M

103.5

R1

ENM

91.0

R1

120

1.137

377

R1

56.14

R1 : x1,NaC

E N : y NaC

E 63.86 kg/min, N kg/min 0.845, x1,NaOH 0.01 0.152 y NaOH

0.147

378

379

14.E1a.

This is difficult part – converting data Basis 1 lb oil-free solids

y oil

ysolvent 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.35 0.3 0.28

0 0.1 0.2 0.3 0.4 0.5 0.6 0.65 0.70 0.72 Note:

ysolids

1.0

x solids

z 0.20 .242 .283 .339 0.405 0.489 0.600 0.672 0.765 0.810

1 z

0.830 0.80515 0.7794 0.74683 0.71174 0.67159 0.625 0.598086 0.56657 0.552486

y oil z

x oil

1 z 0 0.01948 0.044115 0.07595 0.1153 0.16420 0.2250 0.26124 0.303399 0.3222

0 for all streams, Z = lb solution/lb oil free solids.

Plot data on triangular diagram. See Figure 14.E1a, b, c, d, e. b&c. F + S = M1 = 1500

F x oil,F

S yoil,S

1000

M1x oil,M , x oil,M1

0.252

1500

0.168

See Figure 14.E1a, b, c, d, e. Check: Lever Arm Extract E1 ,

y oil,1

Mass Balances:

M 1S

F

2

M 1F

S

1

0.34; Raffinate: x oil,1

0.092 and x solids,1

1500

0.34 E1 +0.922 R 1

R1

E1

Extract: MB:

R 1 , 252

1040.3,

Finish step c) Stage 2: R1 S2

x oil,M 2

. Find tie line through M1.

M2

E1

459.7 lb

1540.3

, R1 0.092

M2 xoil,M

0.062 . Plot M 2 and find tie line through M 2 .

yoil

2

1540.3

0.115; Raffinate: x oil,2 E2

R 2 , 95.7

0.025 and x solids,2

0.115 E 2

0.80 .

0.025 R 2

R 2 904.8 lb, E 2 635.5 lb d & e – Same answer as b & c but R & E are flowrates. f.

0.730

See Figure 14.E1f. 3 stages is more than enough. Need ~ 2

1 3

equil stages.

380

Lines E N R N

1

and E 0 R 1 intersect at .

381

382

14.E.2. New problem in 3rd Edition. Converting data is the difficult part, but is obviously identical to Problem 14.E.1. Basis 1 kg oil-free solids

x solids

y oil

ysolvent

z 0.20

1.0

x oil

1 z

0.830

y oil z 1 z

0

1.0

0.1

0.9

.242

0.80515

0.01948

0.2

0.8

.283

0.7794

0.044115

0.3

0.7

.339

0.74683

0.07595

0.4

0.6

0.405

0.71174

0.1153

0.5

0.5

0.489

0.67159

0.16420

0.6

0.4

0.600

0.625

0.2250

0.65

0.35

0.672

0.598086

0.26124

0.70

0.3

0.765

0.56657

0.303399

0.72

0.28

0.810

0.552486

0.3222

Approximate solution, use Eq. (13-29a) Oil balance:

S Rt

0

x t ,final

x c ,feed

dx t y

S = Mass Solvent, R t Mass raffinate (solids + solute) x = Mass frac. solute (oil) in raffinate y = Mass frac. solute (oil) in raffinate in extract (solvent) a) M is now at saturated raffinate curve. x oil,M

0.21, x solids,M

0.63

Mass balance F + S = M Solids .748F + (0) (S) = 0.63M

M

0.748

F 1187.3 kg R initial 0.63 S 187.3 kg b) Now mixing is from S to a point on raffinate curve. From equilibrium curve in solution to 14.E.1. 383

x oil

y oil

0.21 0.1625 0.115 0.0675 0.02

1 y oil

.54 .498 .40 .28 0.1

Insoluble Solids M.B. Initial 0.748, F = 100, Final 0.81,

1.852 2.0080 2.50 3.57 10.0

R t final

.81 R t final 748 R tfinal 923.5 kg Raffinate is 0.81 solids, 0.02 oil and 0.17 solvent Solvent remaining in raffinate is 0.17 923.5 157.0 kg Needs to be recovered by evaporation. Do Simpson’s rule in 2 parts. 0.21 0.115 1.852 4 2.008 2.50 6 1

0.115 0.02 6

2

Sadded

2.50 4 3.57

.1961

0.4241

10

0.6202

0.6202 R t , but what is R t ?

Eq. (13-29a) assumes R t

Const.

Use average value of R t .

R t ,avg or

Sadded Stotal

1

R t init R t ,final 1187.3 923.5 1055.4 2 0.6202 R t,avg 0.6202 1055.4 654.6

Initial addition + Sadded

Extract amt S

Stotal

Sremain in raffinate

Oil in extract

x F,0.1 F x final,oil R t,final

187.3 654.6

841.9 157.0

841.9

684.9 by solvent

0.252 1000

0.02 923.5

233.5

Total wt extract 684.9 233.5 918.4 yoil 233.5 918.4 0.254

ysolvent

0.746

14.E.3. New problem in 3rd Edition. Solid Matrix is insoluble. Solids = (.748) 1000 = 748 kg. R t not Constant, but Solid is. Solids Rt x Solids

ydS

d R xA

Solids d

xA x Solids

384

x final ,A x Solids

S Solids

d x A x Solids y

x A ,raf ,init x Solids ,raf ,init

Changes limits integration. x oil 0.21, x Solids

x oil

0.63, x oil x Solids

0.115, x Solids

0.21 0.63 .3333

0.705, x oil x Solids

0.115 .705

0.163

x oil 0.02, x Solids 0.81, x oil x Solids 0.02 0.81 0.0247 Numbers for use in Simpson’s rule are from Solution 14.E.2.

.3333 0.163 6

1

0.163 0.0247 6

2

Sadded Stotal

1.852 4 2.008

2.50

2.50 4 3.57

Solids total integral

10

0.3515

0.6173

Total 0.9688 748 0.9688 724.6 kg

initial added 187.3 724.6

911.9 kg

Extract Amount Solvent

Stotal Sraf ,final 911.9 157 Oil in extract = 0.252 (1000) – 0.02 (923.5) = 233.5 Total weight extract 754.9 233.5 988.4

754.9 kg

wt frac solvent = 0.764, wt frac oil = 0.236

385

Chapter 15 Solution Manual Since this is a new chapter, all problems are new. A. Discussion Problems. 15.A1. The mole fraction water is constant but since the temperature within the vessel varies the total molar density Cm varies and the water concentration = Cw = ywCm also varies. Thus, Eq. (1510a) incorrectly predicts molecular diffusion. Equation (15-10b) predicts no molecular diffusion because dyw/dz = 0. B. Generation of Alternatives. 15.B1. For example, one could operate with both inflow and outflow at the bottom of the tube. If flow is controlled with a constant head tank, the height of liquid in the tube will be very close to constant. C. Derivations. 15.C4. Substitute in q = (μ Re)/(4ρ) into Eq. (15-35d) and obtain δ = [(3μ2Re)/(4ρ2g)]1/3. 15.C5. Start with Eq. 15-52a), set vB=0 and solve for yAvA. Then NA = Cm yAvA. Substitute in the expression for yAvA and Eq. (15-52e) for JA. This gives the desired result. 15.C6. This problem is included to show that one can derive the expressions in books. There is a lot of algebra, but the derivation works. First, can expand the derivative,  1   AB 2 (1  2 x1  x12 )     x1 x  [ B  ( A  B ) x1 ]2  Then take the derivative and expand all terms. The denominator becomes [ A  ( A  B) x1 ]3  [ Bx2  Ax1 ]3 and the numerator simplifies to 2 A2 B 2 x2 . Multiply by x1. Q.E.D. * 15.C7. With CMO and y as mole fraction, vmol  y Av A  yB vB  y Av A  (1  y A )vB  0 . Since NA = -NB, CAvA = -CBvB and for an ideal gas Ci = yi Cm. The total molar concentration Cm is constant. Then, vA = -(1-yA)vB/yA (Eq. A) In terms of mass fractions yA =(yA,mass/MWA)/[yA,mass/MWA + (1 – yA,mass)/MWB]. (Eq. B) Substitute Eq. B into Eq. A and simplify. (1  y A, mass ) / MWB vA   vB (Eq. C) y A, mass / MWA

* Then in mass terms vmass  y A,mass vA  yB ,mass vB which after substituting in Eq. C and simplifying * vmass  (1  y A,mass )  ( MWA / MWB )(1  y A,mass )  vB . (Eq. D) * If MWA = MWB, vmass = 0. We can write vB = NBCB = NByBCm = NBCm(1 – yA) (Eq. E) where the y are mole fractions. Substituting Eq. B into Eq. E and then substituting this into Eq. D, we obtain  1  y A , mass   MW A  Cm N B  (1  y A , mass )  (1  y A , mass )     MW B   MW B  *  vmass (Eq. F) y A , mass / MW A  (1  y A , mass ) / MW B * Since yA,mass varies throughout the distillation, vmass is different for each stage.

386

D. Problems. 15.D1. Dprop,water = 0.87E-9 m2/s. Eq. (15-9), J A , z   D AB

dC A

  ( D AB / L )(C A , L  C A ,0 ) . If C A,0 = 1.2 dz kg/m3 is the known value, C A, L can be larger or smaller than C A,0 . For smaller C A, L we have

C A, L =1.2 – (0.2E-5)(0.0001)/0.87E-9 = 0.9701 If it is larger value, then C A, L =1.2 +(0.2E-5)(0.0001)/0.87E-9 = 1.430 15.D2. Taking the ratio of Eq. (15-23c) at the unknown T and at T =298.16, exp[  Eo / (TR )] = 1.52E-09 for T = 335.18K. Flux D (T )  D (298.16) exp[  Eo / (298.16 R )]

J A , z   D AB

dC A dz

  ( D AB / L )(C A , L  C A ,0 )   (1.52  10 9 / 0.0001)(0.9701  1.2)  0.35 10 5

The temperature can be found with Goal Seek from a spread sheet, but one has to trick Goal Seek into working. Multiply the desired and the calculated fluxes by 1,000,000 and have Goal Seek match these two values. 15.D3.a. 0.181cm2/s, b. 0.198 cm2/s, c. 0.0725 cm2/s, d. 0.198 cm2/s 15.D4. a. 0.0875 cm2/s, b. 0.096 cm2/s, c. 0.175 cm2/s, d. 0.096 cm2/s. 15.D5. Use Arrhenius form in Eq. (15-23c) but for mole fraction 0.0332 instead of infinite dilution. Write the equation for both known temperatures and divide one of these equations by the other. The constant Do divides out. Take the natural log of both sides and solve for E/R. The result is  D (T )   1 1  E / R  ln  AB 1  /     D AB (T2 )   T2 T1  The constant Do can be found from the known conditions at T 1 Do  DAB (T1 ) / exp[ E / ( RT1 )] Or from the known conditions at T 2. The results are: E/R = 1348.3, E = 2677.6 cal/mol, DAB (x=0.0332, T=300) = 1.313×10-9m2/s. 15.D6. Same equations as in 15.D5. At 298.16 K for the infinite dilution value set C sucrose = 0. Final results are Eo = 4953.8 cal/mol, DAB(infinite dilution, T = 320K) = 0.925×10-9m2/s. 15.D7. For an ideal solution the term in brackets in Eq. (15-22) is equal to 1.0. Write this equation for two of the xA values with the corresponding diffusivities (e.g., x = 0.0332 with D = 1.007×10-9 m2/s and x = 0.7617 with D = 1.226×10-9m2/s). Then have two equations with the two unknowns: o o o o and DBA . Solve for the two unknowns. Results are DAB = 0.998×10-9 m2/s and DBA = DAB -9 2 1.308×10 m /s. Check results with the other two mole fractions and find that the fit is good. 15.D8. From http://www.engineeringtoolbox.com/ the density of methanol at bp is 750.5 kg/m3 (used a linear interpolation), which means partial molar volume = 1/(density/MW)= 0.0426 kg/m3. Viscosity of water is 1.0 cp = 0.001 Pa s = 0.001 kg/(m∙s). a. With φB = 2.26, DAB = 1.43×10-9 m2/s. b. With φB = 2.26, DAB = 1.56×10-9 m2/s. 387

2 15.D9. Combining Eqs. (15-35b) (15-35d), vvertical ,max,liq  0.5  9  gq /   Assume that the bulk is pure 1/3

water with infinite dilution of ethanol. From Perry’s Chemical Engineer’s Handbook, 8th edition, (p. 2-305) at 1.0 bar (0.1 MPa) water has ρW,m,liq = 55.212 kmol/m3 →ρW,liq=994.64 kg/m3 and ρW,m,vapor = 0.032769kmol/m3 → ρW,vapor = 0.5903 kg/m3. The water boils at 372.76K. At this temperature, from p. 2-432, the viscosity of liquid water in Pa∙s is, W ,liq  exp[52,843  3703.6 / T  5.866 ln T  (5.879 1029 )T 10 ]  2.807 104 Pa  s The viscosity of the vapor at 372.76K is (p. 2-426) W ,vapor  (1.7096 108 )T 1.1146  1.2561105 Pa  s or kg/(m∙s).

Now we can calculate the vertical velocity of the liquid water for q = 7.5×10-6m2/s (remember to use liquid properties). 1/3

 9(994.64)(9.81)(7.5 E  6) 2  vvertical ,max,liq  0.5  9  gq /    0.5    0.130 m / s 0.0002807   A check of the units show they work. The modified Reynolds number (using gas properties) is, 2

Re 

d tube  ( v gas  vliq , y ,max )



1/3



(0.10)(0.5903)(0.81  0.130) 1.256  10 5

 3195.6

/ The gas phase Schmidt number is Sc gas    The viscosity and density were found earlier.  D EW  gas The diffusivity of ethanol and water in the vapor phase at 372.76K and 1.0 bar = 0.98717 atm can be estimated from the Chapman-Enskog theory with the parameters in Table 15-2. This value of DEW = 1.658×10-5 m2/s. Then Scgas = 1.283. Since both Re and Scgas are in the range for Eq. (15-47a), the

modified Sherwood number is, k p d tube ( pB )lm  0.0328(Re) 0.77 Scgas 0.33  0.0328(3195.6).77 (1.283).33  17.79 DAB ptot 15.D10. From the Chapman-Enskog theory DNH3-air = 2.05×10-5m2/s at 318.16K and p = 1.2 atm. Problem 15.D10a, 3rd ed. MW A 28.9 MW B 17 const 1.86E-07 T 318.16 p 1.2 T^3/2 5675.033 sigma A 3.711 sigma B 2.9 sigma AB 3.3055 eos A/kB 78.6 eps B/kB 558.3 eps AB/kB 209.4812 kT/EpsAB 1.5188 Col integ 1.197 Linear interpolation table 15-2 D AB 2.05E-05 D, cm^2/s 2.05E-01 The concentration at z = L is CNH3 (L) = CNH3 (z = 0) + JNH3L/DNH3-air. Results are 0.0002483 kmol/m3 and 0.0002117 kmol/m3. 15D11. D = JL/ΔC = 4.114×10-5m2/s. Set up spreadsheet to obtain this value. Since the collision integral was entered manually, had to do several iterations. After 6 iterations T = 396.2K (see spreadsheet, and note that collision integral does not exactly match the value of kT/εAB.

388

Problem 15.D11, 3rd ed. MW A 28.9 MW B T 396.1642 p sigma A 3.711 sigma B eos A/kB 78.6 eps B/kB kT/EpsAB 1.891168 Col integ D AB 4.11E-05 D desired 4.11E-05 chkB7-B8 chk x E5 Problem MW A T sigma A eos A/kB kT/EpsAB D AB D desired

17 const 1.86E-07 0.9 T^3/2 7885.202 2.9 sigma AB 3.3055 558.3 eps AB/kB 209.4812 1.1069 Linear interpolation table 15-2 1.90E-10 1.90E-05 Goal seek to zero changing B3

15.D11, 3rd ed. 28.9 396.164199034186 3.711 78.6 =B3/F5 =F2*F3*SQRT(1/B2+1/D2)/(D3*F4*F4*D6) 0.00004114

MW B p sigma B eps B/kB Col integ

17 0.9 2.9 558.3 1.1069

const T^3/2 sigma AB eps AB/kB Linear interpolation in Table 15-2

0.0000001858 =B3*SQRT(B3) =0.5*(B4+D4) =SQRT(B5*D5)

chkB7-B8 =B7-B8 chk x E5 =100000*D8 Goal seek to zero changing B3

15.D12*. From http://www.engineeringtoolbox.com/ viscosity is 1.0 cp = 0.001 Pa∙s = 0.001 kg/(m∙s). At 298.16 K, DAB = 1.114 m2/s. Density water = 998.3 kg/m3, viscosity water = 0.001 kg/(ms). Calculate δ = 0.000115282 m, vy,avg = 0.04338 m/s, Re = 19.966. This is a long residence time with Re< 20 so there are no ripples. Shavg = 3.41 and kavg = 3.295E-05 m/s, and 0.000168 kg/s carbon dioxide are absorbed. 15.D13. From http://www.engineeringtoolbox.com/ viscosity is 1.0 cp = 0.001 Pa∙s = 0.001 kg/(m∙s). At 298.16 K, DAB = 1.114 m2/s. Density water = 998.3 kg/m3, viscosity water = 0.001 kg/(m∙s). Calculate δ = 0.0001663 m, vy,avg = 0.090241 m/s, Re = 59.898. This is a long residence time, laminar flow, with no surfactant so there are ripples. Shavg = 5.8 and kavg = 3.89E-05 m/s, and 0.000198 kg/s carbon dioxide are absorbed 15.D14. From http://www.engineeringtoolbox.com/ viscosity is 1.0 cp = 0.001 Pa∙s = 0.001 kg/(m∙s). At 298.16 K, DAB = 1.114 m2/s. Density water = 998.3 kg/m3, viscosity water = 0.001 kg/(m∙s). Calculate δ = 0.0007717 m, vy,avg = 1.9441 m/s, Re = 5989.8. This is turbulent flow with 1300 < Re < 8300. Scliq = 899.2, Shavg = 255.5 and kavg = 0.0003689 m/s, and 0.00188 kg/s carbon dioxide are absorbed 15.D15. From http://www.engineeringtoolbox.com/ viscosity is 1.0 cp = 0.001 Pa s = 0.001 kg/(m∙s). At 298.16 K, DAB = 1.114 m2/s. Density water = 998.3 kg/m3, viscosity water = 0.001 kg/(m∙s). Calculate δ = 0.000115282 m, vy,avg = 0.04338 m/s, Re = 19.966. This is a short residence time with Re< 20 so there are no ripples. Shavg = 9.942 and kavg = 9.61E-05 m/s, 7.851E-09 kg/s carbon dioxide are absorbed. 15.D16. From http://www.engineeringtoolbox.com/ viscosity is 1.0 cp = 0.001 Pa∙s = 0.001 kg/(m∙s). At 298.16 K, DAB = 1.114 m2/s. Density water = 998.3 kg/m3, viscosity water = 0.001 kg/(m∙s).

389

Calculate δ = 0.0001663 m, vy,avg = 0.090241 m/s, Re = 59.898. This is a long residence time, laminar flow, with surfactant so there are no ripples. Sh avg = 3.41 and kavg = 2.28E-05 m/s, and 0.0001165 kg/s carbon dioxide are absorbed. 15.D17. Used a spreadsheet set up to solve Example 15-6. For δ = 0.001 meter one obtains xNH3 = 0.04988, yNH3,surface = .21593, Nwater = 0.5393, NNH3 = 0.0228307. The concentrations are the same as in Example 15-6, but the fluxes are 10× larger. 15.D18. Part a. For two part solution need values at xE = 0.25 and 0.35. The average molecular weights are calculated as in Example 15-5, and are used to determine the average molar densities. The Fickian diffusivities are estimated by interpolating between values given in the Table in Example 15-5. The activity coefficients are determined in the same way as in Example 15-5. Then the Maxwell-Stefan diffusivities are found by the same method. The values are listed below MWavg

DEW, m2/s

 m ,kmol/m3

γE

DEW , m2/s XE = 0.25 25.0 36.28 0.633×10-9 1.9028 1.495×10-9 -9 XE = 0.35 27.8 31.62 0.625×10 1.5553 1.609×10-9 Write Eq. (15-61c) for both intervals. For Δz from xE = 0.2 to 0.3 we obtain (values at xE = 0.2 and 0.3 are in Example 15-5), 36.28(1.495  10 9 )[1.7083(0.3)  2.1582(0.2)] zN E    9.3445  10 9 1.9028(0.25) From xE = 0.3 to 0.4 (interval is over length δ- Δz) we obtain, 31.62(1.609  10 9 )[1.4338(0.4)  1.7083(0.3)] (0.00068  z ) N E    5.7027  10 9 1.5553(0.35) Adding the two equations to remove the unknown Δz and then solving for NE and Δz,, we obtain NE = -2.128×10-5kmol/s and Δz = 0.0004223m Part b. Since the interval Δz is greater than the interval δ – Δz = 0.0002577m, we subdivide the interval from xE = 0.2 to 0.3 into 2 parts. The values needed are given below. MWavg XE = 0.225 XE = 0.275

DEW, m2/s

 m ,kmol/m3

24.3 25.7

37.625 34.99

0.659×10-9 0.624×10-9  z1 N E   5.5371  10 9

γE 2.01976 1.79959

DEW , m2/s 1.482×10-9 1.532×10-9

Equation (15-61c) is now written 3 times:  z 2 N E   3.9846  10 9

(   z1   z 2 ) N E   5.7027  10 9 and solved for the 3 unknowns Δz1, Δz2, and NE. Obtain NE = -2.2389×10-5kmol/s, Δz1 = , 0.0002473, and Δz2 = 0.0001780m. 15.D19 (Optional, Unsteady diffusion) At the average C = 0.001 mol/L Dsucrose  0.5228 105 cm 2 / s . Equation becomes

CA C A,0

 1  erf

z 4(0.5228  10 5 )t

Numerical values of C A / C A,0 are easily obtained with a spreadsheet or with the use of Table 17-7.

390

z, cm 0 0.01 0.05 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.2 1.4 1.6 1.8 2.0 3.0 3.5 3.56 4.00 5.0

t = 1000 s 1 0.9221 0.6249 0.3281 0.0505 0.00335 9.16E-05 1.009E-06 7.61E-12

t = 10000 s 1 0.9753 0.8771 0.7571 0.5362 0.3535 0.2161 0.1220 0.6352 0.0304 0.0134 0.00538 0.00198 0.000206 1.49E-05 7.49E-07 6.2E-10

t = 100000 s 1 0.9922 0.9610 0.9221 0.8449 0.7692 0.6957 0.6249 0.5574 0.4936 0.4346 0.3788 0.3281 0.2406 0.1710 0.1177 0.0784 0.0505 0.00335 6.20E-04 4.99E-04 9.16E-05 1.01E-06

Part b. C  1.0  10 6 when C / C0  5.0 104 , which for t = 100000 s occurs for a thickness of xgel, a gel forms and R increases (probably to l.0) . 17.A8. New problem in 3rd edition. Since there is a gel the retention of the low molecular weight compound also increases. 17.A9. New problem in 3rd edition. Do not invest. Osmotic pressure can often be ignored in UF because with large molecules with high molecular weight the mole fraction is always low even if the weight fraction is high. With low mole fraction the osmotic pressure is low. If there is a concentrated salt with a low molecular weight the mole fraction will be high and the osmotic pressure cannot be ignored. 17.B1. Look at Suk, D.E. and Matsuura, T. (2006) ‘Membrane-based hybrid processes: a review’, Sep. Sci. Technol. Vol. 41, pp.595–626 for additional processes. 17.B.2. New problem in 3rd edition. One possible approach is as follows: Increase stirring to increase the mass transfer coefficient and reduce the wall concentration to prevent gel formation. Then use a permeate in series cascade with recycle of the retentate from the second module in series back to the feed of the first module. The low molecular weight product is the permeate from the second module. The intermediate molecular weight polypeptide product is the retentate from the first module.

17.D1.

PCO 2

15 10

cc STP cm

10

pr

PCH4

0.48 10

10

PCO2 PCH4 31.25 a) Generate RT curve from Eq. 17-6a. pp yp 1 1 yP pr yr 1 yP AB AB

1 10 6 m

tm

cm 2 s cm Hg

pH

12 atm

pP ˆ CO

AB

1

pL 2

ˆ CH

12 76.0

0.2 atm 4

912.0 cm Hg

15.2 cm Hg

1.0, p p p r

0.016666

y P 1.5042 0.5042 y P 31.25-30.25 y P

418

RT

Curve

yP

yr

Op. Eq., FP FIN

0

0

0.1 0.20

0.00515 0.01114

0.30

0.01830

0.40 0.5

0.02721 0.03882

0.6 0.7

0.05504 0.08000

0.8 0.9 1.0

0.12492 0.2349 1.0

y out

y out

PCO2 t ms

cc STP cm

FP

FP

b.

1000

FIN

gmole

1 hr

h

3600 s

0.0002859

cc STP

J CO 2

Fin Fin,1

FP1

2.125

0.32

Fin

0, y P

.15 .32

yP

pr yr

0.402, y out

0.46875

0.25625

0.0276 CO 2 conc. (17-2b)

pP yP

0.088888

L

0.46875

2.125 .10

15.2 cm Hg 0.402

0.0002859

cc STP cm 2 s

gmol s

1.0 gmole

2

cm s 1000 cc 22.4 LSTP

FP y P,CO 2

FP

1 0.32

912 cm Hg 0.0276

1 10 cm cm s cm Hg

Area

0.15 CO 2 mole frac

0.10, y P

Answer (from graph)

2

J CO 2

FP FIN

Plot two arbitrary points:

4

FP

y IN

Slope

10

15 10

y out

y IN

y out

FP FIN

yP

J CO2 J CO2

1 FP FIN

yP

0.32

1.2764 E 8 gmole s cm 2

2.80 10 6 cm 2

0.32 kgmole/hr , Fout

Fin

FP

1 0.32

0.68 kgmole/hr.

1 kgmole/hr

1 2

FP,part a

Stage 1. FP1 Fin1

1 2

0.32

0.16

0.16 1.00

FP 2 , Fout1

0.16 , yin ,1

Fin2

FP1

1 0.16

0.84

Fin ,2

0.15

RT curve is unchanged! 1 .16 0.84 Op. Line: Slope 5.25 0.16 0.16 Find arbitrary points to plot line: 0.15 If y out ,1 0, y p 0.9375 (off graph). 0.16

419

If

y out,1

0.04, y p

y out,1

If

5.25 0.04

0.08, y p

5.25 0.08

Answer (from graph): Stage 2

FP 2

0.16

Fin 2

0.84

0.7275

0.9375

y P,1

0.5175

0.625, y out,1

0.1905,

0.0595

y in2

0.0595

FP2 Fin2

0.1905

yin,2

0.3123

1 0.1905

4.2500 . Plot curve 0.1905 Answer: y P2 0.250, yout 2 0.015 (see graph) Stage 15 10 10 cm 3 STP 912 0.0595 15.2 0.625 0.0006715 1 10 4 cm 2s 1L 1 mol JCO 2 J CO 2 2.9976 E 8 1000 cc 22.4 cm 2s mol 1 h FP1 0.16 1000 0.04444 mol s h 3600s FP1 0.625 Area 1 92.67m 2 JCO 2 ,1 Slope

J CO 2

0.9375

Stage 2: J CO 2 ,2

15 1010 1 10 JCO 2 ,2

4

912 0.015

13.2 0.250

1L

J CO 2

1

0.0001482

6.6161 E 8

3

1000 cm 22.4 lh FP 2 0.1905 840 mol h 3600 s FP 2 y P 2 Area 2 1, 680, 000 cm 2 JCO 2 ,2

1:

cm 3 STP cm 2s

mol cm 2s

0.04445 mol s 168 m 2

It is interesting to compare parts a and b.

Part a:

1 stage

Area

280m 2

y out,CO 2

0.0276 or 97.24% CH 4

y P,CO 2 Part b:

2 stage

0.402

Total Area

260.67m 2

y out

0.015 or 98.5% CH 4

y P1

0.625

y P2

0.250

420

17.D2. a.

yP Slope

1 FP Fin

y in

, yP FP Fin FP Fin .7 2.333 , When y out .3

y out

y out 0, y P

y in

0.2 CO2 , FP Fin

0.2 0.3

.3

0.6667

421

When

yP

y in

0.2

0.286 1 FP Fin .7 RT curve is same as in Problem 17.D1. Draw op line. From graph: y P,CO2 0.53, y r y out,CO2 b.

0, y out

J CO2

J CO 2

15 10

10

pP yP

t ms cm 3 STP cm 76 cm Hg

1 10 4 cm cm 2s cm Hg J CO 2

A

PA p r y r

0.06

0.002148

atm

cm 3 STP 2

cm s

60 atm 0.06 FP

, FP

0.6 mol s 0.53

Fin

Fin

0.6

1000 cm 3 STP 3

1 mol cm STP 0.002148 22.4 L STP cm 2s

L STP

3.3 0.53 mol s

atm

.

3, 254, 000 cm 2

Or 325.4 m 2 . Very sensitive to y P & y r values. Can also calculate J Check:

Fin Fin x in

FP

Fout , 2 FP y P

J CO 2

0.6 Fout

Fout y out , 0 .4

FP ˆJ

J CH 4 . A

Fout

1.4

2 0.2

0.6 0.53

1.4 0.016

.402 , OK

422

17.D3. New problem in 3rd edition Since no concentration polarization x w

J solv

K solv t ms

pr

pp

a xr

R and M.B.

xp

xp

xp

x out 1

xF

xF

x out

Then,

pp

K solv t ms 1

1 R

xp Then

pr

x out ,

xp

J solv

pr

Solve for

xr

1 R x out

415.4

a xr

xp ,

1 R x out , R ,

1

0.22, x F

K solv t ms

33.29, J solv

415.4

0.9804 0.0077

0.0077 0.22 0.78 1 0.9804 0.22

0.0098175

0.0001924

1.1 15.446 0.0098175 0.0001924 33.29 17.D4. Partially new problem in 3rd edition.

13.72 atm.

423

a.

xw

M

exp ( J solv /

xr

1 xP

Mxr

xP

1

pr

3.6 E 3.6 E

K solv

solv ) k

pP

4.625 / 997000 g/m 3

exp

6.94 10

Mxr

xP a

pr

pp

3.6 E

1.069 0.054 75 2

4

3.6 E

1.069 0.054

J solv

t ms

1.069

xP

4 4

5

4

59.895

4.625

a Mxr

Since

73

xP

1.0689 .054

K solv K A , K A

K solv

and

3.6 E

KA t ms

4

59.895

2.29

1 atm

0.0665

g 2

m s atm

K solv t ms

g KA m s atm 0.029 g t ms 2.29 atm 1 m 2s c. Write Eq. (17-37a) for old and new situations – Divide new by old. Obtain 0.0665

2

0.75

k new

k old

Everything else divides out. Since

new

rpm ,

old

k new 17.D5. a.

K solv t ms

J solv pr

RT eq., R

k old

.75

2000

0.000117m / s

1000

1.5 10

3

g

2

cm s

pP

1 102 atm

1.47 10 5 g

1 FP Fin

1 x P x out , Op eq., x P

cm 2 s atm

x out

x in

FP Fin FP Fin Solve simultaneously & obtain Eq. (17-26), which with M = 1 is 1 R x in 0.003 0.05 xP xP 0.000272 , x out 0.091 1 R 1 0.997 0.45 1 R 1 xP x P pr

xr

pP

xP

xr

xP a

1 0.000272 0.091 0.000272 0.000272 102

0.091 0.000272 59.895

b. Plot the RT curve and operating line xP xP a 1 pr pP xr M 1 a 1 xP See graph. Intersection occurs at x r

J solv

K solv t ms

pr

pP

1

, xP

0.0585, x P a M xr

1

3.44 atm

xr

1

x in

0.000752

xP

424

425

J solv

1.47 10 2

5

g

78 atm

cm s atm FP 1 x P , FP A J solv

2 3

2

Fin

3

4

1000 g kg

g cm 2 s

3.36 10 6 cm 2

2

9.91 10 g cm s

Eq. (17-45) can be written as J solv /

k

solv

n xg

336 m 2

n xr

n x r . Slope = k and intercept = k n x g cm

J solv /

solv

0.052 0.037 0.026 0.0134 See Figure. Plot J solv /

solv

Intercept

Stirred cell data:

cP

J solv /

J solv

g L, solv

wt frac

xr

min

n xg

Slope

J solv

0.000991

3.33 kg s

4

Plot J solvent vs

17.D7.

5 kg s

3.33 kg s 1 7.52 10

A 17.D6.

59.895 atm 3 0.0585-7.52 10 -4

dextran 0.012 0.03 0.06 0.135 versus ln xr

0.0185 0.01596

K solv t ms

J solv p

23.1 / (997 g / L )

n xr 4.423 3.507 2.813 2.00

1.159, which is x g

69.25 3.0

23.1

0.314 .

g 2

m s bar L 0.0232 2 m s bar

426

J A ~ J solv c P Mc

With

0.00696 g (m 2 s) . Also, J A

0.0232 0.30 1, 1 R c

Spiral Wound:

JA

Solving for M c

JA

0.00696

c out J solv

10 (0.0232)

J so l v c P

0.03 → R c

0.97

M c cout 1 R c J solv

cP

1.0

c out 1 R c

8 0.030

K solv

J solv

M c cout 1 R c J solv

t ms

p

4.1660 80.8g / (m 2s)

23.1 3.5

Since osmotic pressure is ignored, M c does not effect solvent flux in UF.

yp

1

He H 2

rd

17.D8. New problem in 3 edition From 17-6b, y rHe

He H 2

Check y rH

2

2

H 2 He

b.

b

Pr

1 y PH

2

1 y PH

H 2 He

y rHe Then

Pp

1

H 2 He

pp pr

1 y rH

2

1

0.975

2

y PHe

y r ,He

0.025 0.07656

c

y F,He (1 yP

1

2.8314 .975

1

.9234

0.07656 OK.

0.05 0.07656

1

.07656

3.8314 1 .2 .025

3.8314

y rout He

1

1

10

y FHe

y F,He 1

1 yp

1 y P He

0.5152

Use solution in Eqs. (17-9) to (17-10e) pp .75 a 1 .2 .261 1 1 pr .25

1

Pr

0.2610 , p p p r 0.2 90.8 10 10 .739 0.2 .975 1 0.021397 0.025 .261 .739 0.25 .279475

y rHe

y PH

23.7 10

PHe PH 2

Pp

.739 .2 3

)

.261 .05

b

b2

4ac

2a

Must use minus sign to have positive y p . y PHe

.25

2.3648 .05

1

.25

.25

1.4874

0.0522

1.4874

2.212359 .49377 2

.15763 4.7296

2.3648

1.4874 1.645 4.7296

0.00333

427

y r,He

y F,He

y PHe

1

1

c. Solve RT eq. (17-6b) for y p : y r

0

yr

pp

1

1 yp

y 2p

pr

pp

1

1

c

yr

yr

pr

y rH2

.9234,

1 yr

pr

y rHe

y PHe

y rHe

FP y PH

FP

A

0.025, 2

.06

.739

.80786

.65264

y p,H 2

0.004842 0.1813

0.05,

y H2 IN

.95

.5152

0.975

FIN

pr yr pp yp tm s 51.52 m3 STP h 1h

h

m3

3600 s

90.8 10

10

cm 3

cm 3 STP

14311.11

14311.11

2

a.

yp yr

PH 2 A

s

cm 3 STP

s STP cm

cm 2s cm Hg

A 17.D9.

1 yp

pr

c

y He,IN

1000000 cm 3

Pp y P,H 2

1

.2956 0.05 0.06 0.004842 0.06

51.52m 3 STP

PH 2 Pr y r,H 2

1

.739 .2 1

.80786

y rHe

0.07656 y p,H 2

FP y PH tm s

pp

1

0.01566

4ac

2a y FHe

Use 17-5a written as

FP

1

100m3 STP / h.

FIN

For Part A

pp

b2

Use + sign for positive y p , d.

y p2

0.1478

.06 .261 b

yp

.739 .2

pr

b

pr

0.10001

b

pp

1

.25 pp

1

a

a

.05

3 .00333

555,186 cm 2

.975 1.0 10 4 cm

380 .9234

76 .975

cm Hg

55.52 m 2

Plot the data on a semilog plot in the form of J solv /

solv

J solv L / (m 2 h)

428

J solv

xr

k n

xg

.

From graph, slope When x r

k n xr.

Intercept

18.3 and k

18.3

k n xg

0, J solv / n xg

k solv

2

=J solv

82.9 L m h

23.0 g m 2s

4.53

5.08 m / s

xg

k n xg L

5.08

2

m h

m

2

s

23.0g / (m s)

k n xg

92.8%

The value of xg is very sensitive. b.

There is only one point further out on the ℓn axis. Any error in point is greatly amplified in the least squares regression. Hence, another point in this region would be most useful. The higher the concentration, the better.

17.D10. a) New problem in 3rd edition

2700 800

3.375 , a

b

1

c

yin 1

yp

3.8884

1

Solve RT eq. and op. eq. simultaneously pp 0.3 .5 1 2.375 1.6116 pr 0.7 2

pp pr

yN 1

1

1

1

3.375 .25 3.8884

2

.7

2.375

.3

.25

1

2

.7

.7

.7

3.8884

1.20536

4 1.6116 1.20536

2 1.6116

.5

, use minus sign to have yp between

0 and 1..

429

yp

3.8884

2 1.6116

From op. eq.: y r b)

Since Fˆp y p,A

A

15.1196 4 1.6116 1.205 y IN

yp

1 PA A ˆ A

t ms

1

PA ˆ A p r y r

pp yp

.365295

.7

pr yr

Fˆ p y p,A t ms

.3

0.365295 .25

0.2006

.7

pp yp

ˆ . Since F IN

1 mol s, Fˆp

FˆIN

0.3 mol s

mol 0.365295 1.2 10 4 cm s 3 cm STP cm 1.0 L STP 1 2 L STP 10 3 cm 3 STP cm s cm Hg 22.4 mol

0.3 A 2700 10

10

76 cm Hg atm A

A

c)

Fˆ p y p,A t ms PA ˆ A p r y r

pp y2

, Fˆ p

Fˆ IN

mol s

0.325 1.2 10 4 cm

9.0823 104 cm 2

17.D11. New problem in 3rd edition.

xp

Gelling occurs at a solvent flux of J solv

J solv

Then x gel

solv

atm

0.4 mol s

1 3 cm STP cm 22.4L STP 1.0 L STP 76 cm Hg 2700 1010 2 cm s cm Hg mol 10 3 cm 3STP atm

A

J solv

.5 .365295

6.569 104 cm 2

0.4 A

2.0 .2006

5200

L 2

m day

x out exp (J solv /

997

g

0 since R 0

x IN 1

0.001 .6

0.5 0.325

atm

0.0016667

5200 L/(m 2 day ) which is

day

L 86400 s

solv ) k

1.0, x out

2.0 0.175

60.0

g m2s

0.0016667 exp

17.D12. New problem in 3rd edition p p / p r

1.0 4.5 .

60.00g / (m 2 s) 997000g / m 3 2.89 10 5 m / s

He H 2

PHe PH 2

0.01334

0.261. Use Eq. (17-6b),

430

y p ,He

p

1

He H 2

1 yp

pr

y r ,He

1

1 yp

Eq. (17-7c)

.261

y FHe

y rHE

.2 .254

y p He

y rHe

.1 .254

a)

If x p

x in , cut

.55 , perfectly mixed

(17-27) 1 R 0.55, what value R required. Find R (including concentration

0.00050 &

polarization effect). From Eq (17-27), x p

xp

0.25446

1 R xn

xp

R x IN

1

0.353

17.D13. New problem in 3rd edition .035 NaCl Rejection

1 1 .1 4.5 .261 1 .1

0.1 .261 1

x IN

xp → R

x IN

xp

x IN

x IN

x IN

0.035 0.0005

xp

xp

Rx IN which gives

0.035

.55 .0005

0.035 0.001

0.9935

b)

If

c)

If R° (inherent rejection coefficient with M = 1) for part b is R 0.992, what was value of M that gave R 0.9869 MA 1 M CaseB . R Case B 1 1 R CaseA . Let A be highly stirred RA R M CaseA

xp

0.0016,

M CaseB

17.D14. wB

0.55 , R

xpR

RT curve: y w

Feed

xp

0.035

M CaseA 1 R CaseB

1.0 1 0.9869

1 R CaseA

1 0.992

wB

1

xw

wB

43 (mole frac). Since x w ,IN

B

x IN

141.6

cal

0.9869

1.6375

43 x w

1 xw

1 42 x w

0.10, only need RT curve below 0.10. Create table and plot

xw

yw

0.10 0.08 0.05 0.03 0.01 0.0025 0.001

0.8269 0.78299 0.6935 0.571 0.3028 0.0973 0.0413

kcal

g 1000 cal

0.9 10.5

.55 .001

74.12 g mol

0.1 9.72

10.5 kcal mol

10.42 kcal mol

431

C PB

0.625

1 1000

0.046 , C PW

74.12

CP,in MW feed

xW , F MWW

a) Assume y P and

C PL,in

0.9 0.046

xB , F MWB

1 1000

0.1 0.018

Tout where Tin

0.5 10.5 Tout

30

then,

x in

1

0.018

0.0435 kcal mol C

0.5 9.72

P

molar ratio. Slope Op line

18.016

0.1(18.016) 0.9(74.12)

0.5 to calculate λp

Tin

1.0

68.51

10.11 0.0435 10.11

30

0.129. This is a

6.75 , and op line goes through point (mole fractions)

0.10

0.775. Plot operating line. From graph, y P 0.57, x out 0.129 (mole fraction water). This value of yp is reasonably close to our assumption. x out

0, y P

Fp / FF

( Fp MW p ) / ( FF MWF )

0.031

( MW p / MWF )

432

MWp

y p ,W MWW

y p , B MWB = 0.57(18.016) + 0.43 (74.12) = 42.13

.129(42.13 / 68.51) Area

b)

Cut

Permeate Rate

0.08

x in,w

Then

P

Tin

c)

Tout

Flux

( 0.0791 100 lb h)

0.2 lb h ft 2

39.53 ft 2

0.10

1 cut

Slope

0.0791 in (lb/h)/(lb/h).

0.92

11.5. Find y P 0.68 from graph. cut 0.08 0.32 10.5 0.68 9.72 9.97

C PL,in

x out

P

0.05, x P FP Fin

P

30

0.08 0.0435kcal / (mol o C)

9.97 kcal mol

48.3 C

0.6935 (From RT table or graph).

x in

x out

0.10 0.05

yP

x out

0.6935 0.05

0.3065 10.5

0.6935 9.72

0.0777

9.959

433

Tin

Tout

C PL,in

17.D15. Parts b to h are new in 3rd edition

xP

0, x r,out

Fout

Fin

Fin

FP

Fin Fout 1

0.0777 0.0435

RT curve:

xP

Fin x r ,in

FP x P

Mass balance perfectly mixed Since

30 C

47.8 C

1 R M xr

0.

Fout x r ,out

x r,in FP

0.8 . Then

Fin

Fin Fin Fout 1 x r ,out 0.10 0.125 , Fout 0.8Fin 0.8 Alternate graphical solution gives same result. 6.5

xP

9.959

1 0.8

80 kg h

Op. line

xP

-4

xr

x r,in 0.10 0.125

xr

x r,out

xP

1

RT curve

FP Fin FP Fin

x r ,out

Slope

x r ,in FP Fin

0.8

4 0.2 When x r,out 0, x P

x r,in

0.10

FP Fin

0.2

b. Area = Fp / Jsolv = (20kg/h)(1 L)/0.997 kg)(24h/day)/ (2500 L/m2 day) = 0.193 m2. c. Gel formation occurs when x w = 0.5 and xw = M xout = 0.125 M. M = 0.5/0.125 = 4.0 d. Gel formation occurs when xw = 0.5 and xw = M xout = M xF / (1 – θ’) = 1.2 (0.1)/(1 – θ’) Then 1 – θ’ = 0.12/0.5 = 0.24 and θ’ = 0.76. e. Gel formation occurs when x w = 0.5 and xw = M xout = 1.2 xF / (1 – θ’) = 1.2 xF / (1 – 0.2). Obtain xF = 0.333. xr,out = xF / (1 – θ’) = 0.3333/0.8 = 0.416 f. We have xr,out = xF /(1 – θ’) = 0.20/0.75 = 0.26667. M = xgel / xr,out = 0.5/0.26667 = 1.875. First occurs when Jsolv = 2500 = k ln (M). Obtain k = 3977 L/(m2 day) = 4.603×10-5 m/s. g. M = 1.875 and k = 3977. Since we change the pressures, J changes which will change M. However with constant stirring k is constant. First, assume no gel and calculate J and M. pr pp K Jsolv 2500 L (m 2 kg) L J solv K solv , solv 2083.33 2 t ms t ms p r pp 2.2 1.0 bar m day bar Then, without gel,

J solv

2083.33 3.4 1.0

5000 L m 2 d ay

From Eq. (17-34) M = exp (Jsolv / k) = 3.516.

434

0.5

0.2. Then, x w

3.516 0.2

Mx F 1

0.878, gel forms.

.8

0.2

xF

With a gel, previous work is incorrect. Set R = 1.0, x p = 0, x r

0.25, 1 .8 And from Eq. (17-45), Jsolv = k ln (xgel/xr) = 3977 ln (0.5/0.25) = 2756.6 L/(m2 day) Note: The same answer is obtained in parts g and h if convert to J´solv and use k in m/s. h.

J solv

k n

x gel

.5

k n

xr

Case C, R C

xp

J solv

MB

x IN

1 R C x out

K solv

pr

tms

1.387 152 1.1

m d ay

pr

pp

1 RB

pr

pp

a M C xout

Equation:

log P

log P

0.27027

2446.6 L (m 2 d ay)

0.27027

1.0, p p,B

1.0 .061 14.1 .024 10.96

B

0.0007505]

1.1, p r,B

12.06

3.27

b

18.71 g m 2s

42 , P 10000,

1.6232 0.27875

1.9

4a b 4a b

1.9020 2b b 0.95100 , a 0.16805 logP 0.951

log

.74

xp

b

a log 10, 000

8, log

C

0.0001,

a log 0.0001

log 1.9

0.5

n

0.2

0.01230

15.446[ 3.27 .01230

log 42

0.2, x r

15.2

1 RC

pp

0.26, x F

0.976, M B

1.1, p r,C

17.D17. New Problem in 3rd edition. a) log a log P b , P

b) If

L 2

0.0093 0.26 .74 1 .939 .26 0.0007505

1

1 RC

3977,

Case B, R 0B

.939, p p,C

For Experiment C. M C

x out

3977

xr

17.D16. a) New problem in 3rd edition

b)

k

Gel forms since it did previously,

.16805

0.90309

log 0.16805

0.28509

.951 0.16805

PO2

1.93 Barrers

435

17.D.18. New problem in 3rd edition. Ideal Gas: Vol% = Mole %

PN2

250, PCO2 FP

2700,

PHe

300, PHe 3

0.4 m3 s. FIN

FIN

550

1.0 m s

10 2 cm

10 3

10 6 cm 3 s , FP

3

m 76 cm Hg

Part A

pr

2.5 atm

Part B

pr

76 cm Hg

t ms

0.8 mil

Eq. (17-11d)

10

190 cm Hg

atm

pp

0.4 76

.00254 cm

FP A

30.4 cm Hg

Need to guess value of FP A or of y P ,

Pi t ms p p

Since CO 2 has highest permeability, CO 2 will be concentrated; thus, guess y p,CO2

y r ,CO 2

Then

y p,CO 2 ,guess

1

FP A guess

Then,

PCO 2

y IN ,CO 2 1

t ms y p,CO 2

76 cm Hg

0.002032 cm

mil

Pi t ms p r

K m,i

pp

.4 106 cm3 s

yCO2,IN

0.4 and y IN,CO2

where p r y r.CO 2

0.40 0.4.

p p y p,CO 2

Use FP A in Eq. (17-11d) to calculate all K mi Then check

y IN,i

y r,i

1

K mi 1

1.0 ?

Put in Spread Sheet. Can use Goal Seek to force Results: a. y P,N2

y r,N2 b.

.15037, y P,CO2

.3164, y r,CO2

.54446, yP,He

.3037, y r,He

y r,i

1.0 as change y P,CO2 .

.3351, yP,H 2

.06099, y r,H2

.3189

.27154,

.999885

1.0000766

Same answers for mole fractions since p r p p is same.

yF N2 F,cm3/s pr, cm Hg P N2 P He Fp yp CO2 Fp/A K N2 K CO2

306 2008

HW 8

Problem 2

0.25 1000000 190 0.000000025 0.00000003 400000 change yp to 0.544455884 0.003983763

yF CO2 tm, cm pp, cm Hg P CO2 P H2

yF 0.4 He 0.002032 theta 76 0.00000027 0.000000055

get sum=1 yr CO2 A, cm 2 A, m 2 0.475237299 y r N2 1.792765611 yr CO2

0.05 yF H2 0.4

0.3

0.303696077 100407575.7 10040.75757 0.316417678 0.303696077

436

K He K H2

0.549397235 yr He 0.851323363 yr H2 Goal seek Sum 0.150373483 0.544455884 0.033509684 0.271546029 0.99988508

yp N2 yp CO2 yp He yp H2 sum

0.060993544 0.318969314 1.000076613

17.E1. New problem in 3rd edition For dilute systems J solvent Transfer Eq. (17-7c)

R

1

xp x out

1

J total solution , FIN

Fp

xF

x out

0.022 0.056

FIN

xp

x out

0.00032 0.056

0.00032

0.9943 , J s u c rose

0.056

J solventx p

Fout , Basis: FIN

Fp

1.0

0.6106

J solvent

x

solution

p

Permeate

0.997 0.4 x p 0.997 0.4 0.00032 0.99713 kg L Initial assumption is OK. J solv J solv / solv (3.923 g / m 2 s) / (997 g / L) 0.003935 L / ( m 2 s) permate

(3.923g / m 2 s)(0.00032)

J sucrose b)

Eq. (17-27)

0.00126g / (m 2s)

1 xp

K water

water solute

K sucrose

x p pr

Mxr

pp

xp

Mxr

xp a

1 .00032 0.056 0.00032 w s

0.00032 60.0 1.1

From Eq. (17-16c),

K water pr

Eq. (17-18) c.

pp

a Mxr

59.895 1.0 .056

K sucrose

M xr

R

1 M 1 R

Then RT equation is

xp

1 R x out

Operating Equation is (17-23)

xp

0.0706 g

0.00126 0.056 0.00032

xp

Solution 1.

xp

0.00032

J sucrose

tms

g sucrose

xp

3.923 60 1.1

g water

J solv

tms

K water t ms

0.056 0.00032 59.895

3.131

1

0.6393 x out

1

2.1 1 0.9943

m 2 s atm

0.0226

g sucrose m 2s wt frac

0.98803

0.01197 x out x out

xF

.39

0.036066

Solve RT & operating equation simultaneously. x out

.61

x out

0.05537 , x p

0.022 0.61

0.000663

437

Check

R

pr

t ms

60 1.1

pp

a M xr

K sucrose

Mx r

t ms

3.67g/(m 2s)

0.000663]

J water x p / (1 x p )

0.00243 g (m2 s)

3.67 0.000663 / (1 .000663)

Alternatively, J sucrose

xp

59.895[ 2.1 0.05537

J sucrose

J sucrose

0K

0.98803

x out

K water

J water = 0.0706

xp

1

xp

0.0226[ 2.1 (0.05537) 0.000663]

0.00261

g m 2s

6.9% different

xp xp

Solution 2. RT Eq. (17-21), x r

2.1 1 xr

Simplifies to, Linearize x r @ x p Slope =

3.140 59.895 0.997

0.0003

xr

0.02509

3.140 59.895 .997

1 xp

0.02509 . Note xp = 0.0003 is an arbitrary point.

0.01196

0.01196 x r or x p Solve simultaneouly with Operating Equation xF 1 xp x out 0.6393 x out 0.036066

0.05538 , x p

1

2.1 391.66 x p

Then linear form of RT equation is x p

x out

3.140 .997 60 1.1

x p 186.5 x p 185.391

0.0003

xp

1

0.01196 x out

0.0006623 . Very close to value obtained with retention analysis.

C PL,in

FP

17.E2. (was 16.D11 in 2nd ed.) Eq. (17-59b):

Fin

Tin

Tout , Tin

Tout

85 25

60

P

Stage 1. Assume y P ~ .05 water, 0.95 ethanol P

0.95

For Feed CPL,in

w

0.05

E

0.1 CPL,w 55 C 0.1 4.1915

2290.3 kJ kg (See Example 17-9)

0.9 CPL,E 0.9 2.7595

kJ kg K

2.903

kJ kg K

Where average temperature from 25 to 85 is 55ºC and C P values are from Perry’s 7th, pp. 2306 and pp. 2-237.

FP1

2.903

Fin

2290.3

60

0.0760 and FP1

0.0760 100

7.60 kg hr .

438

Op. line intersects y P

x out

x IN 1

Slope Op. Eq. is, y P

0.10 (water wt. frac.) 1 0.760

0.1

12.16 x out

12.16

0.0760

12.16 x out 1.32 0.0760 If y P 1, x out 0.32 12.16 0.0263 Plot Op. Line on Figure 16-17a and find intersection: y P1 0.66, x out1 x 2,in 0.055 (water values)

Fout1

Fin 2

100 7.60

92.40 kg h

Stage 1 Trial 2. Since yp ≠ yp, assumed, do a second trial.

yP

0.66 water, 0.34 E, FP1

2.903 60

F1in

1892

FP1

0.0921 100

w

0.34

9.21 kg h , Fout1 9.86 , y P

0.0921 0.086

If y P

0.66

0.66 2359

E

0.34 985

1892 kJ kg

0.0921

1 0.0921

Slope

P

100 9.21 91.79 kg h 0.1

9.86 x out

0.0921

9.86 x out

1.086

1, x out

0.00872 9.86 Plot operating line and determine (from graph)

y P1 0.64, x out1 x in 2 value of yp is close to the assumed value of 0.66. Can proceed to stage 2. Stage 2: Estimate y P For x in 2 2

0.50 (water),

0.050, CPL,in

FP2

2.967

Fin 2

1672

Fout 2

60

0.05 4.1915

0.50

.1065 90.66

0.050

1 0.1065 0.1065

1672

E

2.967 9.64 kg h

8.40

8.40 x out 0.4421 0.1065 0.4421 . Draw op. line. Intersection gives y P 0.34

For x out 2 use MB. x out 2 ,

8.40 x out

w

0.95 2.903

8.10 kg hr , Slope

90.66 9.64

0, y P

0.50

0.1065, Fin 2

yP

If x out

P

0.05 water. This

Fin 2 x in 2 Fout 2

FP2 y P2

90.66 0.050

9.64 0.34

81.00

x out 2 ,w 0.0155 or x out,2 ,ETOH 0.9845. This is a close as we can get graphically. 9.21 0.64 9.64 0.34 Mixed Permeate: y p,mix 0.487 wt frac water 9.21 9.64

439

Area

FP

kg 1000 g h kg

J g h m2

, J from Fig. 16-17b based on x out

Stage 1

J

0.8333 g h m 2 , A1

Stage 2

J

0.208 g h m 2 , A 2

9.34 1000 g h 0.8333 9.64 1000

11, 208 m2

46, 346 m2

0.208 Other flow patterns will reduce area. Area is large because of low flux caused by low ethanol permeation rate.

17.F1.

RT eqn., y

x x 1

1 x

x Benz

0.2,

18.3, y

x Benz

0.1,

6.66, y

Operating equation Slope

1

, x Benz

0.3,

18.3 .2 1 17.3 .2

6.66 .1 1 5.66 .1 .9 .1

16.6, y

16.6 .3 1 15.6 .3

0.87676

0.8266

0.4253 . Plot RT equation.

9 . Plot on graph. Find y PBenz ~ 0.844, x out,Benz ~ 0.238

440

P

CPL,in

y Pbenz

benz

1 y Pbenz

x Pbenxw CPLbenz

iP

0.844 94.27

1 x benz,in CPL ,iP

1 0.844 164

0.3 0.423

.7 0.73

105.15 cal g

0.6379 cal g C

441

Tin

Tout

P

C PL,in

50 C

0.1 105.15

66.48 C

0.6379

17.H1. (was 16.G1 in 2nd edition) This is set up for Area being the unknown and cut being known. Problem 17.H1 Fr,in 10000.000000 yin,A 0.2500 cut=Fp/Fin 0.2500 tmem,cm 0.002540 pr,cm Hg 300.0000 pp,cm Hg 30.0000 yin,B 0.5500 P,A 0.0000000200 Fptot 2500.000000 yin,C 0.2000 P,B 0.0000000050 Fr,out 7500.000000 P,C 0.0000000025 Guess values of A or equivalently Fp/A until sum y,r and sum u,p are = 1.00 Fp/A 0.0007059 (this is final result) KA 2.507328 KB 0.7720 KC 0.4015 sum x eq y,r,A 0.181576 y,p,A 0.455271198 y,r,B 0.583244 y,p,B 0.450268647 y,r,C 0.235190 y,p,C 0.094429331 Area, cm2 3541578.1 sum y,r 1.000010 sum y,p 9.999692E-01 These results agree very well with Geankoplis’ results. 17.H.2. New problem in 3rd edition Part a) y p 0.5243, y r,out 0.0610, A b)

y p,avg

0.6193, y r,out

3, 200,152 cm 2

0.0203, A

2, 636,196 cm 2

17.H3. New problem in 3rd edition Counter –current. Shows final guess for theta. Fin, cm3/s 100000 yin 0.209 thetatot PA/tms 0.003905 pr, cm Hg 114 pp, cm Hg M 15 N 100 yroutguess df 0.9 j=N-i+1 Fr yp yr Area Fp Fp/Fr,j-1 yp Areatot yincalc Fincalc Massbal yrout

100 28600 0.173174301 0.144051968 9710.750234 714 0.024356963 0.235015561 824015.8215 0.208999973 100000 9.09495E-13 0.144051968

99 98 29314 30028 0.1738615 0.174547 0.1447613 0.14547 9664.4075 9618.879 1428 2142 0.0475556 0.069677

97 30742 0.1752323 0.1461768 9574.1438 2856 0.0907935

0.714 PB/tms 76 0.2 erroracc

0.00175 0.0000001

96 95 94 31456 32170 32884 0.175916 0.1765981 0.17727918 0.146883 0.147588 0.14829195 9530.183 9486.978 9444.50995 3570 4284 4998 0.110973 0.1302761 0.14875885

442

17.H4. New problem in 3rd edition The spread sheet equations are shown below for part b. Part a agreed with problem 17.D14. Part b answers: yp,W = 0.412, θ = 0.2122, xout,W = 0.0160, θ’= 0.158, Area = 79.0 ft2. Note that if the starting guess for yp,W is too high, Goal Seek will converge on an answer with yp,W > 1, which is obviously not physically possible.

17.H5.

This problem is very similar to Example 17.7. It is easiest to solve on a spreadsheet, which is

shown below. The results are shown in the spreadsheet. New problem in 3rd edition

443

444

17.H6. New problem in 3rd edition The spreadsheet is similar to that for problem 17.H5 and is shown below,

17.H.7. The same spread sheet that was used in problems 17.H5 is used.

445

446

SPE 3rd Edition Solution Manual Chapter 18. New Problems and new solutions are listed as new immediately after the solution number. These new problems are: 18.A3, 18.A16, 18.B4, 18.C4, 18.C14, 18.D3, 18.D8, 18D9, 18D14, 18D15, 18D18, 18D21, 18D24, 18D25, 18.D29, 18.D30, 18.F1, 18.H1-18.H2. Chapter 18 was chapter 17 in the 2nd edition. Most problems from that edition have the same problem number, but the chapter number is now 18 (e.g., problem 17.D6 is now 18.D6). 18.A1. 1c; 2 b; 3a 18.A.2. 1c; 2a; 3b 18.A.3. New problem in 3rd edition. One barrier is lack of knowledge. Most chemical engineers are not trained in use of adsorption, chromatography, and ion exchange. Thus, they do not think of these processes as a potential solution. A second barrier is the simulation tools are not as developed and widely available as the simulation tools for equilibrium staged separations such as distillation. 18.A4. Regeneration is too difficult. 18.A5. In the SMB the mass transfer zone between the two solutes stays inside the cascade. In a chromatograph the MTZ exits the column and must either be completely separated, which requires a significant amount of desorbent, or recycled appropriately. 18.A7. d 18.A.8. New problem in 3rd edition. The LUB approach assumes constant pattern behavior. Linear systems do not have constant pattern behavior. 18.A9. 18.A10.

d e

18.B.4. New problem in 3rd edition. There are obviously many possibilities. One is to develop sorption processes that use an energy separation processes (e.g., pressure or temperature) to produce purge or desorbent from the feed so that a separate purge or desorbent does not have to be added. 18.C1.

T

1

e

Vavailable P

B

18.C4.

rd

e

1

e

1 f cry

1

e

1

P1

clay

P

e

e

e

P4

1

P1

f

f cry

P2

1 f cry f cry

1 cry

1

e P2

f cry

P2 P1

K di Vcol. 1 f cry

f

f cry

f

P2

(same as 18-3b)

New problem in 3 edition. Amount in mobile phase = e (Vol. Col. Segment) Amount in pores = 0 (no pores) Amount exchanged Δz A c cRT Δy K DE No 1 Obtain, u ion

e

c

e

Δz a c Δx cT

term because c RT is equivalent/L

ε e Δz A c Δx c T v int er

ε e Δz A c c T Δx Δz A c c RT Δy K DE v int er Simplify to, u ion (18-44) c RT y 1 K DE x ecT

445

18.C7.

CA

1

C AF

2

z uAt

1 erf

4E eff t u A v inter

12

Sketch of break through:

erf (a)

.9 a

1.164

95%

t final

5%

tw

t st

t1

erf (a)

At 5% point,

0.90 a 1.164

1.164

L 4E

t st u A v

12

Or let

u A t st

2.328 E t st x1

12 st

t , uAx

2 1

uA

12

x2

By definition, Use

u A t final uA v

12

12

L u A t st

v 2.328 E

12

u 1A2 v1 2

x1

2.328

L 2

0.

E u At v

4Lu A

2u A 2.328

Let

L

4E t final

E 1 2 u 1A2 2.328 v1 2

x1

1.164

and at 95%,

12 fin

t , then x 2 t MTZ

t final

t st

x 22

E 1 2 u 1A2 v1 2

2.328 2 E u A v 2u A

4L u A

x12

sign for both (has to be to have positive times).

446

4u

2 A

4u

2 A

x

2 2

x

2 1

2

2.328 E 1 2 u 1A2 v

2.328E 1 2 u 1A2 v

2

12

2

v

v

t MTZ

x

2 2

x

2 1

E uA

4L u A

2.328

12

2

4L u A

v 2

E uA

v

E uA

v

E uA

2.328

2

2.328

v

2.328E 1 2 u 1A2 v1 2

4

2

12

2.328E 1 2 u 1A2

2

12

2.328

2.328 E 1 2 u 1A2

2.328

2

E uA

v

4L u A

4L u A

4L u A

4u 2A 2

If

4L

2.328 E v t MTZ

2 2.328 E

12

v1 2 u A

18C9. New problem in 3rd edition. For Figure 18-7B, In

L

In – Out = Accumulation t vinter A c CT,after x i,after

Out

Accumulation

very reasonable since E is usually small ,

t vinter A cCT,before x i,before

LA c yi,after

yi,before CRT

LAc x i,after CT,after

Note that C RT is constant. After dividing both sides by

v int er x i ,afterC T ,after x i ,beforeC L

C RT y i ,after

y i ,before

x i,before CT,before

t A c , mass balance is

T ,before

L

x i ,afterC T ,after x i ,beforeC T ,before t t For Figure 18-7B with a total ion wave, L u total ion v int er t The first and third terms in the mass balance cancel each other. Thus, L C RT y i,after y i,before 0 t Which requires, yi,after yi,before 18.C10.

447

1 A

v1

2

v2

v2

F

v3

3

u A ,i

C A ,i L i

u A1

M 1A u port

vF

u BL

C B,i Vi

u B2

M 2 B u port

v B,prod

u A3

M 3A u port

vD

u B4

M 4 B u port

v3 v4

v4

B 4

v A ,prod

v1

CB v2

M 2B u port

CA v2

C A v3

M 3A u port

CA v 2

vF

M 3A u port CA M 2B

If all

18.D1.

Rearrange:

vD

(2)

M 3A u port

CB

M 2B CB

v4

M 3A CA

v1 where v 4

u B4 CB

v1 Thus

(1)

M 2B u port

vF

u port

vD

CB

CA vF

Subtract eq. (2) from (1),

Then

CA

u A1 CA

M 4B u port CB M1A u port CA

M 4B u port

M 1A u port

M 4B

M 1A

CB

CA

CB

CA

M 1A CA

M 4B CB

M 3A CA

D

vD

F

vF

Mi

1.0,

M 4B CB

D F

pA

1

qA

q MAX

1 CB

pA

1 CA

1 q MAX K A

1 CB

1 CA

M 2B CB

vf

M 3A CA

1.0

. Plot p A q A vs. p A

448

296 K p/q 135.863 278.679 478.666 696.073 939.619 1116.143 1189.735

p 275.788 1137.645 2413.145 3757.6116 5239.9722 6274.1772 6687.8589

At 296 K Intercept Slope

KA

0.163636

1 q max

, q max

1 q max

0.163636

1 q max K A

80

80

480 K p/q 1786.943 1709.129 1974.657 2309.538 2778.150 3011.134 3122.979

At 480 K

1

Intercept

q max K A

6.1125

0.00204545

p 637.7598 1296.2036 2378.6716 3709.3486 5329.6030 6246.5981 6687.8589

Slope

KA

1380

0.260606

1

q max K A 1 , q max q max

1 q max

0.260606

1 q max K A

1380

3.8372

0.00018884

449

18.D2.

L soln

a = 22 liter soln/kg ads = 22

1 kg

kg ads 1000 g

b = 375 liter soln/g mole anthracene = 375

q max K A,c C A

q

18.D.3. New problem in 3rd edition.

uj

Part a

1

1

e

p

a

0.022

K A,C

2.104

1

Kd

uj

b

2.104

u

c.

u AN

u DN

e

HETP

L g anth

v super

40

5.671 cm min

L u S,DN

4.576

5.567

u S,AN

L N

g ads

K ij

1782 0.00301

4Ru

N1 2

From (18-83)

2

L

g ads.

40 1 0.69

2.104

g anth

Time AN = L/uAN = 25/5.671 = 4.408 min 40 u DN 5.474, time 1 0.69 1782 0.00316

b.

104.33

u S,AN

N 10885

0.002297 cm 2

From (18-81),

2

width at half height 5.54 peak max N To find width in time units, peak max is in time units = retention L u S,AN 4.40864 min , width 0.09946 min

d. time

0.425 width 1/2 height

t

18.D4.

p

e

1 0.69 1782K j

1 mol

0.10456

s

, u AN

g ads

, vinter

1

e

L

mol 178.22 g

v inter

e

40

L

, thus, K A,C

1 K Ac C A

q max

0.022

v sup er

10.0

v int er

0.43

e

23.26

0.042271 min

cm min

v int er

a) u s

1

1

e

p e

us 0 1

Kd

1

1

e e

(18-15c) p

s

Kx T

f

23.26 0.6027 cm min 0.57 .48 1.0 0.57 0.52 2100 17.46 0.43 0.43 684 t br 200 cm 0.6027 cm min 331.8 min

450

b) Assume wall heat capacity is small: v int er u th 1 e 1 1 e 1 p e

u th 1.636 t th,br c) K x

t br M.B.in

10.0

e

23.26 .57 .52 2100 2000

C ps

F

C pf

5.911

cm min

23.26

1.23 g g @80 C , u s 80

200 cm 5.4868 cm min

A c cm 2

331.8 min 0.684 g cm 3

10.0 A c 0.684

331.8 33.84 .0011

Alternate: Eq. (18-24)

C 80 C 0

0.0011 g tol g fluid

33.84 0.0011

2.611 Cconc → C conc us 0 1 u s 80

0.0011 113.92

5.4868 cm min

1.636 2.603 36.45 min , see figure.

1

C 80

s

.43 684 1841 200 cm 5.911 cm min 33.84 min

out Simplifying:

p

1 u th 1 u th

36.45 33.84 C conc

297.96 .0011

1 0.6027 1 5.4868

2.611 1 5.911 1 5.911

0.1255 wt frac.

113.92

0.1253 wt frac . A very considerable amount of concentration occurs.

451

80º, C = 0 z usol (80ºC)

uth

0.0011

C=0 0.0011 331.8 min

0.0011

t Cconc

33.84 36.45 min 0.1255 Cout 0.0011 0 33.84 18.D5.

36.45

vsuper 20 cm min vint er vsuper e 20 0.4 50 cm min For step input w. unfavorable isotherm, get a diffuse wave. v int er Langmuir formula: u s 1 e 1 p 1 e a 1 Kd p s 1 bc e e But now

us 1

b .6 1.01 .54 .4

0.46

0 50 .6 .46 1.124 kg 1.2 .4 liter 1 0.46 c

c,g/l us, cm/min 18.2437 0 16.676 0.25 14.794 0.50 12.565 0.75 9.997 1.00 7.1813 1.25 1.50 4.3499

2

1.81

time, min

2

50 0.93067 1 0.46 c

2

tout = L/us, min 2.741 min 2.998 3.3797 3.979 5.002 6.9625 11.4944

452

18.D6. a)

f

u th 1

1

e

p

f

e

u th

.57 1 .5 .43

1

e

p

e

C ps

.684 2240

.57 .5 920 1.80 .43

50 cm 12.61 cm min v int er

1

e

p

e

c)

1

C pf

If wall effects are negligible, 0.684 2240 30

b) t thermal,br

u s 300K

C p f v int er

Kd

1

1

e e

p

K xy

s

W C pw eAc

12.61 cm min

3.965 min 30 .57 .57 1 .5 1.0 .5 12.109 .43 .43

u s 350K

6.5298 K xy

t br 300K

50 3.0964 min . Exits at c F

3.0964 cm min

4.423 in same eqn.

0.010 .

453

At t = 20, start hot, t br,hot Feed is concentrated. C 350 C 300

50 12.61 20

1 u s 300

1 u th

C 350

0.010 3.2989

t

1 u s 350

20 L u s 350 K

1 u th

23.965 min

1 3.0964

1 12.61

1 6.5298

1 12.61

3.2989

0.032989 g L . This continues until breakthrough at

20 50 6.5298

27.6572 minutes

0.032989

18.D6.

g/L

0.010

0 t

18.D7.

vint er

vsuper

16.1478

15 0.434

e

1

1

e

p

Kd

1

tr

L us

(A) p s

K 4

e

34.56 0.566 0.43

0.566 0.57 1.0

0.434 60 cm 0.3715 cm min

161.49 minutes. Then exits at C F

1

e

e

1

27.6572

34.56 cm min v int er

a) At 4ºC: u s

us 4 C

23.965

0.3715 cm min

1820 0.08943 0.434 161.49 min . Concentration out is zero from t = 0 to t =

0.01 .

454

v int er

b) u th

1.

1

e

p

1

1

e

e

e

C ps

s

C pf

f

WC pw e

A c C pf

f

34.56 17.293 cm min , 0.566 0.43 0.25 1820 0.434 1.00 1000 60.0 17.293 3.4696 min +1200

u th 1.743

t br,th

p

L u th

Eq. A but with K(60º)

u s 60 C

34.56 0.720258 cm min 0.566 0.43 1.743 1820 0.045305 0.434 t br,conc 60 L u s 60 60.0 0.720258 83.3035 min +1200 C=0 60º 60

60º

z

0º 60º C=0

uth Elution time: 0

c high 60

chigh

cF

CF

1 us 4

3.4696

1 u th

0.01 2.6918 0.05783

83.3

C high

1 u s 60

1 u th

1.38839 0.05783

18.D.8. New problem in 3rd edition. Example 18-3: vinter,F

0.3799 cm min , u s v inter,purge,0 C

u th vinter,purge,

6.466 cm min , u s vinter,purge,80 C

If t purge

t purge

0.019796 kmol m3

18.60 and yinter,purge

u s vinter,F, 0 C

u s v inter,F, 80 C

C=0

25.58 18.60

25.58 cm min .

0.3799

0.5225 cm min

4.343 cm min

18.60

4.343 3.158 cm min 25.58 hot purge time and t F is cold feed time, with

t hot wave breakthrough

vinter,purge

18.56 min (from Example 18-3) then breakthrough

equation is u s v Inter,F, 0 C t F u s v Inter,purge,0 C t

thermal,breakthro ugh

120 cm

455

120

tF

0.5225 18.56

290.35 min 0.3799 The next feed input at 290.35 + 18.56 = 308.91 min. This starts a cold thermal wave at v Inter,F , u th v Inter,F 4.701 cm min which breaks through in another 25.53 min for total time to cold breakthrough of 308.91 + 25.53 = 334.44 min. The solute is hot, first at v Inter,purge u s 80 , v Inter,purge

18.60

u s 80 ,v inter,F

3.158 cm min after 18.56 minutes. Next solute step is

4.343 25.58 u s vinter,purge, 80

t Exit Time Solute

18.56

u s,F 80 , vinter,F

120 4.343 18.56 tF

4.343 cm min and then

t purge

t

3.158

120

12.47 min

290.35 18.56 12.47

334.44

Since Exit Time Solute entire time.

t

321.38 min.

breakthrough cold wave, the solute is at 80°C the

Solute exits from 290.35+18.56=308.91 min to 321.38 minutes = 12.47 minutes & it exits at superficial velocity of 8.0 cm min . Mass Balance All solute in = Solute out t F vsuper A c c IN t out vsuper A c c out,AVG

tF

c out ,AVG

t out

290.35

c IN

0.0009 wt frac

12.47

0.02096 wt frac.

This is same as peak concentration in Example 18-3, but greater than x out,avg 0.00748. To have same concentrations need to recycle the material exiting at feed concentration in counterflow system. NOTE: Counter flow system has advantages of not contaminating the product end of the column and typically has less spreading of the zone. 18.D.9. New problem in 3rd edition. a.

u s,feed,M

e

vinter

vSuper e

e

1

e

p

Kd

0.05 m s. vinter

v inter

1

e

1

p

s

RTK A,p

0.5 0.43

0.1163 m s

0.01712 m s

from Eq. (18-27) is same as Example 18-4, M 0.2128 Pressurization Step Feed end (for pressurization) 0.75m (Measured from closed product end) M

z after which is 0.75 0.5584

y M after

0.003

Feed Step u sfeed

0.01712 m s

0.75 m

4.0 atm

0.2128

1.0 atm 0.1916 m from feed and

4.0

0.5584 m

0.2128 1

1.0 7 sec

0.001007 0.11984 m

0.1916 m for pressurization step

= 0.3114 m. Does not breakthrough in first cycle. From 0 to 0.11984 m, concentration is y F .

456

Blowdown. Measuring from closed top, z before

z after

0.4386 m

0.2128

1.0

0.5890 m 4.0 The far end of the feed wave does not get removed from the bed. 0.11984 or 0.75 0.11984 0.6302 m from closed end has z after

1.0

0.6302

0.4386

0.75 0.3114

0.2128

0.8463 m, so it all exits. The mole fraction of this portion is

4.0

y after

y feed

The close wave

1.0

0.2128 1.0

0.003 2.9781 0.00893 4.0 Part of the feed that was pressurized also exits during blowdown. 1.0

0.2128

1.3431 z before z before 0.5584 m from closed 4.0 (product) end. This is 0.75 0.5584 0.196 m from feed end. This gas entered at an unknown pressure between p L 1.0 and p H 4.0. Can calculate this pressure from Eq. (18-28c) This z after

0.75

p before

z before

p after

z after

1

z before

y after,press

0.003

y after,BD

0.001007

0.5584

4.0

4.0

1 0.2128

1.00003 atm

0.75

0.2128 1

0.001007 1.00003 This gas is depressurized to 1.0 & exits column

Exit from Col y

1

After Pressurization Step.

0.2128 1

0.00300 or essentially the feed composition.

4

0.008933 0 .0030 time

Part b. Want z after blowdown

0.75, then z before

z after

p after p before

0.2128

1.0 z before 0.75 0.5584 4.0 from closed end, which is 0.75 0.5584 0.1916 m from feed end. Want the feed to end at this point. During constant pressure feed step, feed travels u s,feed t F 0.01712 t F . Then for pressurization step z after (from feed end)

0.1916 0.1712 t F . From closed end this is

457

0.5584 0.1712 t F 0.5584 0.1712 t F

z after

z before

p before

0.2128

4

0.5584 or t F 0. 1 Thus, need a purge step if have feed step at constant pressure for complete cleanout. 18.D10. a)

pt.10 : z after

y after

0.4,

0.002

Travels,

0.75

p after

0.2128, p before

A

3.0

0.4 0.5

0.000876

1.015128 0.4 m

25.126s 1.0s for blow-down

0.01592 m s

3.0

.48

1 .2128

1.05128 atm

0.2128 1

3.0

b) Start with Arbitrary point at t = 1 sec (end repress) z after

p before

1 0.2128

2.4763 atm , y after

0.002

26.126s

0.48 (.02 from feed end) 3.0

0.2128 1

0.00172

.5 2.4763 Dist. Traveled @ t = 30s: 0.02 + 0.01592 × 29s = 0.48168 m For blow-down: distance from closed end = 0.01832 cm

z after

0.01832

Purge: u M,purge

0.5

0.5

.2128 1.0

0.026824 , y after ,BD 0.00172 0.007048 3.0 3.0 0.01751 m s . Exits bottom column during purge (point 11)

(distance traveled)/upurge 18.D11.

.2128

31 s +

0.5-0.026824 0.01751

58.023s

If repressurize with product, bed remains clean. Feed step is same as to point 3 (at 0.462 m from feed end) on Figure 18-13. Blowdown then pt. 4 (0.056 m from top) and purge exits at pt. 8 (56.36s) Product gas is cleaner (y = 0), but there is lower productivity – less feed per cycle. See Figure.

458

BD 3

4

y=0

y=0

y = 0.0082

8 18.D12. a) The clean bed receiving feed has a shockwave for Langmuir isotherm.

320 cm 3 min

v sup er

vsup er

r2

A c , where A c

6.366 cm min , vinter

vsup er

4 cm

2

50.2654 cm 2

6.366 10.434 14.669 cm min

e

v int er

u sh 1

1

e

p

Kd

1

p s

e

e

q

q after

q before

c

c after

c before

c after

1

e

where c before

50 mol m 3 , q after

q c

0, q before

0.190 50 1 0.146 50

0

1.1446 mol kg

14.669

u sh 1

0.566 0.57 1.0

0.566 0.43

0.434

0.434

t br

L u sh

1.1446

1820

50 cm 0.5843 cm min

0.5843 cm min

50

85.579 min

Outlet concentration is zero until t br then becomes 50. Concentrated solution eluted by dilute soln. Gives diffuse wave for Langmuir isotherm. v u s u diffuse 1 e 1 p 1 e a 1 K p d p 2 1 bc e e

us 1

0.566 0.57 1.0

14.669 0.566 0.43 1820

0.19

0.434

0.434

1 0.146 c

2

1.74336

14.669 193.92 1 0.146c

2

Create Table.

459

18.D13.

A

c

50, u s

3.218, t

L us

50 3.218 15.537 min

c

0, u s

0.07497 cm min , t

c

40, u s

2.491, t

20.071 min

c

30, u s

1.737, t

28.779 min

c 15, u s

0.7052, t

70.898

c

0.2205, t

226.80

5, u s

de xtran, B

L us

666.93 min

fructose (1)

CA v1

u A1

M1u port

CB v 2

u B2

M 2 u port (2)

C A v3

u A3

M 3u port

CB v 4

u B4

M 4 u port (4)

(3)

v F,sup er 1000 cm 3 min , vF 2 40 e 4 CB vF Solve eqs. (2) and (3) simultaneously, u port CB M2 M3 CA v2

CA

v3

1 1

1

e e

v2

v3 v4

KA

1 .6 1 0.23 .4

0.7435 , C B

M1 CA M2 CB

u port

u port

0.97 0.7435 0.99 0.4914

3.03175 cm min

60 3.03175

M 4 u port

1.03 3.03175

CB

0.4914

KB

0.4914

19.791 min

6.1079 cm min ; V2,sup er

0.7435

1 .6 1 .69 .4

L t sw

3.03175

CA

V1,sup er

1

e

3.955 cm min : V1,sup er

1.01 3.03175

Recycle flow

1 1

3.03175

M 3 u port

1.9894 cm 3 min

e

0.4914 1.9894 cm min 0.4914 0.99 1.01 0.7435 L t sw u port

u port

v1

Vol Feed D2 4

v F , v F,super

v1

D2 e

4

1988.176 cm 3 min

3070.15 cm 3 min

4.1184; V3,sup er

2070.14 cm 3 min

6.3547; V4,sup er

3194.19 cm 3 min

1988.176 cm3 min

460

D

V4,sup er

Check:

3194.19 1988.176 1206.0 cm3 min ,

V1,sup er

VD

D

VF

F

V4

V1

VF

F V2

Extract Product

V4

M 1 u port

CB

CA

vF

M2

V3

M1 CB CA CB M3 CA

.97 .4914 .7435 1.2060 , OK .4914 .99 1.01 .7435 V1 3070.15 1988.18 1081.97 cm 3 min 3194.19 2070.14 1124.05 cm3 min

18.D.14. New problem in 3rd edition. From Eq. (18-40c) K K

KK

Anderson’s data:

1.2060

F

1.03 .4914

D

Raffinate Product

M4

M4 CB

D

2.9 1.3

H

H

KK

Li

KH

Li

2.2308

DeChow’s data: K K H 2.63 1.26 2.0873 For the shockwave Eq. (18-46) holds for K+ Since resin is initially in H+ form, x K,before CK,before CT 0 and y K,before

a)

x K,after

CK,after CT

y K,after

CR ,K,after CRT

CR ,K CRT

0.

1.0 1.0 v inter

u sh ,K

y K ,after 1 C RT K DE ,K x K ,after e CT

1

25 0.42

u sh ,K

y K ,before x K ,before

L

44.84 min 1 2.2 1 0 u sh 1 1.0 0.42 0.1 1 0 Same for both sets of data since K K H does not enter into equation when initial and feed contain only one ion. b) C t 1.0, u Sh,K 9.542 cm min , t sh 5.24 min c)

Ct

1.0, x K,before

yK Anderson’s Data: y K ,before

y K ,after

0.2, x K,after

0.85. y K values depend on equilibrium parameter.

K KH x K 1

K KH 1 x K 2.2308 0.2

1

2.2308 1 0.2

2.2308 0.85 1

1.115 cm min, t sh

2.2308 1 0.85

0.3580 0.9267

461

u sh

25 0.42 1 2.2 1 0.42 1.0

1.0

0.9267 0.3580 0.85 0.2

10.662 , t K

L u sh

4.69 min

DeChow’s data:

y K ,before y K ,after

2.0873 0.2 1

0.3148

2.0873 1 0.2 2.0873 0.85

1

0.9220

2.0873 1 0.85

25 0.42 L u sh 4.95 min 10.100 , t K 1 2.2 0.9220 0.3148 1 1.0 0.42 1.0 0.85 0.2 4.69 4.95 % difference 100 5.55% 4.69 d) There is a difference if either initial or feed contains both ions. System with higher K K H had higher shock velocity. u sh

v

18.D15. New problem in 3rd edition. Part a. u sh ,i

y i,after 1 c RT K DE x i,after e cT

1 For both

Na & K ,

x i,after

y i,before y i,after

t center

0 1.0

v

u i,sh 1

y i,after 1 c RT K DE x i,after e cT

y i,before x i,before

For both Na+ and K+: xbefore = 0.4 and xafter = 0.9. For Na+ K Na H xNa (2.0 / 1.3)(0.4) y Na ,before 1 ( K Na H 1) xNa 1 [(2.0 / 1.3) 1](0.4)

y Na , after

x i,before

25 0.42 5.186 cm min 1 2.2 1 0.42 0.5 L u sh 50 5.186 9.64 min

Thus same u sh , u sh

Part b.

x i,before

y i,before

K Na

H

1 ( K Na

H

xNa 1) xNa

(2.0 / 1.3)(0.9) 1 [(2.0 / 1.3) 1](0.9)

0.506

0.933

462

v

u sh ,Na

y Na ,after 1 c RT K DE x Na ,after e cT

1 t Na

L / ush , Na

50 / 5.98

(25 / 0.42) 1(2.2)(1.0) 0.933 0.506 1 (0.42)(0.5) 0.9 0.4

y Na ,before x Na ,before

5.98

8.36 min .

For K+ we obtain,

y K ,before

y K , after

KK

H

1 (KK

H

KK

H

1 (KK

H

u sh ,K

Part c.

(2.9 / 1.3)(0.4)

1) xK

L / ush , K

xK

(2.9 / 1.3)(0.9)

1) xK v

50 / 7.054

(25 / 0.42) 1(2.2)(1.0) 0.953 0.598 1 (0.42)(0.5) 0.9 0.4

y K ,before x K ,before

7.054

7.09 min .

1 c RT dy K DE dx e cT dy Na

K Na

dx Na

1

dx Na

t Na

K Na

1

L u shNA

KK 1

1

0

xK

KK

K Na x Na

dy K

.9

Li

KH

Li

KH

Li

Li

2

1 x Na

25 0.42 1 2.2 1 0.955 .42 0.5

u Na

5.409

1 xK

H

xK

2

1

KK

0.855, u K

Li

Li

KH

KH

Li

Li 2

1 xK

25 0.42 1 2.2 1 0.855 .42 0.5

5.979

H

2.0 1.3 1.538,

u Na

3.477,

t Na

14.38 min

0

KK xK

2

0

dy Na

dx K

K Na

KK

H

2.9 1.3 1 .5

dx Na

1

0.955,

2

2.9 1.3

dy dx

2

9.244 min

dx K

x Na

1 x Na

H

2.0 1.3 1 .5

dy K

x Na

K Na

H

2.0 1.3

dy Na

Part e.

0.953

1 [(2.9 / 1.3) 1](0.9)

v

u 1

Part d.

0.598

1 [(2.9 / 1.3) 1](0.4)

y K ,after 1 c RT K DE x K ,after e cT

1 tK

xK

H

2.9 1.3

2.231,

uK

2.442,

tK

20.47 min

0

.9

463

dy Na

1.538

dx Na

1

dy K

0.538 .9

2.231

dx K

1

0.502,

2

1.231 .9

0.698, u

2

7.159, t Na

Na

uK

6.984 min tK

9.5075,

5.259 min

Part f. The velocities and hence the derivatives are equal. Thus, K Na

dy Na dx Na

1

K Na

KH

Li

KH

Li

Li

2

1 x Na

KK

dy K

Li

dx K

1

KK

Li

Li

KH

KH

Li

Li

1 xK

2

With xNa = xK. The result from a spreadsheet is x = 0.35056 18.D16.

vint er

vsup er

15 0.40

e

MW p f

37.5 cm s

28.9 g mol 50 kPa

1.0 kg

0.5832 kg m 3

3

m kPa 1000 g 298 K mol K q kg toluene kg carbon . Then, shockwave velocity is is in c kg toluene kg air v int er RT

u sh

0.008314

1

1

e

p

1

Kd

1

e

e

e

q y

s f

37.5 cm s

u sh 1

0.6 0.65 1.0

0.6 0.35

0.4

0.4

1500 kg m 3 q 2 0.5832 kg m 3 y 2

q1 y1

37.5 cm s

u sh

For

1.975 1350.308

q2 y2

u sh ,1 : y1

0, y 2

0, q1

u sh 2 : y1 y2 u sh 2

q1 y1

0.0005, q 2

37.5

u sh ,1

At

p

2000 0.0015 1 2200 0.0015

L min

0.47619 0.104976 cm h

0.69767

37.5 0.69767 0.47619 1.975 1350.308 0.0015 0.0005

L min : u sh1t

1 2200 0.0005 0.00002916 cm s

0.47619 1.975 1350.308 0.0005 0.0005, q1 0.47619

0.0015, q 2

2000 .0005

0.00012539 cm s

0.451393 cm h

u sh 2 t 10 h where t is in hours.

464

Solve for

u sh 2 10

t

0.451393 10

13.03 h 0.451393 0.104976 cm L min u sh1 t 0.104976 13.03 h 1.368 cm h Thus, for any column of partial length we will see a single shockwave exit the column. v sup er 21.0 18.D17. v int er 52.5 cm s 0.4 e pV n RT u sh 2

v

Cinit

Since

u sh1

MW n

MW p

28.9 50

V

RT

1000 g kg 0.008314 298

C F , Get 2 diffuse waves v int er

us 1

1

e

p

Kd

1

1 y 0.0010 0.00075 0.00050 2nd wave (0.00050) 0.00025 0.00

1

e

e

us

p

s

e

f

52.5 0.6 0.35

.6 0.65 1.0 0.4 0.4 q u s cm s y 195.31 0.0001991 284.799 0.0001365 453.515 0.00008573 - add 20 hours 832.466 0.00004671 2000 0.00001914

us y

0.5383 kg m 3

0.001

q y

where

q

2000

y

1 2200y

2

52.5 cm s q 1.975+1350.23 q y y t L u s 25 u s

1500 0.583

125,581s = 34.8835 h 183,117.6s = 50.866 h 291,596.6s = 80.999 h 100.999 h 535,250.5 = 148.681 + 20 = 168.681 h 1285937.96 = 357.205 + 20 = 377.205 h

0.00075

0.00025

z

us y

us y 0

0

0.005 t

2

us y

0.0005

465

80.999

34.88 0.001

50.866

0 0.00075 0.0005

c

·

100.99

· 168.88

0.00025 t

18.D.18. Part a.

. New problem in 3rd edition. u S,G 11.12 S cm min is calculated in Example 18-9.

20 0.61 1.0 0.88

u S,F 1 0

8.416 , u

From Eq. (18-93), N

4Ru u S,G

From Eq. (18-78a)

N

Part b.

1

u S,F

2

L v E eff

u S,F

2

9.771

2

229.465

L

2 229.465 5.0 cm 2 min

2N E eff v

114.73 cm

20 cm min

tG

L u S,G

114.73 11.25 10.20 min

tF

L u S,F

114.73 8.416 13.63 min

Part c. Eq. (18-80a),

K Ag

Li

t

L

1

uS

N

13.63 min

t ,F

K AgK

uS,G

0.39

L

18.D19.

377.2

KK

a) Ion wave: u total con

Li

vint er

8.5 2.9

vsuper

1/ 2

,

10.20 min

t ,G

229.465

1/ 2

0.673 min

1/ 2

1 229.465

2.93 , y Ag e

1

3.0 0.4

Breakthrough of ion wave, 50 cm 7.5 cm min

0.900 min 2.93 x Ag 1 1.93 x Ag 7.5 cm min

6.667 min

466

b) Shock wave,

v int er

u sh

y Ag after 1 C RT KE x Ag after e CT

1

before: x Ag

7.5 cm min 1 2.0 1.0 1.0

u sh 1

0.4

1.2

u s,Ag

x Ag

x Ag

0.5, u s

x Ag

0, u s

From spreadsheet:

1.0 .

y Ag

1.4516 cm min , t sh v int er 1 C RT dy KE dx e CT

1

7.5 cm min 1 2.0 2.93 1 1.0 0.4 1.2 1 1.93 x

1.0, u s

1

7.5 12.208 2.93

50 cm

u sh

1.4516 cm min

1 C RT KE e CT 1

1

34.44 min

1

2 Ag

3.0965

2

cm min

u_dif,Ag 3.097163211 2.852615804 2.598940969 2.337763116 2.071284133 1.802351071 1.53450084 1.27196632 1.019637556 0.782936124 0.567638906

2.0 eq L , c T

0.02 eq L , x Ca

K CaK C RT

0.6183 2.0

CT

0.02 75 cm, vsuper

61.83

K Ag -K

K Ag

1 x Ag

K

2

7.5 12.208 1 1.93 x Ag

L

, t out

us

7.5 50 1.8021 , t out 12.208 1.8021 1 3.86 7.5 L 0.5678 cm min , t out 13.208 us xAg 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

Column: L

L

v int er

2

16.147 min

27.745 min

50 0.5678

88.0555 min

t_dif,Ag 16.14380534 17.5277722 19.23860549 21.38796684 24.13961427 27.74154314 32.58388572 39.3091764 49.0370324 63.86217015 88.08416667

Was 18D23 in 2nd edition. Table 18-5, K CaK a.) c RT

x Ag before

1.0

c) Diffuse wave: u s

18.D20.

0 . after: x Ag

y Ag

y Ag before

K Ca

Li

K K Li

0.8 at t

5.2 2

2.9

2

0.6183

0.

shockwave .

20 cm min ,

vinter

20 .4

50 cm min

467

p

Feed:

u sh

0,

0.4, K E

e

v int er , before: x Ca C RT K E y 1 x e CT

from Eq. (18-43) y Ca

u sh

1.0

50 1 2.0 0.971965 1 1.0 0.4 .02 0.8

x Ca

K Ca C RT CT

0.8

0.971965 75

0.16407 cm min , t br

b.) Regenerate: at 500 min → Ion wave at vint er New

0 , after: x Ca

0, y Ca

u sh

50 cm min takes

75 50

457.1 min

1.5 min.

1.2366 y Ca unchanged. Use Eq. (18-43) with new value K Ca C RT CT .

0.9689677 . Obtain diffuse wave. 3

v imter dy Ca 1.2336 1 y Ca 1 x Ca u diffuse where 3 C RT K E dy Ca dx Ca 1 x Ca 1 y Ca 1 C T e dx Ca (Wankat, 1990, Eq. (9-25b)). dy Ca 50 At x Ca 0, 1.2366, u Ca 6.96088 cm min 1.0 2.0 dx Ca 1 1.2366 0.4 1.0 75 cm t out 10.7745 min (slow wave) 6.96088 At x Ca 0.96897, and y Ca 0.971965

dy Ca

1.2336 0.028035

dx Ca

0.03103

3

3

1.96897

0.908386

1.971965

50 9.022 cm min , t out 2.0 1.0 1 0.908386 1.0 0.4 At x Ca 0.5, y Ca 0.534927

75

u Ca

dy Ca

1.2336 0.465073

dx Ca

u Ca

.5

3

18.D.21. New problem in 3rd edition. vF u port , C Tol M2 M3

C Tol

1.5

8.312898 min (fast wave)

0.97013

1.534927

50 2.0 1.0 1 0.97013 1.0 0.4

Cy

3

9.022

75

8.546 cm min , t out

8.546

8.776 min . (in-between)

1 1

1

e e

p

Kd

1

1

e

0.132234 p

K Tol 300

e

468

0.0061 e 2175.2696 300

K Tol 300 K C xy

1

.95 .10007

vF

0.6479 v F 0.6479 cm min , L u port t SW 64.79 cm 1.05 .132234 0.95 .6479 C xy 6.1447 cm min , v3 v 2 v F 5.14476 0.10017 0.6479 C Tol .95 4.6547 0.132234

v2

M 2 u port

v1

M1 u port

v4

M 4 u port C xy

1.05 0.6479

v Tol prod

v2

v1

6.7914 0.10017 6.1447 4.6547 1.4900

v xy,prod

v4

v3

6.7914 5.14476 1.6466

vD Check: vOut 18.D22.

0.10017

1.63627 0.68930 12.1092

u port

E eff

v sup er us

v4

v1

6.7914 4.6547

v tol prod

ED

v xy,prod

u s2

dp 6 k M,c

20 cm 3 min

1 0

v 1

0.0105 e 2115.1052 300 12.1092

8.5972 , K xy 300 K

cm

2

3.1366 , v total in

K

2

1

K

2.035

D F

2.1367

vD

3.1367

vF

, where

6.366 cm min , 15.915

e

2.1367,

dp

1

6 k m,c

k m,c a p

v sup er

v

OK.

15.915 cm min

7.821 cm min

e

E eff

0.15

Eq. (17-69) X

cm 7.821 min

2

0.6 0.69

5.52 min

C

1

CF

2

Argument of erf, a

1

0.4

2

8.063 cm 2 min

z ust

1 erf

4 E eff u s t v int er

12

for step up

200 7.821 t 15.849 t

12

469

Step down:

1

X L, t 8

L u s t 8u s

1 erf

2

12

4 E eff u s t 8 v int er

323.04 7.821 t

Argument of erf , a

15.849 t

Total Solution X X L If t 25.573 min us

X If t

X

1

1

and X

2 L 8u s

2

2

1.998

4 X

31.2835

a

341.89

12

1 2

0.979235 0.0084 cF See also Problem 18.G1.

cF

1 2

0.499 c

1.44473, erf a

12

1 2

50 0.970835

Cinitial

1 .983186

1

e

.998 24.975 .

0.95847

p

Kd

1

CF1 X z, t 17.5

e

1

e

CF1X z, t 28

1.63627 0.68930 12.1092

0.132234 p

K Tol 300

e

8.5972 , K xy 300 K 1

0.0084

48.54

1

0.0061 e 2175.2696 300

C xy (300K)

0.999

0.983186 , X

Cinitial

Tol

K Tol 300 K

c

2.773, erf

0.979236

0.970835 , c

18.D23. Was 18D24 in 2nd edition. Cout

12

532.13

468.663

1.95847

25.0 (for smaller t, can ignore X )

63.96

31.2765

29.575, a

1.69189, erf a

18.D.24. New problem in 3rd edition. vF u port , C Tol (300K ) M2 M3 1 C C

0.50 or c

1 2, a

0, X

c

xy

0 , c cF

7.358

12

279.6

For higher t, X = 1.0,

0.999

Peak at 25.575

12

123.03

0, a

1 1.000

33.575, a

us 1

a

126.792

0.0105 e 2115.1052 300 12.1092

0.10017

470

u port

.90 .10007

vF

1.4812 v F

1.10 .132234

M 2 u port C xy (300K)

v1

M1 u port C Tol (300K)

v4

v2

v1

vD

v4

Check: vOut

v4

v1

13.3084 cm min , v3 0.10017 1.4812 .90 10.0812 0.132234

13.3084 10.0812

v3

v tol prod

v xy,prod

v2

vF

12.3084

16.2655

0.10017 16.2655 12.3084

16.2655 10.0812

148.12 cm

3.2272

1.10 1.4812

M 4 u port C xy (300) v xy,prod

u port t SW

0.90 1.4812

v2

vTol prod

1.4812 cm min , L

3.9571

6.1843,

7.1843 , v total in

vF

D F vD

6.1843 7.1843

OK.

18.D.25. New problem in 3rd edition. Zones 2 & 3 are same as in 18.D.24 since at 300 K u port 0.6479 v F 0.6479 cm min , L u port t SW 64.79 cm

v2

6.1447 , v3

v1

M1 u port CTol 273 K

M1

5.14476

0.5 and M 4

and v 4

M 4 u port C xy 350 K

2.0 (reciprocal values).

K Tol 273K

0.0061 exp 2175.2695 273

17.612

K xy 350K

0.0105 exp 2115.1052 350

4.423

1

C Tol 273K

1

C xy 350K

v1

0.2135

1.63627 0.68930 4.423

0.5 0.6479 0.07259

vTol prod

0.07259

1.63627 0.68930 17.612

v2

4.4627 , v 4

1.6820 , v xy prod

v1

v4

v3

2.0 0.6479 0.2135 0.9260 , v D

v4

v1

6.0707 1.608

D / F 1.608

18.D26. a)

N

u sD

2

4Ru s u sA 1.0 1 5.8

u s,B

, R

0.147059 , u

v

1.5 , u s A 1

1

1.0 s

KA

6.5

0.15385

0.15045

471

4 1.5 0.15045

Need N b) t R ,A

t ,A

CA C A ,max

17689 , L

0.0067873

L

884.45

uA

0.15385

L

1

uA

N

5748.88

95.813 min

t

tR

2

2 t

2

exp

0.05 N

884.45 cm.

95.813 min

12

exp

t,min CA CA,max

12

1

0.7204 min

17689

t 95.813 2 0.7204

2

p

2

90 92 94 95 95.813 96 97 7.27E-15 8.3E-7 0.0421 0.52898 1.00 0.9669 0.2573

CA

0.33 X A L, t

CF

18.D27.

18.D28. a) u p

2

25.0 cm

L t center

35.4 min

b) Large-Scale system

1.0 0.33 X A L, t

1 .55 X A L, t .8t F

0.55 0 X A L, t

0.706 cm min , L MTZ,lab

u pt MTZ

t MTZ, LS

d 2p ,LS D eff

1.0

t MTZ,lab

d 2p ,lab D eff

0.12

t MTZ,LS

Independent of velocity

0.4t F

69.44 2.8

tF

0.706 2.8

1.9774 cm

2

69.44

2

194.44 min

v super u p ,LS

e

u p ,lab

LS

v super e

lab

12

4

9

3

→ u p,Ls

0.706

4 3

0.941 cm min

lab

L MTZ,larg e scale u p t MTZ 0.941 cm min 194.44 min 183.03 cm For frac. bed use = 0.80 & symmetrical pattern, 0.5 183.03 0.5 L MTZ L 457.6 cm 4.576 m , t br t center 1 Frac bed use 1 .8 t center

457.6

L up

486.27 min , t br

486.27

194.44

0.941 2 This is length of feed time if column is completely regenerated. 18D.29. K CaK a.) c RT

K Ca

Li

K K Li

5.2 2

2.9

2.5 eq L , cT

2

t MTZ 2

389.05 min .

0.6183

0.03 eq L , x Ca

0.7 at t

0.

472

K CaK C RT

0.6183 2.5

CT

0.03

Column: L p

Feed:

u sh

90 cm, vsuper

0,

e

shockwave .

51.525

25 cm min ,

0.39, K E

y Ca (1 y Ca )

25 / .39

64.10 cm min

1.0

v int er , before: x Ca C RT K E y 1 x e CT

from Equilibrium,

vinter

0, y Ca

0 , after: x Ca

K Ca K C RT

x Ca

CT

(1 x Ca ) 2

2

0.7

400.75

Solve this for unknown y value. I used a spreadsheet. yCa 0.95128

u sh

64.1 1 2.5 0.95128 1 1.0 0.39 .03 0.7

b.) Regenerate: Ion wave at vint er New

K CaK C RT

(0.6183)(2.5)

CT

1.1

0.22000 cm min , t br

35.0 / 0.39 1.4057 y Ca

old y and with new value K Ca C RT CT 400.75 , find x Ca

90

409.10 min

u sh

89.74 cm min takes

90

1.003 min.

89.74

0.95128 unchanged. Use equilibrium with y Ca (1 y Ca )

2

K Ca K C RT

x Ca

CT

(1 x Ca ) 2

to

0.94251 . Obtain diffuse wave. 3

v imter dy Ca 1.4057 1 y Ca 1 x Ca u diffuse where 3 C RT K E dy Ca dx Ca 1 x Ca 1 y Ca 1 C T e dx Ca (Wankat, 1990, Eq. (9-25b)). dy Ca 89.74 At x Ca 0, 1.4057, u Ca 9.7631 cm min 1.0 2.5 dx Ca 1 1.4057 0.39 1.1 As an alternative can do numerical calculation of derivative. At x = 0, y = 0. x = 0.001, y = 0.001404 and y / x (0.001404 0) / (0.001 0) 1.404 , which is reasonably close.

90 cm

9.22 min (slow wave) 9.7631 At x Ca 0.94251, and y Ca 0.95128 t out

dy Ca

1.4057 0.04872

dx Ca

0.05749

3

3

1.94251

1.95128

0.85169

473

89.74 15.049 cm min , t out 2.5 1.0 1 0.85169 1.1 0.39 From equilibrium, at the arbitrary value x Ca 0.5, y Ca u Ca

dy Ca

1.4057 1 0.55544

dx Ca

3

.5

3

1.5

15.049

5.981min (fast wave)

0.55544

0.95282

1.55544

89.74 2.5 1.0 1 0.95282 1.1 0.39

u Ca

90

13.695 cm min , t out

90 13.695

6.572 min .

This is in-between the other two waves. c. To not have a diffuse wave must have

K CaK C RT

(0.6183)(2.5)

CT

CT

1.0

This requires CT > 1.546. 18.D30. New Problem in 3rd edition. K K

H

KK

Li

KH

Li

DeChow’s data: K K H 2.63 1.26 2.0873 a.) This will be a shock wave since K+ is more concentrated in the feed to the column than it is initially and KK-H > 1. v inter

u sh ,K

y K ,after 1 C RT K E ,K x K ,after e CT

1

Ct

1.0, x K,before

yK

y K ,before y K ,after u sh

tK

0.2, x K,after

1

0.85. y K values depend on equilibrium parameter.

K KH 1 x K

2.0873 1 0.2 2.0873 0.85

1

x K ,before

K KH x K

2.0873 0.2 1

y K ,before

2.0873 1 0.85

0.3148 0.9220

25 0.42 0.9220 0.3148 1.0 0.85 0.2

1 2.2 1 0.42 1.0 L u sh 49.5 min All three times are the same for the shock wave.

10.100 cm/min,

474

b.) This will be a diffuse wave since K+ is less concentrated in the feed to the column than it is initially and KK-H > 1.

v inter

u diffuse,K

1

dy K 1 C RT K E,K dx K e CT

dy K

K KH

At xK = 0.15, dxK

u diffuse,K

25 / 0.42 2.2(1) dy K 1 (0.42)(1.0) dx K

(1 ( K KH

2.0873 1) xK )

2

25 / 0.42 2.2(1) dy K 1 (0.42)(1.0) dx K

1.543

[1 (1.0873)(0.15)]2 59.524 dy 1 5.238 K dx K

59.524 1 5.238(1.543)

6.554cm / min

Thus, at xK = 0.15, tK = L/udiffuse,K = 500/6.554 = 76.29 min. Then at xK = 0.5 we obtain

dy K

K KH

2.0873

dxK

(1 ( K KH 1) xK ) 59.524 u diffuse,K dy 1 5.238 K dx K

2

0.876 [1 (1.0873)(0.5)]2 59.524 10.65cm / min 1 5.238(0.876)

Thus, at xK = 0.5, tK = L/udiffuse,K = 500/10.65 = 46.94 min. Then at xK = 0.8 we obtain

dy K dxK u diffuse,K

K KH (1 ( K KH

59.524 dy 1 5.238 K dx K

1) xK ) 2

0.5970

59.524 1 5.238(0.597)

14.42cm / min

tK = L/udiffuse,K = 500/14.42 = 34.67 min. 18.D31. New problem in 3rd edition. a.

vSuper

10

vint er

10 .4

25,

e

0.4,

L

30.0

475

c RT

2.4,

1.10,

KK

2.9 Na

1

y K ,after 1 c RT K DE x K ,after e cT

b.

u sh,exp t

c.

L MTZ

L t center

u sh t MTZ

L MTZ l arg e scale

1 2.4 1.0 .4 1.1

1

x K ,before

1 0 1 0

7.75 min . , t center ,measured

7.31 min

7.31 7.75

100 6.00% 7.31 30 7.31 4.10 cm min .

4.10 7.57 7.06

Frac. bed use (symmetric wave) d.

25

y K ,before

3.783 cm min, t center,exp ected % error

1.45

2.0

v int er

u sh ,K

u sh,K

cT

L MTZLab

2.093 cm

1 0.5 L MTZ L

d 2p v Super D eff

16 d 2p

l arg e scale

d 2p v Super D eff

0.965 d 2p

Lab

Lab

Lab

200 D eff

L MTZ,Lab

100 D eff

With same beads assume no change in D eff .

L MTZ,larg e scale

16 2 2.093 cm

frac bed use

1 0.5 L MTZ L 1 0.5 66.98 200 t center

Breakthrough start time

v inter,large scale

u sh

u sh ,lab ,exp tl

v inter,lab scale

Breakthrough start time

0.5 t MTZ

L u sh

2580 ft 3 , h=2580/860=3ft.,

End View

0.833

0.5 L MTZ u sh

8.2

[200 0.5(66.98)] / 8.20

18.F1. New problem in 3rd edition. Constraints: w L

wLh

66.98 cm

860 ft 2

p0

20.31min

T max

6 atm

500 C

932 F

88.14 psia

73.14 psig

Weight vessel

Di

ts

L

0.8 D i t s

Seider etal, s

Eq.16 59

h

w Seider etal. (2004), Eq. (16.61)

476

Pd

exp

0.60608 0.91615

Wall thickness

tp

(Eq. 16.60)

s

Relate w to h.:

n p0

96.66 D i

2SE 1.2 Pd

2 13,100 1

2

96.66 psig

3.7057E 3 D i

1.2 96.66

490 lbm ft 3 0.284 lbm in 3 p. 529 (Seider et al, 2004)

cos

cos

90 D

Pd D i

0.0015655

S 13.100 psi p. 529 with SA-387B steel, E = 1.0

where

Weight

n p0

3.7057E 3 D

0.5h

1.5

3

r 0.5w

r w

D w

r

D

D

cos 90

860 D cos 90

cos

D in ft. (In Spreadsheet A cos

cos

1

1

cos

3 D

cos

0.8D

3D

3

1

D

3.7057 E 3 D 490

1

In Spreadsheet, angle is in radius 90 2 D Weight Width L 3.5 33655 1.80 477.0 3.7 31358 2.17 397.1 3.8 30735.9 2.33 368.7 3.9 30325. 2.49 345.1 4 30070.31 2.65 325 4.1 29933 2.7947 307.7 4.2 29889 2.939 292.6 4.3 29918 3.08 279.2 6 35103 5.196 165 8 44837 7.42 115.96 10 56197 9.54 90.15 12 68940 11.62 74.02 14 83144 13.67 62.88 Goal seek L = 60 D = 14.64 ft Weight = 88052.75 Width = 14.333 L = 60 From Seider et al, p. 527: Cp

Cv

(Eq. 16.53) horizontal (Eq. 16.55)

CpL

From p. 531, Fm

Fm C v

Bare module factor, FBm

CBm

0.20294

2

118,323 → Cp

1.0 in 2000

Cp

144, 711 in mid 2000

118323 2724 121047

3.05 for horizontal

Cp Fm

Absorbent: p. 553 Cp

0.04333 n w

2724

1.2 for low-alloy steel, C v

Installed Cost: Calc C p with Fm

0.054266 ft

CPL in mid 2000 (MS = 1103)

exp 8.717 0.2330 n w

1580 D

ts

1.0, 2000

$60 ft 3 , Cp

3.05

1.0 1.0 1.2

60 2580

1

$393, 400

$154,800

477

18.G1. Was 17G1 in 2nd edition. Figures are labeled 17G1.

478

479

480

18.G2.

Was 17G2 in 2nd edition. a.) With QDS with 50 nodes find t center

t MTZ 18.G3.

6.0 3.13

4.52 min 2.87 min

Was 17G3 in 2nd edition.

Find

D F 1.0. D 141.55 E R CA 0.343 and CB 0.219

Eq. (17.31a)

u port

a)

M 2B CB

M 3A CA 141.55 cm 3 min

F

vF e

u port t sw

v1,int er

v1,sup er

Dc

2

4

0.4

10

2

4

4.5057 cm min

4.5057 1 0.219 L u port

CA vint er

Recycle Rate

vF

F.

u A1

2.7295 cm min 1 0.343 50 2.7295 18.32 min

M1u port

2.7295 cm min

2.7295

7.9577 cm min 0.343 0.4 v1,int er 3.18308 cm min

3.18308

10

2

4

250 cm 3 min

Obtained raffinate = 96.6% and extract = 94.3%. b) One approach is to keep a symmetric cycle. Then D = 283.1 and E F E R 212.325 2 Flow optimizer can be used to give t sw ~ 9.1 and Recycle rate ~ 500. Depending on values obtain raffinate and extract > 97%. 18.G4.

Was 17G4 in 2nd edition. Figure below is labeled 17G4.

18G5.

Was 17G5 in 2nd edition. Figure below is labeled 17G5.

481

482

483

18.G6.

Was 17G6 in 2nd edition. a.

k m,a p

1.5 min1 , L

25.0 cm

484

v sup er

20.2 ml min

Eq. (18-66)

2.0 m 2 4

6.366 cm min 6.366 cm min

19.1 1.5 min -1 Satisfied, but close. Thus some bypasses but most undergoes equilibration. 18.G7.

25.0 cm < 4.5

Was 17G7 in 2nd edition. Figure is labeled 17.G7.

485

18H1. New problem in 3rd edition. Spreadsheets with numbers and formulas shown.

486

487

18.H.2. New problem in 3rd edition. The spreadsheets are shown on the next pages. They are based on the previous, but includes both a step up and a step down. Because of the quirk in Excel not allowing negative arguments, it was set up with multiple solution paths. The correct solution occurs when there are numbers. Time, min

15

20

22.5

25

C

0

.0134

1.798

24.96

25.5726 27.5 25.0

30

33.575

42.32 48.52 24.97

35

37.5

11.13 1.114

40 .040

488

489

490