[Phillip C. Wankat] Instructor's Solution Manual -(BookZa.org)

February 13, 2017 | Author: Franco Gates | Category: N/A
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SOLUTION MANUAL

for

SEPARATION PROCESS ENGINEERING. Includes Mass Transfer Analysis 3rd Edition (Formerly published as Equilibrium-Staged Separations) by Phillip C. Wankat

SPE 3rd Edition Solution Manual Chapter 1 New Problems and new solutions are listed as new immediately after the solution number. These new problems in chapter 1 are: 1A3, 1A4, 1B2-1B4, 1D1. A2.

Answers are in the text.

A3.

New problem for 3rd edition. Answer is d.

B1.

Everything except some food products has undergone some separation operations. Even the water in bottles has been purified (either by reverse osmosis or by distillation).

B2.

New problem for 3rd edition. Many homes have a water softener (ion exchange), or a filter, or a carbon water “filter” (actually adsorption), or a reverse osmosis system.

B3.

New problem for 3rd edition. For example: the lungs are a gas permeation system, the intestines and kidney are liquid permeation or dialysis systems.

B4.

New problem for 3rd edition. You probably used some of the following: chromatography, crystallization, distillation, extraction, filtration and ultrafiltration.

D1.

New problem for 3rd edition. Basis 1kmol feed.

.4 kmole E  .4  MW  46   18.4 kg

.6 kmol Water  .6  MW  18 

10.8 kg total  29.2 kg

Weight fraction ethanol = 18.4/29.2 = 0.630

Flow rate = (1500 kmol/hr)[(29.2kg)/(1 kmol)] = 43,800 kg/hr.

17

Chapter 2 New Problems and new solutions are listed as new immediately after the solution number. These new problems are: 2A6, 2A9 to 2A16, 2C4, 2C8, 2C9, 2D1.g, 2.D4, 2D10, 2D13, 2D24 to 2D30, 2E1, 2F4, 2G4 to 2G6, 2H1 to 2H3. 2.A1.

Feed to flash drum is a liquid at high pressure. At this pressure its enthalpy can be calculated as a liquid. eg. h TF,Phigh

c p LIQ TF

Tref . When pressure is dropped the mixture is above

its bubble point and is a two-phase mixture (It “flashes”). In the flash mixture enthalpy is unchanged but temperature changes. Feed location cannot be found from T F and z on the graph because equilibrium data is at a lower pressure on the graph used for this calculation. 2.A2.

Yes.

2.A4. 1.0 Equilibrium (pure water)

yw

zw = 0.965 Flash operating line

.5

2.A4 0

0

.5

xw

1.0

2.A6. New Problem. In a flash drum separating a multicomponent mixture, raising the pressure will: i. Decrease the drum diameter and decrease the relative volatilities. Answer is i. 2.A8.

a. K increases as T increases b. K decreases as P increases c. K stays same as mole fraction changes (T, p constant) -Assumption is no concentration effect in DePriester charts d. K decreases as molecular weight increases

2.A9.

New Problem. The answer is 0.22

2.A10.

New Problem. The answer is b.

2.A11.

New Problem. The answer is c.

18

2.A12.

New Problem. The answer is b.

2.A13.

New Problem. The answer is c.

2.A14.

New Problem. The answer is a.

2.A15.

New Problem. a. b. The answer is

The answer is 36ºC

3.5 to 3.6

2.A16. New Problem. The liquid is superheated when the pressure drops, and the energy comes from the amount of superheat. 2.B1.

Must be sure you don’t violate Gibbs phase rule for intensive variables in equilibrium. Examples: F, z, Tdrum , Pdrum F, h F , z, p F, TF , z, p

F, z, y, Pdrum

F, TF , z, y

F, h F , z, y

F, z, x, p drum

F, TF , z, x

etc.

F, z, y, p drum

F, TF , z, Tdrum , p drum

F, z, x, Tdrum

F, TF , y, p

Drum dimensions, z, Fdrum , p drum

F, TF , y, Tdrum

Drum dimensions, z, y, p drum

F, TF , x, p

etc.

F, TF , x, Tdrum F, TF , y, x

2.B2.

This is essentially the same problem (disguised) as problem 2-D1c and e but with an existing (larger) drum and a higher flow rate. With y = 0.58, x = 0.20, and V/F = 0.25 which corresponds to 2-D1c. lb mole If F 1000 , D .98 and L 2.95 ft from Problem 2-D1e . hr Since D α V and for constant V/F, V α F, we have D α With F = 25,000:

F.

Fnew Fold = 5, Dnew = 5 Dold = 4.90, and Lnew = 3 Dnew = 14.7 . Existing drum is too small. 2 2 Fexisting D exist 4 2 Feed rate drum can handle: F α D. gives 1000 .98 .98 Fexisting 16,660 lbmol/h Alternatives a) Do drums in parallel. Add a second drum which can handle remaining 8340 lbmol/h. b) Bypass with liquid mixing

19

y = .58, V = .25 (16660) = 4150 16,660

25,000

LTotal x

8340

Since x is not specified, use bypass. This produces less vapor. c) Look at Eq. (2-62), which becomes

V MWv

D

3K drum 3600

L

v

v

Bypass reduces V c1)

Kdrum is already 0.35. Perhaps small improvements can be made with a better demister → Talk to the manufacturers. c2) ρv can be increased by increasing pressure. Thus operate at higher pressure. Note this will change the equilibrium data and raise temperature. Thus a complete new calculation needs to be done. d) Try bypass with vapor mixing. e) Other alternatives are possible. 2.C2.

2.C5.

a. Start with

V

zA

zB

F

KB 1

KA 1

xi

Fz i L VK i

Fz i

xi

Then y i

From

Kixi

yi

and let V

L

L F

xi

F L Ki

F L or x i

L F

zi 1

L Ki F

K i zi L 1 Ki F

0 we obtain

K i 1 zi L F

L 1 Ki F

0

20

zi

2.C7.

1 V/F f

0 0

Ki 1

.1 -.09

.2 -.1

V F

1 f

.3 -.09

V F

From data in Example 2-2 obtain:

.4 .5 -.06 -.007

.6 .07

.7 .16

.8 .3

.9 .49

1.0 .77

2.C8. New Problem.

p drum F=L+V

Tdrum

Fz y

x

Lx Vy

Solve for L & V Or use lever arm-rule

z

21

2.C9.

New Problem. Derivation of Eqs. (2-62) and (2-63). Overall and component mass balances are,

F

V L1

Fzi

L2

L1x i,L1

L2

x i,L2

Vyi

Substituting in equilibrium Eqs. (2-60b) and 2-60c)

Fz i

L1K i,L1

L2

x i,L2

Solving,

x i,L 2

L 2 x i,L2

VK iV

L2

x i,L2

Fz i L1K i,L 2

Fz i

L2

VK i,V

L1K i,L1

L2

F V L1

L2

VK i,V

L2

Dividing numerator and denominator by F and collecting terms.

x i,liq 2

Since

yi

K i,V

L2

x i,L2 , y i

2.D1.

a.

V

i 1

K i,L1

V

L2

x i,liq1

0.4 100

Slope op. line See graph. y b.

K i,L1 x i,L 2

L2

1

K i,L1

L2

C

1,

i 1

L2

yi

0

L 1 1 F

1

F V

K i,V

V F

L2

1

V F

C i 1

L2

C

yi

i 1

x i,L 2

1

L2

L 1 1 F

0

0

V 1 F

(2-62)

K i,L1 L 2 z i L K i,L1 L 2 1 1 K i,V F K i,L1 L 2 1 z i

K i,L1

K i,V

L2

L2

1

V 1 F

V F (2-63)

60 kmol/h

L V 3 2, y x 0.77 and x 0.48 600 and L

1

1 , thus,

C i 1

L2

1 zi

x i,liq 2 , we have x i,liq1

x i,liq 2

K i,V

K i,V L 2 zi L 1 1 K i,V F

L2

K i,V

40 and L

0.4 1500

c. Plot x

i 1

K i,L1

C

which becomes

In addition,

1 C

Stoichiometric equations,

Since x i,liq1

1

zi L 1 1 F

z

0.6

900 . Rest same as part a.

0.2 on equil. Diagram and y

V F z 1.2 0.25 . From equil y d. Plot x 0.45 on equilibrium curve.

x

z

0.3. yint ercept

zF V 1.2

0.58 .

22

L

Slope Plot operating line, y e. Find Liquid Density.

MW L Then, VL

xm

V z at z

x

x m MWm MWm

F V

xw

MWw

.2

w

L

v

RT

.2 32.04 32.04 .7914

MW L VL

P MW

v

.8

4 V V F .2 0.51 . From mass balance F 37.5 kmol/h.

x w MWw

m

Find Vapor Density.

1 V F

.8

.8 18.01

18.01 1.00

1 atm 26.15 g/mol

22.51 ml/mol

20.82 22.51 0.925 g/ml

(Need temperature of the drum)

MW v y m MW m y w MW w .58 32.04 .42 18.01 Find Temperature of the Drum T: From Table 3-3 find T corresponding to y .58, x 20, T=81.7 C 354.7K v

20.82

82.0575

ml atm mol K

354.7 K

26.15 g/mol

8.98 10

4

g/ml

Find Permissible velocity:

23

u perm

K drum

L

K drum

v

exp A

v

B

nFlv V

Since V

Wv L

F V

1000 250

15615

WV

L

6537.5

V MW v

A cs

u perm 3600 Thus, D

MW L

nFlv

lb

4

nFlv

20.82

15, 615 lb/h,

v

14.19 3600 8.98 10

4A cs

2.598

14.19 ft/s

250 26.15 454 g/lb 4

2-60

4

0.0744, and n Flv

.925

8.98 10

E

6537.5 lb / h

lbmol

4

4

3

250 lbmol/h,

750

8.89 10

.925 8.98 10

.442

D

250 26.15 L

V

2

0.25 1000

V MW v

WL

.442, and u perm

nFlv

F

F

750 lbmol/h, and WL

Flv Then K drum

C

g/ml 28316.85 ml/ft

3

2.28 ft 2 .

1.705 ft. Use 2 ft diameter.

L ranges from 3 D 6 ft to 5 D=10 ft Note that this design is conservative if a demister is used. f. Plot T vs x from Table 3-3. When T 77 C, x 0.34, y 0.69. This problem is now very similar to 3-D1c. Can calculate V/F from mass balance, Fz Lx Vy. This is V z y 0.4 0.34 Fz F V x Vy or 0.17 F y x 0.69 0.34 g. Part g is a new problem. V = 16.18 mol/h, L = 33.82, y= 0.892, x = 0.756. 2-D2.

Work backwards. Starting with x 2, find y2 = 0.62 from equilibrium. From equilibrium point V plot op. line of slope This gives L V 2 1 V F 2 3 7. F 2

z2

0.51 x1 (see Figure). Then from equilibrium, y1

For stage 1, 2.D3.

0.4

z1

x1

0.55 0.51

F

y1

x1

0.78 0.51

V F 0.6 V L y x z V F V Op. eq. 2 y x 2 3 3 See graph: y 0.55 x M 0.18

a.

z

V

0.78 .

0.148 .

6.0 k mol h, L

4.0

T ~ 82.8 C linear interpretation on Table 2-7 .

24

b. Product

78.0 C

x

Mass Bal: Fz or

0.30,

y

Lx Vy

4.0

0.665,

F V x Vy

10 V 0.3 0.665V

V 2.985 and V F Can also calculate V/F from slope.

c.

V

F 10,

y If y

L V

F

x

0.3

V

z

7

V F

z

0.8, x Then z

3&L

x

0.2985

7

z 0.3

0.545 @ equil 0.3 0.8

7

0.545 0.6215. 3 7 Can also draw line of slope through equil point. 3

25

2.D4. New problem in 3rd edition. Highest temperature is dew point Set

zi

yi .

Ki

K ref TNew If pick C4 as reference: First guess

yi

yi K i

1.0

K ref TOld K bu tan e

.2

.35

yi K i

1.0,

41 C : K C3

T

yi

.2

.35

.45

Ki

8.8

4.0145

.9

4.0145 .6099

yi

.2

.35

.45

Ki

6

2.45

.44

K C4,NEW

3.1,

K C6

0.125

.45

4.0145 T too low K i 3.1 1.0 .125 Guess for reference K C4 4.014 T 118 C : K C3 8.8,

K C4,NEW

0

yi x i

xi

Want

V F

2.45 1.2 .2

.35

.45

Ki

6.9

2.94

.56

.9

0.6099

2.45, T

85 : K C2

6.0,

K C6

0.44

1.20

2.94, T

yi

K C6

0.804

96 C : K C3

6.9,

K C6

0.56

Gives 84 C

26

K C4

Use 90.5º → Avg last two T

yi K i

2.D5.

2.7, K C3

.2

.35

.45

6.5

2.7

.49

6.5, K C6

0.49

1.079

T ~ 87 88º C Note: hexane probably better choice as reference. a) v1 = F2 v2

y1 = z2 z1 = 0.55

y2 1

2

F1 = 1000

V F

L

y1

V1

V1

V1

z Plot 1st Op line.

V

0.45454 & L1

V1

F1

1000

V1 = 687.5 kmol/h = F2

V

c) Stage 2 At x

F

0, y

0.25 ,

z V F

From graph y 2

V2

V F

F2

y1 = 0.66 = z2

y = x = z = 0.55 to x1 = 0.3 on eq. curve (see graph) 0.55 0.80 .25 0.454545 .55 0 .55

L

Slope

L1

F

x1

0.25 2

x2

x1 = 0.30

b)

p1,2 = 1 atm

L

V

687.5

F

1000

1

0.75F

3, y x V 0.25F 0.66 2.64. At y 0, x 2 0.25

0.82, x 2

0.25 687.5

0.6875 0.66. Plot op line

z2

F L

z

z

0.66

L F

0.75

0.88

0.63 . 171.875 kmol/h

2

27

2.D6.

V F = 1.0 kmol/min

T = 50ºC P = 200 kPa

zc4 = 0.45 zc5 = 0.35 Zc6 = 0.20

K i 1 zi 1

0 , First guess V/F = 0.6

Ki 1 V F

1.4 .45

0.2 .35

0.7 .2

1 1.4 .6 1 0.2 0.6 Use Newtonian Convergence

1 0.7 0.6

df k

F

i 1

V k 1

Fk

0.0215

2

K i 1 zi

c

d V F

V

Kc6 = 0.30

L

RR eq.,

f1

Kc4 = 2.4 Kc5 = 0.80

1

Ki 1 V F

2

fk df d V F

28

df1 V d F V

F2

f2

1.4 2 0.45

0.2 2

1 1.4 .6

0215

0.6

2

0.35

1 0.2 0.6

0.7 2

2

0.20

1 0.7 0.6

2

0.570

0.6377

0.570

1.4 .45

0.2 .35

0.7 0.2

0.00028 1 1.4 0.6377 1 0.2 0.6377 1 0.7 0.6377 Which is close enough. yi K i x i zi 0.45 x c4 0.2377, V 1 1.4 .6377 yc4 2.4 0.2377 1 Ki 1 F 0.35 x c5 0.4012, y c5 0.8 0.4012 0.3210 1 0.2 0.6377

x c6

2.D7.

0.20 1 0.7 0.6377

xi

V

zA

zB

F

KB 1

KA 1

KM

5.6 and K P

1.0002

0.3

0.7

F

0.21 1

5.6 1 zM

0.30 0.4012

yi

0.9998

0.3

xP

1 xM

0.1466 V 1 4.6 0.2276 1 KM 1 F 5.6 0.1466 0.8208 0.8534 , y M K M x M

yP

1 yM

0.1792

Use Rachford-Rice eqn: f

Converge on V F

0.1084

0.2276

K i 1 zi

V F

1

Find K i from DePriester Chart: K1

.076, V

Ki 1 V / F

73, K 2

F V F

4.1 K 3

152 kmol/h, L

0 . Note that 2 atm = 203 kPa.

.115

F V 1848 kmol/h .

zi

we obtain x1 .0077, x 2 .0809, x 3 .9113 V 1 Ki 1 F From yi K i x i , we obtain y1 .5621, y 2 .3649, y3 .1048 Need hF to plot on diagram. Since pressure is high, feed remains a liquid h F CPL TF Tref , Tref 0 from chart From x i

2.D9.

, y8

0.21

V

Eq. (2-38) x M

2.D8.

0.3613

0.5705

CPL

CPEtOH x EtOH

CPw x w

29

Where x EtOH and x w are mole fractions. Convert weight to mole fractions. Basis: 100 kg mixture 30 30 kg EtOH 0.651 kmol 46.07 70 kg water 70 18.016 3.885 Total = 4.536 kmol 0.6512 100 Avg. MW 0.1435, x w 0.8565 . 22.046 Mole fracs: x E 4.536 4.536 Use CPL at 100 C as an average CP value. EtOH

C PL Per kg this is

37.96 .1435

18.0 .8565

C PL

20.86

MWavg

22.046

0.946

20.86

kcal kmol C

kcal kg C

h F 0.946 2000 189.2 kcal/kg which can now be plotted on the enthalpy composition diagram. Obtain Tdrum 88.2 C, x E 0.146, and y E 0.617 . For F 1000 find L and V from F = L + V and Fz Lx Vy which gives V = 326.9, and L = 673.1

Note: If use wt. fracs. CPL

23.99 & CPL MWavg

1.088 and h F

217.6 . All wrong.

30

2.D.10 New Problem. Solution 400 kPa, 70ºC

K C3

From DePriester chart Know y i

Kixi ,

xi

zi 1

Ki

V 1 F

5, ,

z C4

35 Mole % n-butane

K C4

1.9,

xi

yi

K C6 1

x C6

0.7

0.3

zi

K i 1 zi 0 z C3 1 z C6 z C 4 .65 z C6 V 1 Ki 1 F zC6 zC6 V For C6 0.7 z C 6 0.7 1 0.7 V V F 1 K C6 1 1 0.7 F F V z C6 0.7 0.49 F 4 .65 z C 6 0.9 .35 0.7z C 6 RR Eq: 0 V V V 1 4 1 0.9 1 0.7 F F F 2 equations & 2 unknowns. Substitute in for z C6 . Do in Spreadsheet. R.R.

2.D11.

Use Goal – Seek to find V F. V 0.594 when R.R. equation 0.000881 . F V z C6 0.7 0.49 0.7 (0.49)(0.594) 0.40894 F L F 0.6 V F 0.4 & L V 1.5 Operating line: Slope 1.5, through y x z 0.4

31

2.D12.

For problem 2.D1c, plot x = 0.2 on equilibrium diagram with feed composition of 0.3. The resulting operating line has a y intercept z V / F 1.2 . Thus V F 0.25 (see figure in Solution to 2.D1) Vapor mole fraction is y = 0.58. Find Liquid Density.

MW L Then, VL

x m MWm xm

MWm

x w MWw xw

MWw

m

w

.2 32.04 .2

32.04 .7914 L

Find Vapor Density.

p MW v

RT

v

.8

.8 18.01 18.01 1.00

MW L VL

20.82

22.51 ml/mol

20.82 22.51 0.925 g/ml

(Need temperature of the drum)

MW v y m MW m y w MW w .58 32.04 .42 18.01 26.15 g/mol Find Temperature of the Drum T: From Table 2-7 find T corresponding to y .58, x 20, T=81.7 C 354.7K

32

1 atm 26.15 g/mol

v

82.0575

ml atm mol K

354.7 K

8.98 10

4

g/ml

Find Permissible velocity:

u perm

K drum

K drum,horizontal

L

v

1.25 K drum,vertical

Since V Wv L

F V

v

V F

B

0.25 1000

V MW v

1000 250

exp A

nFlv

250 26.15 lb lbmol

V

15615

WV

L

6537.5

A cs A Cs

MW L

8.98 10

0.5525

8.98 10

u perm 3600

Or y c

AT

D

2.D14.

xc 1 xc

cp

Raoult’s Law: K C 4

4

4

A Cs 0.2

1.25

20.82

15, 615 lb/h,

2.598

17.74 ft/s

4A T

4

g/ml 28316.85 ml/ft 3

9.12 ft 2

3.41 ft and L 13.6 ft

1.76 .7 1 .76 .7

1

,

cp

0.80418

0.5682

pc

VPC 4 PTot

4.04615 ,

VPC 4

11121 mm Hg

log 10 VPC 6

3.2658 ,

VPC 6

1844.36 mm Hg

1.0

4

0.5525

log 10 VPC 4

xi

nFlv

0.19582 cp

1

E

0.0744, and n Flv

17.74 3600 8.98 10

v

1.824 ft 2 ,

1 yp

3

250 26.15 454 g/lbm

2.D13. New Problem. The answer is ycresol = 0.19582 xp Since x c 0.3, x p 0.7, y p 1 1 xp

yc

4

0.925 8.98 10

V MW v

With L/D = 4,

nFlv

750

.925

0.442, and K drum ,horiz

u perm

D

6537.5 lb / h

L

WL

K drum ,vertical

2

nFlv

250 lbmol/h,

750 lbmol/h, and WL

Flv

C

zi 1

Ki 1 V F

1.0

33

0.3 0.7 1 11121 1844.36 1 1 0.4 1 1 0.4 P P Solve for Pdrum = 3260 mmHg zi xi V 1 Ki 1 F .3 11121 x C4 0.1527, y C 4 K C 4 x C 4 0.1527 11121 3260 1 1 .4 3260 1844.36 x C6 1 x C 4 0.84715, y C6 0.84715 0.47928 3260 Check 1.00019 2.D15.

This is an unusual way of stating problem. However, if we count specified variables we see that problem is not over or under specified. Usually V/F would be the variable, but here it isn’t. We can still write R-R eqn. Will have three variables: zC2, ziC4, znC4. Need two other eqns: z iC4 z nC4 constant, and z C2 z iC4 z nC4 1.0 Thus, solve three equations and three unknowns simultaneously. Do It. Rachford-Rice equation is, K C2 1 zC2 K iC 4 1 z iC 4 K nC 4 1 z nC 2 0 V V V 1 K C2 1 1 K iC 4 1 1 K nC 4 1 F F F Can solve for zC2 = 1 – ziC4 and ziC4 = (.8) znC4. Thus zC2 = 1 – 1.8 znC4 Substitute for ziC4 and zC2 into R-R eqn. K C2 1 .8 K iC 4 1 K nC 4 1 1.8 z nC4 z nC 4 z nC 4 V V 1 K C2 1 1 K iC 4 1 1 K nC 4 F F K C2 1 V 1 K C2 1 F Thus, z nC 4 K C2 1 .8 K iC 4 1 K nC 4 1.8 V V 1 K C2 1 1 K iC 4 1 1 K nC 4 F F Can now find K values and plug away. KC2 = 2.92, KiC4 = .375, KnC4 = .26. Solution is znC4 = 0.2957, ziC4 = .8 (.2957) = 0.2366, and zC2 = 0.4677

2.D16.

0.52091

1 V 1 F

0

1 1

V F

z C1 0.5, z C4 0.1, z C5 0.15, z C6 0.25, K C1 50, K C4 .6, K C5 .17, K C6 1st guess. Can assume all C1 in vapor, ~ 1/3 C4 in vapor, C5 & C6 in bottom V / F 1 .5 .1 / 3 .53 This first guess is not critical.

0.05

34

R.R. eq.

f

K i 1 zi

V F

1

49 .5 1 49 .53 Eq. 3.33

V F

.4 .1

.83 .15

.95 .25

1 .4 .53

1 .83 .53

1 .95 .53

f V F

V F

2

1

zi K i 1

1

1

Ki

V 1 F

2

V/F

1

0.53 and f V / F

calculate

V/F

2

.53 0.157 2.92

x C1

.584 150 1

1

0.157 .

0.584

87.6 kmol/h and L 150 87.6 z C1

K C1 1 (V / F)

y C1 K C1 x C1 50 0.016883 Similar for other components.

0.157

2

where

V

2-D17.

0

Ki 1 V F

.5 1 49 .584

62.4

0.016883

0.844

L F 1.5 V 0.4F 400, L 600 Slope Intercepts y = x = z = 0.70. Plot line and find xA = 0.65, yA = 0.77 (see graph) b. V = 2000, L = 3000. Rest identical to part a. c. Lowest xA is horizontal op line (L = 0). xA = 0.12 Highest yA is vertical op line (V = 0). yA = 0.52. See graph a.

35

d.

V = 600, L = 400, -L/V = -0.667. Find xA = 0.40 on equilibrium curve. Plot op line & find intersection point with y = x line. zA = 0.52 zh 1 zi V xh 2.D18. From x i , we obtain V F Kh 1 1 Ki 1 F Guess Tdrum , calculate K h , K b and K p , and then determine V F .

K1 1 z i

Check: Initial guess: If x h

Kh

Try T

0 ?

.85 then Tdrum must be less than temperature to boil pure hexane

4.8, K p =11.7 .

0.6 1 V 0.85 1.471 . Not possible. Must have K h F 0.8 1 73 C where K h 0.6 . Then K b 3.8, K p 9.9 . 0.6 1 .85 .6 1

V F

Check:

K i 1 zi 1

K1 1 V F

94 C . On this basis 85° to 90° would be reasonable. Try 85°C.

1.0, T

K h =0.8, K b

1

1

0.85

0.706

0.735

8.9 .1

Ki 1 V F

0.6

2.8 .3

8.9 .735 1

.4 .6

2.8 .735 1

.4 735

0.05276

Converge on T ~ 65.6 C and V F ~ 0.57 . 2.D19.

90% recovery n-hexane means 0.9 Fz C6 Substitute in L

F V to obtain z C6 .9

C 8 balance: z C6 F

Lx C6

z C6

or

Vy C6

L x C6 1 V F x C6

F V x C6

1 V F x C6

K C6 Vx C6

x C6 K C6 V F

Two equations and two unknowns. Remove x C6 and solve

z C6 Solve for V F.

.9 z C6 KV F

.93C 6

V

.1

F

.9K C6

1 V F

.1

. Trial and error scheme.

Pick T, Calc K C6 , Calc V F, and Check f V F If not K ref new

0?

K ref Told 1 df T

36

Try

T

70 C. K C4

V

3.1, K C5

.1

F .9 .37 .1 Rachford Rice equation 2.1 .4 f 1 2.1 .231 1

.37

K ref

0.231 . .08 .25

.63 .35

.08 .231 1

.63 .231

.28719

.37

K ref Tnew Converge on TNew

.93, K C6

0.28745 use .28 1 0.28719 ~ 57 C. Then K C4 2.50, K C8 .67, and V F

0.293 .

2.D20. New Problem. The K values are: K E 8.7 , K B 0.54 , K P 0.14 Can use Eq. (2-40), (2-41) or (2-42). If we use (2-42) the R – R eqn

f

F

Then RR eq =

0

K i 1 zi 0 V 1 Ki 1 F 7.7 .2 .46 z B

V

1+7.7 25

K C2

a.)

zB

K C5

0.153

Soln to Binary R.R. eq.

1

0.6078

zA

zB

F

KB 1

KA 1

0.55

0.45

F

.153 1

4.8 1

0.5309

zC2

0.55

V 1 3.8 .5309 1 K C2 1 F x C5 0.8177 , y C5 0.1251 Need to convert F to kmol. Avg MW 0.55 30.07 0.45 72.15

F 100, 000

V

kmol

K drum

L

0.1823,

yC2

0.8749

49.17

2033.7 kmol/h

hr 49.17 kg

V F F 1079.7,

u Perm

b.)

kg

.8 z B

0.86 .25

V

V x C2

zE

0.8764 1.0955z B

0.3499

4.8

1 zB

.86 .8 z B

.46 .25

0.5265 0.51977 z B

0 .5757z B 2.D21.

1

Use z F

L

F V

954.0 kmol/h

v v

To find

MW L

0.1823 30.07

0.8177 72.15

64.48

37

MW V 0.8749 30.07 For liquid assume ideal mixture: V1

x C2 VC2,liq

0.1251 72.15

x C5 VC5,liq

x C2

MW C2

35.33

x C5

MW C5

C2,liq

VL

0.1823

30.07

MW L

64.48

VL

103.797

L

72.15

0.8177

0.54

0.621 g/ml

v

RT atm g 700 kPa 35.33 101.3 kPa mol v ml atm 82.0575 303.16K mol K WL K drum : Use Eq. (2-60) with FlV WV

K drum

kmol 64.48 kg

WV

997.7

WV

881.5 35.33

exp

h

31,143.3

0.621

n 0.2597

2.6612

AC

D

0.18707

0.0010149

n 0.2597

0.009814 1.0 m

s 3.2808 ft

v

0.8111

4A C

m s 3600 s h

n 0.2597 4

2

0.3372

2.6612 ft/s

0.8111 m/s

1079.7

V MWV u Perm 3600

3

n .2597

0.621 0.009814 ft

L

0.2597

0.81458

0.3372

v

31,143.4 kg/h

0.009814

1.877478

0.009814 g / ml

6, 4331.7 kg/h

kmol

64331.7

0.0145229 u Perm

103.797 ml/mol

0.63

MW v

For vapor: ideal gas:

FlV

C5,liq

kmol h

35.33

0.009814

g cm 3

kg kmol kg 1000g

6

10 cm m3

3

1.392 m 2

1.33 m

Arbitrarily

L D

4,

L

5.32 m

38

2.D22.

K i 1 zi

V

f

F

1

K iP 1 z iP V 1 K iP 1 F

Ki 1 V F

Solve for V/F.

1

V

K NP 1

K iP 1 z iP

F

K iP 1 z iP

V F

p tot

log10 VP

K NP 1 z NP

760 mm Hg,

where z iP

z NP

0

1.0

90 C

1580.9

3.011679 20 219.61 1027.256 mm Hg , K iP 1027.256 760 1.35165

VPiP Note: MWiP

8.11778

MWNP .

z iP

F

0.24384 0.5

z iP 1

x NP

K iP

1 x iP

5. 0°C, 2500 kPa Fig 2.12: K M

5.7,

V

0.4095 V 1 F 0.5905; yip K iP x iP

K ethylene

1.43,

K Ethane

0.98,

y NP

0.44653

K C6

0.007

1

V

.47 .4

0.43 .05

F1

1 4.7 .6

1 .43 .6

V

V

F

F

2

.02 0.35 1

f

xi

x ethane

zi 1

Ki 1 V F

0.354, x C6

0.993 0.2

.02 .6

V F

1

zi K i 1

1

1

yi

0.55347

0.6 (equal split ethylene and ethane)

F

Then Eq. (2-38),

0.629

0.24384 0.35165

x iP

First, try

0.5 in both wt & mol frac., as does z NP .

0.35165 0.5

V

Find

K NP 1 z NP

F

1499.2

7.84767

NP

log10 VPiP

Eq. (2-46)

T

V

0

2.75943 90 204.64 574.68 mm Hg , K NP 574.68 p tot 0.75616

VPNP

f

K iP 1

K NP 1 K iP 1

p drum

2.D23.

1

K NP 1 z NP V 1 K NP 1 F

1

.993 6

0.6059

2 2

Ki 1

V F

. xM

0.104, x ethylene

0.502,

0.0108

0.040

1.0001 OK

Ki xi

39

2.D24. New Problem. p = 300 kPa At any T. K C3

y C3 x C3

K C6

y C6 x C6

K’s are known.

Substitute 1st equation into 2nd

K C6

1 x C3 K C6

Solve for xC3,

x C3 K C3 K C3

At 300 kPa pure propane K C3

x C3

at 110°C

x C3

1 K C3 x C3

1 x C3

1 K C3 x C3

K C6

1 K C6

K C6

&

y C3

1.0 boils at -14°C

At 300 kPa pure n-hexane K C6 at -14°C

1 x C3

1 K C6

x C3

Check:

1 yC3

K C3 1 K C6 K C3

K C6

(Fig. 2-11)

1.0 boils at 110°C 1 K C6 1 K C6 0 K C3

0,

1, y C3

y C3

1 1 K C6

1.0

1 K C6 K C3 0 K C3

0

Pick intermediate temperatures, find K C3 & K C6 , calculate x C3 & y C3 .

T 0ºC 10ºC 20ºC 30ºC 40ºC 50ºC 60ºC 70ºC

K C3

1.45 2.1 2.6 3.3 3.9 4.7 5.5 6.4

K C6

0.027 0.044 0.069 0.105 0.15 0.21 0.29 0.38

x C3 1- 0.027 = 0.684 1.45 - 0.027 0.465 0.368 0.280 0.227 0.176 0.136 0.103

yC3

K C3 x C3

0.9915 0.976 0.956 0.924 0.884 0.827 0.75 0.659

See Graph

40

41

b.

L V 0.6 0.4 1.5 x C3 0.3 , V F 0.4, Operating line intersects y x 0.3, Slope 1.5 L F y x z V V F 0.3 at x 0, y z 0.75 V 0.4 Find yc3 = 0.63 and xC3 = 0.062 Check with operating line: 0.63 1.5 .062 0.75 0.657 OK within accuracy of the graph.

c.

Drum T: K C3

d.

y

.8,

y C3 x C3

x ~ .16

0.63 0.062 10.2 , DePriester Chart T = 109ºC

Slope V

L

y

.8 .6

V

x 1

.16 .6

f

F

1.45

1 f

0.45

f

.45

0.69

2.D25. New Problem. 20% Methane and 80% n-butane V Tdrum .50 ºC , 0.40 , Find p drum F K A 1 zA K B 1 zB V 0 f V V F 1 KB 1 1 KA 1 F F Pick p drum

1500 kPa: K C4

13

(Any pressure with K C1 Trial 1

1 and K C4

12 .2

f1

K nC4

.6 .8

1 12 .4 1

1.0 is OK)

0.4

d f Pold

Need lower p drum

0.2178

1 .6 .4

K C 4 Pold

K C 4 Pnew

0.4

1

.2138

0.511

1.0

Pnew

1160

K C1

15.5 .2

f2

.489 .8

1 15.5 .4

K C 4 Pnew

Pnew f3

16.5 1

0.4305

.4863

0.055769

0.511 1

1100

0.541 0.055769 K C1 17.4

16.4 .2 1

.489 .4

16.4 .4

.459 .8 1

.459 .4

0.0159 , OK. Drum pressure = 1100 kPa

42

b.)

xi

yC1 2.D26. New Problem. a)

b) Stage is equil.

K C5

zi 1

0.2

, x C1

Ki

K C1x C1

V 1 16.4 .4 1 F 17.4 0.02645 0.4603

0.02645

Can solve for L and V from M.B. 100 = F = V + L 45 Fz 0.8V 0.2162L Find: L = 59.95 and V = 40.05 y C3 0.8 K C3 3.700 x C3 0.2162

0.2

.2552 0.7838 These K values are at same T, P. Find these 2 K values on DePriester chart. Draw straight line between them. Extend to Tdrum , p drum . Find 10ºC, 160 kPa. 2.D27. New Problem. a.)

VPC5 : log10 VP

1064.8

6.853

0 233.01

VP 191.97 mmHg b.)

VP

3 760

2.2832 ,

2280 mmHg ,

log10 VP 6.853 1064.8 / T 233.01 Solve for T = 71.65ºC c.) Ptot 191.97 mm Hg [at boiling for pure component Ptot d.)

C5:

log10 VP

30 233.01 637.51 mm Hg

VP

K C5 C6:

VPC5 Ptot

log10 VPC6

VPC6 K C6 e.)

KA

yA x A

1064.8

6.853

2.8045

637.51 500 1.2750 1171.17

6.876

30 224.41 187.29 mm Hg

187.29 500 KB

2.2725

0.3746 yB x B

(1 y A ) / (1 x A )

If K A & K B are known, two eqns. with 2 unknowns K A & y A

x C5

1 K C6 K C5

K C6

VP ]

1 0.3746 1.2750 0.3746

Solve.

0.6946

y C5 K C5 x C5 1.2750 0.6946 0.8856 f.) Overall, M.B., F = L + V or 1 = L + V C5 : Fx F Lx Vy .75 0.6946 L + 0.8856 V Solve for L & V: L = 0.7099 & V = 0.2901 mol g.) Same as part f, except units are mol/min.

43

2.D28. New Problem.

V h

D

F L From example 2-4, x H With h D

C, D

0.19, Tdrum

378K, V F

0.51, y H

0.6, z H

0.40

V MWv u perm 3600

v

C

C=4, MWv = 97.39 lbm/lbmole (Example 2-4)

1

3.14 10 3 g mol

v

28316.85ml

454g lbm

ft

3

0.198

lbm ft 3

Example 2.4

u perm

K drum

L

v

,

K horiz

1.25 K vertical

V

From Example 2-4, K vertical

u perm V

V F

0.5541 F

0.4433 , K horiz

0.6960 0.00314

0.00314 lbmol 0.51 3000 hr

1.25 0.4433

0.5541

12

8.231 ft s [densities from Example 2-4]

1530 lbmol hr

lbmol lbm 97.39 h lbmol ft s lbm 8.231 3600 0.1958 3 s h ft 1530

D

h

5.067 ft

4D

20.27 ft 1 Use 5 20 or 5 22 ft drum. 2 2.D29. New Problem. The stream tables in Aspen Plus include a line stating the fraction vapor in a given stream. Change the feed pressure until the feed stream is all liquid (fraction vapor = 0). For the PengRobinson correlation the appropriate pressure is 74 atm. The feed mole fractions are: methane = 0.4569, propane = 0.3087, n-butane = 0.1441, i-butane = 0.0661, and n-pentane = 0.0242. b. At 74 atm, the Aspen Plus results are; L = 10169.84 kg/h = 201.636 kmol/h, V = 4830.16 kg/h = 228.098 kmol/h, and Tdrum = -40.22 oC.

44

The vapor mole fractions are: methane = 0.8296, propane = 0.1458, n-butane = 0.0143, i-butane = 0.0097, and n-pentane = 0.0006. The liquid mole fractions are: methane = 0.0353, propane = 0.4930, n-butane = 0.2910, i-butane = 0.1298, and n-pentane = 0.0509. c. Aspen Plus gives the liquid density = 0.60786 g/cc, liquid avg MW = 50.4367, vapor density = 0.004578 g/cc = 4.578 kg/m3, and vapor avg MW = 21.17579 g/mol = kg/kmol. Since the flow area for vapor = LD and L = 4D, the area for flow = 4D 2. Then the equation for the drum diameter is D = {[(MWV) V]/[ρV uperm (L/D)]}0.5 = {[(21.17579 kg/kmol)(228.098 kmol/h)]/[(4.578 kg/m3)(uperm ft/s)(1 m/3.281 ft)(3600 s/h)(4)]0.5 where the unit conversions are used to give D in meters. The value of u perm (in ft/s) can be determined by combining Eqs. (2-59) and (2-60) for vertical drums with Eq. (2-64a). Flv = (WL/WV)[ρV/ ρL]0.5 = (10169.84/4830.16)[0.004578/0.60786]0.5 = 0.18272 Resulting Kvertical = 0.378887 , Khorizontal = 0.473608, and uperm = 5.436779 ft/s, and D = 0.4896 m and L = 1.9585 m. Appropriate standard size would be used. 2.D30. New Problem. a. From the equilibrium data if yA = .40 mole fraction water, then x A = 0.09 mole fraction water. Can find LA and VA by solving the two mass balances for stage A simultaneously. LA + VA = FA = 100 and LA (.09) + VA (.40) = (100) (.20). The results are VA = 35.48 and LA = 64.52. b. In chamber B, since 40 % of the vapor is condensed, (V/F)B = 0.6. The operating line for this flash chamber is, y = -(L/V)x + FB/V) zB where zB = yA = 0.4 and L/V + .4FB/.6FB = 2/3. This operating line goes through the point y = x = zB = 0.4 with a slope of -2/3. This is shown on the graph. Obtain xB = 0.18 & yB = 0.54. LB = (fraction condensed)(feed to B) = 0.4(35.48) = 14.19 kmol/h and VB = FB – LB = 21.29. c. From the equilibrium if xB = 0.20, yB = 0.57. Then solving the mass balances in the same way as for part a with FB = 35.48 and zB = 0.4, LB = 16.30 and VB = 19.18. Because xB = zA, recycling LB does not change yB = 0.57 or xA = 0.09, but it changes the flow rates VB,new and LA,new. With recycle these can be found from the overall mass balances: F = VB,new + LA,new and FzA = VB,newyB + LA,new xA. Then VB,new = 22.92 and LA,new = 77.08.

45

Graph for problem 2.D30.

46

2.E1. New Problem. From Aspen Plus run with 1000 kmol/h at 1 bar, L = V = 500 kmol/h, WL = 9212.78 kg/h, WV = 13010.57 kg/h, liquid density = 916.14 kg/m3 , liquid avg MW = 18.43, vapor density = 0.85 kg/m3 , and vapor avg MW = 26.02, Tdrum = 94.1 oC, and Q = 6240.85 kW. The diameter of the vertical drum in meters (with u perm in ft/s) is D = {[4(MWV) V]/[3600 π ρV uperm (1 m/3.281 ft)]}0.5 = {[4(26.02)(500)]/[3600(3.14159)(0.85)(1/3.281)uperm]}0.5 Flv = (WL/WV)[ρV/ ρL]0.5 = (9212.78/13010.57)[0.85/916.14]0.5 = 0.02157 Resulting Kvertical = 0.404299, and uperm = 13.2699 ft/s, and D = 1.16 m. Appropriate standard size would be used. Mole fractions isopropanol: liquid = 0.00975, vapor = 0.1903 b. Ran with feed at 9 bar and pdrum at 8.9 bar with V/F = 0.5. Obtain WL = 9155.07 kg/h, WV = 13068.27, density liquid = 836.89, density vapor = 6.37 kg/m3 D = {[4(MWV) V]/[3600 π ρV uperm (1 m/3.281 ft)]}0.5 = {[4(26.14)(500)]/[3600(3.14159)(6.37)(1/3.281)uperm]}0.5 Flv = (WL/WV)[ρV/ ρL]0.5 = (9155.07/13068.27)[6.37/836.89]0.5 = 0.06112 Resulting Kvertical = .446199, uperm = 5.094885 ft/s, and D = 0.684 m. Thus, the method is feasible. c. Finding a pressure to match the diameter of the existing drum is trial and error. If we do a linear interpolation between the two simulations to find a pressure that will give us D = 1.0 m (if linear), we find p = 3.66. Running this simulation we obtain, WL = 9173.91 kg/h, WV = 13049.43, density liquid = 874.58, density vapor = 2.83 kg/m3, MWv = 26.10 D = {[4(MWV) V]/[3600 π ρV uperm (1 m/3.281 ft)]}0.5 = {[4(26.10)(500)]/[3600(3.14159)(2.83)(1/3.281)uperm]}0.5 Flv = (WL/WV)[ρV/ ρL]0.5 = (9173.91/13049.43)[2.83/874.58]0.5 = 0.0400 Resulting Kvertical = .441162, uperm = 7.742851 ft/s, and D = 0.831 m. Plotting the curve of D versus pdrum and setting D = 1.0, we interpolate pdrum = 2.1 bar At pdrum = 2.1 bar simulation gives, WL = 9188.82 kg/h, WV = 13034.53, density liquid = 893.99 , density vapor = 1.69 kg/m3, MWv = 26.07. D = {[4(MWV) V]/[3600 π ρV uperm (1 m/3.281 ft)]}0.5 = {[4(26.07)(500)]/[3600(3.14159)(1.69)(1/3.281)uperm]}0.5 Flv = (WL/WV)[ρV/ ρL]0.5 = (9188.82/13034.53)[1.69/893.99]0.5 = 0.0307 Resulting Kvertical = .42933, uperm = 9.865175ft/s, and D = 0.953 m. This is reasonably close and will work OK. T drum = 115.42 oC, Q = 6630.39 kW,

47

Mole fractions isopropanol: liquid = 0.00861, vapor = 0.1914 In this case there is an advantage operating at a somewhat elevated pressure.

2.E2.

This problem was 2.D13 in the 2nd edition of SPE. a. Will show graphical solution as a binary flash distillation. Can also use R-R equation. To generate equil. data can use x C6 x C8 1.0, and yC6 yC8 K C6 x C6 K C8 x C8 1.0 Substitute for xC6

1 K C8

x C6

K C6 K C8 Pick T, find KC6 and KC8 (e.g. from DePriester charts), solve for xC6. Then yC6 = KC6xC6 T°C

KC6

125 120 110 100 90 80 66.5 Op Line Slope

KC8

4 3.7 3.0 2.37 1.8 1.4 1.0

xC6

1.0 .90 .68 .52 .37 .26 .17

L

1 V F

.6

V

V F

.4

0

.0357 .1379 .2595 .4406 .650 1.0

yC6 = KC6 xC6 0 .321 .141 .615 .793 .909 1.0

1.5 , Intersection y = x = z = 0.65.

See Figure. yC6 = 0.85 and xC6 = 0.52. Thus KC6 = .85/.52 = 1.63. This corresponds to T = 86°C = 359K

b. Follows Example 2-4.

48

MW L VL

x C8 MW x C6

MW

x C8 MW

C6

MW

x C8

C6

C6

MW v

C8

86.17

.52

.659

C8

MW L

99.63

VL

145.98

L

.52 86.17

C8

yC6 MWC6

.48

114.22

42.57

454 g/lbm

.85 86.17

.15 114.22

1.0 90.38 g/mol 0.00307 g/ml ml atm RT 82.0575 359K mol K Now we can determine flow rates V V F .4 10, 000 4000 lbmol/h F pMW v

L

V MW v

F V

4000 90.38

6000 lbmol/h, WL

Flv

L MW L

v

597, 780

0.19135

Wv

L

361, 520

42.57

exp

1.87748 0.01452

u Perm

K drum

L

.81458

2.1995 v

V MW v

A Cs

4A Cs

v

597, 780 lb/h

0.111, nFlv

2.1995

2.1995 0.00101

.18707 2.1995

4

2.1995

2

0.423

42.57 19135 .19135

6.3 3600 0.19135

4 83.33

90.38

6000 99.63

4000 90.38

u Perm 3600

D

3

0.423

v

ft 3

361, 520 lb/h

WL

K drum

lbm

0.19135 lbm/ft 3

v

Wv

99.63

145.98 ml/mol

.703

28316 ml/ft 3

.682 g/ml

yC8 MWC8

.48 114.22

6.30 ft/s

83.33 ft 2

10.3 ft. Use 10.5 ft.

L ranges from 3 × 10.5 = 31.5 ft to 5 × 10.5 = 52.5 ft. Note: This uPerm is at 85% of flood. If we want to operate at lower % flood (say 75%) have u Perm75% 0.75 0.85 u Perm85% 0.75 0.85 .63 5.56 Then at 75% of flood, ACs = 94.44 which is D = 10.96 or 11.0 ft.

2.F1

xB

0

.1

.2

.3

.4

.5

.6

.8

.9

1

yB

0

.22

.38

.52

.62

.71

.79 .85 .91

.96

1

.7

Benzene-toluene equilibrium is plotted in Figure 13-8 of Perry’s Chemical Engineers Handbook, 6th ed. 2.F2.

See Graph. Data is from Perry’s Chemical Engineers Handbook, 6th ed., p. 13-12.

49

Stage 1)

z F1

.4 .4

Intercept Stage 2)

z F2

13

.164

z F3

23

.240

f

.240

Intercept 2.F3.

f .164

Intercept Stage 3)

f

12

13 1.2

y1

23

T

Converge to

T

12

18 C,

T

0 C,

z1 K1

K1

x2

x3

K1z1

K1

0C

x1

1,

.164

13 23 .01

Slope

.480

Dew Pt. Calc. Want Try

.872

.246

2,

13

Slope

Bubble Pt. At P = 250 kPa. Want Guess

23

Slope

z2

12

y2

.240

y3

.461

z3

1

.022

1 K2

.043,

K3

.00095,

.52

K2

0.11,

K3

0.0033,

120.26

1.0

1.93,

50

Converge to T 124 C . This is a wide boiling feed. Tdrum must be lower than 95°C since that is feed temperature. First Trial: Guess Td,1 70 C : K1 7.8, K 2 1.07, K 3 Guess V F

.083

0.5 . Rachford Rice Eq. 7.8 1 .517

fV F

1

.07 .091

6.8 .5

1

V F .6 gives f .6

.07 .5

1

.14

.083 1 .5

.101

V F .56. f 0.56

By linear interpolation

.083 1 .392

.0016 which is close enough for first

trial.

V

V F F 56, zi

xi

1

Ki 1 V F

x1

.1075

y1

.839

Data: Pick

L

x2 y2

.088 .094

44

and y i

x3 y3

Kixi

.806

x 1.001

.067

y .9999

th

25 C . (Perry’s 6 ed; p. 3-127), and (Perry’s 6th ed; p. 3-138)

Tref 1

81.76 cal/g

44

3597.44 kcal/kmol

2

87.54 cal/g

72

6302.88 kcal/kmol

3

86.80 cal/g

114

9895.2 kcal/kmol

at

T

0 C, CpL1

0.576 cal / (g C) 44

For

T

20 to 123 C, CpL3

at

T

75 C, CpL2

25.34 kcal/(kmol C) .

65.89 kcal/(kmol C)

39.66 kcal/(kmol C) . (Himmelblau/Appendix E-7)

Cpv a bT cT 2 propane a = 16.26 b = 5.398 × 10-2 c = -3.134 × 10-5 -2 n-pentane a = 27.45 b = 8.148 × 10 c = -4.538 × 10-5 -3 **n-octane a = 8.163 b = 140.217 × 10 c = -44.127 × 10-6 ** Smith & Van Ness p. 106 Energy Balance: E(Td) = VHv + LhL – FhF = 0 Fh F 100 .577 25.34 .091 39.66 .392 65.89 95.25 297, 773 kcal/h Lh L

44 .1075 25.34

VH v

56 .839 0.94

.088 39.66

3597.4 16.26 5.398 10

6302.88 27.45 8.148 10

0.67 9895.3 8.163 140.217 10

E Tdrum Converge on For

V F

.806 65.89

2

2

70.25

117, 450

45

45 3

45

240, 423

60,101 Thus, Tdrum is too high. Tdrum

57.2 C : K1

0.513, f 0.513

6.4, K 2

0.0027. V

.8, K 3 51.3, L

.054 48.7

51

x1

.137, x 2

.101, x 3

.762,

x1 1.0000

y1

.878, y 2

.081, y3

.041,

y1 1.0000

Fh F 297, 773; Lh L 90, 459; VH v Thus Tdrum must be very close to 57.3°C. x1 .136, x 2 .101, x 3 .762

209,999; E Tdrum

2685

y1 .328, y 2 .081, y 3 .041 V 51.3 kmol/h, L 48.7 kmol/h Note: With different data T drum may vary significantly. 2.F4.

New Problem.

yV Lx

V F

x

Find:

0.4, V

0, y

Fz 4kmol h , L

F

z 2.5 .25 V V = 4 kmol/h, L = 6 kmol/h.

or

y 6,

L V

1.5

L V

x

F V

z

slope

0.625

From the graph, x = 0.19 y = 0.34 Equilibrium is from NRTL on Aspen Plus.

52

FIGURE 2.F.4.

2.G1. Used Peng-Robinson for hydrocarbons. Find Tdrum 33.13 C, L 34.82 and V 65.18 kmol/h In order ethylene, ethane, propane, propylene, n-butane, xi (yi) are: 0.0122 0.0748 , 0.0866 0.3005 , 0.3318 0.3781 , 0.0306 0.0404 , 0.5388 0.2062.

2.G2.

x i yi

Used Peng-Robinson. Find Tdrum 30.11 C, L 31.348, V 68.66 kmol/h. In same order as 2.G1, are: 0.0189 0.1123 , 0.0906 0.3023 , 0.3255 0.3495 , 0.0402 0.0501 , 0.5248 0.1858 .

53

2.G3.

Used NRTL-2.

xM

Tdrum

0.18 and y M

xM

79.97 C ,

0.2475, y M

0.6287 .

Compares to graph with

0.55 . Different equilibrium data.

2.G4. New Problem. COMP METHANE BUTANE PENTANE HEXANE V/F = 0.58354 2.G5.

x(I) 0.12053E-01 0.12978 0.29304 0.56513

y(I) 0.84824 0.78744E-01 0.47918E-01 0.25101E-01

New Problem. Used NRTL. T = 368.07, Q = 14889 kW, 1st liquid/total liquid = 0.4221, Comp Furfural Water Ethanol

Liquid 1, x1 0.630 0.346 0.0241

Liquid 2, x2 0.0226 0.965 0.0125

Vapor, y 0.0815 0.820 0.0989

2.G6. New Problem. Used Peng Robinson. Feed pressure = 10.6216 atm, Feed temperature = 81.14oC, V/F = 0.40001, Qdrum =0. Note there are very small differences in feed temperature with different versions of Aspen Plus. COMP METHANE BUTANE PENTANE HEXANE V/F = 0.40001

x(I) 0.000273 0.18015 0.51681 0.30276

y(I) 0.04959 0.47976 0.39979 0.07086

2.H1. New Problem. The spreadsheet with equations for problem 2.D16 is shown in Appendix B of Chapter 2. The spreadsheet with numbers for i-butane replacing n-butane is below. MC flash, HW 2.G.b., MC flash with ibutane K const. aT1 aT2 aT6 ap1 ap2 ap3 M -292860 0 8.2445 -0.8951 59.8465 0 iB -1166846 0 7.72668 0.92213 0 0 nPentane -1524891 0 7.33129 0.89143 0 0 nHex -1778901 0 6.96783 0.84634 0 0 p T deg R 509.688 psia 36.258 F 150 zM 0.5 z iB 0.1 z np 0.15 znhex V/F 0.602698586 0.25

54

guess KM KiB KnPen KnHex xM xib xnPen xnHex Sum RR M RR nB RRnP RRnHex sum RR

51.86751896 0.926804057 0.175621816 0.05400053 Use goal seek for cell B24 to = 1.0 change B9 0.015793905 0.104615105 0.29812276 0.581601672 1.000133443 0.803396766 -0.007657401 -0.2457659 -0.550194874 -0.000221409

2.H3. New Problem. Use the same spreadsheet as for problem 2H1, but with methane feed mole fraction = 0. Answer: V/F = 0.8625, xib 0.08596648 xnPen 0.203540261 xnHex 0.710481125 KiB 3.886544834 KnPen 1.264637936 KnHex 0.574940847 yib = xib Kib = .33411 and so forth

55

Chapter 3 New Problems and new solutions are listed as new immediately after the solution number. These new problems are: 3A7, 3A10, 3A11, 3C3, 3C4, 3D4, 3D8, 3G2. 3.A7.

3.B1.

Simultaneous solution is likely when one of the key variables can be found only from the energy balances. For example, if only 1 of x D , x B , D, B, FR A dist are given energy balances will be required. This is case for most of the simulation problems and for a few design problems. In some simulation problems the internal equations have to be solved also. a.

x D , x B , opt feed, Q Re b

x D , x B , opt feed, Q C x D , x B , opt feed, S (open steam), sat’d vapor steam All of above with fractional recoveries set instead of x D , x B D, x B , opt feed, L/D b. N, N F , col diameter, frac. recoveries both comp. N, N F , col diameter, FR A dist, Lo D N, N F , col diameter, FR A dist, QR N, N F , col diameter, FR A dist, QC N, N F , col diameter, x D , QC

 or x B 

N, N F , col diameter, S (sat’d steam), sat’d vapor steam, x D  or x B  Many other situations are possible [e.g., 2 feeds, side streams, intermediate condensers or reboilers etc.] 3.C1.

See solution to problem 3-D2.

3.C2.

See solution to problem 3-D3.

3.C3.

New Problem in 3rd Edition. Fmix  F1  F2  D  B

Fmix z mix  F1z1  F2 z 2  Dx D  Bx B (Mole frac. MVC)

z mix 

F1z1  F2 z 2 Fmix

Now solve like 1 Feed Column  Fmix & z mix  . From Eq. (3-3),

z mix  x B Fmix kmol/h. xD  xB B  Fmix  D kmol/h.

D

3.C4.

New Problem in 3rd Edition. See solution to 3D4, Part b.

56

3.D1.

V1 F1

QC

F1z1  F2 z 2  Bx B  D x D

1

Lo

F2

D, x D

V QR

xD  0.85

 z avg  x B  Solve D    Ftotal  xD  xB  Ftotal  F1  F2  1500 kg/h

z avg 

L

x B  0.0001  0.01% 

F1  F2  B  D

D

F1z1  F2 z 2  Ftotal

1000 .60   500  0.10   0.43333 1500

0.43333  0.0001 1500   764.62 kg/h 0.85  0.0001

B  Ftotal  D  1500  764.62  735.38

B, x B

kg h

Mass balance calculation is valid for parts a & b for problem 3G1. a)

Lo  3, Eq  3-14  QC  1  L0 / D  D  h D  H1  D h D is a saturated liquid at x D  0.85 wt. frac. From Fig. 2-4, h D ~ 45 kcal/kg H1 is saturated vapor at x D  y1  0.85, H1 ~ 310 kcal/kg

QC  1  3 764.62  45  310   810, 497 kcal/hour

EB around column.

F1h F1  F2 h F2  Qcol  QC  QR  Dh D  Bh B

h F1  81 C, 60 wt% ethanol  ~ 190kcal / kg; h F2  20 C, 10 wt% ethanol  ~ 10kcal / kg h B (sat’d liquid – leaves equil contact, ~ 0 wt% ethanol) ~ 100 kcal/kg, Qcol = 0 (adiabatic)

QR   764.62  45    735.38 100   1000 190   500 10    810, 497   657, 259 kcal/kg (b) V B  2.5 mass.

57

V (Satd vapor)

Reboiler

QR

L

Sat’d liqd

Lh L  QR  VH V  Bh B L  V  B  L  1838.45  735.38  2573.83

V B  2.5 or V  2.5B  1838.45 B, x B =0.0001 Sat’d liqd

Approximately x B ~ yV ~ x L . Thus h L  h B  100 . HV  640 kcal kg

QR  1838.45  640    735.38 100    2573.83 100   992, 763 kcal / h

EB.

QC  Dh D  Bh B  Qcol  F1h F1  F2 h F2  QR

QC  34407.9  7353.8  190, 000  5000  992, 763  1,146, 001 kcal/h 3.D2.

Column: mass bal: F + S = D + B (1) MVC: Fz + SyS  Dx D  Bx B (2) Note: yS  0 energy bal: Fh f  SHs  QC  Dh D  Bh B (3) Condenser: mass bal. : V1  Lo  D (4)

energy bal.: V1H1  QC   Lo  D  h D (5)

Solve Eqs. (1) and (2) to get:

D

Fz  Fx B  Sx B 100 .3  100 .05  100 .05   36.4 xD  xB .6  .05

Note: Not Eq. (3-3). Solve Eqs. (4) and (5) to get:

QC  D 1  L D  h D  H1 

Substitute Q C into Eq. (3):

L Dh D   F  S  D  h B  Fh F  SH S  1 D D  h D  H1 

From Figure 2-4: h F  8, h D  65, h B  92, HS  638, H1  608 kcal/kg.

L D 3.D3.

36.4  65   163.6  92   100  8   100  638   1  2.77 36.4  65  408 

External balances: F + C = B + D

Fz  Cx C  Bx B  DyD

(1) (2)

QR  Fh F  Ch C  Bh B  DH D

(3)

58

F = 2000, C = 1000, z = .4, x C  1.0, x B  .05, yD  .80,

h F  20 C   30.7,

h C  sat 'd liquid   50, h B  sat 'd liquid   92, H D  sat 'd vapor   327 kcal/kg L  VB Lx N  Vy Re b  Bx B

Around reboiler:

Lh N  QR  VH reb  Bh B For a total reboiler: x N  x B , y N  x N  x B , h N  h B  92 M.B.:

 V  B h

N

 Q R  VH reb  Bh B

QR since h B  h N H reb  h N HReb  617 (saturated vapor at y N 1  0.05 ) Fz  Cx C  Fy D  Cy D Solve Eqs. (1) and (2) for B: B  x B  yD 800  1000  1600  800 Thus B  800 and D  2200 .05  .8 or

From Eq. (3),

V

QR  Bh B DH D  Fh F  Ch C

QR   800  92    2200  327    2000  30.75  1000  50   804,500 cal/h V 3.D4.

QR 804500   1532.4 kg/h H reb  h N 617  92

New Problem in 3rd Edition. F  B D MVC Fz  Bx B  Dy D But given recoveries. Thus, use: Fz F,M (Frac Rec Methanol in distillate)  Dy D,M and

Fz F,W (Frac Rec water in bottom)  Bx B,W

z F,M  0.3, z F,W  1  z F,M  0.7 yD,M unknown, x B,W unknown.

Methanol 29.7  100  0.3.99   Dy D,M If 99% methanol recovered in distillate, 1% is in bottoms 0.3  100  0.3 0.01  Bx B,M Water

68.6  100  0.7  0.98  Bx B,W  2% water in distillate

1.4  100  0.7  0.02   DyD,W

Since  x i  1 and  yi  1,   Dy D,i   D, and   Bx B,i   B Thus,

D  DyD,M  DyD,W  29.7  1.4  31.1 kmol h

B  Bx B,M  Bx B,W  0.3  68.6  68.9 kmol h Check:

B  D  100  F

OK

59

yd,M 

DyD,M D

 29.7 31.1  0.955

x B,M  Bx B,M B  0.3 68.9  0.00435

D  31.1 & L0 D  2. Thus L0  62.2

a)

Reflux liquid is in equilibrium with vapor y D,M  0.955 From equilibrium data (Table 2-7) x M,0 ~ 0.893 (linear interpolation) E.B. Partial condenser: V1H1  Qc  DH0  L0 h 0

b)

V1  D  L0  93.3

V1y1  Dy0  L0 x 0  y1   Dy0  L0 x 0  V1

y1,M   29.7   62.2  0.893  93.3  0.914; y1,W  1  y1,M  0.086  M  35, 270 J [email protected] C  35, 270 kJ / kmol  choose MeOH reference 64.5ºC.

  40, 656 J mol@100 C  40, 656 kJ/kmol  choose water reference 100ºC. The condenser is at 66.1ºC (linear interpolation Table 2-7).

H1   M y1,M   W y1,W  y1,M CP,V,M  T1  64.5   y1,W CP,V,W  T1  100  V1 is at T1 in equilibrium with y1,M  0.914. From Table 2-7 T1 ~ 67.6 C





Assuming only constant & linear T term are important in CP,V eqs., C P,V  CP Tavg . For

67.6  64.5 67.6  100  66.65 C . For water, Tavg   83.8 C . 2 2 J 1000 mol kJ kJ  42.93  0.08301 66.05  48.41  48.41 mol  kmol  1000J kmol C

methanol Tavg 

CPV,M

J kJ  34.04 o mol C kmol C TM  67.6  64.5  3.1 ; TW  67.6  100  32.4

CP,V,W  33.46  0.00688 83.8  34.04 Then

H1   35270  0.914    40656  0.086    48.41 3.1 0.914   34.04  32.4  0.086  H1  35775.5 kJ kmol

Note λ terms dominate.

H D is at yD  [email protected] C

H D   M yD,M   W yD,W  yD,M CP,V,M  TD  64.5  yD,W CP,V,W  TD  100  66.1  64.5  65.3 2  42.93  0.08301 65.3  48.35

CP,V,M  CP,V,M  TM,avg  . Tavg,M 

CP,V,M

100  66.1  83.05 2  33.46  0.00688  83.05  34.03

CP,V,W  CP,V,W  TW,avg  . TW,avg 

CP,V,W

TW  66.1  100  33.9

60

H D  35270 .955   40656  0.045   48.35  0.955 1.6   34.03  0.045 33.9  HD  33682.85  1829.5  73.88  51.91  35534.3 Reflux liquid at 66.1ºC and x M,0  0.893, x W,0  0.107 Reference MeOH 64.5ºC, water reference 100 ºC h 0  CPL,M x M,0 TM  CPL,W x W,0 TW

kJ kmol C h 0   86.85 0.8931.6    75.4  0.107  33.9   149 kJ kmol

CPL,M  CPL,M  Tavg   75.86  0.1683  65.3  86.85

QC  DH D  L0 h 0  V1H1  31.1 35534.3   62.2  149    93.3 35775.5   2, 242, 030 kJ h Overall EB Fh F  QC  QR  DH D  Bh B or QR  DH D  Bh B  QC  Fh F

h B is saturated liquid with x B,M  0.00435 and x B,W  0.99565 Interpolating in Table 2.7 TBot  99.2 C

h B  CPL,M  0.00435 99.2  64.5   CPL,W  0.99565  99.2  100  99.2  64.5  81.86 and CPL,M  75.86  0.1683 81.86   89.64 2 h B  13.53   60.06   46.5 kJ kmo l Tavg,M 

Feed is saturated liquid at z M  0.3, z W  0.7. From Table 2-7, TF  78 C

h F  CP,LM  0.03 78  64.5  CPL,W  0.7  78  100 

Tavg,M   78  64.5 2  71.25 and CP,L,M  75.86  0.1683  71.25  87.85 h F  805.4 kJ kmol Then

Q R   31.1 35534.3   68.9  46.5    2, 242, 030   100  805.4   1,105,116 3.D5.

 3204

 2, 242, 030

 80,536  3, 424, 479 kJ h

Mass Balances: F = D + S + B, Fz  Dx D  Sx S  Bx B Solving simultaneously, B = 76.4 kg/min, D = 13.6 kg/min. Condenser: QC  V1  h 0  H1 

V1  L0  D   L D  1 D  4 13.6   54.4 kg/min From Figure 2-4, h 0  7.7 kcal/kg (x = .9, T = 20°C),

H1  290 kcal/kg (y = .9, sat’d vapor). Thus, Q C = 54.4 (7.7 – 290) = -15,357 kcal/min Overall Energy Balance: Fh F  QR  QC  Dh D  Sh S  Bh B From Figure 2-4,

QR  Dh D  ShS  Bh B  Fh F  QC

h S  61  x S  .7, sat'd Liq'd  ; h F  200  z  .2, 93 C  ,

61

h B  99  x B  .01, sat'd Liq'd  , h D  h o  7.7 Thus,

QR  13.6  7.7   10  61   76.4  99   100  200    15357   3635.3 kcal/min

3.D6.

 z  xB   .4  .002    2500    998 lbmol/h.  .999  .002   xD  xB 

From Eq. (3-3), D = F 

Then B = F = 1502. Condenser: V   Lo  D    Lo D  D  D

QC   h D  H V  D  Lo D  1 With 99.9% nC5 have essentially pure nC5 . Thus, it is at its boiling point.

 h D  H V    C5  11,369 Btu/lbmol. QC   11,369  998 4   45,385, 048 Btu/h

Overall:

QR  Dh D  Bh B  Fh F  QC

Distillate is at boiling point of pure nC5 is at boiling point of nC6

 K C6  1.0 

 K C5  1.0 on DePriester Chart) = 35°C.

Bottoms

 67°C.

Converting to °F: 35°C = 95°F, 67°C = 152.6°F, 30°C = 86°F. Note feed is obviously a subcooled liquid. Arbitrarily, pick a liquid at 0°F as reference. (This will not affect the result and other values can be used.) CPF  x C5CPLC5  z C6 CPLC6

CPF  .4  39.7   .6  51.7   46.9 Btu/lbmol °F

h F  CPF  TF  0    46.9  86   4033.4 Btu/lbmol

Distillate is almost pure nC5 . Liquid at 95°F

h D  CPLC5  TDist  0    39.7  95  3771.5 Btu/lbmol

Bottoms is almost pure liquid nC6 at 152.6°F.

hC pLC6  Tbot  0    51.7 152.6   7889.4 Btu/lbmol

QR   998 3771.5   1502  7889.4   2500  4033.4    45,385, 048   50,861, 491 Btu/h 3.D7.

 z  xB   0.7  0.001    1000    700.4 kmol/h  0.999  0.001   xD  xB 

Eq. (3-3), D  F 

B  F  D  299.6 kmol/h Condenser: Lo   Lo D  D   2.8  700.4   1961.1 kmol/h Only this reflux is condensed since product is a vapor. QC  Lo    where λ is for essentially pure n-pentane.

kmol  Btu  2.20462 lbmol   QC   1966.1 11,369   h  lbmol  1 kmol   Btu 1J J QC  49,154, 204.85  5.18176 1010 -4 h 9.486 10 Btu h

62

From overall balance QR  DHD  Bh B  Fh F  QC Distillate is vapor at b.p. of pure n-pentane (35°C from DePriester chart, K C5  1.0 ) Bottoms is boiling n-hexane (67°C) Conversions: 35°C = 95°F - distillate & Feed and 67°C = 152.6°F - bottoms As reference, arbitrarily choose liquid at 0°F. Feed is subcooled liquid.

CPF  z C5CPLC5  z C6CPLC6   0.7  39.7    0.3 51.7   43.3Btu lbmolo F h F  CPF  TF  0    43.3 95  0   4113.5Btu lbmol

Distillate H D   C5  CPLC5  Tdist  0 

H D  11,369   39.7  95  0   15,140.5 Btu lbmol Bottoms is pure C6 @152.6 F

h B  CPLC6  Tbot  0   51.7 152.6  0   7889.4 Btu lbmol

 kmol  Btu   kmol  Btu  QR   700.4 15,140.5    299.6  7889.4  h  lbmol   h  lbmol   kmol  Btu    2.20462 lbmol   Btu    1000   49,154, 204.85  4113.5     h  lbmol    kmol h     Btu 1J QR  68, 675,167.9  7.240 1010 J h -4 h 9.486 10 Btu 3.D8.

New Problem in 3rd Edition.

F  300,

z E  .3,

z w  .7

98% rec. E in distillate, 81% rec water in bot.

D  Dist.  .98  90   1  .81 300 .7   128.1 kmol/h y DE 

.98 90   0.6885

128.1 B  Bottoms  .02  90   .81 210   171.9 kmol h yD , D, H D b.

Partial Condenser.

Vapor H1 y1 V1

Qc

L0  2, L0  2D  2 128.1  256.2 kmol h. x 0 , L0 , h 0 D x 0 in equilibrium with y 0 , thus from equation data x 0  0.575. Entering vapor y1 (from graph)  0.61

63

V1  L0  D  256.2 128.1  384.3 kmol h. c.

E.B. on PC. V1H1  Qc  DHdist  L0 h 0 . Can use Figure 2-4 by converting mole fracs to mass fracs. Basis 1 kmole. .6885 mol E  MW  46   31.671 kgE Distillate

5.607 kgW 37.28 kg total Mass frac. E = 31.671 37.28  0.8496

.3115 mole W  MW  18  Vapor V1

0.61 mole E  46   28.06 kgE

7.02 kgW 35.08 kg total Mass frac E = 28.06 35.08  0.7999 .39 mole W 18 

Liquid reflux L 0

0.575 mole E  46   26.45 kgE

7.65 kgE 34.1 total Mass frac E = 26.45 34.1  0.7757 ~ 310 kcal kg, H1 ~ 330 kcal kg, h 0 ~ 65 kcal kg 0.425 mole W 18 

From Figure 2-4, Hdist

Q c  DH dist  L 0 h 0  V1H1  128.1

kmol 37.28 kg kcal 310  256.2 34.1 65 hr kg kg

35.08 kg 330  2, 400,517 kcal hr kmol Fh F  QR  Qc  DHdist  Bh B  364.3

Overall EB.

Know Qc  2, 400,517 kcal h and

DHdist  1, 480, 426 kcal h.

To find Fh F and Bh B , need to convert mole frac to wt frac. Basis 1 kmol 30 mole % E: .3 mole  46   13.8 Feed

12.6 total 26.4 kg kmol Mass frac E  13.8 26.4  0.5227 70% W : .7 18  

Bottoms

0.01047 mole  46   .48162

17.811 total 18.293 kg kmol Mass frac E  0.48162 18.28  0.0263 0.98953 mole 18 

From Figure 2-4

h F  satd liqd   70 kcal kg

h B  satd liqd   97

64

QR  DHdist  Bh B  Fh F  Qc

Then

kmol 18.29316 kg 97 kcal 26.4 kg kcal  300 70 h kmol kg kmol kg   2, 400,517   1, 480, 426  305,525  554, 408   2, 400,517   3,632,069kcal h

QR  1, 480, 426  171.9

3D9. New Problem 3rd Edition. B = (xD – z)/(xD – xB)F = [(0.9999 - 0.76)/(0.9999 – 0.00002)](500) = 120

QR  Lh  VH  Bh B and L  V  B

Q R    L  B  h  Vh  V  H  h   V

Assume h  h B .

V   V B  B  1.5 120   180 kmol h. Bottoms is almost pure water.  w  9.72 kcal mol  9720 kcal kmol

QR  180 kmol/h  9720   1.750 106 kcal h

3.D10.

2 atm × 101.3 kPa/atm = 202.6 kPa. Pentane Recovery: 0.995  Fz P  Dx D

D

 0.9951000  0.55   547.6333  0.9993

kmol/h

B = 1000 – 547.6333 = 452.3667

Since

 Pentane Recovery Bot  F zp , 1  .9951000  0.55  0.006079 

Bx B  xB

452.3667

mol frac pentane

Distillate is essentially pure Pentane. Bottoms Pure in Hexane. From DePriester Chart K P  1@ p  202.6 kPa when Tdist  59.5 C

K n H  1@ p  202.6 kPa when Tbot  94 C

 L  QC  1  o  D  h D  H1  D  h D  pure pentane   CPLC5  Tdist  Tref  choose Tref  25 C

For Total Condenser, Eq. (3-14)

kcal kcal  59.5  25  1369.65 kmol C kmol H1  h D   assuming λ is independent of temperature

h D  39.7

Btu  1 lbmol   0.252 kcal  kcal  7680.196     lbmol  0.454 kmol   Btu kmol  kmol kcal Eq. (3-14) is QC  1  2.8 547.6333  6310.5  13,132, 288 h h kcal kcal h B  pure hexane   CPLC6  Tbot  Tref   51.7  94  25  3567.3 kmol C kmol H1  1369.65  11369

Feed is a liquid at 65°C

h F   CPC5 zC5  CPC6 zC6   TF  Tref  h F  39.7  0.55   51.7  0.45    65.25   1804kcal / kmol

65

QR  Dh D  Bh B  Fh F  QC

QR   547.63331369.65   452.3667  3567.3  1000 1804    13,132, 288 QR  13, 692, 081 kcal/h. Note that QC and Q R are relatively close. 3.E1.

Was 3.D8 in 2nd Edition.

QC D, x D  0.990 M.B. F + S = D + B F

Fz M  SyS,M  Dx D  Bx B

p = 1.0 atm

100 kmol h z M  0.6

Since steam is pure, yS,M  0 Know F, z M , x D , x B

S B

x B  0.02

Unknowns S, D, B, Need E.B.

Fh F  QC  SHS  Dh D  Bh B Pick as basis liquid at 0°C, h W  0 & h M  0 (essentially steam table choice) Assume ideal mixtures.

h F  CPavg  TP  Tref  where CPavg  0.4 CPW ,L  0.6 C PM ,L

Felder & Rouseau p. 637 CPW  0.0754 kJ/mol

CPM  0.07586  16.83 105 T kJ/mol

CPM  CPM  Tavg   CPM   0  40  2   CPM  20 C 

CPM  0.07586  16.83 105  20   0.079226 CPavg  0.4  0.0754   0.6  0.079226   0.077696

h F  0.077696  40  0   3.1078 kJ/mol feed = 3107.8 kJ/kmol

Can also use steam table for water

H S is sat’d vapor steam 1 atm,

HS  2676.0

kJ 18.0 kg kg kmol

Steam Table F&R, p. 645 H S =48,168 kJ/kmol

h D is sat’d liquid at x D  0.99 . From Table 2-7, T = 64.6°C

h D  CPavg  64.6  0  where CPavg  0.01 CPW  0.99 CPM

CPM  CPM  Tavg   CPM  32.3 C   0.07586  16.83 10 5  32.3   0.081296

CPavg  0.01 0.0754   0.99  0.081296   0.08124

66

H D  0.08124  64.6   5.2479 kJ mol  5247.9 kJ kmol h B : Since leaving an equilibrium stage, sat’d liqd. 2% MeOH Table 2-7, T = 96.4°C

h B  CPavg  96.4  0  where CP avg  0.98 CPW  0.02 CPM CPM  CPM   96.4  0  2   C PM  48.2 C 

CPM  0.07586  16.83 105  48.2   0.08397 CPavg  0.98  0.0754   0.02  0.08397   0.07557 kJ mol

h B  0.07557  96.4   7.28509 kJ mol  7285.09 kJ mol Q C  do EB around condenser

L  QC   dist V1   dist  Lo  D    dist  o  1 D D  dist  0.99 MeOH  0.01 W  M  35.27 kJ mol &  W  40.656 kJ mol kJ dist   0.99  35.27    0.01 40.656   35.324  35,323,86 kJ kmol mol QC    35,323.86  2.3  1 D  116,568.7D kJ h

Felder & Rousseau:

Plug Q C & numbers into E.B.

100  3107.8  116,568.7D  48,168S  5247.9D  7285.09B

or 310,780 + 48,168S = 121,816.6D + 7285.09B Solve simultaneously with 2 MB. 100 + S = D + B 60 + 0 = 0.99D + 0.02B One can use algebra or various computer packages. Obtain:

E2.

D = 56.33 kmol/h, B = 211.71 kmol/h

S  168.04 kmol/h, QC  6,566, 000 kJ/h.

Was 3.D9 in 2nd Edition. F+S=D+B

F  500

kmol mol  500, 000 h h

Fz  SyS  Dx D  Bx B

2 eq. 3 unknowns Condenser: QC  1  L0 / D  D  h o  H1  Note Eq (3-14) not valid. For enthalpy pick reference pure liquid water 0°C and pure liquid methanol 0°C. Felder & Rouseau: CPMeOH  75.86  0.01683T at Tavg 

64.5  0  3225, CPMeOH  76.4 J mol C. 2

67

Assuming distillate pure methanol, boils at 64.5°C

h D  CPMeoh ,liq  T  Tref    76.4 J/mol  64.5  0   4928.0 J mol H1  h D  1  at 64.5 F   4928  35270 J/mol  40,198 J/mol

QC   4  4928  40198 D  141, 080D J/h where D is mol/h Overall Energy balance: F h F  SHS  QC  DhD  BhB Bottoms is essentially pure H 2 O at 100°C

J  J  h B  CPW ,liq  T  Tref    75.4  100  0   7540 mol C  mol  HS  h B   W  at 100 C   7540  40656 J/mol  48196 For feed. 60 mole % Methanol boils at 71.2°C (Table 2-7).

h F   CPi zi  T  Tref    0.6  76.459  71.2    0.4  75.4  71.2   5413.7 J/mol

Now, Eqs are (1) F + S = D + B or 500,000 + S = D + B (2) Fz  Dx D  Bx B or (500,000) (.6) = 0.998D + 0.0013B

 Lo   D  h D  H1  or QC  141, 080D D  (4) Fh F  Sh S  QC  Dh D  Bh B or (500,000) (54137) + S (48196) (3) QC  1 

+ Q C = D(4928) + B (7540) Solve simultaneously: D = 298.98, B = 1245.5, S = 1044.2 kmol/h Q C = - 4.218 × 10+7 kJ/h 3.F1.

An enthalpy composition diagram is available on p. 272 of Perry’s Chemical Engineer’s Handbook, 3rd ed., 1950.

 z  xB  0.79  0.004 D  25, 000   19, 788.5 kmol/h F  0.997  0.004  xD  xB  Note that N 2 mole fractions were used since N 2 is more volatile. B = F – D = 5211.5 From enthalpy comp. diag. h D  0, H1  1350 kcal/kmol, h B  160, h F  1575 . Eq. (3-3)

Then,

QC  1  Lo D  D  h D  H1    519788.5  0  1350   133,572, 000 kcal/h QR  Dh D  Bh B  Fh F  QC

QR  0   5211.5160    25, 000 1575   QC  95, 030, 000 kcal/h

3.F2.

We will use the enthalpy composition diagram on p. 3-171 of Perry’s 6th edition or p. 3-158 of Perry’s 5th ed.Do for 1 kmol of feed: Conversion of feed from kg to moles. Basis 100 kg 30 kg NH 3 = 1.765 kmol 70 kg H 2O  3.888 Total 5.653 kmol Thus 1 kmol is 100/5.653 = 17.69 kg

68

Will work problem in weight fractions since data is presented that way. 95% recovery: (0.95) Fz = Dx D or, D = (.95) Fz / x D = (.95) (17.69) (.3)/(.98) = 5.15 kg. B = F – D = 12.54 kg

x B   Fz  Dx D  B  17.69 .3   5.15  98   12.54   0.021

From diagram: h D  55, H1  415, h B  150, h F  5 kcal/kg

Eq. (3-14), QC  1  Lo D  D  h D  H1    3 5.15  55  415   5562 kcal/kmol feed and QR  Dh D  Bh B  Fh F  QC

QR   5.15 55  12.54 150   17.69  5   QC  7815 kcal/kmol of feed

G1. a.) Using NRTL. QC  778,863 kcal/h, QR  709,520 kcal/h b.) QC  1, 064,820 kcal/h, QR  995, 478 kcal/h G2.

New Problem in 3rd Edition. ASPENPlus. D = 988, L/D = 3, Peng Robinson, N  40, NF  20 (arbitrary values in Radfrac)

x D  1.000

x DC6  1.211107

x B  0.0013316

QC  4.4426 107 Btu h, QR  4.9852 107 Btu h

69

SPE 3rd Ed. Solution Manual Chapter 4 New Problems and new solutions are listed as new immediately after the solution number. These new problems are: 4A6, 4A13, 4C10, 4C16, 4D6, 4D9, 4D13, 4D15, 4D18, 4E4, 4E5, 4H1 to 4H3. 4A1.

Point A: streams leaving stage 2 (L2, V2) Point B: vapor stream leaving stage 5 (V5) liquid stream leaving stage 4 (L4) Temp. of stage 2: know K y 2 / x 2 , can get T from temperature-composition graph or DePriester chart of K = f(T,p). Temp. in reboiler: same as above (reboiler is an equilibrium stage.)

4A2.

a. Feed tray = .6, z = 0.51 (draw y = x line), yF =0.52, xF = 0.29. b. Two-phase feed. c. Higher

4A6. 4A7. 4A13.

4A14.

New Problem in 3rd Edition. Answer is a. See Table 11-3 and 11-4 for a partial list. New Problem in 3rd Edition. A. Answer is b B. Answer is a C. Answer is a D. Answer is a E. Answer is b F. Answer is a G. Answer is b If feed stage is non-optimum, the feed conditions can be changed to have an optimum feed location.

4B2.

a. Use columns in parallel. Lower F to each column allows for higher L/D and may be sufficient for product specifications. b. Add a reboiler instead of steam injection. Slightly less stages required and adds 1 stage. c. Make the condenser a partial instead of a total condenser. Adds a stage. d. Stop removing side stream. Fewer stages are now required for the same separation. e. Remove the intermediate reboiler or condenser and use it at bottom (or top) of column. Fewer stages, but all energy at highest T (reboilers) or lowest T (condenser) for same separation. Many other ideas will be useful in certain cases. 4C7. Easiest proof is for a saturated liquid feed. Show point z, y D satisfies operating equation. Solution: Op. Eq.

y

Substitute in

y

yD V But

q 1.0, V

Lz D, L

L V x yD , x

L V 1 xB z

L V xB F, L V

y D D Fz Bx B Which is external mass balance.

B QED.

70

Can do similar for enriching column for a saturated vapor feed. 4.C10. New Problem in 3rd Edition. If we consider λ, the latent heat per mole to be a positive quantity, then QR V . With CMO and a saturated liquid feed V V (1 L / D) D , and then

QR / D 4.C16.

(1 L / D) .

New Problem in 3rd Edition. Define a fictitious total feed FT , z T , h T

FT

F1

F1z1

F2 , z T

F2 z 2

, hT

F1h F1

F2 h F2

FT FT Intersection of top & bottom operating lines must occur at feed line for fictitious feed F T. (Draw a column with a single mixed feed to prove this.) This feed line goes through y x z T

with slope where

FT

qT

qT

qT 1

H mix

hT

H mix

h mix

and H mix , h mix are saturated

vapor and liquid enthalpies at feed stage of column with mixed feed.

A

Given p, L/D, saturated liquid reflux, x D , x B

z0

B

opt feed locations, z1 , z 2 , F1 , F2 , h F1 , h F2

z2 y

zT z1 Plot top op line. Plot all 3 feed lines. Draw

xB

line from point A to y = x = x B to obtain

x b.)

Does

qT

q1F1

op. line. Connect pts B & C to get q 2 Fbot. 2

FT

middle op. line.

71

check q T

H mix

hT

H mix

h mix

F1h F1

H mix

F2 h F2 FT

H mix

F1

F2 H mix

h mix

F1H F1

H mix

F2 h F2

h mix FT

where H mix & h mix are vapor and liquid enthalpies on feed stage of mixed column

F1 H mix H mix

qT

h F1

F2

h mix

H mix

h F2

H mix

h mix

FT Usual CMO assumption is λ >> latent heat effects in either vapor or liquid. H mix h F1 H mix h F2 Then q1 and q2 H mix h mix H mix h mix F1q1 F2 q 2 Thus q T if CMO is valid. FT 4D1.

L

a. Top op line: y Intersects y

x

When x

V

xD

x

L V

x D and

L

L D

1.25

V

1 L D

2.25

0.5555

0.9

0, y

b. Bottom op line: y

1

1

L

x

L

xD

V

L

0.4

1 x B , and

Plot – See diagram

L

V B

V B 1

3

V V V B 2 V V Intersects y = x = xB = 0.05 1 0.5 / 2 @y 1 x 0.683 this is convenient point to plot 32 c. See diagram for stages. Optimum feed stage is #2 above partial reboiler. 5 equilibrium stages + PR is more than sufficient.

72

d. Feed line goes from y = x = z = 0.55 to intersection of two operating lines. q Slope 1.0 or q 0.5 . q 1 This is a 2 phase feed which is ½ liquid & ½ vapor. 4D2. New Problem in 3rd Edition. Part a.

y

b. c.

x

zE

.6

Slope

L

F V

1 V/F

.63

1.703 V V V F .37 From Table 2-1, at 84.1° C y .5089 H hF liquid at 20°C and 40 mole % ethanol. q H h The pressure in Figure 2-4 is very close to 1.0 atm, thus it can be used, but must convert to wt frac.

73

Basis 1kmol feed.

.4 kmole E .6 kmol Water From Figure 2-4

H q

Alternate Solution:

40 mole %E

.6 MW

46

0.63 wt frac.

10.8 kg

18

398 kcal kg , h

18.4 kg

total

29.2 kg

75, h F 20 C

10

398 10

1.20 398 75 q 1.2 Slope 6 q 1 .2 40 mole % ethanol boils at 84.1°C (Table 2-1). Then if pick reference as saturated liquid at 40 mole % h F Cp,40%liq 20 84.1

h d.

.4 MW

0,

63 wt%, H

H

40%E

398 kcal kg , h

CPvapor

y E CPEvapor

65, h F

398

kcal kg

C p vapor 120 84.1

y w CPw ,vapor

Assume only 1st and 2nd terms in C P equations are significant. From Problem 2.D9 CPvapor .4 14.66 0.03758T .6 7.88 .0032T kcal/kmol T is C which simplifies to 120

For linear

CPvapor

10.592 0.16952T

C p dT is equal to CPvapor @ Tavg

84.1

Tavg

84.1 120 2 102.05 . Then C Pv ,avg 398

hF

398 15.149

q e.

q

q

f.

Flash

kcal

hF

V F

kg

10.592

12.32 120 84.1

15.147

398 65

333

f, L

13 12 , slope

.7,

12.32

kcal kmol

kcal 1 kmol kmol 27.2 kg

413.15 kcal kg

398 413.15 L L

0.16952 102.05

L

F

0.045. F

12

q q 1

13

L

12 13 12

1 12

L

1 V F

.3

3

V

V F

.7

7

F

13

See graph for feed lines.

74

Graph for 4.D2

75

4.D3*.

a. Basis 1 mole feed. 0.4 moles EtOH × 46 = 18.4 kg EtOH 0.6 moles H2O × 18 = 10.8 kg H2O Total = 29.2 wt frac 18.4 / 29.2 0.63 wt frac EtOH Calculate all enthalpies at 0.63 wt frac. Hv = 395, HL = 65 (from Figure 2-4). hF is liquid at 200°C. Assume Cp,liq is not a function of T. Estimate, 46.1 23 h h L 60 C h 20 kcal C P ,liq .63 wt frac ~ 0.864 T 60 20 80 kg C Then h F

h L 200 q

Hv

CPL 200 60 hF

.864 200 60

h L 60 C

395 167.1

0.691,

q

46.1 167.1

0.691

2.24 Hv hL 395 65 q 1 0.309 b. From Figure 2-4 at 50 wt% ethanol Hv = 446 and hL = 70. Since CMO is valid obtaining both enthalpies at 50% wt is OK. The feed is a liquid h F C P,liq TF Tref CP,liq 250 0

CP,liq CP,EtOH z EtOH CPw z w in Mole fractions Basis 100 kg solution 50 kg EtOH 46.07=1.085 kg/kgmole

50 kg W 18.016 2.775 kg moles Total 3.860 kg moles Avg M.W. 100 3.86 25.91 kg/kgmole Thus, zW = 0.719 and zE = 0.281 CP,liq 37.96 .281 18.0 .719 23.61 C P,liq in kcal kg C hF q

hF

4.D4*. a.

q

Hv

hL

446 70

H CPv 350 50

q q 1

x

z

0.911 . Then,

228 0.58

H

25 300

H hF

L L F where L

L L

25.91

Hv

q q

MWAVG

250 C kg C h F 446 228

Slope

slope

23.61

kcal

H h q q 1 0.6. y

b. c.

0.911

CP

4.D4a

y=x

25 300 z

1.5

0.6 is intersection.

L 0.6F. Then q

feed line

L 0.6F L / F

.5 .6 0.6, and

.7

1.5

F where L

L F 5. q

L F5 L F

1 5 , slope

q q 1

16

76

4.D5*.

h liq

fL

h reflux

3100 1500

h liq

17500 3100

L0

L0 D

1.1

V1

L0 D 1

2.1

H vap

1 fc

L1 V2 Alternate Solution

L1

L1 V2

4D6.

q

qL0

L1

0.524

1.1111 .524

1 f c L 0 V1

For subcooled reflux, Then,

L 0 V1

0.1111

1

0.55

.111 .524

L1

H h0

17500 1500

L0

H h1

17500 3100

1.111

1.1111 L0 L1 D

,

L1

1.111 L0

L1 D L1 D 1 D L1 1.222 0.55 V2 2.222

New Problem in 3rd Edition.

1.111 1.1

D

a) 175

F1

F2

1.2222

B D

85 75 .6 100 0.4 0.1 B 0.9D Solve simultaneously. D 84.375 and B 90.625 kmol hr b) Feed 1. q1 1, vertical at y x z1 0.6 Feed 2.

60% vapor = 40% liquid q 2

Slope feed line Bottom Op. Line

q2

0.4

q2 1

.06

y

L V x

Slope L

Middle

Lx When

x

0, y

F2

L V

0.4 2 3 through y

x

z2

L V 1 x B . Through y

V B 1 V B

0.4

x

xB

32

B V

F2 z 2 F2 z 2

Bx B Bx B

Vy

y

L V

x

F2 z 2

, Slope L V V Also intersects bot. op. line and Feed line 2. Do External Balances and Find D & B. Then V V/B B

Bx B V

2B 181.25

L V B 271.875 At feed 2, L .4F L or L L 0.4F 271.875 40 231.875 V V 0.6F 181.25 60 241.25 L V 0.961 40 9.625 x 0, y 0.126 Plot Middle Op Line. 241.25

77

y

L

V Know that y x Also, L

L

x

F1

1

L

xD V x D and gives through interaction Middle and Feed line 1.

231.875 75 156.875 and V

L V 156.875 241.25

V

241.25 ; thus,

0.65

c) See graph. Graph for 4D6.

78

4.D7*. a. Plot top op. line: slope

L V

.8 , x

L

y

.9. Step off stages as shown on Figure.

V B

2 , x y x B 0.13. Step off stages V (reboiler is an equil stage). Find y2 = 0.515. c. Total # stages = 8 + reboiler Optimum feed plate = 7 or 8 from top. Plot feed line. Goes through x = y = z = .3, and intersection of two operating lines. 9 q gives q = 0.692. slope 4 q 1 b. Plot bottom op. line:

slope

1 1

xD

4.D8*. The equilibrium data is plotted and shown in the figure. q 0.692 and q q 1 9 4

From the Solution to 4.D7c,

a. total reflux. Need 5 2/3 stages (from large graph) – 5.9 from small diagram shown. .9 .462 b. L V min 0.660 (see figure) .9 .236 L V min L D min 1.941 1 L V min c. In 4.D7, L D act

L V 1 L V

.8 .2

L Dact Multiplier Multiplier = 4/1.941 = 2.06

4

L D

min

79

d. Operating lines are same as in Problem 4.D7. Start at bottom of column. Reboiler is an equilibrium contact. Then use E MV AB AC 0.75 (illustrated for the first real stage) Stage 1 is the optimum feed stage. 11 real stages plus a partial reboiler are sufficient.

4D9.

New Problem in 3rd Edition. a) F1 F2 D B

F1z1

F2 z 2

100 F1

Dx B

Bx B

F1 .42

Solve simultaneously, B 113.68, F1 b)

L D L

1 2

,

80 B

F1

B 20

18 .66 80

0.04 B

93.68

L

L D

12

1

V

1 L D

32

3

L

D 40, V L D 120 D V V 120 Saturated Liquid Feed L L F1 40 93.68 133.68, L V c) Top Op. Line – Normal: y Through y Bottom – Normal: y

x

L V x

1.114

1 L V xD

x D , Slope 1 3, y intercept

L V x

L V 1 x B , through y

Also through intersection, F2 feed line and middle op. line.

2 3

x

.66

.44

xB

Feed line F2 slope

L F2

.7

VF2

.3

80

Middle

y

Slope

0, y

Dx D

xD

F1z1

F1z1

80 .66

93.68 .42

0.11212 V 120 d)Opt. Feed F2 stage 1 from bottom, Opt feed F1 , Stage 2. 4 stages + PR more than sufficient. Also,

x

D

(or do around bottom) V V L V . Through intersection feed line F1 and top op. line.

L V x

Graph for 4D9.

81

4.D10*. Operating Line y

L V x

1

L

x D , where

V

Thus, operating line is y = .8x + .192 y a. Equilibrium is x or x1 1 y

L

L D

4

V

1 L D

5

.8

y1 1.79 .76y1

Start with y1 = .96 = x D Equilibrium:

x1

Operating:

y2

Equilibrium:

x2

Operating:

y3

y1

.96

1.76 .76y1

1.76 .76 .96

.8x .192 .8 .9317 y2

x y

1.0 1.0

.192

1.76

.76 .93736

.8x 2 .192 .8 .89476

.9 .9406

0.93736

.93736

1.76 .76y 2

b. Generate equilibrium data from: y

0.9317

.192

0.89476

0.9078

1.76x

1 .76x .8 .7 .8756 .8042

.6 .7253

.5 .6377

.4 .5399

Plot equilibrium curve and operating line. (See Figure). Slope = L/V = .8, y intercept (x = 0) = 0.192, y = x = x D = 0.96. Find x 6 = 0.660.

4.D11. a) Same as 4.D2 part g. q = 1.0668, slope feed line = 15.97. b) Top y L V x 1 L V x D goes through y = x = x D = 0.99

L V

L D 1 L D

Feed line: Slope

0.6969 @ x = 0

q q 1 ,

y

y = (1-L/V) x D = (1-0.6969) 0.99 = 0.300

x

z

0.6

82

Bottom Op line:

L

yV Lx Sy M,S Bx B But y M,S 0 (Pure steam) With CMO B L

V S

B

y

L V

x

L V

xB

y= 0, x = x B . Also goes through intersection of feed line and top op.line. Stages: Accuracy at top is not real high. (Expand diagram for more occupancy). As drawn opt. Feed = #6. Total = 9 is sufficient, c.

L V

min

Slope

L D

0.99 0.57 0.99 0 L V

min

1 L V

Actual L/D is 3.12 × this value.

0.4242 0.4242

min

1 0.4242

0.73684

83

4.D12. L V

y

L V

x

L D 1 L D 1

L V

Bottom slope

34

slope. Top op line goes throug y

xD @ x L V

0, y

.25 .998

B

1245167

S

1044168

1.19

x

xD

0.998

0.2495 From Soln to 3.D9 or from graph. 1.169

Feed line is vertical at z = 0.6. Can also plot top and feed lines, and then find bottom from 2 points y 0, x x B & intersect top & feed . For accuracy – Use expanded portions near distillate & near bottoms. From Table 2-7 from (x = .95, y = .979) Draw straight line to (x = 1.0, y = 1.0) From (x = 0, y = 0) draw straight line to (x = 0.02, y = 0.134) or (x = 0.01, y = 0.067) Opt feed = # 9 from top. Need 13 equilibrium stages.

84

85

4.D13. New Problem in 3rd Edition. a.)

L

b.) See figure.

V

L

See Figure

0.665 0.95 0.30 0.95

MIN

L

L V

0.4385

0.4385

0.7808 D MIN V L 1 L V 0.5615 L L L D 1.5616 c.) 2.0 L D MIN 1.5616 , D V 1 L D 2.5616 L

y intersect 1

V

yD

0.3709 . Top operating line y Goes through y

Bottom y

L V

x

L V

0.6096 L

x

V x yD

1

L

V 0.95

yD

1 xB

Goes through y

x x B & intersection top operating line & feed line. Feed Line: Vertical (saturated liquid, q = 1). Through y x z 0.3 Plot & Step off stages. Optimal feed = 5 below PC. 6 + PC + PR more than sufficient. 0.85 0.025 d.) Slope bottom: See figure for parts c & d. L V 1.941 0.45 0.025 1 1 V B V L V 1.0625 . L V 1 0.941

86

Graph for problem 4.D13.

87

4.D14a.

88

New S.S.

External M.B. S = B Sys = BxB . Since yS = 0 (pure water) xB = 0

B

S 4.D14b.

Two approaches to answer. Common sense is all methanol leaks out and x MA

McCabe-Thiele diagram: This is enriching column with z horizontal feed line is at x

x M,b

y intercept

L V

x

L

1

V

1 L V yD

Bottom operating line

y

0 . Intersection top op. line and

0 , which is also a pinch point. Thus x M,d

4.D15. New Problem in 3rd Edition. Saturated liquid. q Top operating line y

ys

y D , Slope

q q 1

L

L D

2

V

1 L D

3

1 3 0.6885 L V x

1,

0.

0.2295 and y

0 also.

, feed line vertical @ z

x

L V 1 x B goes through y

.3 .

yD

x

xB

And goes through interaction feed line and top operating line. See graph. Optimum feed is stage 2 below partial condenser. Partial condenser + Partial reboiler + 3 equilibrium stages are more than enough to obtain separation.

89

Graph for problem 4.D15.

90

4.D16*.

q

L-L

Hv

hF

F

Hv

hL

Using 32°F = 0°C as reference T, h F

Hv

hL

4033.4 Btu/lbmole.

at feed conditions.

. Approx.

.4 11369 .6 13572 12691 Btu/lbmole For approx. temperature of feed stage, do bubble pt. calc. y1 1 K1x1 K1z1 Pick T = 48°C (~ 40% of way between boiling pts.) K C5 1.5, K C6 .54, K1x1 1.5 .4 .54 .6 K C5 Tnew =

.54 .92

K C5 50 C Hv

=.594, Tnew

1.6,

Note : CP feed,liq Hv q

1.6 .4 .584 .6

.99 Close enough.

, 50 C 122 F

feed

Hv

50 C

K1 x 1

h L 50 C

.92

CPfeed ,Liq 122

46.9 122

5721.8

46.9 is from Prob. 3-D6. 5721.8 12691 18412.8 Btu/lbmole

HV

hF

18412.8 4033.4

HV

hL

12691

1.133

Note: h F is from Prob. 3.D6. 4.D17.

L

Top Op. Line: y

x

L

V L D

V

1 L D

x

L

1

V 7 2

0, y

1

The vertical line at x

xS

9 2 L

x D , goes through y

x

xD

0.9

7 9 2

xD

.9 =0.2 V 9 Plot Top. Step off 2 stages. Find x S ~ 0.81

0.81 is the withdrawal line.

Bot. Op. Line intersects Top at x

xS .

Also know it intersects feed line at x

F

External Balances

Fz

Dx D

x B (unknown)

D B S Don’t know D, B, or x B .

Bx B

Sx S

Feed enters as saturated vapor. Thus q

0&V

F

Bottoms leaves an equilibrium contact, it is saturated liquid L Do flow balances V F 100 V V 100 since S is removed as saturated liquid.

B

91

L

L V

V

7 9 100

L

L S

B

L 62.777 Fz Dx D Sx S

xB

77.777. D

77.7777 15 60

V L 100 77.777

62.7777. L V 22.222 0.9

B

Plot. Op. line Step off stages. 9 is more than sufficient.

4.D18. New Problem in 3rd Edition. Feed F1 : z1

62.7777

62.7777 100 15 0.81

22.222

0.6278 0.444

0.6, saturated liquid, q 1, q / (q 1)

92

Feed F2 : z 2

0.4, 80% vapor hence 20% liquid q q

LF / F

0.2F / F .2

.2

14 q 1 .8 Part a.) Bottom operating line goes through point, y

x

xB

0.04

Max L V to point intersection feed F2 line and equilibrium curve.

L

Slope

1.0 .04

V

V B Part b. L

V L

min

F1

1

1

V

L V 1

1.2326

100

.2 80

116

L L F2

L

V B and V B 1.5 L

1.5B B

L B 116 46.4

D

V

V VF2

100

V 133.6 Check overall balance

116

B

V

L

0.8113

100

L

116

2.2326

.47 .04

max

46.4

2.5 69.6

116

L V

69.6 .8 80

69.6

1.66667

133.6

0.7485

F1

F2

D B

180 133.6 46.4 180.0 OK

To find y D use MVC mass balance

F1z1 or

yD

F2 z 2 F1z1

Dy D F2 z 2

Bx B Bx B

100 .6

D

80 .4 133.6

Actual bottom op. line: y

L V

x

L V

46.4 0.4

0.675

1 xB

L

V B

V B 1

2.5

5

V

V

V B

1.5

3

Goes through y x x B 0.04 , Slope 5 3 2nd point y = 1, x = 0.616 (this was arbitrarily found by setting y = 1.) Plot bottom op. line Dy D F1 z1 L Top Op. line: yV F1z1 Lx Dy D . y x V V Goes through intersection feed line for F 2 and bot. op. line. Does NOT go through y x yD . Since D & F, passing streams, Point z1 , y D is on op. line.

93

Figure for 4D18 4.D19*.

B = 0. Then from external balance F = D + B must have D = F = 1000. Acetone balance becomes Fz Dx D or x D z 0.75 .

L

To predict x B need operating lines. Top: y

V

L

L V

2

V

1 L D

3

x

and y

L

1

V x

xD xD

.75

94

Bottom: L V 1.0 . Thus y = x is operating line. From Figure x B Feed line can be calculated but is not needed.

4.D20*.

0.01 to 0.02

To use enthalpy composition diagram change to wt. fractions. Basis = 1 kg mole Distillate: Weight Fractions: Feed: Weight Fractions: Bottoms: Weight Fractions:

0.8 ETOH = (.8)(46.07) = 36,856 0.2 Water = (.2)(18.016) = 3.6032 Total = 40.459 EtOH = .911, Water = .089 0.32 (EtOH) = (.32)(46.07) = 14.7424 0.68 (W) = (.68)(18.016) = 12.25088 Total = 26.993 EtOH = .546, W = .454 0.04 EtOH = (.04)(46.07) = 1.8428 0.96W = (.96)(18.016) = 17.295 Total = 19.1378 EtOH = .0963, W = .9037

Condenser Energy Balance is V1H1

Lo D

Qc

Qc D ho

H1

Lo h o

Dh D which can be solved for L o D .

1

From chart: h D 54 Kcal/kg and H1 285 Kcal/kg Need D in weight units. Convert feed to weight units. 100 kgmoles Ethanol: .32 46.07 1474.24 kg/hr hr Water: (100)(.68)(18.016) = 1225 kg/hr Total: F = 2699.328 kg/hr

95

F z xB

Then,

D

Then,

Lo D

xD

2699.328 .0963

xB

1489.93 kg/hr

.911 .0963

Q D ho

H1

1

2, 065,113 1489.98 54 285

1 5.0

Now do usual McCabe-Thiele analysis using molar units. Note L o D is the same in mass and molar units. L L L L D Top Operating Line: y x 1 x D and V V V 1 L D

L

5

; x

y

xD

.8, y int ercept x

V 6 Feed Line: Goes through y = x = z = .32 Weight fraction of feed = .546. Then, h f

q

HV

hF

430.7 15

HV

hL

430.7 69

Bottom Operating Line:

y

x

L

V V intersection top operating line and feed line.

L

1

V

15 kcal/kg, H v

1.149 Slope

L

0

xD

1

5 6

.8

430.7, and h1k1

q

1.149

q 1

1.149 1

1 x B . Goes through x

.133 69 .

7.711

y

x B and

From Figure need about 8 equilibrium contacts including a reboiler. Stage 1 above reboiler is the optimum feed stage location.

96

4.D21. Feed 1: q1 Feed 2: q 2 Top:

0 , slope feed line = 0 0.9 , slope

q2

q2 1

L

L D

1.375

V

1 L D

2.375

Goes through y

x

0.9 0.1 L

0.579 , y

V

x D , When x = 0, y

1

Since F1 is saturated vapor, V

Bottom:

x

9 L

1

L V F1

V

xD

xD .

0.40

100 kmoles/hr

L

.1 F2

At feed F2

V

.9 F2

L D B

L V V

V

V

L

L 0.9 F2

0.579 108

0.1F2

100 8 108

62.532

V L 108 45.46 F1 F2 D 100 80 45.46 134.532

But B is saturated liquid. Check L L .9F2

L B 134.532 62.532 0.9 80

134.532, OK

Draw top op line. Intersects with F2 feed line. Then draw bottom op line with slope

L V 1.3453 . Intersection bottom op & q. line gives x B Check

F1z1

F2 z 2

Dx D

B Check External MB 180 F1

F1z1

F2 z 2

F2

x B or x B

D B Dx D

0.09 . 20 36 43.187 134.532

0.095 , OK

45.46 134.53 179.99 , OK

Bx B

56 20.0 36.0 45.46 0.95 134.53 0.095 55.97 OK See McCabe-Thiele diagram: Optimum feed = 5, 7 equilibrium stages (6.65) more than sufficient. fraction ab/ac 0.65

97

4.D22*. Around top of column mass balances are: L D Solving,

L

0.75, y x

0

x

D

yD

Vy Cx W

C

xw V V V For pure entering water, x W 1.0 . With saturated liquid entering, L = C. Then from overall balance, V = D. Thus L/V = C/D = ¾ and D/V = 1.0. Operating equation becomes y 0.75x .92 .75 0.75x .17 Slope

y

V C and Lx Dy D

0.17, y x 1

0.92

yD , y

x

0.68

Not y D

98

4.D23*.

Note L/V ≠ C/D since C is subcooled. Let c = amount condensed. The energy required to heat stream W to the boiling point must come from this condensation. That is, H h c h hW C

c

h

hW

Cp

C

T

C

18 212 100

H h L C c 1.1154C V D c D 0.1154C

17465.4

In addition, C/D = ¾ or D/C = 4/3. L 1.1154C 1.1154

0.1154C

1.1154

0.77 V D .1154C 4 3 .1154 4 3 .1154 This compares to L/V = 0.75 if entering water is a saturated liquid. Very little effect since λ is very large. L L D 3.25 0.7647 V 1 L D 4.25 Goes through

y

x

When

x

0, y

1 L V yD

L

L

Bottom Op Line:

y

V

yD

x

0.85

V

0.20

1 xB

Through y x x D 0.05 and intersects top op line @ feed line Opt. Feed is #4 below partial condenser – see diagram. Need 6 equil stages + P.C. (an equil. contact) Note – Commercial columns usually operate much closer to minimum reflux ratio and have many more stages. b.)

L V

0.85 0.376 min

0.85 0

0.558 ,

L V

L D

min

1

Min

L V

1.26

Min

c.) Total reflux 5 stages + PC sufficient or 4 3 4 equil contacts + PC = 5 3 4 eq. contacts ab 7.6mm where fraction 0.74 or .75 ac 10.3mm

99

100

4.D25.

a. 99.9% methanol is essentially pure. Pure MeOH boils 64.5°C. 1 fc Lo D L Eq. (4-66) 1 where f c CPL TBP Treflux V2 1 1 f c L o D For pure MeOH, CPL

CPL

fc

0.07586 16.83 10 5 T , average (40 + 64.5)/2 = 52.25°C

0.084654 kJ gmole ,

0.084654 24.5 35.27

MeOH

35.27 kJ mole , TBP

0.058804 ,

1.058804 1.2

L1 V2

64.5

1

1.058804 1.2

0.5596

101

b.

4.D26.

L

L D

1.2

0.5454 or 2.59% more reflux with 24.5°C cooling!

V 1 L D 2.2 subcooling not important. a) 50% feed: q

L L

F, L

L

amt vaporized

Usually

L F 20

1 F q 1 20 0.05 , Slope = L F 0.0476 20 20 q 1 1 20 1 1.05 35% feed: Saturated liquid – vertical feed line. Plot both feed lines. The one with lowest intersection point with equilibrium curve will normally control V B q

L

min

L V

Find

L V

max

. Then V B

1 0.1

slope

max

b)

0.46 0.1

L V

L V L

Bot. y

V y

L

x 1,

x y

V L V

x

3

V L

min

1 V

2.497 ,

V B

B

1 min

2.497 1.0

0.6681

2.0043

min

V B

V B 1

3.0043

V

V B

2.0043

L V x B or x

1

max

V

1 x B . Goes through y 1

L V

x 1

1.49892

slope

x B with total reboiler L V 1 xB / L V

1

1

1 0.49892 0.1 0.6338 1.49892 Intersects feed line with 50% feed first. Middle operating line: Do mass balance around bottom of column. Mass balance intersects streams L & V (in column), F50% and B. yV L x F50% z50% Bx B y

L

x

F50 z 50

Bx B

V V Intersects bottom operating line & 50% feed line. @ x 0, y (F50 z 50 Bx B ) / V , Slope

q 50

For 50% feed,

L V

0.05 (L L ) / F50 .

External balances: 250 F1 F2 D B and F1z50 F2 z35 Dy D Bx B Find D = 103.333 and B = 146.666 moles/min Since V B 2.0034, V 293.96 and L V B 440.63 Then from q 50% : L

V

L

F50

B

q 50 F50

L

0.05 100

445.63 100 146.666

440.63

398.964 , Slope

445.63 L V

1.11698

y intercept (x = 0), y

[ 100 .5 146.666 0.1 ] / 398.964 0.0885 Top Intersects feed line for 35% feed and middle op. line and goes through y x y D 0.85

102

EQ

Stages

0.75 = E ML =

actual change liquid

actual

op

change at equilibrium eq. change

Figure for 4D26

103

4.D27*.

Top Op. Eqn:

Feed: q

y

L V

x

1

L

L D

1

V

1 L D

2

L

L

L F

L V

xD

, y intercept

1 F 4 F

1

L V

L

xD q

.46, y

x

xD

.92

54

5 , y x z .48 14 Bx B L Middle: V B L S , V y Bx B L x Sy s y s 0 or y x V V Does not intersect at y x x B or at y 0, x x B . Does intersect top op. line at feed line. Need another point. L V B V B 1 1.5 L L Bottom: 3, y x x B 0.08 y x 1 xB , V B V B .5 V V The steam is another feed to the column: Sat’d vapor q = 0, q/(q-1) = 0, y = x = z = ys = 0. Middle Op. Line intersects this steam at bottom op. line (see figure). F

5 4 , slope

q 1

This problem is a two-feed column with the lower feed (steam) input at a non-optimum feed stage. Otimum feed is 3rd above partial reboiler. Need 5.6 equilibrium stages plus PR.

This problem was 4.D35 in 2nd edition. L L Stripping Section: y x 1 xB, y x xB V V

4.D28.

0.02 . Feed line is vertical

104

L V

1.0 0.02 max

0.81 0.02

V B 1 Op line y

1.24,

1, x=

1.5 L V

V B

1 xB

V B

5800 / 0.675

min

L

6.25, L V

1

.16 .02

1 V

L V

max

1 1

0.24

V B

V B 1

6.25 1

V

V B

6.25

min

L V 1.16 Overall Balance: 10,000 = F = D + B 6000 Fz Dy D Bx B D .695

D

V

4.167

1.16

0.865 . Intersection Op & feed lines is y D

B .02

6000 .695D .02D 200

8592.6 kgmoles/day, B 1407.4 . Need a use for impure distillate.

Figure for 4.D28

105

4.D29.

L D

3,

L

3 4,

L

1

1

xD

V V 4 Step off 3 stages on top op line. Find x S on middle op. line

0.9

0.225

0.76 . Point on top op line at x S

0.76 is also

xD

3

xs, S = 15

V´ F = 100



F=D+B+S

6

z = .6

Fz

Dx D

Bx B

Sx S

Solve simultaneously D = 50.125, B = 34.875 10 x6 = .1 In top:

L = 3D = 150.375 and V = L + D = 200.500

Since saturated liquid withdrawn, V´ = V = 200.500 and L´ = L – S = 150.375 – 15 = 135.375

L V

Middle op. line slope Middle Op line: yV

135.375 200.5

L x Sx S y

L V

x

0.6752

Dx D Sx S

Dx D V

Feed z = 0.6, 20% vapor = 80% liquid Feed line slope q q 1 0.8 0.4

, when x

0, y

11.4 45.1125 200.5

0.2819

q = 0.8. 4

106

4.D30*.

a. Subcooled Reflux:

c

1 3

Lo

500 ,

Lo

Lo D

3

V1

1 Lo D

4

L1 V2

Substituting in values, we have Step off two stages.

x2

, L1

4 Lo 3 1 V1 Lo 3 L1 1.0 L2

0.62

1 14

xS

4 3 1

Lo

c and V2

V1

c

L o V1 1 Lo 3 V1 1 4 54

5

, Top op. line y

x

xD

0.92

yS

107

Mixed feed to column: F S Solving for mixed feed, z M

FM

h F (sat’d vapor), h FM

External Balances: F = D + B, Fz

Lo

3D

Middle: L

1500, and L

L S

FM z M

0.52667

Energy balance for mixed feed, Fh F Since H s

1500 , Fz SyS

4 3

Lo

SHS HS Dx D

FM H FM

h F (sat’d vapor), and q FM

0 (horizontal).

Bx B , Solving simultaneously D = 500 & B = 500

2000 (subcooled reflux), V

2000 500 1500 , V

V

L D

2500

L

1500

3

V

2500

5

2500 , Slope

Intersects Top Op. line at x S

.6

Plot Bot. from y x x B to intersection feed line and middle Step off stages (see figure). Need 4 8/9 equilibrium stages. b. Mass balances for mixed feed injection: V FM

V

V

FM

V

2500 1500 1000 , V B 1000 500

2

4.D31*. Was problem 4.D36 in 2nd edition. Solution is trial & error. Need to pick L/V. Final answer shown in figure. L .63 .385 0.245 .389 V .63 0.63 L V L 0.389 .636 1 L V D 0.611 Note feed stage is not optimum.

108

Figure for problem 4D31 External Balance: F = B = 50, and Fz

4.D32*. 4.D33*. a.

L V

.75 .452 min

L

b.

L D

act

L V min

0.4 .

0.659

1 L V

1.318. L V

0, x

min

L D 1 L D

Top operating Line through y Bottom through y

z

0.397 (tangent pinch)

.75 D

Bx B . Thus x B

yB

x

0.569

yD

0.75

0.1 and intersection feed and top operating lines.

L L F .25F, q 5 4, slope q q 1 5 Optimum feed is 3rd from bottom. Need 9 real stages plus partial condenser (see figure). c. From figure slope of bottom operating line L V 2.025 Feed:

Since saturated steam and CMO valid, B S L V Also have mass balances, S + F = B + D SyS Fz Bx B Dy D ys 0 Solve 3 eqs. simultaneously. S = 760 lbmoles/hr = 13,680 lb steam/hr.

109

4.D34*.

4-E1.

Trial and Error, Feed: L

L F F 2, q

Find (L/V)min (see diagram) 0.95 0.613 L V min 0.95 0

3 2, Slope

3.

Figure for 4.D34

0.3547 ,

L V min

L D

min

1 L V min

0.5497

110

L D

L

2

1.0994 ,

D

act

min

External M.B. F = D + B, Fz or Eq. (4.3) D

z

xB

xD

Dx D

F

xB

L V

L D act

Bx B

0.6 0.025 0.95 0.025

100

D L L

V V´

62.162

L

h

, B = F – D = 37.838

L D D

68.030

V = L + D = 130.192

F

At feed V L″

V

kgmol

L



V″

0.5237

1 L D

P=

xP

L

V (sat’d liquid)

L F 168.030

B

Top op line y 1st middle

L V

x

y

1 L F x D normal . At x = 0, y = 0.4525

L

x

L

1 x B looks like usual bottom! V V Goes through y x x B , and intersection top & feed line. Slope

L

168.030

1.2906 130.192 At pump-around return, V V 130.192 L L P 208.030, L V V

208.030 130.192 1.5979

At pump-around removal, V V 130.192 , L L P L 168.030 Check at bottom L V B or 130.192 168.030 37.838 , OK L L Bottom Op line y x 1 x B , Same as first middle!!! V V Step off P.R. stage 1 & 2 above. x P is liquid from stage 2, x P 0.335 . Vertical line at x P is withdrawal line for pump-around and it is feed-line for return of pump-around. 2nd middle op line slope L V intersects x P withdrawal & feed line where bottom & 1st middle intersect.

111

Using MB:

yV

When

y

4-E2*. Feed:

q

Px P 0, x

Bx B P

Bx B

L V 40

x

P V

xP

Bx B V 32838 0.025

.335 0.0690 L L 208.030 208.030 Draw, 2nd middle – Step off stage 2 & start 3. 3 is op loc. for feed and where pump-around is returned. Need PR + 6 equil stages. (Actually 5 and a large fraction)

L L F

xP

Lx, y

, L

L 1.5F, q

3 2 , slope

q q 1

Bottom op. line: Since steam is saturated vapor S V and B L Thus, 1 S B L V 1.2 . Operating line goes through y 0, x

3

xB

0.015

Middle op. line: V B Side L S or V L S B Side V y Bx B Side x side L x SyS Since

yS

0 this is, y

L

x

Bx B

Side x side

V V Side stream is removed as a saturated liquid so q = 1. Step off two stages (see figure) and find x side 0.0975

112

Find slope: V L Side B

V

S, B L Side Side B 1 0.4 1

S

SB

0.833

1.68

slope

Draw in the middle operating line. Step off 4 stages. Trial and error to find x D for final result).

.85 (see figure

4-E3*. This column has 4 sections. The exact shape is not known ahead of time. Plot top operating line L L D 1.86 L L 0.650, y x 1 xD V 1 L D 2.86 V V

1

L

xD .35 .8 .28, y x x D 0.8 V Step off 8 stages and find x S 0.495 yS . Feed line for this vapor is horizontal. Feed line for feed to column is vertical at z = 0.32. From figure the feed is injected below the liquid withdrawal and above vapor stream from intermediate reboiler. Can now calculate flows in each section of columns. Overall Balance: Fz Dx D F D x B

113

D Flows: L

L D

D

F z xB xD

.3 1000

xB

.78

385

716.1, V

L D 1101.1 V

L

L S

258.8, L V

L

L

L

V ,V

V

S

643.8

0.235

F 1258.8, L V

1.143

L V 1.955 (this is a check) To plot: From stage 8 draw line of slope L´/V´. From intersection of first intermediate operating line and feed line draw line of slope L″/V″. Draw line from intersection of second intermediate operating line with line y y s to y x x B 0.02 . Check if slope L V 1.955 . Optimum feed is 10th below condenser while vapor from intermediate reboiler is returned on 11 th stage. Need 12 ½ stages. Note: Small differences in stepping off stages may change column geometry.

4.E.4. New Problem in 3rd Edition. External Balance. a) F D B, Fz Dx D Bx B

D b)

z xB xD

xB

F

.25 .025 .9 .025

V B 1.0, V

B

100

25.7 kmol h , B

74.3 kmol h

74.3

L V B 148.6, L V 2.0 At feed, amount condensed = C = F/10 = 10 1 L L F C L F F 148.6 110 10 V V C 74.3 10 64.3

38.6

114

At stage 2

L

L L R and V

L

L0

64.3

21.4

L

21.4

V LR

64.3 L L

y

c) Top Op. line:

V

0.333 1 3

38.6 21.43 17.17 kgmoles hour

L V

LR

x

D V

D

L0

L

xD

y

V

x, y

x

L

1

V

LR

D

V

L

xD

xD

x D sin ce V

L

LR

D

Envelope for top

V

L

LR

L

Plot

Middle:

y

Bottom:

y

V L V x

y

x

Slope

Slope

x

1 L V x D . Slope

L V 1 3, y x

38.6 64.3

x x D 0.9 L 0,y 1 xD V

0.600 y

x

xD

0.6

0.9

L V 1 xB

xB

L V

V B 1

2 V B Now have somewhat redundant information. Can plot bottom. Intersection bottom and feed line should also be on middle. – Or use this pt to find middle or bottom op. line. From graph: Opt. Feed = #4. Need 6 stages + P.R.

115

Graph for 4.E.4.

4.E.5. New Problem in 3rd Edition. Bot. Op. Line: External M.B.: F

D

y

V B

At feed stage: L 687.5

V

937.5

V B

V

V

x

V

L V

D B and Fz

250 kmol day , B L

L

L

V

1

2.25

9 4

9

V B

1.25

54

5

1 xB, y Dx D

x

Bx B ,

750 kmol day , V

x B , Slope D F

z xB xD

xB

V B B

L V .3 0.10

.2

1

0.90 0.10

.8

4

1.25 750

937.5 kmol day

937.50 750 1687.5 kmol day V

V

937.5, L

L

F or L

L F 1687.5 1000

687.5

0.733333

116

MB:

L x Dy D

V y, D

V

L , y

L

Top op. line:

y

L V

x

L

1

S

L V

487.5, V

yD

V

1

L

yD V V Goes through y x y D , and intersects feed and bot. op. lines. At side withdrawal: 687.5 L L , V S V or V 937.5 200 737.5 Op line by intermediate condenser: L D V S Dy D Sx S L L x Dy D V y Sx S or y x V V 687.5 Find from intersection L V op line @ y yS plus slope 0.932 or intersection 737.5 top op line and x x S At side stream feed point: L

x

737.5. Thus,

Can draw, yint ercept

L

487.5

V

737.5

1 .661 .9

0.661

0.305

117

Graph for 4.E5.

4-F1.

970.33

L

V

Since bottoms are very pure h B

HS H Equil

v

h water @ 212°F

1381.4 1192

(in column)

Extra heat 189.4 Btu/lb S

B

Since y ≈ x, MW are same

118

Must vaporize material in column. extra heat MW 189.4 v S S MW 970.33

B

L v, V

0.1952S

S v

L

B v

BS vS

2 0.1952

V

S v

1 vS

1 0.1952

If super heat not included L V

BS

1.837

2 , which is incorrect.

4-F3*. An approximate check is to compare molar latent heats of vaporization. Data is available in Perry’s and in Himmelblau. a. See Example 4-4. b. isopropanol λ = 159.35 cal/g. MW = 60.09, λ = 9.575 kcal/mole. Water λ = 9.72 kcal/mole. CMO is OK. c. CMO is not valid. AA 5.83 kcal / gmole, W 9.72 d. nC4. λ = 5.331 kcal/mole, MW = 58.12. λ = 0.0917 kcal/g nC5. λ = 6.16 kcal/mole, MW = 72.15. λ = 0.0854 kcal/g Constant mass overflow is closer than constant molar overflow. e. benzene. λ = 7.353 kcal/mole, MW = 78.11, λ = 0.0941 kcal/g toluene. λ = 8.00 kcal/mole, MW = 92.13, λ = 0.0868 kcal/g CMO is within about 6%. 4G1.

a*. Answer should be close, but not identical, to result obtained in Example 4-4.

4G2.

Was 4.G4 in 2nd edition. Used Peng-Robinson. QC

QR

44, 437,300 Btu/hr,

49,859, 400 Btu/hr.

Optimum N F

17 & N

27 (Total condenser is #1)

x D 0.9992 and x B 0.00187 . 4G3*. See answers to selected problems in back of book. 4.H.1. New Problem in 3rd Edition. Use VBA program in Appendix B of Chapter 4. xd 0.995 xb 0.011 F 250 z 0.4 L/D 3 q 0 feed stg 4 partial reboiler total condenser Ethanol-water Prob. 4H1. VLE 6th 5th 4th 3rd 2nd 1st constant -24.75 85.897 -118.03 82.079 -30.803 6.6048 0 yeqatxint 0.584177 yint 0.4 xint 0.2016667 stage x y 1 0.011 0.069033 2 0.039445 0.217358 3 0.112146 0.451872 4 0.227091 0.607193 5 0.477924 0.769849 6 0.694798 0.867005 7 0.824341 0.919132 8 0.893842 0.954863 9 0.941484 0.979257

119

10 11

0.974009 0.992216 0.991288 0.996557

4.H.2. New Problem in 3rd Edition. The VBA program is in the result is: xd 0.7 xb 0.0001 F 1000 L/D 6.94 q 0 partial reboiler total condenser Ethanol-water VLE 6th 5th 4th 3rd 2nd 1st -47.949 161.42 -212.43 138.68 -46.65 7.9322 yeqatxint 0.099219 yint 0.1 xint 0.0135447 stage x y Reflux rate too low Reflux rate too low

Appendix B of Chapter 4. With L/D = 6.94 z 0.1 feed stg 28 Prob. 4H3. constant 0

With L/D = 6.95 the result is given below. With feed stages below 85 the feed stage was too low. xd 0.7 xb 0.0001 F 1000 z 0.1 L/D 6.95 q 0 feed stg 85 partial reboiler total condenser Ethanol-water Prob. 4H3. VLE 6th 5th 4th 3rd 2nd 1st constant -47.949 161.42 -212.43 138.68 -46.65 7.9322 0 yeqatxint 0.100057 yint 0.1 xint 0.0136691 stage x y 1 0.0001 0.000793 2 0.000194 0.001538 3 0.000295 0.002338 4 0.000404 0.003197 5 0.000521 0.004117 6 0.000646 0.005102 7 0.000779 0.006154 8 0.000922 0.007277 9 0.001075 0.008472 10 0.001237 0.009742 11 0.00141 0.011089 12 0.001593 0.012515 13 0.001786 0.014021 14 0.001991 0.015608 15 0.002206 0.017276 16 0.002433 0.019025 17 0.00267 0.020853 18 0.002919 0.022758 19 0.003178 0.024739 20 0.003447 0.026791 21 0.003725 0.02891 22 0.004013 0.03109 23 0.004309 0.033327 24 0.004613 0.035613 25 0.004924 0.037941 26 0.00524 0.040302 27 0.005561 0.042689 28 0.005885 0.045091

120

29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78

0.006211 0.006538 0.006865 0.00719 0.007513 0.007831 0.008144 0.00845 0.00875 0.009041 0.009323 0.009595 0.009858 0.010109 0.01035 0.010579 0.010797 0.011004 0.0112 0.011384 0.011558 0.011721 0.011874 0.012018 0.012151 0.012276 0.012392 0.0125 0.0126 0.012693 0.012779 0.012858 0.012932 0.012999 0.013062 0.01312 0.013173 0.013222 0.013267 0.013308 0.013346 0.013381 0.013413 0.013442 0.013469 0.013494 0.013516 0.013537 0.013556 0.013573

0.0475 0.049907 0.052301 0.054674 0.057017 0.059321 0.061579 0.063782 0.065925 0.068002 0.070008 0.071938 0.07379 0.075561 0.077251 0.078857 0.08038 0.08182 0.083179 0.084459 0.085661 0.086787 0.087841 0.088826 0.089743 0.090598 0.091392 0.092129 0.092812 0.093445 0.09403 0.094571 0.09507 0.095531 0.095955 0.096346 0.096705 0.097036 0.09734 0.09762 0.097876 0.098112 0.098328 0.098526 0.098708 0.098874 0.099027 0.099167 0.099295 0.099412

121

79 80 81 82 83 84 85 86 87 88 89 90 91 92 93

0.013589 0.013604 0.013617 0.013629 0.013641 0.013651 0.01366 0.013665 0.013706 0.014018 0.016416 0.034534 0.155188 0.490181 0.646064

0.099519 0.099618 0.099707 0.09979 0.099865 0.099934 0.099997 0.100032 0.100305 0.102401 0.11824 0.223718 0.516574 0.652848 0.724878

4.H.3. New Problem in 3rd Edition. The Spreadsheet is: xd 0.7 xb 0.0001 F 1000 z 0.17 Multiplier 1.05 q 0.5 feed stg 17 partial reboiler total condenser Ethanol-water Problem 4.H4. VLE 6th 5th 4th 3rd 2nd 1st constant -47.949 161.42 -212.43 138.68 -46.65 7.9322 0 L/Dmin 1.687377 L/Vmin 0.62789 L/D 1.771746 L/V 0.639217 stage x y 1 0.0001 0.000793 2 0.000229 0.001812 3 0.000418 0.003309 4 0.000696 0.0055 5 0.001104 0.008697 6 0.001698 0.013331 7 0.002559 0.019994 8 0.003797 0.029452 9 0.005555 0.042644 10 0.008006 0.060585 11 0.01134 0.08415 12 0.015719 0.113686 13 0.021208 0.148522 14 0.027681 0.186647 15 0.034766 0.224911 16 0.041877 0.259918 17 0.048382 0.28916 18 0.057276 0.325158 19 0.113592 0.469947 20 0.340102 0.575494 21 0.505222 0.659654 22 0.636883 0.719124 Because the multiplier is close to 1.0, this answer is very sensitive to the data fit used. One solution to the coding is the following VBA program: Option Explicit

122

Sub McCabeThiele() ' Find minimum reflux ratio assuming it occurs at feed plate. Then ' L/D actual = L/D min times Multiplier. Steps off stages from the bottom up. ' Assumes that the feed stage is specified. Sheets("Sheet2").Select Range("A8", "G108").Clear Dim i, feedstage As Integer Dim D, B, xd, xb, F, z, q, LoverD, LoverV, x, y, xint, yint, yeq As Single Dim a6, a5, a4, a3, a2, a1, a0, L, V, LbaroverVbar, LoverDmin As Single Dim LoverVmin, LoverVdelta, Multiplier As Single ' Input values from spread sheet xd = Cells(1, 2).Value xb = Cells(1, 4).Value F = Cells(1, 6).Value z = Cells(1, 8).Value Multiplier = Cells(2, 2).Value q = Cells(2, 4).Value feedstage = Cells(2, 8).Value ' Fit of equilibrium data to 6th order polynomial to find y. a6 is multiplied ' by x to the 6th power. a6 = Cells(5, 1).Value a5 = Cells(5, 2).Value a4 = Cells(5, 3).Value a3 = Cells(5, 4).Value a2 = Cells(5, 5).Value a1 = Cells(5, 6).Value a0 = Cells(5, 7).Value ' Calculate intersection point of two operating lines and use this to find ' minimum L/D and L/V. Initialize LoverV = 1 LoverVdelta = 0.00001 Do LoverV = LoverV - LoverVdelta xint = ((-(q - 1) * (1 - LoverV) * xd) - z) / (((q - 1) * LoverV) - q) x = xint yint = LoverV * xint + (1 - LoverV) * xd ' Equilibrium y at value of x intersection. When yint=yeq, have minimum L/V and L/D. yeq = a6 * x ^ 6 + a5 * x ^ 5 + a4 * x ^ 4 + a3 * x ^ 3 + a2 * x ^ 2 + a1 * x + a0 Loop While yint < yeq 'Print intersection and equilibrium values. LoverVmin = LoverV + LoverVdelta LoverDmin = LoverVmin / (1 - LoverVmin) LoverD = Multiplier * LoverDmin LoverV = LoverD / (1 + LoverD) Cells(6, 2).Value = LoverDmin Cells(6, 4).Value = LoverVmin Cells(6, 6).Value = LoverD Cells(6, 8).Value = LoverV ' Calculate flow rates and ratios. D = ((z - xb) / (xd - xb)) * F L = LoverD * D

123

V=L+D LbaroverVbar = (LoverV + (q * F / V)) / (1 - ((1 - q) * F / V)) ' Step off stages from bottom up. First stage is partial reboiler. Initialize x = xb i=1 ' Loop in stipping section stepping off stages with equilibrium and operating eqs. Do While i < feedstage y = a6 * x ^ 6 + a5 * x ^ 5 + a4 * x ^ 4 + a3 * x ^ 3 + a2 * x ^ 2 + a1 * x + a0 Cells(i + 7, 1).Value = i Cells(i + 7, 2).Value = x Cells(i + 7, 3).Value = y i=i+1 x = (y / LbaroverVbar) + (LbaroverVbar - 1) * xb / LbaroverVbar Loop ' Calculations in enriching section continues to Loop While y < xd. Do While y < xd y = a6 * x ^ 6 + a5 * x ^ 5 + a4 * x ^ 4 + a3 * x ^ 3 + a2 * x ^ 2 + a1 * x + a0 Cells(i + 7, 1).Value = i Cells(i + 7, 2).Value = x Cells(i + 7, 3).Value = y i=i+1 x = (y / LoverV) - (1 - LoverV) * xd / LoverV If x < 0 Then Cells(i + 7, 4).Value = "Feed stage too low" Exit Do End If If i > 100 Then Cells(i + 7, 6).Value = "Too many stages" Exit Do End If Loop End Sub

124

SPE 3rd Ed. Solution Manual Chapter 5 New Problems and new solutions are listed as new immediately after the solution number. These new problems are: 5A15, 5C1, 5D1, 5D2, 5D9, 5D10, 5H1 to 5H5. Problems and solutions from the first edition that were not in the second edition are: 5D6, 5D8, 5D11-5D13, 5E1. 5.A6.

Ethane is less volatile than methane so it decreases toward distillate. At bottoms it is more volatile than propane and butane, so must decrease towards bottoms. Thus ethane concentrates within column. 5.A7.

1. c; 2. c; 3. a (saturated liquid feed); 4. c; 5. B

5.A9.

Pure LK cannot be withdrawn because LNK is present. Pure LNK can be removed at distillate if all LK is removed in side stream. However, recovery of LNK will be < 100%.

5.A13.

If z HNK F cross.

(frac. rec. HK in bot) z HK F, then there is more HNK in bottoms and curves

5A15. New Problem in 3rd Edition. a. a, b. d, c. b, d. c, e. b. 5.C1. New Problem in 3 rd Edition. Start with equilibrium equation, yi,j = Ki,jxi,j and multiply right hand side by Kref,j /Kref,j. One obtains yi,j = αi-ref,j Kref,jxi,j. Then 1 = Σ y = Σ (α i-ref,j Kref,jxi,j) . Since Kref,j is constant, bring it outside the summation and solve for Kref,j = 1/ Σ (αi-ref,j xi,j). This is Eq. (5-29).

125

5.D1. New Problem in 3rd Edition.

Dx M ,d

a)

Dx B,d

1.00 2, 000 0.19

2280

0

0

Dx E,d

0.978 2, 000 0.31

Dx P ,d

1 0.994 12, 000 0.27 D

3638.16

x i,d D

19.44 5937.6 kg/h

b) assume NK (Methanol and n-butanol) do not distribute (all methanol in top and all butanol in bottom).

wt frac. in distillate: M = 0.38399, E = 0.61273, P = 0.00327, B = 0.0.

wt. frac Bx M ,B

0

Bx B,B

1.0 12, 000 0.23

Bx E ,B

1 0.978 12, 000 0.31

Bx P ,B

0.994 12, 000 0.27

x i,b

0

B

0

2760.0

0.4553

81.84

0.0135

3220.56

0.5312

x i,b B 6062.4 kg hr

Check B+D=F, OK. 5.D2.

New Problem in 3rd Edition.

y4

yi P

Raoult’s Law Antoine’s Eqn.

log10 VP

Dew Point condition is

xi

yi P VPi

0.30, y 5

xi

A

0.5, y 6

0.20, P

760 mmHg

VPi x i B T C

1

(from Raoult’s Law)

126

B

(from Antoine’s Eqn) T C Combine with Dew Point condition y5P y6P y4P 1 B B B

VPi

10

A4

10 A

4

T C4

10

0.30 760 10

6.809

5

A5

T C5

10

A6

6

T C6

0.50 760

935.86 T 238.73

10

6.853

0.20 760

1064.8 T 233.01

10

6.876

1171.17

1

T 224.41

Using Goal Seek in Excel, T = 41.3ºC 5.D3.

Assume that ethanol is HK. Then assume that HNK’s are totally in the bottoms. x M,dist .99, x E,dist .01 2195.6 = (.998) (0.22) (10,000) = Dx MD

D

2195.6

2195.6

2217.78 and B x MD .99 Bottoms: MeOH: .0021 (.22) (10,000) = 4.4 4.4 x Mbot 0.0006 7782.22 1.0 .18 10, 000 x n propanol,bot 0.2313 7782.22 1.0 .13 10, 000 x n butanol,bot .1670 7782.22 x EtOH,bot 1 x MeOH,bot x n p,bot x nbut,bot 0.6011 5.D4.

a. .99 F z C5

F D

7782.22

D x C5,dist (1), and .01 Fz C5 =B x C5,bot (2)

.98 F z C6,bot

Bx C6,bot (3), and .02 F z C6 D z C6,dist (4) Assume all heptane in bottoms Fz C7 Bx bot,C6 (5), x dist,C7 0 (6) Take Eqs. 1, 4 & 6: .99 (1000) (.4) = Dx C5d .02 (1000) (.3) = D x C6d 0 = D x C7,dist

Dx l,dist b.

D

402 kg moles/hr

B = F – D = 1000 – 402 = 598 x C7,dist 0

x C5,d x C6,d

.99 1000 .4

0.9851 402 1 0.9851 0.0149

127

x C7,bot

1000 .3 1.0 598 .98 1000 .3

x C6,bot

c.

5.D5.

0.5017

0.4916 598 x C5,bot 1 .5017 .4916 0.0067 L = (L/D)D = (2.5) (402) = 1005 V = L + D = 1005 + 402 = 1407 At feed stage: L = L + .6F = 1005 + 600 = 1605 V = V - .4F = 1407 – 400 = 1007

Assume all methanol and ethanol in distillate. Dx MeOH,dist 0.55 100 0.01 150

Dx EtOH,dist

0.21 100

Dx prop,dist

0.993

Dx bu tan ol,dis

0.03 150 0.23 100

1 0.995

56.5

25.5

0.26 150

0.01 100

61.57

0.70 150

0.53

D 144.1 B Check:

F1

F2

D 105.9

Bx Pr op,bot

1 0.993

Bx but,dist

.23 100 .26 150

0.995 .01 100

0.7 150

0.434

105.47

Check = 105.90

Mole fractions:

x M ,bot

0 , x E ,bot

x but,bot

1 x prop,bot

0.9959

Dx MeOH ,dist

56.5

x M ,dist x E,dist x p,dist

x But,dist

0, x prop,bot

D 144.1 25.5 0.1767 144.1 61.566 0.427 144.1 0.53 144.1 0.0037

Check

= 1.000

0.434 105.90

0.0041

0.392

OK

5.D6. This is 8.D1. in 1st ed.

128

129

5.D7.

Assume 100% recovery C 2 , & propylene in distillate. Assume 100% recovery pentane & hexane in bottoms. Comp.

Distillate

C2 Propylene: n-C3 n-C4 C5 & C6

0.3 (1000) + 0.02 (1500) 0.006 (1000) + 0.001 (1500) (0.991) [1000 (0.45) + 1500 (0.249)] (0.02) [1000 (0.154) + 1500 (0.40)] 0

Bottoms flow rate = F1

x B,C2

0, x B,propylene

F2

D

x B,C5

0, x B,C3

1000 0.09

= 330 = 7.5 = 816.0885 = 15.08 = 15.08 D= = 1168.7

.009 1000 0.45

1500 0.18

1331.33

xD 0.2824 0.0064 0.06983 0.0129 0.0

2500 1168.7 1331.3 kg/h

x B,C4

5.D8.

kg/h

1500 0.249

1331.3 .98 1000 .154 1500 .40 1331.3 0.2704,

x B,C6

0.0056 0.5550

1500 .15 1331.31

0.1690

8.D.6. in 1st edition. Assume all benzene is in the distillate.

130

131

5.D.9. New Problem in 3rd Edition.

132

5.D10. New Problem in 3rd Edition. At the bubble point

x C5

.40 , x C6

.60 , p

yi

760 mmHg , y C5

1.0

y C6

K 5 x C5

K 6 x C6

1.0

133

or

VPC5 Ptot

x C5

0.40 10 TºC 20 51

VPC6 Ptot 6.853

1.0 , VPC5 x C5

x C6

1064.8 T 233.01

0.60 10

(C5 term) x (.4) 441.03 × .4 = 176.4

Final result is:

6.876

VPC6 x C6

Ptot

1171.17 T 224.41

+

1270.095 × .4 = 508.04 +

760

C6 term x (.6) 121.387 × .6 =

SUM 249.23

420.28 × .6 = 252.17

760.21

5.D11. Was problem 6.D2 in 2nd edition SPE. Since x i known, want

yi

1

K i p BP x i

K ref p old

Find new pressure from, K ref p new

Kixi

Use ethane, or n-pentane as reference. First guess: Try K C5 1.0, p 115 kPa

K NC7

0.13

Ki xi

K C2

29 0.1

K rep p new

Ki xi K C5 p new

29

1.0 .35

0.13 .55

1.0

~ 0.3 p new 3.32 8.0 0.1 0.3 .35

3.32 p too low.

440 kPa , K NC7

0.042 0.55

0.042

K C2

8.0

0.927

0.3

.032 p new 400 kPa, K nC7 0.045 K C2 0.927 K i x i 8.7 0.1 0.32 0.35 0.045 0.55 1.004

8.7

Answer (within accuracy DePriester Chart) = 400 kPa 5.D12. Was problem 6.D3 in 2nd edition SPE. a. Highest B.P. Temp. is pure n-octane. K C8 b. Lowest B.P. Temp. is pure n-hexane. K C6

1.0, T 174 C

1.0, T 110 C

5.D13. Was problem 6.D6 in 2nd edition SPE. Let 1 = n-butane, 2 = n-pentane and 3 = n-hexane. p = 101.3 kPa. Bubble Point: First guess. K1 1 at T 1 , K 2 1 at T 36 , K 3 1 at T 68 .

Tfirst K1

z1T1 3.6, K 2

.2

1

1.08, K 3

.5 36 0.36.

Choose 2 as ref. Eq. (6-14) is: K 2 Tnew

Tnew

29 C. K1

2.7, K 5

Eq. (6-14) is: K 2 Tnew

Tnew

28.8 K1

K1x1

38

.2 3.6

1.8 1.368

0.26, and

0.789 1.013

2.7, K8

.3 68

Kixi

.5 1.08

.3 .36

1.368

0.789 1.013

0.779

0.255, and

Kx i i

1.006. OK. TBP

28.8 C.

5. E1. This is 8.E4. in 1st edition.

134

135

136

5.H1. New Problem in 3rd Edition. Same program as 5.H5 except do not list y values as distillate. Different input in spreadsheet – see below. Ternary Distillation: Constant relative volatility. Step off stages from bottom up. Use whole stages. System has A = LK, B = HK and C = HNK C5H9 alpha Alpha CA-B 3.58 alpha B-B 1.86 B 1 feedstage 8 zA 0.35 z B 0.4 z C 0.25 epsilon (values for 0.00000001 N loop( F 200 q 1 L/D 6 convergence 100 df(HNK frac frac rec B in dist 0.996 guess frac rec C bot 1 recovery) 0.9 frac rec A in dist 0.961 D 67.59011 B 132.4099 L/V 0.857143 Lbar/Vbar 1.279858988 Mass balance xAdist 0.995264 xBdist 0.004734 xCdist 1.56E-06 values Mass balance xAbot 0.020618 xBbot 0.601768 xCbot 0.377614 values

137

Stage by stage calculations i xA yA 1 0.020618 2 0.041225 3 0.072406 4 0.11662 5 0.175066 6 0.245778 7 0.322624 8 0.396885 9 0.515565 10 0.633439 11 0.738112 12 0.822076 13 0.883953 14 0.926687 15 0.954865 16 0.97287 17 0.984143 18 0.991109

0.046992 0.0869 0.143487 0.218289 0.308791 0.407144 0.502187 0.584094 0.685129 0.774848 0.846817 0.899855 0.936484 0.960636 0.97607 0.985732 0.991702 0.995362

xB yB xC yC 0.601768 0.7125981 0.377614 0.688364 0.7538807 0.270411 0.720619 0.7419438 0.206975 0.711292 0.6917342 0.172088 0.672062 0.6158891 0.152873 0.612801 0.5274175 0.141421 0.543675 0.4396808 0.1337 0.475123 0.3632905 0.127992 0.42305 0.2920855 0.061385 0.339977 0.2160685 0.026583 0.251291 0.1497869 0.010597 0.173962 0.098934 0.003962 0.114634 0.0630978 0.001413 0.072825 0.0392226 0.000488 0.044971 0.0238835 0.000164 0.027075 0.0142529 5.45E-05 0.015839 0.0082925 1.77E-05 0.008886 0.0046363 5.57E-06

0.240409437 0.159219617 0.114569647 0.089976459 0.075319919 0.0654388 0.058132319 0.05261578 0.02278591 0.009083172 0.003395923 0.001211305 0.000418127 0.000141178 4.69552E-05 1.54307E-05 4.9941E-06 1.56158E-06

Calc frac recovery C in bottoms 0.9999979 j 5.H2. New Problem in 3rd Edition. The spreadsheet is, Ternary Distillation with Constant relative volatility. Step off stages from top down. System has A = LNK, B = LK and C = HK alpha A-B 2.25 alpha B-B 1 Alpha C-B 0.21 feedstage zA 0.25 zB 0.35 zC 0.4 epsilon (values for F 100 q 1 L/D 0.3 N loop( convergence frac rec B in dist 0.9 frac rec C in bot 0.97 df(LNK frac recovery) Guess: frac rec A in dist 1 D 57.66689 B 42.35298 L/V 0.230769 Lbar/Vbar xAdist 0.43295 xBdist 0.546241 xCdist 0.020809 Mass balance values xAbot 0.001251 xBbot 0.082639 xCbot 0.91611 Mass balance values Stage by stage calculations i xA yA xB yB xC yC 1 0.229688 0.43295 0.65203 0.546241 0.118282 0.020809168 2 0.180904 0.386044 0.601681 0.570654 0.217416 0.043302907 3 0.16005 0.374786 0.537147 0.559034 0.302804 0.066179941 4 0.147137 0.369973 0.486906 0.544142 0.365957 0.085884852 5 0.13893 0.366993 0.453607 0.532548 0.407464 0.100458716 6 0.133981 0.365099 0.433372 0.524864 0.432647 0.1100371 7 0.062603 0.208987 0.425676 0.631573 0.511721 0.159440004 8 0.021494 0.097273 0.308019 0.619528 0.670486 0.283199293 9 0.004909 0.032934 0.146011 0.435383 0.84908 0.531682943 10 0.000766 0.006976 0.044919 0.181823 0.954315 0.811200627 Calc frac recovery A in distillate

0.998702 j

Note x1 = xbot

3

6 0.0001 10 0.9 1.565105

Note y1 = xdist

2

138

5.H.3. New Problem in 3rd Edition. (L/D)min = 0.26761 by trial and error using spreadsheet in Table 5.A1.. 5.H.4. New Problem in 3rd Edition. Ternary Distillation with Constant relative volatility. Step off stages from top down. System has A = LK, B = sandwich and C = HK 5.G.e. alpha A-B 1.4 alpha B-B 1 Alpha C-B 0.7 feedstage zA 0.25 zB 0.35 zC 0.4 epsilon (values for F 100 q 1 L/D 5 N loop( convergence frac rec B in dist 0.583 frac rec C in bot 0.995 df(LNK frac recovery) Guess: frac rec A in dist 0.95 D 44.10818 B 55.89182 L/V 0.833333 Lbar/Vbar xAdist 0.532853 xBdist 0.462613 xCdist 0.004534 Mass balance values xAbot 0.026781 xBbot 0.261129 xCbot 0.71209 Mass balance values Stage by stage calculations i xA yA xB yB xC yC 1 0.447934 0.532853 0.544443 0.462613 0.007623 0.004534307 2 0.378937 0.462087 0.609404 0.530804 0.011659 0.007108533 3 0.325116 0.40459 0.658055 0.584939 0.016829 0.010471366 4 0.284385 0.359739 0.692247 0.625481 0.023368 0.01477984 5 0.254167 0.325796 0.71427 0.653975 0.031563 0.020228923 6 0.231958 0.300615 0.726286 0.672327 0.041757 0.027058086 7 0.215598 0.282107 0.730061 0.68234 0.054342 0.035552935 8 0.203353 0.268473 0.726901 0.685486 0.069746 0.046040694 9 0.193893 0.258269 0.717703 0.682853 0.088403 0.058877558 10 0.18623 0.250387 0.703059 0.675188 0.110711 0.074425299 11 0.179649 0.244001 0.683384 0.662985 0.136967 0.093014635 12 0.173652 0.238516 0.65905 0.646589 0.167298 0.114894592 13 0.167915 0.233519 0.630501 0.62631 0.201584 0.140170911 14 0.162254 0.228738 0.598352 0.602519 0.239393 0.168742359 15 0.156599 0.224021 0.563437 0.575729 0.279964 0.200250121 16 0.150964 0.219308 0.526798 0.546633 0.322237 0.234059083 17 0.145428 0.214612 0.489617 0.516101 0.364955 0.269286831 18 0.140099 0.209999 0.453098 0.485116 0.406803 0.304884695 19 0.135091 0.205558 0.418339 0.454684 0.446571 0.339758041 20 0.100543 0.157965 0.402362 0.45154 0.497094 0.39049513 21 0.071479 0.116121 0.372448 0.43219 0.556074 0.45168881 22 0.048123 0.080918 0.329669 0.395957 0.622209 0.523124245 23 0.030232 0.05263 0.276757 0.344144 0.693012 0.603226557 24 0.017191 0.03096 0.2177 0.280057 0.765109 0.688982619 Mass balance: fraction stage, A, B, C calculated at bottom, % error B 0.264616 0.026781 0.261129 0.71209 6.65242E-09 Calc frac recovery A in distillate 0.940127 j 4

19 1E-10 100 0.8 1.21119

Note y1 = xdist

Part d. Fractional recovery of B in distillate. is 0.725. Note that B goes through a maximum of close to 1% on stages 7 and 8. Ternary Distillation with Constant relative volatility. Step off stages from top down. System has A = LK, B = trace sandwich and C = HK 5.G.e. Part d. alpha A-B 1.4 alpha B-B 1 Alpha C-B 0.7 feedstage 19

139

zA 0.38 zB 0.02 zC 0.6 epsilon (values for F 100 q 1 L/D 4 N loop( convergence frac rec B in dist 0.725 frac rec C in bot 0.995 df(LNK frac recovery) Guess: frac rec A in dist 0.99 D 39.37044 B 60.62956 L/V 0.8 Lbar/Vbar xAdist 0.95555 xBdist 0.03683 xCdist 0.00762 Mass balance values xAbot 0.00626 xBbot 0.009071 xCbot 0.984668 Mass balance values Stage by stage calculations i xA yA xB yB xC yC 1 0.934659 0.95555 0.050434 0.03683 0.014907 0.00761993 2 0.909255 0.938837 0.064694 0.047713 0.026051 0.013449322 3 0.878109 0.918514 0.079128 0.059121 0.042762 0.02236485 4 0.839847 0.893598 0.092985 0.070669 0.067168 0.035733649 5 0.793216 0.862987 0.105202 0.081754 0.101582 0.055258739 6 0.737611 0.825683 0.114471 0.091527 0.147918 0.082789801 7 0.673775 0.781199 0.119471 0.098942 0.206753 0.119858439 8 0.604361 0.73013 0.119295 0.102943 0.276345 0.166926476 9 0.53382 0.674599 0.113888 0.102802 0.352293 0.222599678 10 0.467334 0.618166 0.104227 0.098476 0.428438 0.283358088 11 0.409234 0.564978 0.092025 0.090748 0.498741 0.344274648 12 0.361849 0.518497 0.079126 0.080986 0.559026 0.400517103 13 0.325379 0.480589 0.066982 0.070666 0.607639 0.448744495 14 0.298551 0.451414 0.056436 0.060951 0.645013 0.487634893 15 0.279454 0.429951 0.047786 0.052515 0.67276 0.517534464 16 0.266162 0.414673 0.040972 0.045595 0.692866 0.539732259 17 0.257043 0.40404 0.035754 0.040143 0.707203 0.555816779 18 0.250839 0.396745 0.031838 0.035969 0.717324 0.567286133 19 0.246633 0.391781 0.028939 0.032836 0.724428 0.575383054 20 0.193371 0.320667 0.029598 0.035058 0.777031 0.644274899 21 0.145301 0.251 0.029111 0.03592 0.825588 0.713080242 22 0.105056 0.188125 0.027585 0.035283 0.867359 0.776592117 23 0.073452 0.135485 0.025264 0.033287 0.901284 0.831228269 24 0.049874 0.094147 0.022436 0.030252 0.92769 0.875601733 25 0.032959 0.063306 0.019353 0.026552 0.947688 0.910141552 26 0.02117 0.041182 0.016207 0.02252 0.962623 0.93629802 27 0.013123 0.025762 0.013125 0.018405 0.973752 0.955833007 28 0.00771 0.015236 0.010183 0.014373 0.982107 0.970390508 29 0.004108 0.008156 0.007421 0.010525 0.988471 0.981318291 Mass balance: fraction stage, A, B, C calculated at bottom, % error B 0.402462 0.00626 0.009071 0.984668 5.53216E-07 Calc frac recovery A in distillate 0.990012 j 16

1E-10 100 0.8 1.308

Note y1 = xdist

5.H.5. New Problem in 3rd Edition. Ternary Distillation with Bubble point Calc. Step off stages from bottom up. System has A = LK, B = HK and C = HNK feedstage 5 zA 0.3 zB 0.3 zC 0.4 epsilon 1E-08 F 100 q 1 L/D 8 N loop 100 frac rec B in bot 0.997 guess frac rec C bot 1 df 0.9

140

frac rec A in dist K const. aT1 nB=A -1280557 nPen=B -1524891 nHex=C -1778901

aT2 0 0 0

0.995 aT6 7.94986 7.33129 6.96783

Trebguess ap1 -0.96455 -0.89143 -0.84634

500 ap2 0 0 0

p, psia ap3 0 0 0

14.7

D xAdist xAbot

29.940006 B 0.9969938 xBdist 0.002141 xBbot

70.05999 L/V 0.003006 xCdist 0.42692 xCbot

0.88889 Lbar/Vbar 1.26 2.2E-07 0.57094

stage 1 2 3 4 5 6 7 8 9 10

xA 0.002141 0.0086951 0.0291201 0.0853111 0.214736 0.4743433 0.7600825 0.920961 0.9779073 0.994401

xB 0.42692 0.613018 0.714099 0.717438 0.617813 0.483106 0.233736 0.078451 0.022047 0.005596

xC 0.57094 0.37829 0.25678 0.19725 0.16745 0.04255 0.00618 0.00059 4.6E-05 3.4E-06

yA 0.0103991 0.0361347 0.1069354 0.270011 0.5324156 0.786406 0.9294091 0.980028 0.9946891 0.9986729

Distillate mole fracs = y values Calc frac recovery C in bottoms

yB 0.6614042 0.7887656 0.7929732 0.6674458 0.4297616 0.2080991 0.0700679 0.0199309 0.0053079 0.0013269

yC 0.3282 0.1751 0.10009 0.06254 0.03782 0.00549 0.00052 4.1E-05 3E-06 2.2E-07

T 582.283 570.622 561.854 551.835 536.481 513.748 498.351 491.769 489.686 489.103

KB 1.549247 1.286692 1.110453 0.930318 0.695617 0.430752 0.299774 0.254057 0.240759 0.237137

0.9986729 0.00133 2.2E-07 0.9999998 j 3

Option Explicit Sub Ternary_bottom_up_BP() ' Ternary distillation with constant alpha. Frac recoveries of LK and HK given. ' There is a HNK present and its frac rec in bottoms is guessed. Sheets("Sheet1").Select Range("A18", "I150").Clear Dim i, j, k, feedstage, N As Integer Dim AaT1, AaT6, Aap1, BaT1, BaT6, Bap1, CaT1, CaT6, Cap1 As Double Dim F, fracBbot, fracCbot, q, LoverD, LoverV As Double Dim LbaroverVbar, D, B, L, V, Lbar, Vbar, Eqsum, fracAdist As Double Dim xA, xB, xC, yA, yB, yC, zA, zB, zC, xAbot, xBbot, xCbot As Double Dim DxA, DxB, DxC, BxA, BxB, BxC, xAdist, xBdist, xCdist As Double Dim fracCbotcalc, difference, epsilon, df As Double Dim T, p, Tinit, KA, KB, KC, sum As Double 'Input data AaT1 = Cells(9, 2).Value AaT6 = Cells(9, 4).Value Aap1 = Cells(9, 5).Value BaT1 = Cells(10, 2).Value BaT6 = Cells(10, 4).Value Bap1 = Cells(10, 5).Value CaT1 = Cells(11, 2).Value CaT6 = Cells(11, 4).Value Cap1 = Cells(11, 5).Value feedstage = Cells(3, 8).Value F = Cells(5, 2).Value

141

q = Cells(5, 4).Value LoverD = Cells(5, 6).Value zA = Cells(4, 2).Value zB = Cells(4, 4).Value zC = Cells(4, 6).Value fracBbot = Cells(6, 3).Value fracCbot = Cells(6, 6).Value fracAdist = Cells(7, 4).Value epsilon = Cells(4, 8).Value N = Cells(5, 8).Value df = Cells(6, 8).Value p = Cells(7, 8).Value Tinit = Cells(7, 6).Value ' The For loop (remainder of program) is to obtain convergence of guess of ' fractional recovery of A in distillate. For j = 1 To N ' Calculate compositions and flow rates. DxA = F * zA * fracAdist DxB = F * zB * (1 - fracBbot) DxC = F * zC * (1 - fracCbot) BxA = F * zA * (1 - fracAdist) BxB = F * zB * fracBbot BxC = F * zC * fracCbot D = DxA + DxB + DxC B = BxA + BxB + BxC xAdist = DxA / D xBdist = DxB / D xCdist = DxC / D xAbot = BxA / B xBbot = BxB / B xCbot = BxC / B L = LoverD * D V=L+D LoverV = L / V Lbar = L + q * F Vbar = Lbar - B LbaroverVbar = Lbar / Vbar ' Print values of flowrates and mole fractions Cells(13, 2) = D Cells(13, 4) = B Cells(13, 6) = LoverV Cells(13, 8) = LbaroverVbar Cells(14, 2) = xAdist Cells(14, 4) = xBdist Cells(14, 6) = xCdist Cells(15, 2) = xAbot Cells(15, 4) = xBbot Cells(15, 6) = xCbot ' initialize (reboiler =1) and start loops i=1 xA = xAbot

142

xB = xBbot xC = xCbot T = Tinit ' Calculations in stripping section: equilibrium then operating. Do While i < feedstage ' Bubble point calculaton For k = 1 To 10 KB = 1 KA = Exp((AaT1 / (T * T)) + AaT6 + (Aap1 * Log(p))) KB = Exp((BaT1 / (T * T)) + BaT6 + (Bap1 * Log(p))) KC = Exp((CaT1 / (T * T)) + CaT6 + (Cap1 * Log(p))) yA = KA * xA yB = KB * xB yC = KC * xC sum = yA + yB + yC KB = KB / sum T = Sqr(BaT1 / (Log(KB) - BaT6 - (Bap1 * Log(p)))) Next k ' Print values Cells(i + 17, 1).Value = i Cells(i + 17, 2).Value = xA Cells(i + 17, 3).Value = yA Cells(i + 17, 4).Value = xB Cells(i + 17, 5).Value = yB Cells(i + 17, 6).Value = xC Cells(i + 17, 7).Value = yC Cells(i + 17, 8).Value = T Cells(i + 17, 9).Value = KB ' Bottom operating line i=i+1 xA = yA / LbaroverVbar + (1 - (1 / LbaroverVbar)) * xAbot xB = yB / LbaroverVbar + (1 - (1 / LbaroverVbar)) * xBbot xC = yC / LbaroverVbar + (1 - (1 / LbaroverVbar)) * xCbot Loop ' Calculations in enriching section Do For k = 1 To 10 ' Bubble point calculation KA = Exp((AaT1 / (T * T)) + AaT6 + (Aap1 * Log(p))) KB = Exp((BaT1 / (T * T)) + BaT6 + (Bap1 * Log(p))) KC = Exp((CaT1 / (T * T)) + CaT6 + (Cap1 * Log(p))) yA = KA * xA yB = KB * xB yC = KC * xC sum = yA + yB + yC KB = KB / sum T = Sqr(BaT1 / (Log(KB) - BaT6 - (Bap1 * Log(p)))) Next k ' Print values Cells(i + 17, 1).Value = i Cells(i + 17, 2).Value = xA

143

Cells(i + 17, 3).Value = yA Cells(i + 17, 4).Value = xB Cells(i + 17, 5).Value = yB Cells(i + 17, 6).Value = xC Cells(i + 17, 7).Value = yC Cells(i + 17, 8).Value = T Cells(i + 17, 9).Value = KB ' Test for feed stage too low If xA < 0 Or xB < 0 Or xC < 0 Then Cells(i + 18, 3) = "Feed stage too low" Exit For End If i=i+1 ' Test for too many stages, which may mean reflux rate is too low. If i > 100 Then Cells(i + 18, 2).Value = "Too many stages" Exit For End If ' Top operating line xA = yA / LoverV - ((1 / LoverV) - 1) * xAdist xB = yB / LoverV - ((1 / LoverV) - 1) * xBdist xC = yC / LoverV - ((1 / LoverV) - 1) * xCdist ' Test for calculations being done. Loop While yA < xAdist ' Fractional recovery of C based on stage-by-stage calculation. fracCbotcalc = 1 - (yC * D) / (F * zC) difference = fracCbot - fracCbotcalc If Abs(difference) < epsilon Then Exit For fracCbot = fracCbot + df * (fracCbotcalc - fracCbot) ' Test if have convergence of fractional recovery of C. Next j Cells(i + 19, 1).Value = "Calc frac recovery C in bottoms" Cells(i + 19, 5).Value = fracCbot Cells(i + 19, 6).Value = "j" Cells(i + 19, 7).Value = j Cells(i + 18, 1).Value = "Distillate mole fracs = y values" Cells(i + 18, 5).Value = yA Cells(i + 18, 6).Value = yB Cells(i + 18, 7).Value = yC End Sub

144

Chapter 6 New Problems and new solutions are listed as new immediately after the solution number. These new problems are: 6A1, 6A5, 6D3, 6D4, 6G4-6G7. 6.A1. New Problem in 3rd Edition. Trial and error. Try a feed stage and determine the distillate and bottoms mole fractions of the key components. Repeat for additional feed stages. The feed stage that produces the best separation is the optimum feed stage for this value of N. 6.A5. New Problem in 3rd Edition. Trial and error. Pick an N that you think is close (a systematic method to do this is described in Chapter 7). Find the optimum feed stage. If you need more separation to meet specifications, increase N, and if you exceed specifications, try decreasing N. For an initial estimate of the optimum feed location for the new N, assume that the ratio (Optimum feed stage)/(total number of stages) is constant. Continue process until specifications are met or slightly exceeded. 6.C1. a.

With a side stream, mass balance is, Vj y j L j x j Sx S Vj 1 , y j 1 L j 1x j

Fjz j

1

where L j is the flow into the stage below, L j (6-4) to (6-6) are unchanged. Note that the L j b.

Now L jh j

Lj

Vj

1

Fj h Fj

1

L j x j , Eqs.

input into the matrix will be changed.

L jh j S jh j in the E.B. and j

D

j

Fk

k 1 k Substituting into EB we find

DEj

x j . Since L j x j Sx S

L j S, x S

Qj

D hj

j 1 k

1

1

Sk , L j

j

hj

1

SK h j

k j

1

k

1

Vj

1

1

j 1

D

FK h j

k

1

j 1 k

1

j 1

Fk

k

FK h j

1

Sk

1

SK h j

6.C2. Partial condenser mass balances is: Dy1 L1x1 V2 y 2 F1z1 This becomes, DK1 V2 K 2 2 F1z1 1 1 L1 L2 Thus,

B1

1

DK1 L1

, C1

(6-7)

K 2 V2 L2

, and D1

(6-8)

F1z1

Note that only B1 differs. 6.D1.

For n-pentane from Example 6-1, T = 60°C, K C5 Matrix:

L3 1825, L 4 450 B, V2 j = 1 (total condenser),

V3

1.05, L1 V4

L2

825 kmole/hr,

1375.

145

1.05 1375

K 2 V2

C1

L2

j

2, C 2

B2

1

825

j

1

K 3 V3

1.05 1375

L3

1825

V2 K 2

2.75, A 2

L2

j 3, C 3 B3

L4

450

1.791, A 3

L3

1.75

1

2.75

0

6.D2.

1,C5

p = 5 atm: z C2

0 0.791

1

0

1.791

0

to

0.08, z C3

0.33, z C4

yi

1, K C4

5.4 0.08

yi

1.7 0.33

12 , K C2

0

2,C5

0

3,C5

350

4,C5

0

0.49, z C5

0.10

yi

1.0.

506.5 kPa 0.47, K C5

0.47 0.49 Kixi

0.14

0.14 0.10

K C3 20

4.6, K C4

1.7

1.37

1.237

0.35, K C5

1.237

0.10

0.368 0.4521 0.1715 0.0 1.0016 OK

Propane Matrix Analysis: K C3

D3

Fz c3

V2

V3

L1

0

1.7, K C4

K C3 Tnew

Tnew

350

1. DePriester Chart.

5.4, K C3

Need lower T.

0

0

1,C5

x i . Want

Pick C 3 as ref. 5 atm 101.3 kPa atm Try T = 20°C. K C2

1, D 4

0

4.208

As sat’d liquid & for bp calculate z i 1 Guess: Want K C3

1000 0.35

3.208

1

1.67, D1

825

are in Example 6-1.

4,C5

st

Fz C51

4.208, A 4

L4

1.67

L1

550

3.208

1, D 3

1 V4 K 4

1

0

1.05 1375

V3 K 3

D

1

0.791

1, D 2

K 4 V4

4 Reboiler , B 4

Values for

1.75 , B1

1.37, B

330, D1 V4

L1

D L3 L 2 F Total Condenser (1):

D

V5

D2 V6

1025

2025

L4

F D 1000 410 D4

L1 L2

D5

D6

590

L6

0

D 1435 L3

L5

146

B1 1 D L1 1.40, C1 Stage 2.

A2

Stage 3. A 3

1, B2 1, B3

V2 K 2

1

L3

Stage 4:

A4

1, B 4

1

Stage 5:

A5

1, B5

1

Reboiler (Stage 6). A 6

V4 K 4

L5

1, B6

2.918

0

1

1435 1.37

L3

2025

V4 K 4

1435 1.37

L4

2025 V5 K 5

V6 K 6

1435 1.37

V6 K 6 L6

590

0

0

0

0

-0.9708

0

0

0

0

0

0.9708

0

0

1

0

0

0

-1

0

0

0

0

0.9708 0.9708

3.32

=4.32

L6

1.9708

1.918

0.9708

L5

=1.9708, C 5

1

1025

V3 K 3

=1.9708, C 4

L4

Mass balance matrix. 1.40 -1.918

-1

=1.9708, C 3

V5 K 5

1.37 1435

=2.918, C 2

L2

V3 K 3

1

K 2 V2 L 2

1.9708

0.9708

0

1.9708 -3.32 -1

4.32

6.D3. New Problem in 3rd Edition. p = 5 atm = 506.5 kPa z C2 0.08, z C3 0.33, z C4 0.49, z C5 0.10 As sat’d liquid & for bp calculation at z i Result is: Tbp

12 , K C2

4.6, K C3 =1.37, K C4

n-butane Matrix Analysis: K C4

D3

Fz c3

V2

V3

490, D1 V4

V5

x i . Calculation is same as in 5.D11 to obtain T.

0.35, B

D2

V6

D4

L1

0.35, K C5

F D 1000 410

D5

D6

A2

Stage3. A 3 Stage 4.

1, B 2 1, B3

A4

1

1, B 4

1

V2 K 2 L2

V3 K 3 L3 1

=1.49, C 2

=1.2480, C 3

V4 K 4 L4

L6

D 1435

L1

Stage 2.

590

0

D 1025 L 2 , L3 L 2 F 2025 D Total Condenser (1): B1 1 D L1 1.40, C1 K 2 V2 L 2 L1

0.10

=1.2480, C 4

L4

L5

0.35 1435

V3 K 3 L3

1025

1435 0.35 2025 1435 0.35

V4 K 4 L4

2025 V5 K 5 L5

0.49 0.2480 0.2480

0.2480

147

Stage5: A 5

1, B5

V5 K 5

1

L5

Reboiler (Stage 6). A 6

1, B 6

Mass balance matrix. 1.40 -0.49

V6 K 6

1

L6

1435 0.35

V6 K 6

=1.2480, C 5

L6

=1.8513

0

0

0

0

0

0

0

0

0

-1

1.49

-0.248

0

1

1.248

0.248

1

0

0

0

0

0

-1

1.248 -0.851

0

0

0

0

-1

5 60

300

6.D4. New Problem in 3rd Edition. L

1.248

L D D

V L D 360 Saturated liquid feed: V V

360; L

0.8513

590

L F

0.248

0 1.851

400

1

L1 V1

V1

L, L 2 0, V2

L, L3

L, L 4

V, V3

B

V, V4

F D

40

V

2

L2

V3 3 yi

V4

Bubble Pt. Set z i F

1.0 or

Ki xi

L3

xi

yi

Ki xi

1.0 M

3.58,

E

2.17,

NP

1,

NB

0.412

4

b.

yi

xi

i

xi

Eq. (5-30), zi i .3 3.58 Then and

y nP K nP

z NP

i

.25 2.17 NP

zi y x

i

nP

.35 1.0

0.1 .412

2.0077

0.35 1.0

0.1743 2.0077 0.1743 0.4981 0.35 148

KM c.

M NP

K nB

Matrix for n-butanol Stage 1. A1 , B1

1, B 2

L2

300

V4 K 4

C3

F3z nBut

Stage 4. A 4

1.2 1 0 0 d.

y 1 j

2

j=3

1.8468

40

10

1

V4 K 4

2.8468, C 4 , D 4

B 0

0.2463 1.2463

0.1847

1

1.1847

0

1 1.2, V21

V12

B2

A2

D2

C 2 V11

V13

B3

V23 V33

D3

A3

0

2

0

3

10

4

0

1

1

.2463

1.2 1.2463 1

V12

0

1

V13

10

1.8468 1.04105

.20525

.20525 0

0.1847 0.1539 1.2 V3 2 1.1847 1

A3V22

C3 V12

0

0, V31

V3

A 2 V2

V32

1

2.8468

B1

0

0 1.8468

V11

V22

0

1.1847

360 0.2052

1 B4

0.2052

0

0.1847, D 2

400

L3

100 .1

0.2463, D1

360 .2052

V3 K 3

1

K NP

1.2463

L2

L3

L4

D3

V2 K 2

V3 K 3

1, B3

nB NP

1 0.2 1.2,

360 0.2052

1

1.0809, K n-B

0.1

V2 K 2

C2

Stage 3. A 3

1.7832; K E

0.2052, z nB

1 D L1

C1

Stage 2. A 2

K NP

1.04105

0

0.1539

1.0308

1 0 1.0308 9.7014 1.7740

149

V14

j=4

B4

V24

D4

V34

e.

VP

V3

3

A 4 V23

C4 V13

2.8468

V14

0

1.7740

1 9.7014

1.0728

1.0728 9.0428

not needed. 9.0428 (bottoms flow rate)

V24

N

3NB

V23

V33

4

9.7014

2 NB

V22

V33

3

0

0.1539 25.743

3.9619

1NB

V22

V31

2

0

.20525 3.9619

0.8132

VP

VP

nP

Ptot

K nP 760

1.7740 9.0428

n 200

760

0.4981 760

378.5 mmHg

5.2983, n 378.5

1 66.8 273.16 Linear Interpolation: 5.9362

TbP

5.2983

273.16

25.743

in mmHg

Need to interpolate VP data. We know n VP

6.F1.

1

4,NB

K nP

Raoult’s Law

A4

1T 5.9362, n 400

2.9415E 3

5.99146

1 82.0+273.16

2.9415E 3 2.8163E 3 5.2983 5.99146

351.74 or TbP

2.81563E 3

2.9415E 3

0.00284

78.6 C.

Plots of vapor pressure are available in Maxwell (see Table 2-2 for reference) while tabulated values are in Perry’s K VPi p tot . Dew point calculation on feed gives 245.7°F. Overall Mass Balances: D = 30, L = 5D = 150 V = L + D = 180, V V F 180 100 80 , L L 150 , B = F – D = 70 First Trial Values Stage T L V KB KT Kx 245.7 70 = B 80 2.307 1.042 4 0.534 245.7 150 180 2.307 1.042 3 0.534 245.7 150 180 2.307 1.042 2 0.534 1 245.7 150 30 = D 2.307 1.042 0.534 Stage 4 3 2 1 Stage 4 3 2 1

C -2.705 -2.8404 -2.8404

Benzene B 3.705 3.8404 3.8404 1.2

C -1.191 -1.2504 -1.2504

Toluene B 2.191 2.2504 2.2504 1.2

A -1 -1 -1 A -1 -1 -1 -

D 0 35 0 0 D 0 40 0 0

ℓ 7.9886 29.5978 57.058 135.6569 ℓ 27.1651 59.5188 61.5875 64.1742 150

Stage 4 3 2 1 6.G1.

6.G2.

C -.6103 -.6408 -.6408

Xylenes B 1.6103 1.6408 1.6408 1.2

A -1 -1 -

D 0 25 0 0

ℓ 22.7361 36.6120 21.1971 11.3193

Using Peng-Robinson. Aspen-Plus solution:

Stage

T1°C

L kmol/h

V

C4

C5

C8

1

38.31

825

0

x1 y1

0.360 0.6568

0.6013 0.3424

0.0386 0.00083

2

69.16

557.3

1375

x2 y2

0.0993 0.3601

0.4499 0.6013

0.4508 0.0386

3

107.02

1533.9

1107.3

x3 y3

0.03355 0.2288

0.1731 0.5251

0.7934 0.2461

4

140.92

450

1083.9

x4 y4

0.00436 0.04568

0.0429 0.2271

0.9528 0.7272

1. What VLE package did you use? Peng- Robinson. 2. Report the following values: Temperature of condenser = - 2.77 oC Temperature of reboiler = 79.97 oC Distillate product mole fractions C2 0.3636, C3 Bottoms product mole fractions C2

1.2 E 13, C3

0.6360, C4

0.0004

0.000492, C4

0.9995

3. Was the specified feed stage the optimum feed stage? Yes No If no, the feed stage should be: a. closer to the condenser, b. closer to the reboiler. 4. Which tray gives the largest column diameter (in meters) with sieve trays when one uses the originally specified feed stage? Tray # 28 Diameter = 0.792 m. 5. Which components in the original problem are the key components? LK = Propane, HK = butane 6. Change one specification in the operating conditions (keep original number of stages, feed location, feed flow, feed composition, feed pressure, feed temperature/fraction vaporized constant) to make ethane the light key and propane the heavy key. What operating parameter did you change, and what is its new value? D = 20 Temperature of condenser = - 31.54 oC Temperature of reboiler = 50.87 oC Distillate product mole fractions C2 0.9955, C3 0.00448, C4 1.32 E 07 Bottoms product mole fractions C2

0.00112, C3

0.4364, C4

0.5625

151

6.G3.

For column 1 report the following: a. Final value of L/D 1.8 b. Split fractions of ethanol (distillate) 0.9999 and n-propanol (bottoms) 0.9913 c. Mole fractions in bottoms 1.70 E-5, 0.00871, 0.9913 d. Mole fractions in distillate 0.4545, 0.5383, 0.00714 For column 2: a. Optimum feed location in the column. 18 b. Mole fractions in bottoms 0.00689, 0.9800, 0.0131 c. Mole fractions in distillate 0.9917, 0.0083, 0.0

6.G4. New Problem in 3rd Edition. 1. Temperature of condenser = 389.9_ K. Temperature of reboiler = __547.4 K Qcondenser = _-772260____cal/sec, Qreboiler = _____912459__cal/sec Distillate product mole fractions: B= 0.23529, T= 0.76471, BiP = 0.12E-08_________ Bottoms product mole fractions:_B = 0.5 E-10, T = 0.67 E-08, BiP= 1.0000_________ 2. Was the specified feed stage the optimum feed stage? Yes No x If no, the feed stage should be: a. closer to the condenser, b. closer to the reboiler. a__ (Note: Do minimum number of simulations to answer these questions. Do not optimize.) 3. Which tray gives the largest column diameter with sieve trays when one uses the originally specified feed stage? Aspen Tray #__16______Column Diameter =______2.28____meters [Use the default values for number of passes (1), tray spacing (0.6096 m), minimum downcomer area (0.10), foaming factor (1), and over-design factor (1). Set the fractional approach to flooding at 0.65. Use the “Fair” design method for flooding.] 4. Which components in the original problem are the key components (label light and heavy keys) _____LK = toluene, HK = biphenyl_____________________________________________ 5. Change one specification in the operating conditions (keep N, feed location, feed flow, feed composition, feed pressure, feed temperature or fraction vaporized constant at original conditions) to make benzene the light key and toluene the heavy key. Also increase the reflux ratio to 4.0. What operating parameter did you change (not including the reflux ratio), and what is its new value? D = 40________ Temperature of condenser = _368.9____ K, Temperature of reboiler = 407.7____ K Distillate product mole fractions: _B = 0.9283, T = 0.07173, BiP = 0.8 E-19________ Bottoms product mole fractions: _B = 0.01793, T = 0.79457, BiP = 0.1875_________ 6.G.5. New Problem in 3rd Edition. 1. Temperature of condenser = _121.07___ K. Temperature of reboiler = _166.23___ K Qcondenser = ____-757506.6____cal/sec, Qreboiler = ______1058466.75____cal/sec Distillate product mole fractions:__B = 0.9779, T = 0.22070, pxy = 0.6004 E-05__ Bottoms product mole fractions:___B = 0.0055189, T = 0.55698, pxy = 0.43750___ 2. Was the specified feed stage the optimum feed stage? Yes No x If no, the feed stage should be: a. closer to the condenser, b. closer to the reboiler. a (Note: Do minimum number of simulations to answer these questions. Do not optimize.) 3. Which tray gives the largest column diameter with sieve trays when one uses the originally specified feed stage? Aspen Tray #____24______Column Diameter =______2.28___meters 152

[Use the default values for number of passes (1), tray spacing (0.6096 m), minimum downcomer area (0.10), foaming factor (1), and over-design factor (1). Set the fractional approach to flooding at 0.7. Use the “Fair” design method for flooding.] 4. Which components in the original problem are the key components (label light and heavy keys) ________benzene = LK, toluene = HK______________________ 5. Change one specification in the operating conditions (keep N, feed location, feed flow, feed composition, feed pressure, feed temperature or fraction vaporized constant) to make toluene the light key and p-xylene the heavy key. What operating parameter did you change, and what is its new value?__D=260______ Temperature of condenser = _142.2____ K, Temperature of reboiler = _183.98__ K Distillate product mole fractions: _B = 0.30769, T = 0.68850, Pxy = 0.003805________ Bottoms product mole fractions: __B= 0.3177 E-06, T = 0.007066, Pxy = 0.99293_____ 6.G.6. New Problem in 3rd Edition. Part a. L D 27 . b. L D 60 . c. D = 147, S = 453 (liquid) distillate mole fracs: E = 0.99007, B = 0.00993, P = 0.5 E-9 side stream mole fracs: E = 0.0009845, B = 0.98930, P = 0.000854 bottoms mole fracs: E = 0.7 E-14, B = 0.00043, P = 0.99957 d. distillate : E = 0.89146, B = 0.10854, P = 0.556 E-8 side: E = 0.041845, B = 0.95794, P = 0.000218 bottoms: E = 0.1 E-14, B = 0.0001095, P = 0.99989 Since vapor mole fraction ethane > liquid mole fraction (ethane is LK), have more ethane in vapor side stream. e. The separation of n-pentane and n-butane is much more difficult than between ethane and nbutane. Thus side stream purity is less. Also feed has lot more pentane than ethane, which makes side stream below feed less pure. 6.G.7. New Problem in 3rd edition. 1. Report the following values: Temperature of condenser = _373.28____ K. Temperature of reboiler = ___411.75___ K Qcondenser = _-829828_____cal/sec, Qreboiler = ____1012650_____cal/sec Distillate product mole fractions: M = 0.59998, E = 0.36184, NP = 0.038177, NB = 0.3087E -05 Bottoms product mole fractions: M= 0.2042E-04, E = 0.03816, NP = 0.46182, NB = 0.50000_ 2. Was the specified feed stage the optimum feed stage? Yes No X If no, the feed stage should be: a. closer to the condenser, b. closer to the reboiler. Answer a (Note: Do minimum number of simulations to answer these questions. Do not optimize.) 3. Which tray gives the largest column diameter with sieve trays when one uses the originally specified feed stage? Aspen Tray #_____18_____Column Diameter =___1.77____meters [Use the default values for number of passes (1), tray spacing (0.6096 m), minimum downcomer area (0.10), foaming factor (1), and over-design factor (1). Set the fractional approach to flooding at 0.7. Use the “Fair” design method for flooding.] 4. Which components in the original problem are the key components (label light and heavy keys) ______________LK = ethanol, HK = n-propanol_______________________

153

5. Change one specification in the operating conditions (keep N, feed location, feed flow, feed composition, feed pressure, feed temperature or fraction vaporized constant) to make methanol the light key and ethanol the heavy key. What operating parameter did you change, and what is its new value?_____D = 60____ Temperature of condenser = __368.66__ K, Temperature of reboiler = _404.23___ K Distillate product mole fractions: M = 0.97858, E = 0.021417, NP = 0.155 E-07, NB = 0.1 E-10_ Bottoms product mole fractions: M = 0.0091787, E = 0.27654, NP = 0.35714, NB = 0.35714__

154

Chapter 7 New Problems and new solutions are listed as new immediately after the solution number. These new problems are: 7.A1, 7.A4, 7.D2, 7.D10, 7.D11, 7.D14, 7.D21, 7.G1. 7.A1. New problem in 3rd edition. f. none of the above. 7.A.4. New problem in 3rd edition. a. estimate fractional recoveries nonkeys at total reflux. 7.C4.

Use yi, j

1

K i x i, j 1 . Then substituting into Eq. (7-20), we have Vmin K i x i, j

which is

L min x i, j

Vmin K i x i, j 1

L min x ij

1

Vmin K i K HK

1

Dx i,dist

Dx i,dist where K i

1

Kx i,dist , or L min x i, j

L min

c

Total flow rate L min is L min

i 1

L min x ij

L feed

7.C5.

VF

For saturated vapor feed have

L min

i

i

(A)

1 i

i

(B)

1 i

(C)

c i 1

L min

1/

i 1

Dx i,dist 1/

Fz i

c

i

(7-33 analogue)

i L

i

i

(7-29 analogue)

L

F . For binary system Eq. (7-33) is, z

1

1

Vmin K

L

From Eqs. (A) and (C) we have L min

i

Dx i,dist V K i 1 min HK L min

L min L min Add Eqs. (A) and (B), and use external mass balance, qF

K HK . Rearranging,

c

1

Vmin K HK L min

Vmin K1

L

i

Dx i,dist Vmin K HK L min

1

Bx i ,bot

L min

By a similar analysis obtain,

Let

L min x ij

1

1 1

2

1

z2

2

Clearing fractions we obtain 1

2

z

2

z

1 1

2 2

1

After some algebra this

z

2

Solutions are, For sat’d liq’d

1 1

0 or VF

1

z

2 2

2

1

z

1 1

2

2

0

z2

0 . Clear fractions and equation is linear.

155

x 1 x

n 7.D1.

dist

x 1 x

a. Eq. (7-16), N min

n

bot

.992

.008

.014

.986

10.36

n AB n 2.4 This includes the partial reboiler. Eq. (7-40a) gives, x 1 x

n

z 1 z

N f ,min b. Saturated liquid:

Vf

dist

n

B

B

zB

T

B

,

zT

0

T

1.0 2.4

zB

zB

T

5.97

n 2.4

AB

After clearing fractions and solving for

.008

.4 .6

feed

0 . Eq. (7-33) becomes

T

.992

n

zT

2.4 .4

1.0 .6

1.53846

Which does lie between the α’s of the keys. To use Eq. (7-29) we need D. From mass balances (Eq. (3-3)).

z

D Eq (7-29) is:

x bot

x dist

Vmin

B

x bot

.4 .014

F

.992 .014

Dx Bdist

T

Dx Tdist

B

Vmin

L min c.

L D Using Eq. 7-42b,

1.1 L D N

3.9468 kg moles/hr

.

T

2.4 3.9468 .992 2.4 1.53846 Vmin D 6.9013 .

min

10

1.9234 , x

1.0 3.9468 .008 1 1.53846 L D min 1.75

L D

L D

min

L D 1

10.848

0.0598

N min

0.5563 N 1 .5563 N min Solving for N, N 24.6 (includes reboiler) 1 .5563 N F,min 5.97 From Eq. (7-40b), N F N 24.6 14.2 N min 10.36 Try stage 14 from top for feed stage. 7.D.2.

New problem in 3rd edition.

p

5 atm, z C2

Saturated liquid and for bp. Calc., z i

0.08, z C3

x i . Want

Pick C 3 as reference (this is arbitrary). 5 atm 1st Guess: Want K C3

1 (light key), K C4

yi

101.3

0.33, z C4

0.49, z C5

0.10

1.0 .

kPa

506.5 kPa atm 1 (heavy key). Use DePriester Chart.

156

20 C , K C2

Try T

yi

1.7 , K C4

5.4 , K C3

5.4 0.8

1.7 0.33

0.47 , K C5

0.47 0.49

0.14 0.10

0.432 0.561 0.230 0.014 1.237 K C3 20 1.7 Need lower T. K C3 Tnew 1.237 Kixi Tnew

12 , K C2

yi

n N MIN

b)

C2 C4

0.35, K C5

C4

HK be reference.

0.10

1.37 0.35

C3C4

3.914

0.997 0.998 0.003 0.002

12.01874

n 3.914

0

i

C4 C4

1.0,

3.914 0.33

13.14

i

3.914

Solve for φ. Find φ = 1.74 (Note i

VMIN

0.1 0.35

C5 C4

0.286

VMIN . In Eq (7-33) divide through by F. 13.14 0.08

zi

8.808 (includes PR).

1.36456

4.6 0.35 13.143,

Sat’d liquid feed VMIN

Eq (7-29)

1.37

0.368 0.4521 0.175 0.0 1.0016

For remainder, let

a)

4.6, K C4

0.14

1.0

3.914

LK HK

Dx i,dist

1.0 0.49

0.10 0.286 0.286

HK HK

)

. Assume all C 2 in distillate & all C 5 in bottoms

i

Dx i,C2

Dx i Dx i

80, Dx i,C5 C3

C4

0

0.997 1000 0.33 1 0.998 1000 0.49

D VMIN

c)

13.14 80

329.01 3.914

13.14 1.74 L MIN VMIN

L MIN

Eq (7-42b)

N N

Dx i,d

0.98

409.99

0.98 1.0

0 683.23 kmole h 3.914 1.74 1.0 1.74 D 273.24; L D MIN 0.6664

1.15 L D

Ordinate Gilliland

329.01

MIN

0.7664

0.7664 0.6664 1.7664

N MIN

0.05662

0.545827 0.591422 0.05662

N 1 0.5608 8.808 1 0.5608

0.002743 0.05662

0.56079

21.33 (include PR)

157

Dx LK Dx HK

n N F,MIN Eq. (7-40b)

n

N F,min

NF

N min

N

1 N min AB

x 1 x

n 7.D4.

x 1 x

N min

a.)

dist bot

log

xA xB

n

n3.914

.552

NF

xA xB

d

xA xB

d

xA xB

bot

bot

.01773 0.98227

y

L D

L D

13

.36 .64

1.287

y*

10.02

0.9915 y* 0.9915 0.6 x

1

1 x

(L/V) min = 0.534,

c.) Abscissa =

.545 .455

.9915 .0085

(L/V)min = * z = .6

bot

AB

1 N min

n AB n 2.4 b.) Feed is saturated liquid, feed line is vertical.

y*

xA xB

d

log

xA xB

4.552

21.33 11 (approximate)

8.808

7.D3. At total reflux use Fenske Eq. (7-11). N min

AB

329.01 0.98 0.33 0.49

n

LK HK

log

Rearrange, log

dist

z LK z HK

0.7826 x z 0.6

L D

L V min

1

L V

1.144 min

2.2286 1.144

0.336 L D 1 3.2286 From Eq. (7-42b) N N min 0.002743 0.545827 0.591422 0.336 0.3553 ordinate N 1 0.336 N min ordinate 10.022 0.3553 N 16.1 1 ordinate 1 0.3553 10.022 0.3474 From fitted curve ordinate = 0.3474, N 16.2 1 .3474 Error = 25-16 25 36% low. Aspen Plus equilibrium data is not α = 2.4. Note that α = 2.24 may be a better fit. min

158

log 7.D5.

Fenske Eq. is: N min

xD 1 xD

xB 1 xB

log

30,

7.D6.

N min

1.30 and x D xD 1 xD

log Fenske Eq.: N min

xD 1 xD

0.984, this is x B

xB 1 xB

log

2.4 .4

x 1

1 x

xD

L V

xD

min

Then,

x

.01 .99

.993 .616

z

.993 .4

1.15 1.746

0.636 ,

0.4 .

L D

L V 1 L V

min

1.746 . min

2.01

act

L D

Gilliland Correlation: Abscissa

L D

2.01 1.746

min

L D 1 3.01 N N min N 10.82 Ordinate 0.557 N 1 N 1 Clear fractions, and find N = 25.3 (including partial reboiler).

7.D7.

p

5 atm. From the solution to problem 6.D9: Tbp

K C2 Let C 4

4.6, K C3

1.37, K C4

HK

reference.

z

Eq. 7-33, 0

i i i

Want

LK HK

0.35, K C5

0.10

0.02 0.008

8.7121 1.36456

4.6 0.35 13.143,

C2-C4

0.10 0.35

C4 C4

6.38

1.0

0.286

Vmin

13.14 0.08 13.14 3.914

.0878

12 C

n 3.914

C5 C4

Sat’d/liquid feed, Vmin

3.01

3.914 0.35 0.98 0.992

C3 C 4

a) Eq. (7-15), Including PR N min

C4

.264

1.37

HK be reference.

n

b)

10.82

2.4

.616

1.4 .4

y*

L D

1

0.229

.993 .007

log

log

Determine y in equilibrium with feed z

y*

xB 1 xB

xD 1 xD

Solving for x B , we obtain: x B Since N min

xD 1 xD

N min

or

3.914 0.33 3.914

1.0

HK HK

1.0 .49 1.0

. Converge to

0.10 0.286 0.286 = 1.74.

159

Dx i

C2

Dx i

C4

Fzi

0.98 1000 0.33

C3

.008 1000 0.49

13.14 80

Eq. (7-29), Vmin

80, Dx i

323.4 3.914

3.92, Dx i

C5

323.4 0

3.92 1.0

0 671.05 13.14 1.74 3.914 1.74 1.0 1.74 D Dx i,d 407.32, Lmin Vmin D 263.73, L D min

c)

L D 1.2 L D N

Eq. (7-42b),

N min

0.002743

0.545827 0.591422 0.073

0.073

0.073

0.5402

15.05, incl. PR. Nfeed ~ 9 1 0.5402 a. Can do this graphically, or can calculate slope of a line from y x x D .992 to intersection of feed line and equilibrium, or use Underwood. Easiest to calculate slope. Feed line y z F .4 . Equilibrium: x

y y

V

c.

min

N min

.992 .3755

1.2

22.83 N

L

.958 ,

xB 1 xB

D

L V min

L D

27.4 , Abscissa

L D

22.83

1 L V

.992 .005 .008 .995 n 1.11

n This is 95.9 stages plus partial reboiler.

L D

.3755

.4 1.11 .6

.992 .4 xD 1 xD

n b.

.4

1 y

L

min

L D 1

96.9

27.4 22.83 27.4 1

.161

N min

or N 181.9 which includes partial reboiler. N 1 This separation would probably not be done by distillation. LF Fz Feed 80% liquid, L F .8F, , Slope VF .2F. Feed line: y x VF VF Ordinate

7.D9. a.

0.777 0.647 1.777

N 1 0.5402 6.38

N

7.D8.

0.777 . Ordinate Gilliland

min

0.647

See Graph.

.47

L V

L D b. c.

N min

6

L D

L

Min top op line is tangent.

3 4

min

1

min

L V

min

V

slope min

0.5175 1 .5175

.8 0.386 0.8 0

8 2

4

0.5175

1.0725

eq. contacts. See graph.

1.05 1.0725

1.1262 . Abscissa, Gilliland Correlation is

actual

160

L D

L D

min

1.1262 1.0725

0.053666

2.1262

2.1262

L D 1 Ordinate ~ 0.63 from graph. From eq. (7-42b), Ordinate

0.02524

0.545827 0.591422 0.02524

(agrees with graph). N min ordinate 6.75 0.6396 Then N 20.5 1 ordinate 1 0.6396 Need 20 eqs. contacts + P.R. N F,min 6 20.5 18 N F,min from graph = 6. N F N N min 6.75

0.002743 0.02524

0.6396

(7-40b)

7.D10. a) New problem in 3rd edition. Eq. (7-15)

n N MIN

FR E ,dist

FR B,bot

1 FR E ,dist 1 FR B,bot n

EB

161

0.989 0.998

n EB

b)

N MIN

0.011 0.002

N MIN

13.14

n 13.14

4.159 is known..

PB

2.5756

FR B,bot

Dx D

4.159

PB N MIN

1 FR B,bot c)

4.159

3.91. N MIN

Eq. (7-17) FR P ,dist

10.7114

PB

3.91 0.998 3.91 0.002

4.159

0.3677

FR i,dist Fzi

i

Ethane

Dx DE

0.989 100 0.3

29.67

Propane

Dx DP

0.3677 100 0.33

12.134

n-butane

Dx DB

0.002 100 0.37 3

D Also accept D = 0 since total reflux.

i 1

0.074

Dx i,d

41.878

kmol h

7.D.11. New problem in 3rd edition.

D 200

zn

0.35

z iP

0.4

z NP

2

V1

0.25

1 B Use Underwood Eqns. – Case A Assume LNK (propane) is all in distillate. b)

Vfeed

F 1 q

Eq. (7-33). F 1

F since q i

F zi

Dx p,dist

Fz p

20

0

where φ is between α’s of two keys (B and H)

i

1.0 > φ > 0.2. Equation is, 2.04 0.2 1.0 .35 1.0 2.04 1.0 Solving for φ obtain φ = 0.62185.

0.20 0.45 0.20

162

Then

Vmin

Dx B,dist

Dx i,dist

. Find D from fractional recoveries.

0.99 100 .35

34.65

Dx p,dist

20

Dx H ,dist

1 0.98 100 .45

0.9

D

VMIN

L min 7.D12.

2.04 20

55.55

1.0 34.65

0.2 0.9

2.04 0.62188 1.0 0.62188 0.2 0.62188 VMIN D 64.4314 and L D min 1.1599

A = benzene (LK),

AB

2.25, FR A,dist

B = toluene (HK),

BB

1.0, FR B,bot

C = cumene (HNK),

CB

0.210

0.98 0.99

n a. Use Fenske eqn. at total reflux. N min

FR A ,dist

1 FR B,bot

1 FR A ,dist

FR B,bot

n 0.98 0.02

n N min

0.01 0.99

FR A ,dist

AC N min

where

AC

AC

1 FR A ,dist

AB

10.47

n 2.25

N min

FR C,bot

119.98

AB

2.25

CB

0.21

10.71

10.47

10.71 FR C,bot 0.98 10.47 10.71 0.02 (We can also substitute into Eq. (7-17)). N min AC

FR C,dist

FR B,bot

Feed is sat. vapor. q

0,

100 Find

Vfeed

0.21 N min

CB

1 FR B,bot b. Underwood equations – Case B analysis

1 . All cumene goes to bottoms.

F 1 q

.99 .01

F 100 ,

10.47

0.21

Vfeed

2.25 40

1.0 30

0.21 30

2.25

1.0

0.21

8.1 10

10.47

C i 1

i

12

0

Fz1

1

1.6516.

163

C

Vmin

Dx A,dist

100 0.4 0.98

i 1

From mass balance, L min

100 0.3 0.01 1.0 39.2

2.25 1.6516

1 1.6516

1.25 2.71

min

L D

ordinate = 0.46. With N min

xA xB

Underwood:

Vf

V V

Dx B,d

Dx T,d Vmin

L min

dist

D

39.6

min

2.71

feed

AB

n 2.25 N feed N

gives N feed

1 .9899

0.0101

5.30.

10.25. Use stage 10 or 11.

.99 .02 99 , FR C 1 FR C 0.0204 .01 .98 log 99 0.0204 log 4851 5.438 log TC log 1 0.21

2.5 .25

0, 0

1.526 or 0.3374. Use

zA zB

0.9899, xB 39.6 0.9899 0.4 n 0.0101 0.3

min

N min

107.2

0.155 . From Figure 7-3 the

min

39.2

min

1 FR T

L D

n

min

N min Use Fenske eq. FR T

L

min

x A ,dist

7.D13.

39.6

3.39

L D 1

n

N feed

0, D

10.47, we find N = 20.24.

To find N feed , we need N feed

N feed

i,dist

0 146.78

D 146.78 39.6 107.2 ,

abscissa for Gilliland correlation

N feed

Fz1 FR

0.4, Dx C,dist

2.25 39.2

Vmin

L D 1.25 L D

where Dx i,dist

1

39.2, Dx B,dist Vmin

c.

Dx i,dist

i

2.5

1.0 .30 1.0

0.21 45 21

0.3374 as it is between keys. Vmin

Fz B

3 i 1

1

Dx i,d 1

25 (assume all benzene in dist.)

.99 Fz T 2.5 25

29.7, Dx c,d 1.0 29.7

0.02 Fz c

0.9, D

.21 9

2.5 0.3374 1.0 0.3374 0.21 0.3374 V D 16.64 and L D min 0.2993

55.6

72.24

164

N

Gilliland: Ordinate

N min

9 5.438

0.3562

N 1 10 Abscissa ~ .29 (original Gilliland) or .36 (Liddle) L L L If use 0.29 have, 0.29 0.29 D D min D

If use Abscissa = 0.36,

L

.29 0.2993

D L

1 .29 .36 .2993

D

.83 1.03 which are quite different. Safer to use

1 .36

higher value. If 2.25, N min is same. Underwood Eq. gives BT

Vmin

72.68 , L min

29.40 44.78 1.492

Which is 2.7% different than for

BT

1.4666 or 0.3367. Use 0.3367.

V D 17.084 and L D min

0.3073

2.5.

7.D14. New problem in 3rd edition. Use Gilliland correlation to find the value of the minimum reflux ratio, (L/D)min = 1.4

FR B,dist FR C,bot

n 7-D15.

1 FR B,dist

Fenske: Eq. (7-15), N min

N min TC

1 FR C ,bot

FR tol,dist

Dx d ,tol

N F,min

BC

where

FR C ,bot

n

1 FR C,bot

n N min TC

Eq. (7-17), FR T ,dist

1 .21

0.8238 167

Dx d ,benz

0.9992 397

Dx d ,cum

0.0001 436

x LK x HK

dist

n

LK-HK

.9992 .9999 .0008 .0001 2.25 n .21

6.89

1 TC

.21

6.89

1 .21 .9999 .0001

n

z LK z HK

0.8238, and FR tol,bot

6.89

x dist

137.57 396.68

0.2568 0.7418

.0436 D n

0.1762

534.294

0.0008

.7418 .397 .000815 .0436 n 2.25 .21

1.94 .

Underwood: Use a Case C analysis since toluene is a sandwich key. 3 2.25 397 1.0 167 .21 436 1Fz1 Eq. (7-33): 0 VF is, 0 i 1 2.25 1.00 .21 1 1.216 and 0.3373 which lie between α’s.

165

Eq. (7-29):

Vmin

3

Dx i,dist

i 1

Write for For

i

2.25 396.88

becomes Vmin

1.0 Dx tol

2.25

i

.21 0.436

1.0

.21

1.216 and for 0.3373. Obtain 2 eqns and 2 unknowns: Vmin and Dx tol,dist .

1.216 , Vmin

.3373, Vmin

863.525 4.629 Dx tol . For

466.15 1.509 Dx tol .

Solving simultaneously, Vmin

563.84, Dx tol

64.740

D

Dx i,d

396.88 64.74 .436

L min

Vmin

D 101.79 and L D

L D L D min

462.056 min

0.2203

1.2 .2203

0.445 L D 1 2.2 N N min Ord. .245 (Original Gilliland) N 1 Obtain N = 9.45 (includes reboiler) N F,min 2.91 N 9.45 2.66 (Use stage 3) Estimate N F N min 6.89 Gilliland Abscissa

n 7.D16. a. Fenske: N min b. Underwood:

x

x

1 x

dist

Vfeed

EP

2.1 .6

1.0 4

2.1

1.0

To find D: D

1 x

bot

n Fz E

z xD

1.0 F z p 1.0

EP

1

.99 .992 .01 .008 n 2.1

n

,

0 or

xB

F

xB

, zE

.6, z P

1.44 . Use

.6 .008 .99 .008

1000

1.0 .008

D

2.1 1.44

1.00 1.44

c. Use Gilliland: Ordinate

N

N min

Sounds harder than it is: 0

f

Vmin

L

D 1285.3,

D

2.13 min

.30 12.69

VF F

F

3.132

0.558 N 1 31 L D L D min Abscissa ~ 0.8 (Original Gilliland), L D L D 1

7.D17.

V V

602.85 . Then

2.1 .99

1888.12, L min

V feed

1.44 which is between 1.0 and 2.1

Vmin

Vmin

.4,

12.69

i i

zi

,

tol-xy

3.03,

-

xy xy

2.4

1

166

Expand & Solve for

, 0

tol

z tol

xy

tol

xy

Result is linear,

xy

tol

tol z tol

i

1 z tol tol z tol

Dx i,dist

z

D

,

3.03 3.03 .1

0.3

.9

3.03 3.03 .3

0.5

.7

3.03 3.03 .5

.7

.5

3.03 3.03 .7 .3

.9

3.03 .996 3.03

3.03 3.03 .9

.1

3.03 2.51870

1 2.51870

3.03 .996

.004

3.03 1.88316

1 1.88316

D

.004

3.03 1.503722

1 1.50372

3.03 .996

.004

3.03 1.25155

1 1.25155

90.2834

1.071806

54.93685

3.03 .996

70.0405

1.25155

Vmin

L min

.004

49.7976

1.503722

, L min

1

3.03 .996

29.5547

1.88316

0.988

Vmin

9.3117

2.51870

xB

1

xy

z .008

F

1.0 .004

D

z tol 0.1

D

xB

xD

i

Vmin

sin ce

tol

xy z xy

Vmin

Then

z xy

77.6383

D D

3.03 .996

.004

3.03 1.071808

1 1.071808

98.0683 117.739

D

129.08

100 D

L

45.625

D min 4.8997

48.0836

1.6269

48.2707

0.96934

47.6985

0.68101

38.7990

0.429

Check for z = 0.5.

Slope

xD

L V xD

L

y*

V y

y*

.5 = z

min

y*

x D .5

min

where

xF 1

1 xF

0.75186

0.996 0.75186 x

As z

0.996 0.5

L

L

L V

.4922

D

V L

1 L V

1 4922

min

, although L D min , Vmin , thus Qc

Q R,min

min

Vmin

0.4922

0.96934 Perfect

as expected.

also.

167

FR A ,dist FR B,bot

n

1 FR A ,dist

Fenske: Eq. (7-15): N min

7.D18.a.

n

Where A = propane, B = butane,

1 FR B,bot AB

1/ .49

AB

2.04 . (Note value α.)

.9854 .8791

n

.0146 .1209

N min

8.7

n 2.04

For N F,min assume no LNK is bot and no HNK in distillate D = .229 + (.9854) (.368) + (.1209) (.322) = .631 .9854 .368 .1209 .322 x prop 0.575, x C4 .631 .631

x C3 x C4

n N F,min

z C3 zC4

dist

n

.575 .0617

n F

.368 .322

2.94

0.713

C3-C4

Underwood Eqns. (Case A.) 0

0.0617

Vfeed

1

Fz1

for 1.0

.49

1

0

9.92 .229

f

9.92

L min

1.0

9.92 .229

0.6213 , Vmin

Find

1.0 .368

N

.081 0.10

.49

1.0 .363

.10

.49 .0389

9.92 .6213 1.0 .6213 .49 .6213 D 1.057 .631 .426, L D min 0.676

Vmin

Gilliland Correlation (Fig.7-3): abscissa Ordinate

.49 322

N min

L / D (L / D) min L/D 1

1.057

0.33

.32 (Original Gilliland, ~.36 Liddle). Find N = 13.24 (14.13 Liddle) N 1 b. With N = 20, ordinate to Gilliland correlation is, N N min 20 8.7 0.538 N 1 21 Abscissa = 0.1. Since L D min 0.676, solve for L/D = 0.862. c.

FR C 6

FR C 3

dist

1

where

FR C 6

C6 C3

0.0156 1 0.0156

N min C6C3

bot

FR C 3

0.10 dist

N min C6 C3

bot

0.10, FR C3bot

1 FR C3dist

0.0156, and N min

8.7.

8.7

0.10

8.7

0.00000013 , FR C6

bot

0.99999987

168

For all practical purposes all C6 in bottoms at total reflux. d.

FR C3dist

0.999, L / D 1.5, FR B,bot

n N min

1) 2)

0.8791

FR C3dist FR C 4 bot 1 FR C3dist n

.999 .8791

n

1 FR C 4bot

.001 .1209

12.47

0.713

C3C 4

For D assume all LNK in dist, No HNK in dist D = 0.299 + (0.999) (0.368) + (0.1209) (0.322) = 0.6356

f

Now

0

9.92 .229

1.0 .368

9.92

.49 .332

1

.081 .1

.49

.1

Which is same [φ depends only on feed & α’s]. Thus, same φ = 0.6213

Vmin

L min 7.D19.

9.92 0.229

1.0 0.368

0.49 4689 ??

9.92 .6213

1 .6213

.49 .6213

1.0709 0.6356

Use Figure 7-3. Ordinate

L D

N

0.4345

N min

25 11

N 1

L D

L D min

1.0709

0.6848. Very little change.

0.5385

26

0.08 with L D 2.2286 . L D L D 1 Abscissa approximated between original & fitted curves.

Then Abscissa

7.D20. a) Distillate

Dx Bdist

x dist Find

Fz B

min

5, Dx Tdist

1.0 becomes

Fz T

5

15

D

D

min

1.97

15 ,

0.57895 0.07018 1.0

D = 57.001 kmoles/hr, B = 100 – D = 42.999

n b) Can use Fenske eq. (7-11) or alternatives. N min

AB

xylene cumene

Xylene balance, Fz

x x,bot

35

xA xB

dist

xA xB

bot

n

AB

A

xylene

B

cumene

K xy

K xy K tol

xy

0.330

K cum

K cum K tol

cum

0.210

57.00 0.57895

0.0465, x cum,bot

1 .0465

1.57143

42.999 x x,bot 0.9535

169

n N min

0.57895 0.07018 0.04650 0.9534

11.35 n 1.57143 This is # equil contacts at total reflux. Dx x ,dist 57.001 0.57895 c) Alternative: FR xy,dist Fz x 35

1 FR C,bot

Dx C,dist

57.001 0.07018

Fz C

45 n

Use Eq. 7-15.

N min

0.088896 , FR cum,bot

0.91110

FR B,bot

1 FR A dist n

n N min

FR A dist

0.94288

1 FR B,bot AB

,

A

xylene

B

cumene

0.94288 0.91110 0.057122 0.088896 n 1.57143

11.35

7.D21. New problem in 3rd edition. Assume all ethanol in distillate and all n-butanol in bottoms.

Dx E ,dist Dx i

Fz E

100 .3

30

Fz ip Frac Rec iP dist

P,dist

Dx n

P,dist

Fz nP

Dx n

B,dist

0

100 .25 .986

1 Frac Rec nP dist

24.65

100 .35 .008

0.28 0

D

Dx i,dist

54.93

xE,dist = xE,dist/D = 0.5461, xi-P,dist = xi-P,dist/D = 0.4488, xn-P,dist = xn-P,dist/D = 0.0051

Bx E ,bot Bx i

P,bot

0

0 Fz iP 1-Frac Rec iP dist

100 .25 .014

0.35

100 .35 0.992

34.72

Bx n

P,bot

Fz nP Frac Rec nP bot

Bx n

B,bot

Fz n

B

100 0.10

10.0 B = 45.07

x i,bot

Bx i,bot

B

xi-P,bot = Bxi-P,bot /B = 0.0108, xn-P,bot = Bxn-P,bot /B = 0.7704, xn-B,bot = Bxn-B,bot /B = 0.2188

170

FR ip,dist FR nP ,bot

n b.

Fenske eq. (7-15)

1 FR iP ,dist

N Min

n N min

n

iP nP

.986 .992 .014 0.008 n 1.86

This includes PR

x iP x nP

n

N F,MIN Eq. 7 40a , N F,MIN c.

1 FR nP ,bot

n 8733

9.0748

0.62058

0.62058

z iP z nP

dist

n

14.62

0.4488 0.25 0.0051 0.35 0.62058

n

iP-NP

7.76

Underwood Equation: Assume NKs do NOT distribute: Case A. i Fz i Eq. (7-33) Vfeed i

Vfeed

For saturated vapor

z

F divide (7-33) by F. 1

i i

, which becomes

i

1

E nP

zE

iP nP

ENP

1

z iP

nP NP

iP NP

3.58 0.3

z NP

NB NP

NP NP

1.86 0.25

z NB

NB NP

1 0.35

0.412 0.10

3.58 1.86 1 0.412 Solve for φ between α values of keys. LK = i-propanol, HK=n-propanol. 1.0 1.86 . From Goal Seek on spreadsheet 1.48648 i

Then from Eq. (7-29) Vmin

Dx i,dist

where Dx i,dist

Thus, want

values from part a.

i

VMIN

L MIN d.

L D 1.1

VMIN

3.58 30

1.86 24.65

1.0 0.28

0 173.47 3.58 1.48648 1.86 1.48648 1.0 1.48648 D 173.47 54.93 118.54 , L D MIN 118.54 54.93 2.16

L D

Min

Gilliland abscissa, x

or

L D

L D

1 MIN

L D 1

x

L D L D MIN

L D L D MIN

L D MIN L D 1 1 L D

1

1 L D MIN

1 1.1 1 1 1.1 L D MIN

1.1 1 1 1.1 2.16

1

0.0683

171

N

From Eq. 7-42b,

Assuming 7.F2

N MIN

0.5456 N 1 N 1 0.5456 0.5456 N MIN

NF

N F,MIN

N

N MIN

NF

7.76

33.4

33.4

N includes PR

17.7 or Stage 18 below total condenser.

14.62

Equilibrium data is available in a variety of sources such as Perry’s Handbook. Data used here is from Perry’s (3rd ed.), p. 574. a) Need to obtain avg. α from equilibrium data. yN2 1 x N2 0.1397 0.9615 x 0.0385, y 0.1397, 4.055 1 y N2 x N2 0.8603 0.0385

x N2

0.4783, y

x N2

.9190, y N2

n Fenske N min

xD

.9770,

4.01 3.744

avg top&bot

b)

0.7893 (needed for part b)

x 1 x x 1 x n

1/ 2

dist

=

.9770 .0810 .0230 .9191

3.744

3.875

n

bot

.998 .002 .001 .999 n 3.875

AB

9.685

z

where x* is in equilibrium with feed y z 0.79 xD x * From equilibrium data x* ~ 0.48. L V min L 0.40 .998 .79 0.66667 L V min 0.40 , D min 1 L V min 1 0.40 .998 .48

L V

min

L / D 1.1

c)

L/D

Gilliland Correlation: abscissa

N

0.7333

min

L D

L D

1 L D

min

0.06666 1.7333

0.0385

N min

0.6 N = 25.7 including PR. Need 25 equil. Stages N 1 7.G.1. New problem in 3rd edition. a. At total reflux N MIN 9 Original correlation, ordinate

b. L D

MIN

0.92

172

Chapter 8 New Problems and new solutions are listed as new immediately after the solution number. These new problems are: 8.A1, 8.A2, 8A7, 8A12, 8.D1, 8D6, 8D12, 8.D13, 8D15, 8D17, 8D20, 8.D22, 8D23 to 8.D25, 8.E1, 8G1-8G5, 8.H3. 8.A1. New Problem in 3rd edition. a. 2-pressure distillation 8.A2. New Problem in 3rd edition. b. extractive distillation 8.A7. New Problem in 3rd edition. If there are volatile and non-volatile organics, a single equilibrium contact gives an organic layer that contains no non-volatiles. Extra stages do not increase the separation. If there is entrainment, a second stage may be useful. 8A.12. New Problem in 3rd edition. Steam distillation is normally operated with 2 liquid phases in the still pot and in the settler after the condenser. There is usually no reflux. Azeotropic distillation is normally operated with one liquid phase in the column and in the reboiler, but with 2 liquid phases in the condenser and settler. One of the liquid phases is refluxed to the azeotropic column. 8.C2.

y org x org org w in water

p org x org in w

yw x w

At solubility pt. x w in w

pw x w

in w

.975 and x org in w

H org x org x org

H org

VP

VP

w

xw xw

.025, x org in org

Vapor pressures (Perry & Green, 1984). N-butanol: T = 70.1°C 84.3°C l00.8°C VP = 100 mm Hg 200 400

w

VP

org

VP

w

x org x org

in org in w

.573

117.5°C 760 mm Hg

Water: T = 100.8°C, VP = 782 mm Hg, T = 84.3, VP ~ 421.8 mm Hg VPorg VPorg 400 200 at 84.3°C: 0.474 , at 100.8: 0.5115 VPw 421.8 VPw 782 For

org w

use average between 92 and 100°C. Can linearly interpolate at T = 96°C,

VPorg / VPw

0.501 , w org in w

From y w

w o

1

org w in w

1

xw

w o

org w in w

0.501 .573 .025 11.483 1 11.483

0.0871x w

1 xw

1 0.9129x w

generate equilibrium curve,

xw

1.0

.995

yw

1.0

.9495 0.8961

.990

At constant x w , the calculated y w

0.0871

.985

.980

0.8512 0.8102

.975 0.7726

y w ,exp eri min tal . Difference at x B

approximately .7726 .752 .752 100

0.975 is

2.74% .

These equations work better for mixtures which are more completely immiscible.

173

8.D1.

New Problem in 3rd edition.

Top Op. Eq.,

y

L V x

1 L V xD, xD

L

L D

4

V

1 L D

5

xB

0.11 (from diagram). Need 2 equil. Stages.

.8 , y intercept x

0

1

L V

xD

.975

.2 .975

.195

Graph for problem 8.D1. 8.D2. The columns are sketched in the Solution to Problem 8-C2. B1 is butanol phase and B 2 is water product. Two equilibrium diagrams are shown. a. F B1 B2 , Fz B1x B 1 B2 x B2

174

B1 b.

z

x B2

x B1

.28 .995

x B2

L

Col. 1. Bottom Op: y

V

Feed: 70% Liquid, q = .7, Top y

L V

x

V

L

x

V

1

q

7

.3

0.995 .436 0.995 0

min

0

1256.55

1 x B1 , Intersects y

.7

q 1

F B1

3

x

x B1

0.04 .

. Intersects y = x = z = .28.

x B2

V

L

y intercept x

Note that reflux is x 0

L

1

From Figure 8-D2a:

3743.45 , B2

.04 .995

1

L

x B2

V

0.562 ,

L V

1.23 .562

0.69

0.307

0.573 . Optimum feed stage = 3. Need about 3 stages + partial reboiler. L

Stripper (Column 2): y

L V

V

V 2

B V

L

x2

V

2

V B

1

2

V B

2

1 x B2 1.132 .132

8.57

Construction is shown in Figure 8-D2b. Need 1 2/3 eq. contacts or P.R. + 2/3 equil. stage.

175

8.D3.

a)

y = 0.4, x = 0.09 from graph.

B) V/F = 0.3

L

20 = Fz = Vy + Lx

20 = 0.4 V + 0.09 (100 – V) V 11 .31 35.48 kg moles / h L = 64.52 kg moles/h F zw x w 100 0.99 0.999

yw

xw

1 V/F

.7

7

V

V V/F .3 3 L F 7 .99 y x z x V V 3 0.3 See Enlarged figure [Be careful with scales] y w 0.969, x w 0.999

100 = F = V + L

V

F V

0.969 0.999

Use Table 8-2 to find

= 30

V, Tdrum @ y w

kg moles/hr hr

0.40, xw

0.09 , L = 70 and Tdrum ~ 108 C

176

8.D4.

Compositions x

100

0.975 [Shown on Figure 8-D3a]

L and 88 = Fz = L x

L

F z x

L x

100 .88 0.573

= 76.37 kmoles/hr, L 23.63 kmoles/hr x x .975 0.573 a) Water conc. W is 0.975. 200 = F = W + B W = 200 – B Water balance: (200) (.8) = Fz = W(0.975) + B (0.04) Solve: B = 37.433 kmol/hr, W = 162.567 kmol/hr V B V B 1 5 L L b) L V , Bot. Op. Eq. y x 1 xB . V V B 4 V V

L

8.D5.

F

0.573 , x

Goes through y

x

xB

177

Plot operating line: If y = 1, x

1

0.25 .04

1.25 PR + 2 stages mores than sufficient. (see graph) 0.75 0.04 c) L V Slope 1.332 min 0.573 0.04 V 1 1 L V 3.012 min L V L V 1 0.332

0.808

178

8.D6.

New Problem in 3rd Edition.

F

D, x dist Reflux

L

V

B, x bot 8.D.6 Part a)

F D F .65

D

z

F

V

b)

B .975D .02B

xB

xD

xB

100

.65 .02

V B B 136.124, Reflux

L

L V V

170.155.

Or V F D L 136.124 100 65.969 170.155 L V B V B 1 4 1 1.25 V V V B 4

c)

L

y Goes through

V

y

L

x

x

V

1 xB

xB

Calculate arbitrary point at x .6 y 1.25 .6 .25 .02 See Figure: Need 2 stages + PR d)

65.969

.975 .02

What is V B

MIN

?

V B

.745

V MIN

L V

1 L V

MAX

1

3.0894 On graph.

179

180

8.D7.

y W x W in organic

yA

y A x A in organic

x A in organic

W A in organic

0.9636 0.372 W A in organic

8.D8.

0.0364, y W

1 0.0364

0.628, x W in organic =1-0.628=0.372

44.69

0.0364 0.628

Convert wt frac to mole frac. MW C8 H14 O 72 14 16 102 and MW water Basis 1000 kg 0.994 wt frac. ether:

988 kg W = 54.889 kg moles 12 kg E = 0.118 kg moles x W in organic .998

y = 0.959 ether:

41 kg W = 2.278 kg moles 959 kg E = 9.402 kg moles y W 0.155

y W x W in org W - E in org

.195 .033

yE x E

z(wt) = 0.004 water:

V

min

1

1

y equil w feed xD min

L V

7.026

7.026 0.022 1 x

L V

L D

1

min

.805 .967

4 kg W = 0.222 kg moles 996 kg E = 9.765 kg moles z W 0.022

x

xD

L

18

6 kg W = .333 kg moles 994 kg E = 9.745 Total = 10.078 x W in organic 0.033

0.012 wt frac. ether:

y in equil w feed

0.9636

z

0.138

6.026 0.022

.998 .138 .998 .022

7.467;

min

Generate following equilibrium data using

L D

0.882

11.20; act

w E in org

L V

0.918 act

7.026 :

xW

0

0.01

.022

.033

yW

0

0.066

0.137

0.195 181

Top Op. line:

y

L

x

V Where L/V = 0.918, x D Bottom: From y

x

1

L

xD V .998 , and y = intercept = (1 - .918) (.998) = 0.082

x B to intersection of feed line and top operating line.

xB Obtain x B

0.0004 wt frac.: 0.4 kg W = 0.022 kg moles 999.6 kg E = 9.800 kg mole 0.0023

See plot in Figure: Optimum feed is top stage. Need 4

8.D9.

Convert to Mole fractions: MW C6 H14 O 72 14 16 102; MWwater

3 5

equil. Contacts.

18

Basis for all conversions is1000 kg soln. Top Layer Separator = 0.994 wt frac. ether 6 kg W = 0.333 kmole 994 kg ether = 9.745 kmol Total = 10.078 kmol 0.333 Mole frac. x w in org 0.033 10.078

182

xD

0.998

0.033 Bottom Layer separator is

988 kg W = 54.889 kmol

= 0.012 wt frac. ether

12 kg E = 0.118 kmol Mole frac. x W in org

z = 0.02

0.998

xD

41 kg W = 2.278 kmol Vapor into Condenser is

959 kg E = 9.402 kmol

yazeotrope = 0.959 wt frac. ether Mole frac. y W

y W x W in org W E in organ phase

Feed is y W

in org

L D

act

1 .195

7.026

1 0.033

0.02 mole frac water = z. In equil. With feed:

xD min

0.195 0.033

y E x E in org

y

x *f L V

0.195

xD

0.02 1 y

z x

0.998 0.02

* f

0.988 0.002896

2 L D min

114.36 , L V

Plot on graph, and plot top op. line: x 0, y 0.00868. x 0.04, y

0 0

0.9913 .04

L V

L

0.9828 ,

D

min

L D act

1

0.01 0.066 0.00868 x

x

1

L V

57.18 ,

0.9913

L D

Top Op. y L V x 1 L / V x D,W through y When x = 0, y = 0.00868 7.026x Eq. Data. y . Generate curve, 1 6.026x

x y

0.002896

7.026 6.026 0.02

x D,W

0.022 0.137

0.998 .

0.033 0.195

0.04833 0.03, y .9913 0.03

0.00868

x W ,bot is at intersection y = z = 0.0208 top op. line, x W,bot 0.0123 Step off stages from top down. 1 equil stage is sufficient. But with this very high reflux rate consider alternatives.

183

8.D10.

a.

VP

C10

x C10

VP

W

xW

760

Assume the water layer is pure, x W 1.0. Try 95.5°C, VPC10 60, VPW 645.7. (.99) (60) = 645.7 = 705.1. Too low. Try higher temperature. The attached plot of VP vs. T allows estimation of vapor pressure. (Note: a plot of log (VP) vs 1/T will be easier to interpolate and extrapolate.) 97.0°C: VPC10 63, VPW 682.07 , (.99) (63) + 682.07 = 744.4 97.5°C: VPC10 T = 97.6 gives VPW

65, VPW

694.57 , (.99) (65) + 694.57 = 758.9

697.1 which will be too high. Thus T = 97.5°C is close enough.

184

b.

nw n org

p tot

VPorg x org VPorg x org

760

65 .99

10.81

65 .99

This is significantly less than in Example 8-2 where 296.8/4.12 = 72.04 mol decane are used. Difference is due to higher n-decane concentration in liquid.

x F,org 0.9, 95% recovery → 5% left. Octanol left = 0.05 (.9) (1.0) = 0.045 kmol/h Nonvolatiles in bottoms = 0.10 kmol/h

8-D11.

octanol water

x oc tan ol in org

xF

0.045 1.10 0.045

0.3103

W steam

185

a) Water VP can be fit to log10 VP T = 95.5

log 0 645.67

T = 100 log10 760

B

A

A

A

273.16 T B

over short ranges T. T in C, VP is mmHg.

273.16 95.5 B 2.8808 273.16 100

2.8100

A B 368.66

(1)

A B 373.16

(2)

To solve for A and B, subtract 1 from 2 B B B = 2164.42 0.07080 0.00003271B 368.66 373.16 B A 2.8808 8.68105 373.16 Now find T for which p tot VPW x W VPO x O 760 mm Hg where x W 1.0, x O 0.3103 On Spread Sheet find T = 99.782°C VP O x O 19.075 b) y O 0.025098 0.3103 0.007788 p tot 760 754.072 yW x 1.0 0.99220 760 a) Moles octanol = F z O .95 1.0 0.90 0.95 0.855 kmol/h b) Moles water

nW

Check Eq. (8-18): n W

n org

yW y org 0.855

19.075 0.3103

8.D12. New Problem in 3rd edition. All cases

a)

D2

40

1.0 .65 .65 .55

D1

280

.35 .01 D1

B1

0.9922 0.007788 760

60, B2

108.93

19.075 0.3103

40 , D 2

B2

108.93

x P,B2

x P,dist1

x P,dist1

x P,dist 2

140

D1 140 40 180 . b)

0.855

Total feed Col 1 = 246

1400

1440.

Total feed Col 1 = 1500

186

8.D13. New Problem in 3rd edition.

F1

D, x dist

V

L

V

L

F2

B, x bot Part a. F1

F2 D B Water: F1z1 F2 z 2

100 80

D B

Dx dist Bx bot 100 .84 80 .20 Solve simultaneously, D = 99.25 and B = 80.75 V b) V B 121.125, L V B 201.875 B V

Since feed 2 is saturated liquid L L F2 121.875

c)

Doing Mass balance around top V y F1z1 y

L

Lx

Dx dist

Dx dist

F1z1

V

D 0.975

B 0.04

V 121.125

Doing Mass balance around bottom V y Bx bot y

V

L V

x

Lx F2 z 2

F2 z 2 Bx bot

V

These two equations are equivalent.

Slope

L

y

d) Bot. op. line:

L V

V

V B

V

L

x

V

V B

1 x B . Goes through y 1

V B

L

x

xB .

5 3 . Plot Bot Op. line.

121.875

1.0062 V 121.125 At intersection F2 feed line and bot op. line (at x .2, y 0.306667 ) with slope 1.0062 2 stages + PR is more than sufficient (See graph). Op. line above feed 2:

Slope

187

Graph for 8.D13. 8.D14. Figure is on next page.

F

Part b.

B1

B1 c)

D2

F

B2 ,

Fz

z x B2 x B1 z x b2

x B2

x b1 x b1

D1

D2

B1x EB1

0.85 0.006

F

0.992 0.006

x b2

x d1

x d1

xd2

B2

100

B2 x E 100

B2

85.60 , B2

F B1

14.40 kmol/h

0.85 0.992

0.006 0.449

0.006 0.992

0.449 0.75

21.196

35.596 kmol/h

188

D2

D1

xE

101.3 kPa

1333

F

zE

0.75

0.449

0.85

kPa

ethanol

Water Ethanol

99.4 mole % water

99.2 mole % B1

B2

8.D.15. Part a) New Problem in 3rd Edition. p org VPoc tan ol x oc tan ol =Ptotal where x oc tan ol is mole fraction octanol in organic phase. At 0.05 atm and boiling T, porg From Antoine equation,

log10 VP

oc tan ol

6.8379

0.05 atm. 38 mmHg

1310.62

T 136.05 T 129.8C, VPoctanol 80.905 mmHg

At Since p org

p org

38

0.470 VPbenzene 80.905 Average mole wt solids and non volatile organics can be calculated. Basis 100 kg mol octanol 15 130.23 0.470 15 85 mol octanol mol non-volatiles 130.23 MW

0.470

b)

38 mmHg, x octanol,mole

15

85

15

MW 654.04 130.23 MW 130.23 95% recovery is true on both mass and mole basis. Distillate octanol flow rate 0.95 100 0.15 14.25 kg h. Since MWoc tan ol

130.23, this is 14.25 130.23

In waste there are 0.05 15

0.109 kmol h.

0.75 kg hr octanol and 85 kg h (organics + solids), or

85.75 kg h total. Wt frac octanol 0.75 85.75 0.00875. 0.75 130.23 Mole frac. octanol in waste 0.0424 0.75 130.23 85 654.04

189

c)

For equilibrium in still pot VP

oct

x oct in org

VPw 1.0

The still pot is perfectly mixed; thus x oct in org Since water boils at 100°C when P

p tot

x oct in waste

760 mmHg .

0.0424 mole frac.

760 mmHg, T < 100°C.

Eq. (8-15) becomes VPoct 0.0424

VPw 1.0 760 Substituting in the Antoine equations for octanol and water and solving with a spread sheet, T = 99.97°C. VPoct 19.27 mmHg and VPw 759.18 mmHg. d)

n oct

From Eq. (8-18),

VPoct x oct

nw

p tot

VPoct x oct

From spread sheet n oct n w Since

n oct

0.109 kmol h, n w

water

101.27

This is a lot of steam! 8.D16.

0.001076 n oct 0.001076 101.27 kmol h

kmol 18.016 kg kmol

1824.5 kg h water in distillate.

kmol

Distillate 1: 0.997 EtOH, 0.0002 solvent. Calculate x d1,W Distillate 2: 0.999 water, 0.00035 solvent. Calculate x D2 E F = 100, x F,E

0.81, sat'd liq'd, x F,solv

1 .9972

0.0028

1 0.99935

0.00065

0

Find D1 , D 2 , M where Makeup is pure solvent. 0 Water: x W M Fz W D1x D1W D 2 x D2W 0 Ethanol: Mx E,M Fz E D1x d1,E D 2 x d 2,E Ethylene Glycol:

Mx Esolv

1.0

Fz solv

Solving water & ethanol balances obtain: D 2 From Ethylene Glycol balance, M 81.2316 0.0002

D1x d1,solv

Dx d 2,solv

18.7913 and D1

81.2316 kmol/h.

18.7913 0.00035 0.02282 kmol/h Can also use overall balance instead of EG bal. Then M D1 D 2 F 18.7913 81.2316 100 0.2290 , OK 8.D17. New Problem in 3rd edition. Since everything now exits the bottoms, B = S + F, and x A = FzA/(S+F), xB = FzB/(S+F), xsolvent = S/(S+F). 8.D18.

Ethanol Product: Water Product:

F 100, z E

0.997E, 0.0002 solvent, 0.0028 water 0.9990W, 0.00035 solvent, 0.00065 ethanol

0.20 z W

0.80

190

Water bal: x WM M

Fz W

PE x EP,W

E bal:

x EM M Fz E

EG

0 x M,solvent M Fz solv

PE x EP,E

M 8.D19.

(B)

PW x WP,E

PE x EP,solv

Solve A & B for PE & PW :

(A)

PW x WP,W

PW x WP,solv where x M,solvent

80.0240 kmol/h, PE

PW

80.0240 0.00035

F H

Ethanol:

Fz E 0 B1x E,bot 2 B2 x E,bot 2 0 Fz H H 1.0 B1x H,bot1 B2 0

Hexane:

B2 where H = makeup hexane.

Solving simultaneously, B1 8.D20. New Problem in 3rd edition. a.

1000 0.8094

0.03201 kmoles/h

do M.B. in wts.

M.B. around System. Since everything in wt. units

B1

20.0074 kmol/h

20.0074 0.0002

Overall:

1.0

8000.04, B2

1000

F

Ex E,Ethanol prod

2000.04 and H

0.08 kg/h.

E W

Wx E,wprod

809.4 0.998E 0.0001W 808.3 E 811.0 kmol h 0.9979 W F E 1000 811 189.0 b)

V

V

Fx WF

boilup ratio

L L

Ex W ,Ethanol prod

1000 0.1906

y w ,1 V E Pr od

811.0 0.002

0.300 629.93 811.0

629.93

0.777

V E Pr od. 629.93 811.0 1440.93 L F sat 'd liquid feed 1440.93 1000

440.93

If CMO strictly valid then, L

reflux

440.93

Can also estimate L from _ settler

Pentane flow rate in V1

Ethanol flow rate in V1 Ethanol lost in Water Product.

Pentane flow rate

y P,1V1

629.93 0.6455

406.62

191

Ethanol flow rate in V1

y E,1V1

629.93 0.0555

E in V1 E lost

x E,reflux

L from settler to Col1

Ethanol lost in water product

Lfrom _ settler _ calculation

W prod x E in Wprod

189.0 .0001

1. CMO not totally valid 2. There is some water in reflux 3. K dE value may be incorrect.

Ethanol returned to distillation column V1 y E1 WPr od x E,W Pr od 34.96 0.0189 Using average estimate for L 0 Then 34.91

0.0189

441.6

Match not perfect because:

c.

34.96

440.93 441.6 2

34.19 kmol h.

441.3

441.3 x E in pentane x E,pentane

Then since assume K d

0.0792

x E,Re flux,pantane _ layer

1, x E,Water layer

0.0792

192

d.

x E,water

0.0792

V1 1

V W

0.5

V

L L

283.5

V

94.5

y E1 y E1 W

0.5W

W V

283.5 kmol h.

3

L 0.0792

W .0001 V

283.5 .0792

189.0 .0001

94.5

189.0

x E,W Pr od

94.5 kmol h.

0.237

0.0001

8.D21.

193

L

Bottom:

V

V B 1

1.5

L

L

1 xB .Goes through y V B 0.5 V V Feed line = Horizontal (q = 0). Through y = x = z = 0.4 Top. MB: yV Lx Dx D and V L D y

3, y

L

x

x

x

x B with slope = 3

L

1

x D goes through y x x D 0.975 V V Intersects Feed line where bottom op line does. Opt. Feed #1 above reboiler. 3 equilibrium stages + PR is sufficient. 8.D22. New Problem in 3rd edition.

D, x dist Reflux

L

V

B

F Part a. F

D Part b. yV

y

D B & Fz z x bot x dist

x bot

F

Dx dist

Bx bot

0.20 0.08 0.975 0.08

100

13.41 kmol/h and B

Lx Dx dist L V

x

D V

x dist

Substitute in D

Points on operating line: y

x

x dist

F, L

B, thus slope

L V

L

V L to obtain y

0.975 and x

Alternative point is at feed line (y = z = 0.2) & x

V

86.59

x bot

1

0, y intercept

L V 1

x dist L V

x dist

0.08

B F 86.59 100

1 0.8659 0.975 0.1307 Part c. Need 2 Stages. See graph. Part d. Pinch at feed line intersection with equilibrium is at x

V

x

0.8659

y intercept

0.02 .

194

Figure for problem 8D22.

195

8.D23. New Problem in 3rd edition.

Water phase

XNM = 0.086

N.M. Phase

NM

xw = 0.312

W

F

Water Product

Nitro Methane Product Part a. External balances F

NM Pr od

WPr od

NM :

F .25

NM Pr od

.25

.01

.98 .01

NM Pr od .98

100

WPr od .01

24.74, WProd

75.26

a. W Column:

z = .25, horizontal feed line L L Top y NM x NM 1 x NM ,bot ,col NM Mass balance through top of W column V V and around col. NM. Can easily show that

196

y NM x NM x NM,bot,col NM But do not know L/V so cannot plot yet. Bottom operating line looks familiar:

y NM

x NM

x NM,bot,col w

V

L

x

V

1 x NM ,bot col w

0.01

L

V B 1

54

V

V B

14

col w

L

y NM

5

Can plot bottom operating line. Arbitrary point: x

0.2, y 5 0.2 .01 .96 Now can plot top operating line from intersection of bottom operating line and feed line to point y NM x NM x NM,bot,col w 0.01 See graph. Need PR + ~ 1 2 stage. Build PR + 1 stage. yw

b. NM Column is a stripping column:

L V

To plot,

V 1 B V B

col NM

yw

xw

xw

.3, y w

Need PR + ~ 1 c. W col. Want V. V

1 3

3

B

0.02 1

.3

3

.02

0.3933

xw

0.3 is arbitrary point

stages.

100 V, V= V

1 x w ,bot ,col NM

3

V

B

B V to cond. from Wcol

NM col want V. V=

L V

4

x w,bot,NM _ col 4

L V xw

B

3 NM Pr od

1

WProd

1

75.26 18.81 4 4 118.81 kmol hr 3 24.74

74.22 kmol hr . To condenser.

197

Graph for 8.D23.

198

8.D24. New Problem in 3rd edition. From Equilibrium, y bu tan ol Overall Mass Balance: 100 F V B Butanol MB: 100 0.025 V .092

2.5

.092 V

0.092 at x bu tan ol

0.004

B .004

F V .004

2.1 .088V V 23.864, B 76.136 kmol hr This problem can also be solved graphically, but using basic mass balances is easier. 8.D25. New Problem in 3rd edition. Part a)

VPbenzene x benzene where x benzene is

porg

mole fraction benzene in organic. At boiling T, p org From Antoine equation, log10 VP At T

93 C, VPbenzene p org

1.0 atm. 1211.033

6.90565

benzene

T

220.790

1112.44 mmHg . Since

760, x ben,mole

p org

760

0.683 VPbenzene 1112.44 Average mole wt solids and non volatile organics can be calculated. Basis 100 kg 0.683

0.683 b)

Moles benzene Moles benzene + Moles non-volatiles

20

80

20

80 .683

20

78.11

MW

78.11

MW

78.11

1 .683

20 78.11 20 80 78.11 MW

MW

673.2

90% recovery is true on both mass and mole basis. Distillate benzene flow rate 0.9 100 0.2 18.0 kg h . Since MWbenzene 78.11, this is 18 78.11 0.230 kmol h In waste there are 2.0 kg/h benzene and 80 kg/h (organics + solids), or 82 kg/h total. Wt frac benzene 2 82 0.0244 2 78.11 Mole frac. benzene 0.1773 2 78.11 80 673.2

c)

For equilibrium in still pot VP

b

x b in org

VPw 1.0

The still pot is perfectly mixed; thus, x b in org Since water boils at 100ºC when P boils at 80.1ºC, but mole fracs low.

p tot

x b in waste

760 mmHg

0.1773 mole frac.

760 mmHg, T 100 C . Benzene is more volatile and

Antoine equation for water: log10 VPw

8.68105

2164.42 273.16 T 760

VPb 0.1773 VPw 1.0 Substituting in the Antoine equations for benzene and water and solving with a spread sheet, T 92.0411 C . VPben 1082.5 mmHg and VPw 568.1 mmHg. Eq (8-15) becomes

199

d)

n ben

From Eq. (8-18),

VPben x ben

nw

p tot

VPben x ben

From spread sheet n ben n w

n ben

Since

0.337876

0.230 kmol hr, n w

water

0.6807

n ben 0.337876

kmol 18.016 kg kmol

e. To vaporize benzene condense moles water

kmol

n ben

ben

0.6807 kmol hr

12.264 kg h water in distillate.

.

w

This occurs at 92.0411 C

365.1911K; From Perry’s table 2-237, H

hg

h

hf

H

h kJ kg

T = 360

886.7

498.7

388

T = 370

898.6

518.1

380.5

5.1911

0.0911 w

5

x 380.5 388

2265.67 2278.3

Moles water condensed

2278.3

388

384.1

kJ

30, 002 kJ kmol. 10 kg Note: 8th edition, Table 2-193 is very slightly different after unit conversion. T = 360 2663 384.7 2278.3 Water Table 2-352. T = 370 2671 405.88 2265.67 Linear interpolate

2277.8 kJ kg

41,037 kJ kmol

0.230 30, 002

kg h water in waste

0.1682 kmol h water (in waste) 41037 0.1682 18.016 3.029 kg h

200

8.E.1. New Problem in 3rd edition.

Water phase

XNM = 0.086

N.M. Phase

F2

xw = 0.312

NM

W

Water Product

Nitro Methane Product

250

Part a.

F1

F2

NM balance

PNM

PW

F1z1NM

F2 z 2 NM

PMN x

NM Prod

Pw x

NM mol frac

135.5 8 127.5 135.5 PNM Pw b.

F1

PNM

PNM .98

250 PNM

w Pr od NM mol frac

Pw 0.01

0.01

.97PNM

2.5

133 .97 137.11 kmol hr 250 137.11 112.89

Column W – Use y NM vs. x NM (water phase) plot. Top operating line

y NM

L V

Bottom Operating Line y NM

L V

L V

V B

V B 1

43

V

V B

13

water col

water col

x NM

1 L V

x NM

L V

V

4. L

V

wcol

water col

F2 z NM 2

x NM

NM Pr od

Vwcol

1 x NM

Water Pr od

V

B col w 37.63 V . B B col w 150.52 L F1

201

y

Top IS NOT from

x

x NM

0.98 to intersection feed and bottom operating line.

NM Pr od

Instead from intersection of feed and bottom operating line with slope

L V

L F1

V

50.52 37.63 1.32 25 .

Optimum feed is top stage. Need PR + 1 stage. c.

Column NM. Top y Bottom

yw

L V L V

yw

xw

NM col

xw

NMcol

F1z1,NM

1 L V NM x w col

xw

L V

xw

1 xw

NMcol

in WM Prod.

L

0.02.

V

W,W Prod.

V B 1

2.00 V V B Draw bottom operating line. Top is through intersection bottom operating line and feed line F 2 . Slope L V 0.906 (see item d). Need 2 stages + PR. Optimum feed is stage above PR. d.

Column W: PW

V

Bcol w

NM Pr od

112.89

V B Bcol w

1 3 112.89 L

V Bcol w

Saturated liquid feed: V Column NM:

37.63

V

37.63

V B

hr. PNM 1

50.52.

Bcol NM

137.11

2.0

V B

V

L F

kmol

V B 1.0, L V

L

150.52.

V B B 137.11 kmol hr,

L

2.0 137.11

274.22

V 137.11, L L 150 124.22 , L V 0.906. Minimum boilup rate NM column gives combination bottom & top operating lines to go through

Saturated liquid feed V e.

reflux point: y W

0.5, x w

0.312 .

From bottom operating line intersection with feed line

y INTER

L

x

z2

0.15 is

L

1 x W ,NM Prod V V Slope of top operating line to reflux point is L F2 L 0.5 y INTER V

V B Guess V B

z2

0.312 0.15

V

1

L V

L V 1

V

.

Calc V & L & L V

Check is two calculated values L V are same.

V

V B B 137.11 V B .

Calc y int er

Calc L V

L

V

B

L F2 V

202

Spreadsheet.

V B

MIN

0.6105,

L V

0.846

Graph for Problem 8.E1.

203

8.E2.

Balances at mixing point for F & R. To Butanol Column: Overall: FT F R Water:

FT z T,W

100 R

Fz W

Rx W,reflux

R .573

z T,W

FT z T,W

30 0.573R

30

100 R External balances: 100 = W + B, water: 30 = 0.995W + 0.02 B Solve simultaneously: W = 28.72 & B = 71.28 kmol/h Butanol Col:

FT

V B 1.90, V 1.90B 135.432 L y

R zT

V B x

FT

206.712,

x W,butonal

F

L V 1.5263

0.02

206.712 100 106.712

106.712 0.573

30

0.4409 206.712 Vertical feed line at z T intersects bot. operating Line at y = 0.67 (see graph) Water Col.

V B

0.1143, y

L V L V x

V B

V B 1

V

V B

9.748

L V 1 x B,water,watercol, y

x

x

B,W watercol

0.995

See graph, y leaving column = 0.8

204

205

8.E3.

Basis: 1000 kg sea water (1 h): 965 kg water kmol 18.016 kg

35 kg NaCl kmol 58.45 kg

53.5635 kmol, x F,W

0.98894

0.5988

kmol, x F,salt 0.011056 Total 54.1623 Water Condensate = (0.60) (53.5635) = 32.1381 kmol/h = n W Water Remaining 53.5635 – 32.1381 = 21.4254 21.4254 kg moles W Waste Water is 0.9728 mole frac. water 21.4254+0.5988 salt a. In still, organic phase is pure decane. VPC10 VPW x W p tot 760 mmHg where x W

0.9728 .

Try T = 99°C. VPC10 ~ 68, VPW 733.2 mm Hg 68 + (0.9728) (433.2) = 781.27 mm Hg, which is too high. Converge to T ~ 98.2°C. 707.27 0.9728 nW yW pW VPW x W b. Distillate: 10.4225 n org y org p org VPorg x org 66 1.0 (This calculation is at 98°C, not 98.2, but will be close.)

206

n C10 8.F1.

32.1381 kmol water/h 10.425 mol water/mol organic

3.0819

kmol h

C10 in distillate

V.P. Data n-nonane (p. 3-59 Perry & Green, 1984)

VP = 20 T = 51.2

40 66

60 75.5

100 88.1

200 107.5

400 128.2

See Solution problem 8.D10 for plot. MWnC9 1984), nonane enthalpies are,

128.25 .

hliquid 671.3 KJ/kg 722.5

Hgas 998.2 1036.5

T 360 K 380 K

760 mm Hg 150.8°C

From p. 3-268 (Perry & Green,

Water VP is given in Problem 8-D10 and on p. 3-45 of Perry and Green (1984). a. Try 95.0°C. VPC9 127 mm Hg, VPW 633.9 . Assume water is pure. Pressure: (.99) (127) + (1.0) 633.9 = 759.6. Close enough and lucky! b.

p tot

nW n org

VPorg x org

760 127 .99 127 .99

VPorg x org

5.045 mol water/mol nonane

c. Need to calculate the energy required to vaporize the nonane. T = 273 + 95 = 368 K. By linear interpolation for pure nonane: h1 ~ 691.78, Hgas ~ 1013.52. nonane 321.74 KJ/kg Table 3-302 of Perry and Green (1984): h liq,W

397.36, H vap W

2667.8,

W

2270.44 KJ/kg

mol water condensed

C9

321.74 KJ/kg 128.25 kg/kmol

mol C9 vaporized

W

2270.44 KJ/kg 18.016 kg/kmol

b. Now, VPC9

.020

VPW

1.009

760. Temperature will be higher.

Try 99°C: VPC9 149 mm Hg, VPW 733.24 (149) (.020) + 733.24 = 736.22 which is too low. Try 99.9ºC: VPC9 154, VPW 757.29 154 (.020) + 7570.29 = 760.37. Close enough. The low nonane conc. reduces nonane partial pressure and operation is much closer to 100°C. p tot VPorg x org 760 154 .02 nW 245.75 n org 154 .02 VPorg x org Need lot more steam!

207

8.F2. n-nonane

water F = 1000 95% n-nonane organic waste steam water

a)

Basis: 1 hour All junk in feed (0.50 kmol) is in bottoms

Organic Bottoms is 0.95 n-C9 0.5 junk 1.45 (see part C) .95 x C9,bot ,org 0.65517 1.45 b) Still T. p W p org p tot 102.633 kPa 770 mm Hg pW

porg

VPW (T)x W where x W

K C9 T x c9,org,bot

K C9 T

1

0.65517 770 mm Hg

Procedure: Guess T, determine VPW & K C8

VP

W

504.481K C9

check if pressure eq. is valid

770 mm Hg?

0.16 (DePriester Chart) 652.62 80.717 738.34 Need higher T T = 97°C, VPW 682.07 and K C9 0.17 682.07 + 85.762 = 767.83 slightly low, but close enough

Try T = 96°C

VPW

504.481 K C9 T

657.62, K C9

c) 9.50 kmol n-nonane × .90 = 8.55 kmol n-nonane in distillate. 9.5 – 8.55 = 0.95 kmol n-C9 in bottoms n org p org 85.762 n org 8.55 d) Eq. (8-18) 0.12567, n W 68.031 kmol water nW p W 682.07 0.12567 0.12567 e) EB simplies to

W

n W,condensed

n W ,condensed

nonane C9 W

n org,dist. where λ’s are at 97°C = 370 K.

n org,dist.

8.55 kmol

C9 W

208

From Perry’s 6th ed. Table 3-268 or 7th ed. Table 2-292, nonane λ’s are: h g h f , @ 360 K, 998.2 671.3 326.9 kJ/kg

380 K, 1036.5 722.5 326.9 314 @ 370 K, λ ≈ 320.45 kJ/kg 2 MW C9 128.258 . Then at 370 K, kJ 128.258 kg

41,100.3 kJ/kmol kg kmol Water 370 K: (Perry’s 6th Ed., Table 3-302). kJ 18.016 kg kJ 2671 405.8 2265.2 40,809.84 W kg kmol kg 41,100.3 n W ,cond 8.55 8.611 kmol water 40,809.84 C9

320.45

314.0 kJ/kg

8.G.1. New Problem in 3rd Edition. 1. Final makeup solvent flow rate _________0.02__________ kmol/h. 2. Final value solvent recycle rate (B2) __1400___kmol/h and L/D in col 1 _0.100_. 3. Final values of flow rates D1 _140.0_, B1 _1460.02_, and D2 __60.02___ kmol/h. 4. Mole fractions in stream D1 _Pyr=0.0084259, W=0.99157, Bisphenol=.49E-10_ 5. Mole fractions in stream D2 _Pyr = 0.98001, W = 0.019654, Bisphen = .000333___ 6. Mole fractions in stream B1 _Pyr=0.040287, W= 0.0008079, Bisphen = 0.95890_ 7. Mole fractions in stream B2 (solvent recycle stream) Pyr = .526E-8, W = .2E-12, Bisphenol = 1.0000__ 8. Heat load in cooler on solvent recycle line__-0.15216E8___ cal/s. 8.G2. New Problem in 3rd edition.

Aspen Plus Residue Plot 4.0 atm using NRTL

Pressure can have major effect on VLE for non-ideal systems. Compare T-xy diagrams for acetone MEK at 1.0 and 4.0 atm. Also compare residue curves for acetone-MEK-MIBK at 1.0 & 4.0 atm.

209

210

211

8.G3. New Problem in 3rd edition. a. Final reflux ratio column 1___0.01_ and final reflux ratio column 2 _0.01_______. If these values are not 0.01 you are not finished with Part B. b. Flow rates furfural product ___166.0___ kmol/h and water product ___34.0__ kmol/h. c. Boilup rate in column 2 _____8.0________ kmol/h. d. Mole fraction furfural in furfural product ____0.99816___& mole fraction water in water product ___0.99102____. e. Flow rate of distillate from column 1 _____42.10_____ kmol/h. f. Column 1 condenser temperature __370.3___K, & column 1 reboiler temp. __433.59__ K. g. Outlet temperature of decanter ____375.2______ K. h. Molar ratio of water phase/total liquid in decanter _____0.8393_____ 8.G.4. New Problem in 3rd edition. Column 1: a. Bottoms product mole fraction acetonitrile______0.99915______ b. Distillate flow rate ___240____ kmol/h and bottoms flow rate ___170____kmol/h. c. Distillate mole fraction acetonitrile ____0.67910___________ . Column 2: a. Distillate flow rate ___210_______ kmol/h, and reflux ratio ___1.2____. b. Bottoms product mole fraction water______0.99517______________ . c. Distillate mole fraction acetonitrile _____.77542__________ . 8G5. New Problem in 3rd edition. Results are residue curves and profiles of mole fraction vs plate location. For an equal molar feed, N = 10 does not give the desired purity even if L/D = 10. N = 50 does work with L/D = 2, but not for L/D = 1.0. 8.H1. Part b. Was 8.D12 in 2nd edition of SPE. Use Eq. (8-25b) with 2.4, BB 1.0, BC 0.21. AB A = benzene, B = toluene, C = cumene. Results from Spreadsheet: Stage: Reboiler: x A 0.0003 x B 0.0097 1 0.003298 0.04443 2 0.03137 0.176085 3 0.18019 0.42145 4 0.46126 0.44952 5 0.70274 0.28536 6 0.85421 0.14453 7 0.93403 0.06585 8 0.97145 0.028535 9 0.98791 0.012091 10 0.99493 0.005074 11 0.99788 0.00212 12 0.99912 0.000885

x C 0.990 0.95227 0.79255 0.39836 0.08923 0.01189 0.001265 0.000121 1.102 E-5 9.802 E-7 8.637 E-8 7.580 E-9 6.641 E-10

8.H2. Was 8.D13 in 2nd edition of SPE. Use a spreadsheet with Eq. (8-30) as recursion equation. Result is shown in Figure. The VBA program was given in Example 8-3. The results obtained for the starting conditions given are: k xA xB xC 1 0.990 0.001 0.009 100 0.9763 0.0017 0.0220 200 0.9431 0.0029 0.0534 300 0.8630 0.0049 0.1331

212

400 450 475 500 600

0.6740 0.5044 0.3946 0.2696 0.00042

0.0077 0.0089 0.0092 0.0089 0.00095

0.3183 0.4867 0.5962 0.7214 0.9986

Results for other starting conditions are shown in the figure.

213

Figure for problem 8H2.

8H.3. New Problem in 3rd edition. The spread sheet including the first 10 time steps and time steps 600 to 610, and the VBA program are listed. Part a

214

Simple distillation calc (residue curves) with BP calcs. aT1 aT6 ap1 -1166846 7.72668 -0.92213 -1280557 7.94986 -0.96455 -1481583 7.58071 -0.93159 -1524891 7.33129 -0.89143 -1778901 6.96783 -0.84634

iB nB iP nP nhex Residue curve calc. x1iB sumx time step 1 2 3 4 5 6 7 8 9 10

0.98 1 xiB 0.98 0.979883 0.979764 0.979645 0.979525 0.979405 0.979283 0.979161 0.979037 0.978913

600 601 602 603 604 605 606 607 608 609 610

0.009754 0.009586 0.009418 0.009253 0.009089 0.008927 0.008766 0.008608 0.008451 0.008296 0.008142

0.071054 0.069204 0.067391 0.065615 0.063874 0.06217 0.060502 0.058869 0.057271 0.055708 0.054179

h x1nB T1guess,R xnB 0.01 0.010032 0.010065 0.010097 0.01013 0.010162 0.010195 0.010228 0.010261 0.010294 0 0 0 0 0 0 0 0 0 0 0

0.01 0.01 500 xiP 0 0 0 0 0 0 0 0 0 0

0.919192 0.92121 0.923191 0.925133 0.927037 0.928903 0.930732 0.932524 0.934279 0.935997 0.937678

N x1iP p,psia xnP 0.01 0.010085 0.010171 0.010258 0.010345 0.010433 0.010522 0.010612 0.010702 0.010793

1000 0 14.7 x nHex 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0

542.3988 542.71 543.0165 543.3183 543.6155 543.9078 544.1955 544.4783 544.7563 545.0295 545.2978

epsilon 1E-09 x1nP 0.01 x1nHex 0 TR 472.0604 472.0642 472.068 472.0718 472.0757 472.0797 472.0836 472.0876 472.0916 472.0956

Option Explicit Sub Residue_Curve_BPcalc() ' K value data for nbutane,ibutane, ipentane,npentane and nhexane included. ' Only want 3 for residue curve. Thus, set x values = 0 for 2 components. ' The reference component is nbutane. Dim i, N, j As Integer Dim h, epsilon, xiB, xnB, xiP, xnP, xnHex As Double Dim T, p, aT1iB, aT6iB, ap1iB, aT1nB, aT6nB, Ap1nB As Double Dim aT1iP, aT6iP, ap1iP, aT1nP, aT6nP, ap1nP, aT1nHex, aT6nHex, ap1nHex As Double Dim KiB, KnB, KiP, KnP, KnHex, Ksum, chksum, inside As Double Dim yiB, ynB, yiP, ynP, ynHex As Double Sheets("Sheet1").Select

215

Range("A15", "G1045").Clear aT1iB = Cells(5, 2).Value aT6iB = Cells(5, 3).Value ap1iB = Cells(5, 4).Value aT1nB = Cells(6, 2).Value aT6nB = Cells(6, 3).Value Ap1nB = Cells(6, 4).Value aT1iP = Cells(7, 2).Value aT6iP = Cells(7, 3).Value ap1iP = Cells(7, 4).Value aT1nP = Cells(8, 2).Value aT6nP = Cells(8, 3).Value ap1nP = Cells(8, 4).Value aT1nHex = Cells(9, 2).Value aT6nHex = Cells(9, 3).Value ap1nHex = Cells(9, 4).Value h = Cells(11, 4).Value N = Cells(11, 6).Value epsilon = Cells(11, 8).Value xiB = Cells(12, 2).Value xnB = Cells(12, 4).Value xiP = Cells(12, 6).Value xnP = Cells(12, 8).Value xnHex = Cells(13, 8).Value T = Cells(13, 4).Value p = Cells(13, 6).Value For i = 1 To N j=i+1 Do KiB = Exp((aT1iB / (T * T)) + aT6iB + (ap1iB * Log(p))) KnB = Exp((aT1nB / (T * T)) + aT6nB + (Ap1nB * Log(p))) KiP = Exp((aT1iP / (T * T)) + aT6iP + (ap1iP * Log(p))) KnP = Exp((aT1nP / (T * T)) + aT6nP + (ap1nP * Log(p))) KnHex = Exp((aT1nHex / (T * T)) + aT6nHex + (ap1nHex * Log(p))) Ksum = KiB * xiB + KnB * xnB + KiP * xiP + KnP * xnP + KnHex * xnHex KnB = KnB / Ksum inside = aT1nB / (Log(KnB) - aT6nB - (Ap1nB * Log(p))) T = Sqr(inside) chksum = Ksum - 1 Loop While Abs(chksum) > epsilon Cells(13 + i + 1, 1).Value = i Cells(13 + i + 1, 2).Value = xiB Cells(13 + i + 1, 3).Value = xnB Cells(13 + i + 1, 4).Value = xiP Cells(13 + i + 1, 5).Value = xnP Cells(13 + i + 1, 6).Value = xnHex Cells(13 + i + 1, 7).Value = T yiB = xiB * KiB

216

ynB = xnB * KnB yiP = xiP * KiP ynP = xnP * KnP ynHex = xnHex * KnHex xiB = xiB + (h * (xiB - yiB)) xnB = xnB + (h * (xnB - ynB)) xiP = xiP + (h * (xiP - yiP)) xnP = xnP + (h * (xnP - ynP)) xnHex = xnHex + (h * (xnHex - ynHex)) Next i End Sub

217

Chapter 9 New Problems and new solutions are listed as new immediately after the solution number. These new problems are: 9.A4, 9.A5, 9C1, 9D1, 9.D5, 9.D8, 9D11, 9D13, 9D18, 9D19, 9D21, 9D22, 9D25, 9E2, 9.H1. 9.A4. New Problem in 3rd edition. Answer is g. 9.A5. New Problem in 3rd edition. Answer is c. 9.A6.

b. The same.

9.B1.

Multi Stage F, x F , x Davg , N, L / D, Treflux , P

Single Stage F, x F , x Davg , P

F, x F , x Wf , N, L / D, Tr , P

F, x F , x D tot , P

F, x F , Wfinal , N, L / D, Tr , P

F, x F , x wf , P

F, x F , D tot , N, L / D, Tr , P

F, x Wf , Wfinal , P

F, x F , x Davg , x Wf , N, Tr , P

x F , x Davg , D tot , P

F, x F , x Davg , x Wf , L / D, Tr , P

x F , x Wf , Wfinal , P

x F , x Davg , D tot , N, L / D, Tr , P

x F , x Wf , D tot , P

x F , x Davg , D tot , x Wf , N, Tr , P

F, x F , x Davg , x W

x F , x Davg , D tot , x Wf , L / D, Tr , P

etc.

x F , x Davg , Wfinal , N, L / D, Tr , P x F , x Wfinal , Wfinal , N, L D, Tr , P etc. 9.B2.

a. Replace the column with one containing more trays or more packing. b. Retray or repack existing column. c. Run a batch in several steps. For example, take the feed and operate so that the desired bottoms concentration is met. Collect all the distillate and use this as the feed for a second batch. Operate so that the distillate for this run meets specifications. The bottoms from this run can be used as feed for a 3rd run, or it can be mixed with the next feed batch. An alternate is to first collect distillate of desired purity. Then collect distillate which does not meet purity requirements while bottoms is reduced to the desired purity. The material not meeting requirements is then mixed with fresh feed for next batch. Other operating variations are possible. d. Hook up two batch stills in series – Either to run 1 batch or to run separate batches (second still takes distillate from first as the charge). e. See if product specifications can be relaxed. f. Reducing the pressure increases the relative volatility and may help. However, one must watch for earlier flooding.

9.C.1. Rayleigh eqn xF

Wfinal

F exp x W,fin

d xW xD

xW

218

Because x D

constant, can integrate analytically. xF

xF

d xW xD

x w ,final

Wfinal

n xD

xW

F exp

Wfinal

xD

xF

xD

F

x W ,final

xD xD

xD xD

xF x W ,final

xF x W ,final

D

Wfinal x W,final

xD

F

xD

Wfinal

Fx F Solve for

n x w ,final

n

F

Mass balances are

xW

D xD

xF x W ,final

Thus results are identical. 9.C2.

External balances over entire cycle Fz Dx Dfinal Wx wAvg and F = D + W a. Ignoring holdup on stages and in reboiler - Out = Accum in accumulator is - x w dD

x w dD

which becomes,

dD

Rearrange,

Ddx D

dx D

D

D final

dD

x Dfinal

D F

D

xD xF

xW

dx D xw

xD

xD

d Dx D

x D dD and integrate,

which is,

n

D final F

x D final

xF

dx D xD

xw

b. Assume CMO and draw mass balance envelop around bottom of the column.

L

y

y

L V L

V

V

x

x

B

Lx

Vy Bx w

B

xw V L

V

1 xw

219

xF

rd

Wfinal

9.D.1. New Problem in 3 Edition. Eq. (9-9)

F exp x W,final

From Simpson’s rule

0.1 0.00346

Area

0.00346 0.05173 0.1

y

4 x

0.00346

y-x

0.03096 .34 0.4416

y-x

1

6

From equilibrium curve (Table 2-1). x y

dx

y

1 x

y

x 0.005173

x

0.1

1 y-x 36.366 3.469 2.9273

0.027498 .28827 .3416

The y values are found by linear interpolation of data in Table 2-1. For example, at x = 0.00346, linearly interpolated first 2 data pts Table 2-1. 0.170 y x for x 0.19. 0.019 0.170 y 0.00346 0.03096 For x = 0.00346, 0.019 For y at x = 0.1, y = 0.4375 +[ (.4704 - .4375)/(.1238 - .0966)](.10 - .0966) = 0.4416 [Alternatively, could fit equilibrium data to constant α.] 0.1 0.00346 Area 36.366 6

Wfinal

0.5 exp

D total

0.5 .2125 F xF

x DAvg

9.D2.

Rayleigh equation is Wfinal

0.8555

0.2125 kmol .5 .1

D avg .75

0.8555

0.2875 kmol

Wfinal x W ,final

F exp -

2.9273

.2125 .00346 0.2875

0.1714

dx

y-x Most of values of 1/(y – x) are listed in Example 9-1. From Table 2-7 can easily generate values for x = .55: y = 0.805, y – x = .255. (y – x)-1 = 3.92. The mid-point for Simpson’s rule is at x = .65. Then from Eq. (9-12) and values in Example 9-1, .75 dx .2 3.92 4 5.13 6.89 1.044 x 6 .55 y Wfinal

x Davg

100e

Fx F

.55

1.044

35.20, D total

75

Wfinal x final D total

F Wfinal

Operating equation is y

64.8

35.2 0.55 64.8

A graphical integration counting squares gives x Davg 9.D3.

4 3.469

L V x

1

220

L V

0.859

0.861.

x D where L V

L0 D 1 L0 D

0.65

This is y = 0.65 x + 0.35 x D We have two equil. stages (stillpot and one in the column). From McCabe-Thiele diagram we can get the values of x D which are related to x W . Pick x D and get x W from figure. From this we can generate the following table (only two values are shown in Figure).

xD 0.90 0.895 0.85 0.837 0.720 0.70

x w ,final

xW

Raleigh Equation:

D total x Davg

3.205 3.077 2.193 2.096 1.754 1.748

Wfinal

xF

F

x wfinal

100 39.7

Wfinal x wf

0.36 ]

3.077 4 2.096

6

F Wfinal Fx F

0.588 0.570 0.394 0.360 0.150 0.128

.57 .15

dx w xD

1/(xD-xW)

[Midpoint x w

Simpson’s rule (Eq. (9-12)). xf

xW

dx xD

xw

, Wfinal

1.754

Fe

0.925

0.925

39.7 kmol

60.3 kmol

100 0.57

D total

39.7 0.15 60.3

221

0.847

9.D4.

Wfinal

2.0 kg moles, x F

0.8, x wf

0.4 . Find F, x DAvg , D total x

1

y(equil)

y 1

16.66

y x

4.76

7.143 .4

.6

.8

x

0.80

0.86

16.666

0.70

0.80

10

0.60

0.74

7.143

0.50

0.67

5.882

0.40

0.61

4.76

Can use Simpson’s Rule (Eq. 9-12) or evaluate numerically. xF dx 0.4 4.76 4 7.143 16.666 x 6 x Wfin y

Wfinal

F

Rayleigh eqn

xF

exp

D total

x Wfin

2.0

F Wfin

Fx F

x D AVG

exp

dx y x

3.3333

Wf x Wfin

56.063 kmol

56.068 0.8

2.0 .4

54.063 Wfinal

a)

3.3333

54.063 kmol

D total

9.D.5. New Problem for 3rd Edition

F exp

xF x w ,fin

dx y

x

Can use Simpson’s rule, eq. (9-12) with equilibrium values from plot. 1 1 f x w ,final 2.7397 y x 0.028 0.645 0.28

f

x w ,fin

f xF

xF

1

2

y x

1 0.705 0.52

0.40

1

1

y x

0.735 0.52

0.52

x

3.2787

4.65116

222

0.815

xF

xF

dx

x w ,fin

y x

0.52 0.28

3exp

y D,avg b. Settler:

6

f x w ,fin

2.7397 4 3.2787

6

Wfin

x w fin

0.8202 D V,tot D1

DV,tot y D,avt

Use eq. (9-13)

n

WF F

6

20.5056

1.321 , DV,tot

F Wfin

0.8202

1.6790

0.7088

D2

1.6790

D1x 2

D1

D2 x B

Solving simultaneously, D 2 9.D6.

f xF 0.24

4.65116

3 0.44033

Fz Wfin x w fin D V,tot

4f x w avg

D2

1.6790 0.7088

0.567 and D1 1 1

n

1.112

x W ,F 1 x F x F 1 x W ,f

223

0.573 D1 0.975 D 2

n

1 xf 1 x W ,f

a) F = 1.3, x F n

.6, x Wf

WF

1

1 3

1.4

WF 1.3

Fx F

.3 4

WF

.4

n

.6 7

1.3 0.6

0.3036 0.3

0.6914

1.3 0.3036

0.2335

Wf

0.81725

0.81725 0.3

3.5 0.8125 0.6, x DAVG 0.75

2.0, x F

0.8948 0.5596 1.4544

.7

0.3036 kmol

Wfin x Wf

3.5 0.6

x D AVG

Since

2.4

F Wfin

b) Now Wf 3.5

F

n

0.2335

x DAVG

c)

.3,

F Wfin x D AVG

Fx F

0.6914

x D AVG

Wfin

Wfin x Wf ,

F

x D AVG

xF x W ,f

.

Then Eq. (9-13) becomes

n

0.75 .6 0.75 x Wf

L D 1/ 2, L V

L D

1

1 L D

3

x D AVG

xf

1

x D AVG

x Wf

1.4

F x D AVG

0.5025 . Wfinal

Solution is x w fin 9.D7.

n

x D AVG

xf

xW

0.56

0.11 0.10 0.09

Simpson’s Rule

dx

xF x Wfin

xD

xW

.6 1 x Wf

1 x Wf

1.212 kmol.

.75 0.5025

x Wf 6

1 xD

xW 1

xD

xW

, determine

dx

xF x Wf

xD

xW

f

1/.45 = 2.2222 Interpolate 2.361 1/.40 = 2.50

0.06 0.05 0.02

xF

4

n

slope . Pick series x D values. Plot enriching section op

xD

0.44 0.38 0.26

x Wf .4

2.0 .75 0.60

x Wf

line. Step off two stages. Find x W . Calculate

0.49

n

1 .38

2.6316

1 .33 3.0303 1/.24 = 4.1666

f x WF

0.10 0.02 6

224

0.02

4f x W

4.1666 4 2.6316

0.06 2.361

f xF

0.10

0.22739

WFinal x DAvg

4.0 exp Fx F

0.22739

Wf x Wf

4 0.7966

4 .1

F Wf

3.1864 0.02 4 3.1864

9.D8. New Problem 3rd Edition. a) Op. Eqn.

L

y

v

x

4 L L 1 xd , x d .8@ y D D 5 From McCabe-Thiele plot x w final ~ 0.075 L V

10

b)

F

Wfinal

10 .4 Substitute in for

D tot

D tot

F xF

Wfinal , 4.0

3.1864

1

L V

xd

x , y Intercept

Wfinal

D tot x D

0.075 10 D tot

.8 D tot

.8 .075

225

4.483

Wfinal

1

L V

xD

.2 .8

10 D tot

Wfinal x w final

4.0 10 .075

0.4133

5.517

0.075 Wfinal .8 D tot

.16

9.D8. Figure

9.D8. Part C. Trial & Error to 3 stages ending at x F

y INtercept

0.645

1

L V

xD

1

0.4 (See figure) L V

.8

226

1

L V

9.D9.

a.

L

.645

V

.8

.19375,

.80625

L

L

L V

.15375

D

V L

1 L V

1 .19375

Initial

Mass balances: F

F xF

D total x D

D total

W or 20

D total

W x Wf or 8 .975 D total

0.24

W

.28 W

Solving simultaneously: D total 3.453 and W 16.547 Can also use Rayleigh equation to obtain same result. (Use of the Rayleigh equation for this type problem is illustrated in the Solution to Problem 9-D14.) b. Vapor in equilibrium with x Wf must be within the two phase region. x Minimum is when y

x . This is x Wf ,min

227

0.21 . See graph.

y

x

9.D10.

L

L

x D , Slope, L V L D V V Plot on McCabe-Thiele graph for series of x D values. Op. Eq., y

Simpson’s rule,

1

1 xD

xW

dx W

xF x W ,fin

x

xd

at x W

or

Wfinal

n

10 e

xW

x D,avg

Wfinal F

1.2491

.52, x W

0.52 0.20 6 0.32

Rayleigh eq.

xF

6 xF x W ,fin

xd

xW

xW

xF

Wfinal

xD

2.67

10 0.52

D total

228

xW

x F x W ,fin 2

xD

1.2491

F Wfinal

2.8677 0.2 7.132

.36

1

Fe

2.8677 and D total

Wfinal x Wfinal

xf ) / 2

4

5.95 4 3.70

dx W

15

.20, and (x Wfin

1 xD

10 0.28677

Fx F

x Wfin

1 LD

7.132

0.648

xW

x W ,fin

xD

xW

0.70 0.65 0.60 0.50 0.40 0.30

0.55 0.405 0.25 0.094 0.055 0.035

x D-x W 0.15 0.245 0.35 0.406 0.345 0.265

229

1/(x D-x W) 6.666 4.082 2.857 2.463 2.899 3.774

1

xW

xD xW 5.95

x F = 0.52 xF

x Wfin

3.70

= 0.36

2 x W ,fin = 0.20

2.67

S

9D11. New Problem in 3rd edition. Eq. (9-17)

x butanol Inital

W

dx pot y

x butanol final

Note W water Equation is in terms of butanol. 1.0 dx but pot S Simpson’s rule – need y at x pot W y but .6

x but

x water

y water

y but

1.0 .8

0 .2

0 0.565

1.0 0.435

.6

.4

0.70

1.0, .8, .6 butanol.

0.30

.4

3.333 4 2.299 1.0 6 0.9019 W 1.804 kmol.

1 y but 1.0 2.2999 3.333

0.90191

S More accurate if done in 2 steps. Thus add points below: x but

xw

yw

y but

.9 .7

.1 .3

.42 0.66

.58 .34

1

0.8 .8

.6

.2 6 .2 6

2.299 4 1.7241

1

3.3333 4 2.9412

2.299

Total Area = .91975. S = 1.8395

230

0.33985 0.5799

1 y but 1.7241 2.9412

xF

9.D12.

Wfinal

dx

F exp -

y x

x Wfinal

xF

xF

dx

x Wfinal

y x

x Wfin

dx y

D tot

0.32 x

6

4

y x

0.48, x Wfinal

x W final

1

1

y x

2

xF

0.16, x avg

x Wfinal

1

0.16

0.36

0.20

0.32

0.545

0.225

4.444

0.48

0.66

0.18

5.555

5.555

Fx F

1.51

(y-x)

y x 5.0

, Wfinal

Wfin x Wfin

3.0 exp

3.0 0.48

D tot

y (from graph)

y-x

0.7 0.61 0.37

0.3 0.37 0.29

0.4 0.24 0.08 Simpson’s Rule xF 0.4 0.08 x w ,fin

F exp

Two Liquids.

0.93826

F xF

x D,AVG b)

3.333 4 2.7027

6

D total x L

L x

0.573 and x L

D total

L x

0.662 k/moles

0.662 0.16

0.571

Wfinal

n

F

xf

x w final

dx y

x

1 y x 3.333 2.7027 3.448

3.448

0.93826

D total

F Wfinal

3.1305

Wfinal x w ,final

1.51

2.338

9.D.13. New Problem for 3rd Edition. a) Rayleigh equation: x

xF

0.32

2

y (from eq. data)

2.338 x DAVG ,

Wfinal

y x

x Wfinal x F

x

5.0 4 4.444

F Wfinal

1

6 xF

xF

x Wfinal

4.8695

0.6057 0.975

L

D total

D total x DAVG

231

L

D total

L .573 L .975

D total .6057

D total .6057 .573

L 9.D14. a.

p org

pw

0.3963

0.975 0.573

L

D total .3963

VPC10 x C10 , and assuming water is pure, p W

760 , p org

4.4732

VPW .

VPC10 x C10 VPW 760 Vapor pressure data for C10 was shown in solution to Problem 8.D10. Guess 99.5ºC. VPW 746.52, VPC10 ~ 70.5 mm Hg 7 + 746.5 = 753.5 < 760 At 100ºC VPC10 ~ 70.5 and VPW 760 . 7.05 + 760 = 767.05 > 760 6.5 By linear interpolation: T 99.5 .5 99.74 ºC 13.55 b. Use Mass balances. Initially 9 moles n-decane, 1 mole non-volatile a .1 Final: a mol n-decane where 1 ; thus, a .111 mol 1 a .9 Wfinal 1.111 mol (Water free) D total

F Wfinal

10 1.111 8.889 kmol

Alternate Solution: Raleigh Eq. with xF

xD

1: Wfinal

F exp x W,final

c.

nW

Wfinal

F exp

D total

F Wfinal

D org

p tot

n .1

dx 1 xW

n .9

F exp

n 1-x W

xF x W ,final

.111 F 1.111

8.889. Same result as mass balance.

VPorg x org in org VPorg x org in org

Should really calculate numerically from integral for most accuracy D

nW

p tot

VPorg x org in org VPorg x org in org

0

dn org

However, estimate at final conditions with VPndecane ~ 70.25 (from part a)

nW

760

70.25 .1

n org 70.25 .1 This should be a good estimate. 9.D15.

107.185

Column is similar to figure in Solution to 9.D13, but with 1 stage in column. a) For finding Wfin & D tot don’t actually need to step off stages. Just want to make sure x Wfinal is obtainable – I checked this at total reflux – It works. Use Mass balances: F x F D tot x d Wfin x Wfin Substitute in F

D tot

Wfinal

232

Then,

Wfinal

Solution is Wfinal

F xD xD

xF

100 0.975 0.48 0.975 0.08

x Wfin

55.307 kmol, D tot

F Wfinal

b) Need to draw operating lines until: initial

44.673

2 stages gives x Feed .

final 2 stages gives x W,final . Then L/V = slope. Initial – There will be a pinch at point reflux is returned. y xd y int ercept 0.975 0.41 L V Initial Slope 0.579 xD 0 0.975 0 Final: A few trials resulted in final result. y xd y int ercept 0.975 0.17 L V final Slope 0.826 xd 0 0.975 0

233

9.D16. a. Need L/V so that 3 stages go from x F error) was used to find this).

L V

.84 .57 a

.84 0

.4 to x D

.3214 and

b. Need L/V so that 3 stages go from x wfinal trial-and-error was used to find line.

L V c.

F

D total

F x feed

.84 .13 b

Wfinal D total x D

.84 0

10

D total

Wfinal x wfinal

.8452 and

0.84 . This is line a in Figure (trial-andL V

L D

a

1

0.08 to x D

D

L V

b

1

a

0.84 (see line b in Figure). Again, L V

L

.4737

a

5.4615

b

L V

b

Wfinal

4 .84 D total

.08 Wfinal

Solving simultaneously, Wfinal = 5.789 kmol and D total 4.211 kmol. The Rayleigh equation could be used, is not needed, but gives the same result.

x init

9.D17. Eq. (9-17),

S W

1.0 . Start 1 kmol and keep 1 kmol. Add water as boil. x W ,initial

x W ,final

dx tan k y

Note balance is on original solvent, methanol.

Use equilibrium data from Table 2-7. Generate table of methanol mole fractions:

x

1.0 0.611 0.222 0.11611 0.01

y

1.0 0.830 .588 .450 0.07

1/y

1.0 1.20489 1.6722 2.222 14.9254

Use Simpson’s rule in two steps. Step 1 (x from 1.0 → 0.222)

234

0.778

1.0 4 1.20489 1.6722 0.97143. 6 Step 2 (x from 0.222 → 0.01) 0.216 1.6722 4 2.222 14.9254 0.9005. 6 Total = 1.87194 = S/W with W = 1. rd

n

9.D.18. New Problem 3 Edition Eq. (9-14)

y D ,final

F D final

y D ,final

F D final

yF

dy y x

dy

exp

y

yF

x

F

D final

y D ,final

dy y x

exp yF

Read x values from equilibrium diagram or interpolate from Table 2-1. y x y-x 1 0.1 0.3 0.5

Area

Ethanol MB:

x C,avg

F yF

0.008 0.045 0.155

0.092 .255 .345

0.5 0.1 6

10.87 4 3.92

0.5

D final

exp

C total

F Dfinal

2.90

1.963

0.0702 kmol

1.963

0.4298 kmol

Dfinal y D,final

C total x C,avg

D final y D,final

0.5 0.1

C total

F yF 0.0702 0.5

0.4298

9.D19. New problem in 3rd edition. VPw x w VPoct x oct Ptot , or in mm Hg, 526.123 1.0 9.D20. Was 9.D18 in 2nd edition. x F,C5 0.35 & x W,final,C5

y x 10.87 3.92 2.90

0.05 : x C5,AVG

10.964 .6

0.20, x C8,AVG

235

0.00346 mole frac ethanol

532.7

0.80, p 101.3 kPa.

yi

B.P.

Ki xi

1.0 . For average mole fractions, the BP calculation converges to T =

84º with K C5 3.7 and K C8 0.30 from the DePriester charts. The close enough to estimate α. K C5 3.7 12.33 C5 C8 K C8 0.30 Eq. (9-13),

n

Wf

1

F

11.33

0.35 0.95

0.5573, Wfin

Wfin F D total

0.05 0.65

n

F Wfin

Fx F

x D,Avg

Wfin x W ,fin

L

x

1

V x D & slope

0.98 which is

0.5847

0.95

0.5573 1.5

1.5 0.8359

0.8359 kmol

0.6641 kmol

1.5 .35

D total

9.D.21. New Problem for 3rd Edition.

y

0.65

n

yi

0.8359 0.05

0.7276

0.6641

L D 1.0

L D

L V

1 L D

12

L

x D , but x D varies. Thus, plot series operating lines of arbitrary V 1 2. With a total of 21 equilibrium contacts there will be a pinch where the operating line

intersects the equilibrium curve. This intersection is x W for this x D value.

xD

.7 .6 .5 .4 .3 .2 Want to integrate from x W,final Plot

1 xD

xW

x W ,final xF

Area

xW

xD

.067 .05 .038 .027 .018 .01

.633 .55 .462 .373 .282 .19

xW

0.02 to x F

1 xD

xw

1.5798 1.818 2.1645 2.6801 3.5461 5.26316

0.06 and want middle point at x W

vs x W and find values.

0.02,

0.06,

1 xD

xW

1 xD

xW

0.06 0.02 6

3.23; x W,Avg

0.04,

1 xD

1.65

3.23 4 2.1

1.65

236

0.08853

xW

2.1

0.04 .

Wfinal

F exp

Area D total

2.5 exp F

Wfinal

0.08853 0.2118

2.288 x D,Avg

237

F xF

Wfinal x W ,final D total

0.492

9.D.22. New Problem for 3rd Edition. t OP 9.D23.

Prelim. Calc. Feed; Avg MW

MWavg

D tot

0.1 MW

0.1 46

1000 kg 20.8 kg mol L

L D

toperating = __1.49 to 1.50

QR

E

0.9 MW

0.9 18

h

water

20.80 kg/mol

48.0769 kmol

23

2 5 0.4 V 1 L D 53 All op. lines have slope 0.4. Can draw op. line to x W . Ten stages will go from x D to x W because have large number of stages. Thus, do not need to step off stages.

238

From Graph can create table of 1/(xd – 1/xw) versus xW.

xD

xW

xd

xw

1

0.665 0.10 0.630 0.08 0.499 0.052 x Avg

0.565 0.550 0.447

xd xw 1.770 1.8182 2.237

0.440 0.278 0.140 0.057

0.400 0.258 0.130 0.053

2.500 3.876 7.692 18.868

0.040 0.020 0.010 0.004

239

Simpson’s Rule: x F

0.1, x W,final

xF

dx W

x Wfin

xd

0.004, x W,Avg

xF

x Wfin

xW

xd

6

WFinal F

D total

x DAVG

xF

exp x Wfin

F WFinal

Fx F

1

6 0.096

0.104 2

xW

4 x Wfin

18.868 4 2.237

dx W xd

xW

0.052

1 xd

1 xW

1.770

0.62289; Wfinal

x AVG

xd

xW

0.4734

48.0769 0.62289

48.0769 29.9466 18.1303 kmole

WFinal x Wfin

4.80769

D total

29.9466 0.004 18.1303

240

xF

0.2586

29.9466

9.D24. Was 9D22 in 2nd edition. a) F D total Wfin and Fx F =D total x D

Fx F

D total

D total x D

x Wfin +Fx Wfin

0.62 0.45 0.85 0.45

3.0

Wx Wfin D total

xF

x Wfin

xD

x Wfin

1.275 kmol, Wfin

F

F Dtotal

1.725 kmol

b) Want operating line where 2 equil. contacts gives x w fin 0.45 . y

L

L

xD . V V Surprisingly, with 2 contacts T & E not needed. – Start stepping off stages from top & from bottom simultaneously. The intersection point must be on op. line as is y x x D . L V

Slope

0.85 0.44 0.85 0

0.482,

Figure for 9.D24. 9.D25. New Problem in 3rd edition. Mix together F F1 F2

2.5

241

L

L

L V

D

V L

1 L V

x

0.932

1

xF

F1x F1

F2 x F2 xF

w FINAL

n

F

xF

xF

x FIN 6

0.8 2.5

dx

xw final

x w Final

2.5 y

1 y

xF

x AVG

x

0.32

x

y

x w ,FIN

y

1 x

x w ,AVG

Wfinal

3.333 4 2.666

D

area

Fe

F Wfinal

Then from 0.2 to 0.1 with F

1.151,

x

WFIN1

Now F x

y

WFIN1

1

y x 0.1 0.4 3.3333 0.15 0.51 2.7777 0.2 0.575 2.6666

F2 0.1 6

Wfinal2

0.6166

Fx F

x DAVG

WFIN x w FIN

.2

2.666 4 2.703

6

Ftot x tot

3.030

0.55026

WFINAL1 = F1e-area = 1.5 .5769 = .865 D1

F1

WFinal,1

0.635, x D,AVG ,1

0.865 1.0 1.865,

xF

3.3333 4 2.7777

2.666

Fe D2

area

F

1.865 0.7518 Wfinal2

Wfin 2 x Wfin

2.5 .32

242

.865 .2

0.635

0.285185

1.4022

1.865 1.4022 1.4022 .1

1.09775

D total Higher distillate mole fraction.

1.5 .4

0.2

Total D total Total x D,AVG

0.578

D

D1

F2

y x 0.2 0.575 2.666 0.3 0.67 2.703 0.4 0.73 3.03

x w ,F

1.349 kmol

1

y

x

Values are from methanol-water equilibrium data.

2.817

2.5 0.5398

First go from 0.4 to 0.2 with F1

Part b)

y

1

y x 0.1 0.4 3.333 0.21 .585 2.666 0.32 .675 2.817

6

0.21

2

4

x

.22

x w FIN

0.46275

.635 .46275 1.09775 0.6010

Part c) Go from 0.4 to 0.1 for F1 1

x y 0.1 0.4 0.25 0.62 0.4 0.73

0.3

y x

6

3.3333 2.7027 3.030

Wfinal

3.3333 4 2.7027 1.5 exp

.8587

3.030

0.8587

0.6356

F1

For F2 go from 0.2 to 0.1. Same as 2nd part of Part b.

Wfinal

area

F2 e

0.28518

1.0 .75187

0.75187

F2

Wfinal total

1.3874

Wfinal

Wfin ,

F1

D total

F1

F2

Wfinal tot

1.11255

F2

Differs from b – Numerical error!

Ftot .32

x DAVG

1.3874 .1

0.59436 1.11255 Should be same as part b. There are numerical errors in use of Simpson’s rule. More accurate for .4 to .1 is .4 → .2 (Area = .55026) + .2 →.1 (Area = .285185) Total Area = 0.835445, Wfinal1 F1 e area 1.5 .433681 0.65052 Then

1.40239 , D total

Wfin total

2.5 1.40239 1.09761

2.5 .32

x D AVG

1.09761

Same as for Part b. 9.D.26. New Problem for 3rd Edition.

L D

4, L V

final

1.40239 .1

a&b) final

L D

45

0.60109

2 3, L V

1 L D

53

0.4

0.7 . Step off 2 stages.

0.7 draw op. line slope = 0.8. Two stages gives

0.11 (See graph (labeled 9.D.c))

x w final

Also, draw a few lines with x D 2 stages. b) Generate Table x D , x w , 1 x D Plot

23

0.8

For Part b, draw op. lines with slope 0.4 for arbitrary x D For Part a, From x D

L D

1 xD

Find 2 areas

xw

0.7 and L V between 0.4 and 0.8 . Find x w values with

xw

vs x w . Note there is a break in curve at x D 1. from x F

2. from x w

0.6 to x w

0.185

0.185to x w final

Area 1. Simpson’s rule. x w avg

0.6 0.185 2

243

0.110

0.3925 .

0.7 x w

0.185 to 0.110

1

From graph

Area 1 Area 2.

0.6 0.185 1 xD

6 xw

xD

x w avg

3.56 4 2.37

1.942

1.0363

curve is straight line. Thus, Area = width x Avg height

1.6949 1.942

0.185 0.110 Total area = 1.1726. Rayleigh eqn.,

c)

2.37 .

xw

n

2

Wfinal F

x w final

1.1726

Wfinal

100 e

D total

F Wfinal

x D,AVG

Fx F

dx w

xF

xD

xw

0.13638

1.1726

3.0955

6.9045

Wfinal x Wfinal D total

6.0

3.0955 0.11 6.9045

244

0.8197

L/D 2/3 2/3 2/3 2/3 2/3 -4.0

xD

xw

0.9 .85 0.8 0.75 0.7 0.7 0.7

0.65 0.48 0.36 0.253 0.185 0.145 0.110

x w ,final

245

1 xD xw 4.0 2.703 2.273 2.012 1.942 1.8018 1.6949

9.D.25. New Problem for 3rd Edition. From the methanol-water equilibrium data, the following table can be obtained.

246

xM 1.0 .8 .6

yM 1.0 .92 .825

1 yM 1.0 1.08696 1.21212 x pot ,Initial

S W

x pot ,final

dx MeOH y MeOH

.4

Simpson’s rule:

1.0 4 1.08696 1.2121 6 0.4373W 0.8747 kmol

S

0.4373

9.E1. octanol water

xF pot

0.90, F=1.0 kmole

1 0.95 .9

Final octanol in pot

1.0 0.1 0.1

Nonvolatiles in pot

x oct,W,final

steam

log10 VPW

0.045 kmol

0.045 0.145

0.3103

2164.42

8.68105

, T C, VPW mmHg 273.16 T 0.3103, x W, in W 1.0, T 99.782 C from

a) Final conditions x oct, in org

problem 8.D11. Initial conditions: x oct,in org

0.90, x W in W

solution

1.0

x i VPi 1.0 atm 760 mm Hg From spread sheet find T = 99.377ºC b)

Wfinal,org

c)

Dorg

F

1 z 1 x Wfin

F Wfinal

0.8550

D org

nW

d) Eq. (9-24)

p tot

Estimate VPoct at average T

p tot VPoct

VP VP

0

nW

1.0

oct

1.0 0.9 1 0.3103

n org oct

x oct

x oct

dn oct

(99.782 99.377) / 2 D org

0

dn oct x oct

0.1450

D org

dn org 0

247

p tot VPoct

99.5795 , VPoc tan ol D org

0

dn oct x oct

D org

18.87 mm Hg

to

F 1.0, n org

F W, x org

dn org

W

W

1

Step-by-Step integration,

n org

F

1

x org

x org

1 x feed

1

0.1 F W

dn org

dn org

x org

dn oct

Avg x org

x org,avg 0

1.0

.9

0.1

.1

.9

.8888

0.2

.1

.8

0.875

0.3

.1

.7

0.8571

0.4

.1

.6

0.8333

0.5

.1

.5

0.8000

0.6

.1

.4

0.750

0.7

.1

.3

0.6666

0.8 0.855

.1 0.055

.2 0.145

>

0.8944

>

0.8819

>

0.8661

>

0.8452

>

0.81666

>

0.775

>

0.7083

>

0.58333

0.50 0.3103 >

0.11181 0.11339 0.11546 0.1183 0.12245 0.12903

0.40517

0.14118 0.17143 0.1357 1.1588

nW

p tot VPorg

Porg

O

dn org x org

D org

760 18.8666

1.15882

0.8550

45.826 kmol

e. Continuous had 108.93 kmoles water/kmole organic fed. The continuous always operates at lowest octanol mole fraction in liquid & thus y oct is always at lowest value. Thus, requires more water to carry over octanol then the batch operation. 9.E2. New Problem in 3rd edition. Parts a & b. See solution to problem 8.D25. mol B 78.11 x benz mol B 78.11 80 673.2 c. Find T from Eq. 8-15 With Spread Sheet: SPE, Problem 9E5. Solution for temperature T deg C 92.04234 Do step by step Antoine VP values VPW A,B,C 8.68105 2164.42 273.16 2.754418 568.0906 VPben A,B,C 6.90565 1211.033 220.79 3.034461 1082.583 X ben 0.17727 xw 1 ptot 760 Eq8 -5E-05 Goal seek B6 to zero changing B2 massbeninit 20 massbenfin 2

248

dmorg-sum dnw/dnorg 9-23 d)

18 massben-still 2.960203 dnw

2 dmassorg 0.037898

1

Use Spread Sheet for each time dn W dn org Eq. 9-23 step. Did the addition of steps off-line in table below. nW dn W for n org still init 20 78.11 n benz 2 78.11 Step-by-step integration.

still final

Set dm org

1 kg. Values from spreadsheet. Mass benz still, 777 (kg)

dm org kg Initial T

0→1 1→2 2→3 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 → 18 18

final T d)

20 → 19 19 → 18 18 → 17 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3→2 2

dn W divide all values by 78.11 kmol .664431 .677266 .691548 .707537 .725558 .746073 .769462 .79657 .828276 .865852 .91108 .966543 1.036129 1.12596 1.246275 1.415548 1.670846 2.098857 2.960203

TºC from spreadsheet 75.99 76.27 76.577 76.91 77.269 77.66 78.0996 78.58 79.12 79.715 80.39 81.156 82.03398 83.0487 84.2358 85.6435 87.339 89.422 92.04

For accuracy add time steps 1 to 17 + (time steps 0 & 18)/2. This is then identical to the use of average values of dm for every step.

Steps_1 to 17 0.5 Steps _ 0

17.2794168

18

1.8122565

n W 19.0916733 78.11 18.016 4.4034 kg water Compared to continuous, nw = 12.3, batch requires less because benzene mole fraction is higher in batch operation for most of batch. If we take final cut and used that value for entire operation, 18.016 dn W 18 kg benzene overhead 12.3 same as continuous. 78.011 e) To vaporize (n ben ben ) / w varies since T is different. For accuracy, do for each dm org and find λ values at these T. x d ,fin

9.E3.

Simpson’s Rule: xF

dx d xd

xB

0.13 6

1 xd

249

4 xB

x d .63

xd

1 xB

x d 0.565

xd

xB

x D 0.50

Generate table from graph 1 xd xB xd xB 0.3 0.36 3.0303 0.2 0.597 2.5189 0.1 0.565 2.15 0.04 0.500 2.004

x init

9.E4.

Wfinal

F exp x W ,fin

0.13 6

3.0303 4 2.15

D final

Fe

1 e

B

F Dfinal x B,avg

0.2954

2.004

0.2954

7.442 gmole

2.558 mole. F xF

Dx D,fin

5.0 4.6886

B

2.558

dx y x

T & E since dilution effects F & x init . Dilute with 5 kg water → Start with 6, x init Dilute with 4 kg water → Start 5, x init Dilute with 3.5 → Start 4.5, x init

16

15

0.1667

0.2

1 4.5

0.2222

Dilute with 2.75 → Start with 3.75, x init

1 3.75

0.26667

Dilute with 1.75 → Start with 2.75, x init

1 2.75

0.364

250

0.1218

For each dilution want to integrate using Simpson’s rule until find Wfinal need values of 1 y x at x init , x avg , and x final are determined in the following table.

1.0. Thus,

0.01 for each dilution. These values

Table of Values for Integrations Dilute 5 kg Water 4 kg Water 3.5 kg Water 1.75 kg Water 2.75 kg Water Integrations: Dilute 5 kg:

Wfinal Dilute 4 kg:

Wfin Dilute 3.5:

Wfin Dilute 1.75:

Wfin Dilute 2.75:

Wfin

x 0.16667 0.08833 0.01 0.200 0.105 0.01 0.2222 0.11611 0.01 0.364 0.1870 0.01

y 0.537671 0.38708 0.067 0.579 0.428 0.067 0.598 0.450 0.067 0.706 0.563 0.067

y-x 0.3710 0.29875 0.057 0.379 0.3230 0.057 0.376 0.334 0.057 0.342 0.376 0.057

1/(y – x) 2.695412 3.34724 17.544 2.63 3.095975 17.544 2.6596 2.994 17.544 2.923977 2.6596 17.544

0.2666 0.138 0.01

0.636 0.493 0.067

0.369 0.355 0.057

2.7100 2.8169 17.544

Use Simpson’s rule for each addition. 0.15667 2.695412 4 3.34724 17.544 0.87809 6 6 exp -0.87809 2.4934 . Value is too high. Want 1.0 kg.

0.19

32.5579 1.031 6 5 exp -1.031 1.783 too high 0.2122

32.1796 1.138 6 4.5 exp -1.138 1.442 too high .354

31.106 1.833 6 2.75 exp -1.835 0.4389 too low .2566

31.52 1.348 6 3.75 exp -1.348 0.9740

Close to desired 1.0 kg. Thus, 2.75 kg water. The final still pot is 99% water so have (.99) (.974) = 0.964 moles water remaining. Moles of water distilled off is 2.75 – 0.964 = 1.786. 9.H.1. New Problem in 3rd edition. This problem is challenging for students because they must first derive the forms of the equations they need to use.

251

A. Define. The system is the simple still pot shown in Figure 9-1. Find Wfinal, D, xA,Wfinal, and xA,dist,avg. B. Explore. At first it may appear that the problem in Part a is under specified since there are now five unknowns. However, in specifying the problem based on the fractional recovery of benzene in the distillate we have added the equation for the definition of fractional recovery of A in the distillate. This equation is most conveniently written as, FzA (1 – Frac. Rec. A in distillate) = Wfinal xA,Wfinal

(9-35a)

which becomes, Wfinal = FzA (1 – Frac. Rec. A in distillate)/ xA,Wfinal

(9-35b)

C. Plan. If we write Eq. (9-13) for A and substitute in Eq. (9-35b) we obtain Eq. (9-36),

0

1 AB

1

n

x A ,W ,final 1 x A ,F

1 x A ,F

n

x A ,F 1 x A ,W ,final

1 x A ,W ,final

n

z A 1 Frac. Rec. A.dist x A ,W ,final

Part a. In a spreadsheet Eq. (9-36) is easily solved for xA,Wfinal using Goal Seek. Then W final can be determined from Eq. (9-35b). Then DTotal is determined from Eq. (9-11) and xA,dist,avg is determined from Eq. (9-10) written for component A or from the fractional recovery. Part b. Now solve Eq. (9-36) for frac. rec. of A in distillate using Goal Seek. For both parts a and b can use fractional recovery values and DTotal to find xA,dist,avg = FzA(Frac Rec. A in distillate)/ DTotal D. Do It. Because Eq. (9-36) for xA,Wfinal is nonlinear, it is easiest to solve this problem with a spreadsheet and use Goal Seek to solve Eq. (9-36). The spreadsheets are shown below. Part a F 5 zA 0.37 The 0.37 is in cell D2 alpha AC 10.71 frac rec A in distillate 0.75 xA,Wfin 0.143185 9-36 term 1 -1.25687 term 2 -0.3075 term 3 0.436926 Eq 9-36 -1.7E-05 Use Goal seek Wfinal from 9-35b 3.230095 D total 1.769905 xAdist,avg 0.78394 Part a. Use Goal Seek for cell B8, setting it equal to zero by varying cell B5 (xA,Wfin). Part b. Use Goal Seek for cell B8, setting it equal to zero by varying cell C4. Part b F 5 alpha AC 10.71 frac rec A in distillate xA,Wfin 0.05 9-36 term 1 -2.41222

zA

0.37

The 0.37 is in cell D2

-0.41074

term 3

0.930081 term 2

0.658942

Eq 9-36 -0.00023 Use Goal seek Wfinal from 9-35b 2.586992 D total 2.413008 xAdist,avg 0.713073 Part a. Use Goal Seek for cell B8, setting it equal to zero by varying cell B5. Part b. Use Goal Seek for cell B8, setting it equal to zero by varying cell C4 (frac rec A in distillate).

252

SPE 3rd Edition Solution Manual Chapter 10 New Problems and new solutions are listed as new immediately after the solution number. These new problems are: 10A4, 10A5, 10A7, 10A12, 10A17, 10C4, 10C5, 10D13, 10D19, 10D21, 10G110G4. 10.A9.

10.A13.

A good packing will have: good contact between liquid and vapor, high surface area, low pressure drop, inert, inexpensive, and self-wetting. Marbles have low surface area, poor contact and relatively high ∆p. Figure 10-25 shows that if viscosity increases the ordinate increases and ∆p/foot increases.

10.A14. a.Want low F since this gives low ∆p. b. F decreases as size increases. c. Ceramics have much thicker walls than metal or plastic. Ceramics are used in corrosive environments. 10.A16. 1. a. fewer; 2.b. larger; 3. a. lower 10.A.17. Answer is c. 10.B1.

10.B3.

The trick is to have maximum and minimum positions of the valve with a larger area for vapor flow at the maximum position. a. Use a cage to prevent excess movement. b. Use feet. c. Have a flap that moves up and down. d. Use a spring to provide force and maximum position. e. Sliding valve controlled by an external feedback mechanism. f. Two flaps to make a roof. Many other ideas can be generated. Some possible candidates: Bottle caps Bent bottle caps Natural sponge Synthetic sponge

SOS pads Scotch brite pads Steel wool Cooper cleaning pads

Miscellaneous junk Broken crockery or glassware

Plastic coated wire Tin foil - crushed

Cut up tubing (Tygon) Glass tubing (broken)

String - balls lines tied together

Crushed beer cans Cut up - crushed beer cans

lines twisted lines stretched taught

Coal

Rope

Egg shells Styrofoam packing material Old seat cushions

frayed rope Rope tied into bow ties Porous rock pieces

253

10.C1.

Nuts/bolts/screws/nails Metal filings Wood shavings Kindling Left-over redwood

Pumice Ash from Mt. St. Helens Ashes from coal stove Pieces of cement block Pieces of brick

Staples

Pop-tops

Window screens, rolled up Chicken or barbed wire

Old watch bands - twisted Bent wire coat hangers

Christmas Tree Ornaments Corn cobs

Cookie cutters Combination of the above

L/V constant since L/D const. Thus L/G const. In F v only

n

changes. From Perfect Gas

G

P

MWv . Thus, F v increases as P . However, curve is almost v RT flat in this range and Cs,bflood ~ constant.

Law

G

MWv

u flood

Then

V Thus,

L D

10.C2.

Dia

u flood

P

since

P

1 D and D

F

.5

1

G G

L D

V

Then, F

L

0.2

20 /

Dia 2

.5

C sb flood

F

xD

z

xD

F 1

xB

L

so V

G

.

F

F P

P

P , and exponent = ½

See derivation in solution to Problem 10.C1.

L

V

D

1 D

L D

1 F

xD xD

z xB

Then from Eq. (10-16)

L 4F 1 D

Dia

Since Dia 10.C3.

F1/ 2 , Dia

1/ 2

3600 p fraction u flood

L D 1

Plot points based on ∆p and

xD z xD xB

1/ 2 1/ 2

L

G

G

L

on Figure 10-25. From ordinate values calculate F

for each point. Use an average value.

254

10.C4. New Problem in 3rd edition. Part of the operating lines become closer to the equilibrium curve. Thus, for the same separation more stages are needed. Fortunately, this effect is often small. 10.C5. New Problem in 3rd edition. You can show this by proving that the minimum reflux ratio (Figure 10-18A) or the minimum boilup ratio (Figure 10-19A) must increase compared to the base cases. These ratios increase because cooling the entire feed (Figure 10-18A) or heating the entire feed (Figure 10-19A) automatically changes the feed line and moves the minimum reflux (Figure 10-18A) or minimum boilup (Figure 10-19A) operating lines towards the y = x line. This means larger minimum external reflux ratio or larger minimum boilup ratio. 10.D1.

K C6

y/x

and

K C7

1 y

Solve simultaneously. x

1 x

1 K C7 K C6

K C7

KC6

T 149 C 169 171

KC7

1.0 1.3 1.34

193

K C6 x

Bounds are K C6

Pick T and generate equilibrium curve. o

, y

.72 .75 -

-

1.0

x

y

1.0 .483 .424

1.0 .63 0.568

0

0

Average temperature is T = 171ºC = 444.1 K. y x .568 .424 1.786 1 y 1 x .432 .576 Viscosity equation and terms are given in Example 10-1. 1 1 log10 C6 362.79 .935, T 207.09

log10

n

C7

MIX

436.73

1

1

T

232.53

.5 n .116

1.0 and K C7

.5 n .127

.895,

C6

0.16 .

C7

0.127 .

2.107,

MIX

1.0 .

0.122.

From Eq. (10-6) for 0.217, E o 0.730 . The higher pressure results in higher temperatures and lower viscosities. This increases the predicted column efficiency by 24%. 10.D2.

T v

98.4 273.1 371.5 K (almost pure n-heptane at bottom). 1 100.2 pMW 0.684 62.4 0.205 lb/ft 3 , L RT 1.314 371.5

42.68 lb/ft 3 , 20

Need L V . First, find y at intersection of operating lines. L .999 y Top Operating Line Slope .8 V .999 .5 y .999 .8 .499 0.5998 Then,

L

0.5998 0.001

V

0.5 0.001

1.20

255

L

L MWL

L

G

V MWv

V

Fv

1.2 . This is at bottom where MWL ~ MWv .

0.5

L

v

G

L

0.5

.205

1.2

0.08317

42.68 0.2

From Figure 10-16, Csb

u flood

0.32, C sb

42.68 0.205

0.29

0.32

20

0.205 Saturated liquid feed V V

0.2

12.5

0.29

20

0.5

4.19 2500. Use η = .90 1/ 2

4 2500 1.314 371.5

D

12.35 ft

.90 3600 1 .75 4.19

Somewhat larger. Would design at bottom of column. Use a 12 ft. diameter column. 10.D3.

New 12’ dia col. First, redo entrainment calculation.

L

0.5

0.205

0.0832 , L 1.2 V 3000 V 42.68 For Fig. 10-17 need % flood. In problem 12.D2 designed for 75% flood.

D Use 12 feet:

1.2, F g

1/ 2

const

11.78 , const1/ 2

75% flood

% flood

e

1.2

1/ 2

10.202 /12.0. % flood

0.022 3000

L

A total

12.0

0.9,

A active

v0

t tray

weir

Dia

0.726 ,

0.726 12 ft

weir

90.48 ft 2 , A hole

0.078 in, 3 16 holes, d 0 t tray

v

9.05

h

p ,d ry

Lg

3067.48

4 113.1 ft 2

113.1 1 .2

VMW v 3600

10.20

1 0.9 113.1 11.3 ft 2

Ad Table 10.2,

2

1/ 2

72.3% . Then Fig. 10-17, ψ = 0.022.

67.48 , L e

1 1 .022 This is reasonable amount.

14 gauge,

11.78 .75

37.51 ft s , C0 0.003 37.51

3067.48

2

0.1 90.48 9.05 ft2

2.4

0.759 (unchanged from Ex. 10-3)

0.205

100.2 7.48 42.68

8.71 ft

60

59.87

1 0.01

42.68

0.759

2

2.086 in

897.8 gpm

256

Abscissa

Lg

897.8

2.5 weir

8.71

Parameter

w

Dia

4.01

2.5

0.726, Fweir Fig.10 20

Eq. (10-26),

h crest

0.092 1.03

With 1″ gap,

A du

8.71 12

h du

Eq. (12-27),

h dc

897.8

2.083 in

8.71

0.726 ft 2 2

897.8

0.56

1.03

2/3

4.248 in

449 0.726

2.086 2 2.083 0 4.248 10.417

h dc,aereated

10.417 0.5

11.3 10.417 3600 42.68

t res

3067.48 100.2 12 0.040 12.5

Weeping, Eq. (10-32), h Eq. (10-31), LHS

h

h

2.086 0.0625

2 2.083 0

=4.904 This is OK.

0.0625 in

42.68 3 16

p,dry

x

OK, but close to distance between trays.

20.83

2.148

4.083

RHS 0.10392 0.25119 4.083 0.021675 4.083 LHS > RHS Operation is slightly marginal with high downcomer backup. 1.5 Increase apron gap to 1.5 inches: A du 8.71 ft 1.089 ft 2 12 h du1.5 gap

h dc

0.56

v o,bal

Given Wvalve

2

449 1.089

p,valve

C v Wvalve 2g K vA v

0.08 lb,

Av

Cv 1 12

1.888 in

16.11 inch, OK 1/ 2

(10-36)

v

g

32.3 ft s 2 , K v,closed

Pressure drop in terms of inches of liquid of density

h

0.7682, OK

2.086 2 2.083 1.888 8.05 in

h dc,areated 10.D4.

897.8

2

Wvalve

Kv

v

Av

2g

L

L

0.02182 ft 2 ,

33C v L

1.25,

K v,open

5.5

:

v 02 v

0.1917 lb ft 3 ,

L

41.12 lb ft 3

257

1/ 2

1.25 .08 2 32.2

v o,bal,closed

6.83 ft sec

33 0.2182 .1917 1/ 2

1.25 .08 2 32.2

v o,bal,open

16.73 ft sec

5.5 0.218 .1917

At balance point, h

1.25 0.8 lb

C v Wvalve

p,valve

AL

0.2182 ft 3 41.14 lb ft

L

0.1115 ft closed:

h

33 .1917 p,valve

2.39 10 open:

h

3

v 02 ft

5.5 .1917 p,valve

4

10.D6.

6.83

2.87 10

v 02 for v 0

2 32.2 41.12

3.98 10 10.D5.

1.338 in liquid

v 02 for v 0

2 32.2 41.12

3

v 02 ft

2

v 02 inches

16.73

4.78 10

3

v 02 inches

Do calculation at total reflux. From a McCabe-Thiele diagram (not shown). Total # Contacts = 4.2 N = Total – 1 (P.R.) = 3.2 Length 1 HETP 0.31 m N 3.2 From Fenske eq. and definition of HETP

HETP n

z

x

x

1 x

n n

x 1 x

1 x

dist AB

x dist

1 x

bot

n bot

.987 .013

.008 .992

9.150

z = 3.5 meters. Obtain: a. α = 2.315, HETP = 0.321 b. α = 2.61, HETP = 0.367 c.

AVG

2.315 2.61

1/ 2

2.4581 and HETPAVG

0.344

Can also use McCabe-Thiele diagrams although the solution shown is easier. 10.D7. Current: F v 0.090, 12 spacing, Ordinate 0.2 U nf const., const. 0.2 U nf 0.2 6.0 New: L Vnew 1.11 L V old , then L G new 1.11 L G old , F v,new 1.11F v,old 0.0999 Trays Spacing 24″, Ordinate ~ 0.32, Ordinate = U nf × Const.

258

U Nf 10.D8.

At

Ordinate

0.32

Const

0.2 6

Fv

0.5, Csb,f 1

0.12

0.2

Uf1

C sb

L

Since

nRT,

20

0.5

V

10.D9.

G

0.12

0.5

L

20

RT

G ,new

0.5

1

0.5

0.25; C sb ~ 0.18

4

U f 1,new

C sb,new

G ,old

C sb,new

old

0.18

U f 1,old

C sb,old

G ,new

C sb,old

new

0.12

4

3 2

2

3

3 6 18 ft/s

Mass Balance:

D F D

z xB

0.6 0.01

0.59656 x D x B 0.999 0.01 596.56 kmol/h, B 403.44

L V

0.6, V L D V D 0.6V D 596.56 V 1491.47 kmol h 0.4 0.4 L V D 491.41 kmol h

At top:

At top of col.

V

WL WV 1491.41

kmol

L V since pure MeOH, same mol. wt.

0.6

32.04 kg MeOH

lbm

2.046 lbm

h kmol kg Assume ideal gas. Top of column is essentially pure MeOH. lb 1 atm 32.04 n MWM p lbmol MWM v 3 V RT ft atm 0.7302 607.79 R lbmol R where pure MeOH boils at 64.5 C L

At top

1 0.2

MWv p

G ,old

U fl,new

6.0

G G

n

MWv

G

Fv

New Condition:

L

0.12

G

pV

0.2

6.0

G

20

9.6 ft s

MeOH

h

0.07219

lbm ft 3

1.8 R

607.79 R K 1 kg 2.2046 lbm 28317 cm 3 0.7914 g cm 3 1000g kg ft 3

24 0.0773 64.5

337.66K

105, 346

49.405

lbm ft 3

19.0

259

1/ 2

WL

Fv

WG

0.07219

.6

G L

0.28

0.2

U flood

C sb

20

0.2

19

V

frac u flood 3600

ft 3600 s h s Use either 10 ft (slightly higher frac flood) or 11 ft – (lower frac. flood). 0.90 0.07219 lbm ft 3

D = 10.27ft

3.0, y=

Ref. Bonilla (1993).

L Dact

.9 0.66667 .9 .4

2 L D min

x 1+

-1 x

1.75 ,

L V

0.875D , L avg

Generalize Llow

min

L D

7.25

0.4

x, y feed

L

L V

0.46667

V L 1.75

1 L V

1 .46667

1 L D

act

L V

min

1.75D , Llow

Lavg M where L D

n MWv V

0.875

0.636

2.75

actual

or Llow D

L D

min

0.875

0.5 Lavg

M

10.D11. a. Since Liquid & Vapor have the same mole fractions L G G

23

0.3273

At Minimum (Pinch Point), Llow V

L low

0.75

, at z

0.46667 , L D

1 L V xD

pV = nRT,

7.25 ft s

4 105, 346 lbm h

D

min

0.07219

4V lbm h V

L V

49.405 0.07219

0.28

20

V

Use Eq. 10-14 (Modified), D

10.D10.

0.02294

49.405

Fig 10-16 with 18″ tray spacing: Csb L

1/ 2

p MWv RT , R

L D

min

L V 45.6 cm3 atm gmoles-1

o

R1

260

MWv

.8 46

1 ATM

G

V

L D

L V

.2 18

40.4

3D

6750 lb day , L

4500 6750 .6667 1/ 2

L G

G

45.6 cm3 atm mol 1 R 1

40.4 g mol

H2O

2 F

G

L

1 .82

97

3

1.393 10

3

.82

1/2

.001393 g cm3

4500 lb day

=0.0275

1 g cm3

L

(Table 10-3) .2 G F2 F

G

2 2250

460 R

L G

.6667 1.393 10

L

F = 97 Ordinate This is,

GF

2D

176

.82 g cm3

1 .82

.197 , from Figure 10-25 (flooding line).

gc

.52

0.2

1 g cm 3 62.4 lb ft 3

g cm 3 .82 g cm 3

0.5216 lb (s ft 2 ) 8.64 10 4 s day

2

.197

32.2

45067 lb (day ft 2 ) , G

.75 GF

FIND AREA AND DIAMETER FOR 75% OF FLOODING

AREA

V

.75 G

45067 lb Day ft 2 .75

6750 lb Day

.19970 ft 2

D 2 4 .19970 ft 2 , D .5042 ft 6.05 in COLUMN DIAMETER 6.05 5 8 9.68 which is probably OK. PACKING DIAMETER b. From Fig. 10-25, G 2 F

G

2

97

0.2

1 .82

1.393 10

G

c.

(

G

.52 3

L

0.2

1 62.4

D 2 4 .3503,

L G will be the same; thus

D

2

67500

4 1.998

D

.6679 ft

0.3503 ft 2

8.01 inches

1/ 2

L

G

G

L

V .75 6 ,

.0275

1.927 10 4 lb (day ft )

19267 lb (day ft 2 )

AREA

L

0.036

0.2230 lb (s ft 2 ) 8.64 104 s day 6750 lb day

area

G

2

.82 32.2

AREA

Area

1/2

g c ) .036 at L G

V

will be the same, and G will be the same.

3D as before

.75 .521 8.64 10 4

67,500 lb day

1.998 ft 2 10

earlier value

1/ 2

1.59 ft

19.14 inches

261

10.D12. a.

y

L V x 1 L V x D . When x 0, y 1 L V x D 0.1828 . See figure. Need 2 equilibrium stages. Stop where feed line and operating line intersect. HETP 5 / 2 2.5 ft, x B ~ 0.065

b. M is at x in

y1

.43 L

L M

6.13

L

L M

1.58

L

L

Within accuracy of graph,

3.88

L

L

V L L L L If try a shorter column with same feed won’t work. L L .8 and must adjust column. L V

1

0.8

10.D.13. New Problem in 3rd edition. Saturated vapor feed in problem 10.D.9 has minimum L/V = (0.999-0.6)/(0.999-0.22)=0.5122. This is (L/D)min = 1.05. The actual L/V = 0.6, which is an L/D = 1.5. Thus, the multiplier M of the minimum was M = (L/D)/(L/D)min = 1.5/1.05 = 1.43. For a saturated liquid feed (L/V)min = (0.999-0.825)/(0.999-0.6) = 0.4361, which corresponds to (L/D)min = 0.7733. If we use the same multiplier, L/D =1.43(0.7733) =1.106 and L/V = 0.525. z xB D 0.6 0.01 Mass Balance: 0.59656 F x D x B 0.999 0.01 D 596.56 kmol/h, B 403.44 . These are same as in 10.D9. At top:

L V

0.525, V

L D

V 0.525V

D

262

V

L

D

596.56

1255.9 kmol h 0.475 0.475 V D 659.4 kmol h

WL WV

At top of col.

V

1255.9

L V since pure MeOH, same mol. wt.

kmol

32.04 kg MeOH

2.046 lbm

h kmol kg The density and surface tension calculations are the same as in 10.D9. Assume ideal gas. Top of column is essentially pure MeOH. lb 1 atm 32.04 n MWM p lbmol MWM v 3 V RT ft atm 0.7302 607.79 R lbmol R where pure MeOH boils at 64.5 C L

MeOH

337.66K

Fv

G

WG

L

0.2

U flood

C sb

0.07219

.525

Fig 10-16 with 18″ tray spacing: Csb L

20

V

49.405

0.07219

lbm ft 3

49.405

lbm ft 3

1/ 2

0.02007

0.28 19

0.2

0.28

20

V

h

19.0

1/ 2

WL

lbm

1.8 R

607.79 R K 1 kg 2.2046 lbm 28317 cm 3 0.7914 g cm 3 1000g kg ft 3

24 0.0773 64.5

At top

82, 330

0.525

49.405 0.07219 0.07219

7.25 ft s

4V lbm h

Use Eq. 10-14 (Modified), D V

frac u flood 3600

4 82, 330 lbm h

D

9.08ft ft 0.90 0.07219 lbm ft 0.75 7.25 3600 s h s Probably use 9 ft, which is a slightly higher fraction of flooding. This compares with 10.27 ft for the saturated vapor feed. The smaller diameter column will be less expensive. With a saturated liquid feed and CMO, the vapor flow rate in the bottom of the column is the same as in the top, V = 1255.9 kmol/hr. For problem 10.D9 with a saturated vapor feed, V V F 1491.47 1000 491.47 . Since QR V, 3

QR ,liquid _ feed

(1255.9 / 491.7)QR ,vapor _ feed

2.55QR ,vapor _ feed .

Thus, in this case there is a significant energy price for reducing the column diameter by this method. 10.D14. D

z xB xD

xB

F

0.4 0.0001 0.998 0.0001

1000

400.7415 , B 1000 D

599.258 kmol/day

263

or

V

V

V

1202.225

At bottom, L

L D 1 D

3 400.7415

kmol 1 day

1h

1202.225 kmol day 0.013915 kmol s

day 24 h 3600 s

V B 1202.225 599.258 1801.483 kmol day

L

1801.483

1.49846 V 1202.225 Bottom of column is essentially pure water. Also y boilup Thus

L G

V in

L

G is lb (s ft 2 )

L V 1.49846

lb

kmol 18.016 kg 2.20462 lb

0.013915

s

density water at 100 C

s

kmol m

3

nRT

where 100ºC

n G

V

Fv

K

L

G

G

L

59.83

3

lbm ft 3

1.0

L

MWw

RT

1.8 R

373.16K

35.31454 ft

1 atm

p

MWw

m3

kg W

0.55268 lb s

1 kg

kg 2.20462 lb

958.365

Data from Perry’s, 7th ed., p. 2-92.

pV

xB

18.016 3

ft atm lbmol R

0.7302

lbm ft 3

0.0367

671.688 R

671.688 R

1/ 2

1/ 2

0.0367

1.49846

0.03713

59.83

From Fig. 10.25, Ordinate at flooding = 0.18 0.18

G flood

F

w

Area

F = 33,

32.2, F

110 Table 10 3

1/ 2

0.15147

0.2

1/ 2

0.3892

lbm s ft 2

0.26 cp from Perry’s p. 2-323.

0.80 G flood

0.80 0.3892

V in lb s G actual lb s ft

Diameter b.

, gc

110 1.0 0.26

100

G actual

1/ 2

gc

0.18 0.0367 59.83 32.2

G flood Where

G L 0.2

0.55268 2

4 area

0.8 0.3892 1/ 2

Diameter Intalox plastic

4

Dia

0.31136 1.77505 ft 2

1.77505 Fint Berl

FBerl

1/ 2

1.503 ft 1/ 4

1.503

33 110

1/ 4

1.112 ft

264

lbm ft 3

Fv

10.D15.

WL

v

Wv

L

0.03713 from prob. 10.D14

From Fig. 10-16 with 12″ tray spacing , Csb,flood

0.21

0.2

K

Csb

where σ = surface tension water at 100ºC.

20

Perry’s, 7th ed., p. 2-306 @ 373.15 K,

K u flood

K

L

4V

0.0367

Dia

lb s

10 5 dynes

1m

1N

100 cm

0.21 1.2411

59.83 0.0367

0.206

v

v

0.2

0.21 58.9 20 v

Dia

0.0589 N m

58.9

dynes cm

0.2606

10.52 ft s , u act

.8 10.52

8.416 ft s

Eq. (10-14) modified for units. V is from Problem 10.D14.

lb ft u act 3 ft s

4 0.55268 0.85 0.0367 8.416

1.637 ft . This can be compared to 1.5 ft for packed.

Tray columns with this small a diameter are seldom used in industry. 10.D16.

F1/4 . F1

From Eq. (10-44). Diameter Diameter (3 )

98 and F3

Diameter (1 )

22 Table 10-3

1/ 4

F3

14.54

F1

22

1/ 4

10.0 ft

98

Can also repeat entire calculation which is a lot more work. 10.D17.

At the bottom of the column have essentially pure n-heptane. Then, following Example 10-4, we have. p MW 1 100.2 0.205 lb ft 3 v RT 1.314 371.4 Need L V . Since L V operating lines. Then L V

.8

1 y

where z 1 z 0.6 0 1.2 . .5 0

L

L MWL

G

V MWv 1/ 2

L

v

G

L

1.2

.5, we have y

1.2 1.0

0.6 at intersection of

1.2

0.205 1.684 62.4

1/ 2

0.084

Figure 10-25, Ordinate = 0.05 at ∆p = 0.5

265

0.05 .205 0.684 62.4 32.2

G The value 0.9595 is

V

V

.2

98 .9595 .684 0.205 at 98.4ºC, and

Water

Water

0.375

. Since feed is a saturated liquid

L

0.6944 lbmol/s.

0.6944 100.2

VMWv

Area

G

0.375

185.5 ft 2

1/ 2

D 4 Area 15.37 ft This is somewhat larger than in Example 10-4. Therefore design at bottom. G

G flood

10.D18.

L

gc

ordinate

F

0.2

Assume changing p changes only

G

& ordinate. Then take ratios

G flood ,new

G ,new

G flood ,old

G ,old

ordinate, new ordinate, old

G ,new

p MW

RT

G ,old

p MW

RT

new

p new

old

p old

4 . Assumes small change in

T (in Kelvin). T set by boiling conditions (Vapor Press) not by ideal gas law. Fv

0.5

L

G

G

L

,

0.5

F v ,new

G ,new

p new

F v ,old

L ,old

p old

0.5

2.0 , F v,new

2F v,old

0.4 .

New Ordinate Value ~ 0.5, old value ~ 0.09

G flood ,new

p new

ordinate, new

p old

ordinate, old

G flood ,old

4

0.05 0.09

0.5

0.75

10.D.19. New Problem in 3rd edition. Saturated vapor feed in problem 10.D.9 has minimum L/V = (0.999-0.6)/(0.999-0.22)=0.5122. This is (L/D)min = 1.05. The actual L/V = 0.6, which is an L/D = 1.5. Thus, the multiplier M of the minimum was M = (L/D)/(L/D)min = 1.5/1.05 = 1.43. For a saturated liquid feed (L/V) min = (0.9990.825)/(0.999-0.6) = 0.4361, which corresponds to (L/D)min = 0.7733. If we use the same multiplier as in 10.D9, L/D =1.43(0.7733) =1.106 and L/V = 0.525. This is the slope we use in the top section for the 2enthalpy feed. In the middle section of the column at minimum reflux conditions the slope of the middle operating line is L / V (0.825 0.6) / (0.6 0.22) 0.592 . The external mass balances still gives D

Lmin

( L / D) min D

0.7733(596.56)

the saturated liquid feed V

V and L

596.56 kgmoles/hr, B

461.3 and Vmin L

Lmin

D

403.44 . At minimum reflux

461.3 596.56 1057.86 . At

Fliquid . Thus,

[( L Fliquid ) / V ]min ( L / V ) min 0.592 and Fliquid ,min 0.592V L 165.06 . Since the total feed rate is 1000, the fraction liquefied is 0.16506. The same fraction can be liquefied at the finite reflux ratios. Thus, Fliq 165.06 and Fvap 834.94 .

266

At top use saturated liquid reflux ratio L V

0.525, V L D V 0.525V D D 596.56 V 1255.9 kmol h 0.475 0.475 This is the same as for problem 10.D13 and the remainder of the calculation of the diameter is identical to that calculation. The result of the calculation at the top of the column is 4 82, 330 lbm h

D

9.08ft ft 0.90 0.07219 lbm ft 0.75 7.25 3600 s h s We now need to calculate the vapor flow rate in the bottom. Assuming CMO, in the middle section V V 1255.9 . In the bottom section, 3

V

V

Fvap

1255.9 834.94

420.96 .

Since QR V , QR ,2 enthalpy _ feed (420.96 / 491.7)QR , vapor _ feed 0.86QR , vapor _ feed . Thus, in this case the two-enthalpy feed design results in the same reduction in diameter as liquefying the entire feed, and it has energy savings compared to the vapor feed. However, the two enthalpy feed system will require more stages than the other systems. A complete economic analysis is required to determine the most economical system. 10.D20.

Use Fig. 10-16 to find C sb . Gas is N 2 . Liquid is ammonia. Since system very dilute, treat as pure ammonia liquid & pure N 2 gas.

L

WL kg h

L kmol h

G

WG kg h

V kmol h MWV

0.61

L

100 cm

1 kg 3

cm 1000 g

m

27.36

RT

L atm mol K

16.642

236

28.08

16.642

610 kg m 3

3

175 atm 28.02 g mol 0.08206

17.03

3

pMW v G

Fv

g

MWL

1000L m

253.2 K

3

kg 1000 g

236.0

kg m3

1/ 2

10.35 610 Off chart. Extrapolate using Eq. (10-10e).

log10 Csb

0.94506 0.70234 log10 10.35

log10 Csb

1.891

20

Assume

u flood D

0.2

v

2

0.01286

1.0. Then

0.01286

610 236 236

4 V MW V v

0.85,

Csb

0.22618 log10 10.35

u op 3600

236.0 kg m3 , u op

0.0162 ft s , u op ,V

0.75 uflood

100 kmol h , MWv

0.01215

28.02 kg kmol

0.01215 ft s , Need to watch units

267

4 100 28.02

D

=1.155 m

0.85 236.0 0.01215 3600 1 3.2808 ft

3.79 ft

Probably use 4 ft diameter column – (standard size) Using larger diameter helps take into account the uncertainty in extrapolating to find C sb . 10.D.21. New Problem in 3rd edition. The mass balance and flow rate calculations are the same as for problem 10.D14.

D

z xB xD

xB

0.4 0.0001

F

V or

400.7415 , B 1000 D

1000

0.998 0.0001

L D 1 D

V

1202.225

3 400.7415

kmol 1 day

599.258 kmol/day

1202.225 kmol day

1h

0.013915 kmol s day 24 h 3600 s L/V = 2/3. Top of column is close to pure methanol Thus L G L V 0.66667 G is lb (s ft 2 )

V in Pure MeOH boils at 64.5 C L

MeOH

lb

kmol 32.04 kg 2.20462 lb

0.013915

s

s

1.8 R

337.66K

K

0.7914 g cm 3

kmol

1 kg

0.98290 lb s

607.79 R 1 kg 2.2046 lbm 28317 cm 3

1000g

kg

ft

49.405

3

lbm ft 3

59.83 / 49.405 1.211 Assume ideal gas. Top of column is essentially pure MeOH. W

L

1 atm

n MWM

p

V

RT

v

Fv

MWM

G

WG

L

lb lbmol

3

ft atm 0.7302 lbmol R

1/ 2

WL

32.04

.66667

0.07219

0.07219

607.79 R

lbm ft 3

1/ 2

0.02548

49.405

From Fig. 10-25, Ordinate at flooding = 0.20 G flood

G flood Where

methanol

G actual

0.20

G L 0.2

F

gc

1/ 2

, gc

32.2, F

0.20 0.07219 49.405 32.2 110 1.211 0.34

64.5 C

110 Table 10 3 1/ 2

0.2146

0.2

1/ 2

0.4633

lbm s ft 2

0.34 cp from Perry’s (8th ed.) p. 2-449.

0.80 G flood

0.80 0.4633

0.3706

268

V in lb s

Area

0.98290

G actual lb s ft

2

Diameter

0.3706 1/ 2

4 area

2.6521 ft 2

4

2.6521

1/ 2

1.838 ft

Note that this is larger than the calculation of 10.D14 at the bottom of the column. Thus, do calculations at top of column. b.

Diameter Intalox

F = 33,

Dia

plastic

10.E1.

D kmol

op time Then

V

L

1.838

FBerl

Berl

where D 18.1303 kmol ,

D kmol hr D

1/ 4

Fint

V 0.4V

L

2

D

3

1/ 4

33

1.360 ft

110

, L V

0.4

D

0.6V and t op

0.6V Use Fig. 10-25 or Eq. (10-39a) to find flooding at the end of the operation at bottom of column. kg MWliq 2 L lbm s ft L kmol h kmol 2 kg G lbm s ft V kmol h MWvapor kmol At end of operation at bottom of column x 0.004, y 0.036 (pinch)

L G

MWavg,liq L

0.004 46

18.128

0.4

0.381

19.023

0.996 18.016

18.128 & MWvap

0.036 46

0.964 18.016

19.023

62.4 lbm ft 3 (Essentially pure water). Boils at ~ 100ºC = 373 K p MW G

Then

1.0 atm v

RT

FV

19.023 3

atm ft 1.314 K lbmol 1/ 2

L

G

G

L

0.381

0.038806

373 K

0.038806 62.4

lbm lbmol

lbm ft 3

1/ 2

0.009501

From Eq. 10-39a.

log10 ordinate

log10 ordinate G

2

From Table 10-3, F

0.6864 & ordinate

0.2059 F

70,

1.6678 1.085 log10 0.009501

G L 0.2

0.29655 log10 F v

2

0.2059 agrees with Fig. 10 25

gc

1.0, g c

32.2,

water

100 C ~ 0.26 c p (Perry’s 5th ed., 3-213)

269

G flood From Eq. (10-41), V

70 1.0 0.26

s

12

Then 10.F1.

lb l bmol

s ft 2

2

0.19635 ft 2 and G act

0.7 G flood

0.0038646 19.023 lbmol 3600s 0.453593 kmol V 0.0038646 6.310665 kmol h hr s h 1.0 lbmol D 18.1303 kmol t op 4.7883 h 287.3 min. 0.6V 0.6 6.310665 yM 1 x M x M 1 yM

0.134 0.98 0.02 0.866

. From Equil. data

0.979 0.05

7.582 , Top 7.582 2.454

0.95 0.021

(Note 40% MeOH is probably wt%) Estimate: n mix x M n M x W n W Feed is 60% M 40% W n mix 0.60 n 0.28 0.40 mix

4.31 0.306

2.454

4.31

Column temperature varies from 64.5º to ~ 98.2 ºC. 64.5+98.2 Avg T= 81.35 C 2 from Perry’s 7th ed., T = 81.35ºC, p. 2-323. liquids

M

n 0.35

0.28 cp,

1.1837 ,

W

mix

0.35 cp

0.306

1.3195

From O’Connell’s Correlation, Fig. 10-14, Eq. (10-6): E o

s ft 2

s kmol

Geometric avg

Then

lbm

lbm

G

4 0.19635 0.53488 0.7

lbmol

Need average

Bot :

MWvapor

0.53488

0.2

Area ft 2

lbmol

Area V

0.5

0.2059 0.03698 62.4 32.2

0.52782 0.27511 log 10 1.3195

Eo

45% 0.044923 log 10

1.3195

2

49.5%

If conservative use 45% 10.F2.

To use O’Connell’s correlation (Fig. 10-14), need α and viscosity of feed. KM yM x M yM 1 x M . Used Table 2-7 for values. MW K W yW x W x M 1 yM Can estimate a geometric average at bottom, feed & top 0.134 0.98 0.729 0.6 7.582 , MW ,feed x .4 4.035 M W ,bot 0.02 1.866 0.4 0.271

270

0.979 0.05

1/ 3

2.454 ,

4.2184 0.75 0.021 Averages can be calculated many other ways. The feed is saturated liquid. From Table 2-7, T = 75.3ºC Viscosities from Perry’s, p. 2-323, W 0.39cp & MeOH 0.30 Note: (MeOH, 40% probably refers to wt % - p. 2-322 Perry’s) Estimate n mix x1 n 1 x 2 n 2 MW ,top

n

mix

.4

avg

n 0.30

10.F3.

z xB F

F

.6 n 0.39

4.2184 0.351 Overall Plate effic. = 43.7% Then

bot

Feed

T

1.0465 ,

mix

0.351

1.481

0.30 0.01 100

36.71 B F D 100 36.71 63.29 lb mol h xD xB 0.8 0.01 At top of column L = D(L/D) = 73.4 and V = L + D = 110.1 L F F3 L Stripping section L L qF where q 4 3 F L 206.7, V L B 143.3 Feed line has slope 4/3 and goes through y x z .3 . Top operating line has slope L/V = D

0.667 and goes through

y

x

xD

0.8 . Bottom operating line goes through

y x x b 0.01 and the intersection of top operating line and feed line. McCabe-Thiele solution is shown in Figure. Optimum feed is 8th from top. Need 8 7/8 equilibrium stages plus partial reboiler.

271

Overall Efficiency. For O’Connell Correlation, need

yE 1 x E 1 yE x E x

0.019, y

x

0.3273, y

x

.7472, y

AVG

and

Feed

Tcol

. Using Table 2-1 we find α and following mole fraction.

0.170 .981

0.170 :

.830 .019 .5826 .6727

.5826 :

.4174 .3273

.7815 .2528

.7815 :

.2185 .7472 1/ 3

AVG

1

2

10.575 2.87

1.210

3.324

3

Can estimate μ from p. 99 Ethyl Alcohol Handbook at z .3 .523 wt. frac., 0.55 cp. Thus αμ = 1.83. From O’Connell Correlation E o .42 .

N

8.875 .42

Height 22 18

18

21.1 . Thus, need 22 stages plus partial reboiler

disengagement

48 (bottom sump)

38.5 ft

4V MW V

Diameter Calculation Dia v

flooding fraction u flood 3600

Use average values of parameters in stripping and enriching sections.

MW v

MW

eth

stripping section: 18.25

yeth MW

MW

W

y W for both MW V and MW L .

21.5

WL

L MW L

206.75 20

4135 lb h

WV

V MW V

143.46

2869.2 lb h

20

0.96225 g ml 60.07 lb ft 3 Bottom Bottoms T =100ºC = 672ºR L

P MW V

V

Fv

WL WV

V

L

Enriching section:

RT

1.0

4135 2869.2 26.4

MW

20

0.7302 672 R

0.04076 60.07

0.04076 lb ft 3

0.0375, Csb

0.28 .

40.4

WL

L MW L

73.42 35

WV

V MW V

110.13 35

2569.7 3854.55 lb h

0.766 g ml 47.92 lb ft 3 Distillate Distillate T = 82ºC = T = 639.6ºR; L

V

P MW V

RT

1 35

0.7302 639.6 R

0.05472 lb ft 3

272

Fv K

Csb

WL WV 20

0.2

V

2569.7 3854.55

L

0.05472 47.92

0.0225, Csb

0.28

, ft/sec. σ, surface tension in dyne/cm. 57th ed. Hdbk of Physics + Chemistry, F-45. Bottoms, σ ~ 46 dyn/cm, Middle, σ ~ 25 dyn/cm, Top σ ~ 18.6 dyn/cm

u flood

0.20

stripping:

K

0.28 46 20

enriching:

K

0.285 18.6 20

K

L

V

V

0.33075

enriching: u flood

0.2809 Dia.

0.2809 ft s

60.07 0.04076

V

12.693 ft s

0.04076

47.90 0.05472

8.31 ft s

0.05472 4V MW

V

enriching: V MW V

0.20

, ft s

stripping: u flood

stripping: V MW V

0.33075 ft s

0.90

V

0.75 u flood

V

3600

1/ 2

143.46 lbmol h

20

0.04076

lb ft 3

70392.5

110.13 lbmol h

35

0.05472

lb ft 3

70441.37

Diameters: stripping section: Dia = 1.7 ft and enriching section: Dia = 2.1 ft Probably use 2.5 ft diameter since there is little if any cost penalty. 10.F4.

Numbers from solution of Problem 10.F3 are used. V lbmol s MWV lb lbmol Cross Sectional Area G lb (s ft 2 ) Bottom:

L G

L V

Top:

L G

L V MWL MWV

1” metal Pall rings: F

48,

MWL MWV

0.15,

1.44 ~ 1

1.44

0.667 ~ 1

0.667

0.15

ψ = Density of water/density of liquid At Top: 61 47.92 1.27 At 81ºC ,

w

0.35 cp,

E

y= x d

At top

0.8, n

MWv

y MWE

1/ 2 G

L 2

G F G

0.078 F

G

L 0.2

gc

x1 n

MIX

.8 n .45

MIX

1 y MWw

0.05472 lb ft 3 ,

G

L G

0.45 cp , n

L

G

L

gc

V 110.13 mol h

2

.2 n .35 , and 0.8 46

MIX

.2 18

0.43 40.4

1/ 2

0.02254

0.078 from Figure 10-25

0.078 0.05472 47.92 32.2 48 1.27 0.43

x2 n

47.92

0.667 0.05472 47.92 0.2

1

0.2

1/ 2

0.358 lbm (s ft 2 )

0.0306 lbmol s

273

D2 4

Area

D

4V MW v

1/ 2

V MW v , or G

4 0.0306 40.4

G

1/ 2

0.358

2.096

Probably use 2 or 2.5 foot diameter columns. The calculation at the bottom of the column gives a smaller diameter. HETP N Height of Packing, or 1.2 ~ 10

12.0 ft

10.G.1. New Problem in 3rd edition. The result from Wankat (2007a) is listed in the following Table: Results for distillation of vapor feed 5 mole % methanol, 95 mole % water. Distillate is 0.9543 mole fraction methanol and bottoms is 0.9976 mole fraction water. Tray spacing = 0.4572 m. Base case conditions are listed in Tables 1 and 2 in Wankat (2007a). When two trays are listed, they have the same diameters. The decrease in volume and increase in QR are compared to the base case. FL NF,V 0(base) 10

NF,L --

Two-enthalpy feed: 500 11 6 500 12 6 600 12 6 750 13 6 750 16 7 1000 (all liquid) 9

N 20

dia A Vol tray QR 2.84 6.33 55.0 2 1065

Qc, total -12,330

decr Vol --

20 22 20 20 26 20

2.08 2.07 1.89 1.56 1.56 1.20

-12,340 -12,330 -12340 -12,360 -12,330 -13,680

46.5 % 0.5 % 40.9 % 0 55.8 % 0.5 % 69.6 % 2.3 % 60.3 % 0 82.1% 127 %

3.38 3.38 2.80 1.92 1.91 1.13

29.4 32.5 24.3 16.7 21.8 9.8

2 2 2 2 2 2

1070 1065 1070 1090 1065 2415

Intermediate condenser: FWithdr NF,V NV,with NL,ret N dia A Vol tray 300 11 10 6 20 2.41 4.56 39.6 2 450 11 10 6 20 2.22 3.88 33.7 10/11

QR 1067 1067

Qc, total -12,340 -12,340

Two-enthalpy feed (FL = 600 kmol/hr) plus one intermediate condenser: Fwithdr NF,V NF,L NV,with NL,ret N dia A Vol tray QR,total Qc 100 12 6 5 5 20 1.69 2.23 19.4 2 1073 -12,340 80 12 6 13 6 20 1.72 2.33 20.2 2 1079 -12,345

incr QR --

decr Vol incr QR 28.0 % 0.2 % 38.6 % 0.2 % decr Vol incr QR 64.7 % 0.8 % 63.2 % 1.3 %

Two-enthalpy feed (FL=600 kmol/hr, NF,V =12, NF,L=6, N=20) plus two intermediate condensers: Fwthd1 NVwth1 NLret1 Fwthd2 NVwth2 NLret2 dia A Vol tray QR,total Qc decr Vol incr QR 100 5 4 80 13 8 1.50 1.77 15.4 2/6 1081 -12,350 72.1 % 1.6 % Two-enthalpy feed (FL = 680 kmol/hr) plus one intermediate condenser: Fwithdr NF,V NF,L NV,with NL,ret N dia A Vol tray QR,total Qc 100 12 6 5 5 20 1.51 1.79 15.5 6 1081 -12,350

decr Vol incr QR 71.7 % 1.6 %

With constant Qc and QR, the two-enthalpy feed with FV = 750 and N = 26 appears to be best. 10.G.2. New Problem in 3rd edition. Results are from Wankat, P. C., "Balancing Diameters of Distillation Column with Vapor Feeds," Ind. Engr Chem. Research, 46, 8813-8826 (2007).

274

Table 1. Simulation conditions and results for base cases. F 1000 kmol / h, D = distillate flow rate kmol/h, N = number of trays + condenser + reboiler, tray spacing = 18 inches = 0.4572 m, operation at 80% of flooding, dia = maximum calculated diameter in m, tray = tray at which max diameter occurred. A = max calculated column area in m 2 , Vol = column volume in m 3 = A(N – 2 +1) (tray spacing) where N – 2 is the number of trays and +1 is for disengagement space for liquid and vapor, Q R and Q c = reboiler and condenser duties in kW, p cond condenser pressure

p pressure drop in psi/tray, N feed = optimum feed stage, and condenser is stage 1. Note this solution has a p, each stage. Thus, solution slightly different than students’ solutions. in atm,

N Feed N Dia A Vol tray Q R Q c , L1 D D p cond Ethanol (10 mole %). Water (90 mole%) Vapor Feed Base Case: 23 26 2.61 5.35 85.6 2 902 -10,700 8.0 125 1.80

p 0.1

Table 2. Diameters calculated for standard distillation base cases listed in Table 1. Vapor flow rate Vj and liquid flow rate L j are in kmol/hr, diameter is in m, area is m 2 .

Tray 2 23 24 35

Ethanol-water, vapor Vj Lj Dia 1125 997 2.61 1082 949 1.99 74.3 950 0.71 79.4 956 0.71

Area 5.35 3.11 0.396 0.396

Table 3. Simulation conditions and results for a distillation column separating a vapor 10 mole % ethanol, 90 % water feed (see Table 1 for base conditions). Partial condenser is stage 1. N F,V and N F,L are optimum feed locations for vapor and liquid portions of the feed, respectively. Decrease in column volume Vol (equal to change in area when the number of stages is unchanged) and increases in Q R are compared to the ethanol-water base case (Table 1). For both runs y D,E N Dia A Vol Tray Q R N F,V N F,L FL 0(base)

FL 600 Qc,total

23

--

36

2.61

2.24

35.8

2

902

N N F,V N F,L 23 17 36 Qc,col Qc,feed condenser

Dia

A

Vol

Tray

1.69

2.24

35.8

2

QR 902

0.7901 and x B,W

0.9986.

Q C,col -10,700

Q C,tot -10,700

Decr Vol 58.2%

Incr Q R 0

10.G3. New Problem in 3rd edition. Part a. S Dia = 2.032 m. Distillate mole fractions (vapor) = 0.22222 Eth, 0.77765 Propane, 0.12383 E-03 B, and 0.28503 E-9 pentane. Bottoms mole fractions = 0.14843 E10 Eth, 0.10132 E-03 Propane, 0.81808 Butane and 0.18182 pentane. Other values are in Table for 10.G4. Part b. Worst backup is 0.232 m on stages 30 and 31. Maximum weir loading is 0.0204 m2/s on plate 31. Part c. Same mole fractions, same Qc and QR. Max backup 0.1614 m in panel A on stages 29 to 32. Maximum weir loading is 0.01183 m2/s on plate 31 of panel A which is acceptable. 10.G4. New Problem in 3rd edition.

275

V feed L feed Qc kW QR kW Max Dia Stage yD,C4 xB,C3 Kmol/s Kmol/s m max dia 0* 1 pass .1(NF=16) -1463 2827 2.032 31 .000124 .000101 0 part d .1(NF=15) -1463 2827 2.032 30 .000115 .0000942 0* 2 pass .1(NF = 16) -1463 2827 2.032 31 .000124 .000101 .01 .09 -1463 2600 1.956 31 .000145 .000119 .02 .08 -1463 2373 1.876 31 .000199 .000163 .03 .07 -1463 2147 1.792 31 .000300 .000246 .04 .06 -1464 1921 1.905 32 .000525 .000429 .05 .05 -1466 1696 1.650 18 .00112 .000915 .06 .04 -1472 1474 1.600 18 .003188 .002609 b. Change N=41 NF,liq=18 NFvap=21 N .03 .07 -1463 2146 1.793 34 .0000817 .0000668 * Values from problem 10.G3. Part c. Tray rating program with Dia = 1.793 m and defaults for tray spacing (0.6096m) & for DC clearance (0.0373m) obtain 0.2207 m backup on tray 34, which is acceptable. Maximum weir loading is 0.01887 m2/s on tray 34 which is acceptable. Part d. Shown above, plus maximum backup is 0.2320 on plate 31 (acceptable) and maximum weir loading is 0.0204 m2/s on plate 31, which is marginal.

276

SPE 3rd Edition Solution Manual Chapter 11 New Problems and new solutions are listed as new immediately after the solution number. These new problems are: 11.A19-11.A22,11.B6, 11.C3, 11.D5,11.D8, 11.G3-11.G4. 11.A3. As feed temperature increases

L D

MIN

increases, hence L/V increases and Q c increases.

L V increases and Q R decreases. The number of stages probably decreases. (See Figure 7-3. The abscissa increases as L D

min

and L/D increase. Thus ordinate drops.

Since N MIN is constant, N decreases). 11.A.6. New Problem for 3rd edition. b. A liquid side-stream between the feed and the distillate. c. A vapor side-stream between the feed and the bottoms. 11.A.9.

1 point for b. 2 points for c.

If the feed rate is consistently one half the design capacity, the entire economy of scale will be lost. In addition, distillation columns probably operate at lower than design efficiencies.

11.A11. Column 2 would have to be at a lower pressure than column 1. 11.A.12. New Problem for 3rd edition. The heuristics to not do would include items such as: 1. Remove dangerous, corrosive, and reactive components last. 2. Use distillation only for very difficult separations (α < 1.10). And so forth. 0.33 X,T 11.B4. Use heuristics. Hardest separation is xylene-cumene, x ,c 1.57 0.21 C,T Toulene is most abundant. Use heuristics to: Remove toluene early Do hardest separation last One-by-one in overhead If use equil-molal splits: B

T

X

F C Can also derive coupled systems.

277

B

X

F T C

11.B5.

Add in Heuristic 9 for sloppy separation. T

T

B

B X

F

OR

F

X C

C

11.C.1. New Problem for 3rd edition. Take the log of both sides of Eq. (11-2). log (cost A/cost B) = (exponent) log (size A/size B) log (cost A) – log (cost B) = (exponent) [log (size A) – log (size B)] exponent = [log (cost A) – log (cost B)]/ [log (size A) – log (size B)] 11.D1.

NOTE: This solution requires the solution to Problem 10-D1. Estimate α at feed composition x = z = 0.5 1 K c7 a&b. For binary, x c6 , y c6 K c6 x c6 . Use Fig. 2-11. K c6 K c7 T = 168ºC: K c6 1.29, K c7 0.71, x c6 0.50, y c6 (Guess was aided by solution to Problem 10-D1) .645 .5 1.817 .355 .5

n N MIN

L V

min

0.645

.999 .001 .001 .999 23.13 stages n 1.817 x D y 99 .645 0.7094 x D z .999 .5

278

L

L V

D c.

L D

Gilliland: x

d.

L D

2.441 min

4 2.441

MIN

L D 1

0.3703 23.13

N 1

0.37027

1 0.3703

Fig. 11-1.

D2

5.91 m 2 . Column Vol

4

cos t

$700 m 3 ,

7 bar, Eq (11-5), Fp

2.25 1.82 2.3 1.0

Qc

$844, 000 .5

10.71 0.00756 7

2.30

$ 209,800

4196 50 1 1

total

$1,054,000

500, B 500, L V

Vh D gives, Q c

$4196 tray

7 2.743

.5

L VH v

$131,110

710 5.91 m 2

C P, tray

10.71 0.00756p

69ºC (bp) = 156.2ºF. F 1000, D Energy Bal:

187.3 m 3

3

pD

C BM.tray

Fig.11-9.

m

187.3 m 3

area H

$700

C P,tower

vol Tray cos t Fig. 11-2. $710 m 2 , area Eq. 11-8. CBM,tower 131,110

11.D2.

37.3 contacts or 36.3 stages

N act N eq E o 36.3 0.73 49.7 or 50 stages 24” tray spacing × 50 trays = 100 ft + 4 disengagement H = 104 ft = 31.7 m m D 9 ft 2.743 m 3.2808 ft Tray area

700 kPa

0.3117

5

N N MIN

Eq. (7-42b):

N

1 L V

min

0.8, L D

2000 lbmol h, V

V ho

4 2500

Hv

h D is pure boiling hexane, H v is vapor. Thus, h D

Hv

Qc

2500

Pick 25ºC as basis.

hD

3.39 107 Btu h

13,572

QR

C PL ,c 6 69 25

hB

13,572 Btu/lbmol

c6

Dh D

Bh B

1.8 F C

C PL,c7 98.4 25 1.8

Fh F 51.7

QC 44 1.8

4094.6

50.8 73.4 1.8

Feed is a saturated liquid. From Example 10-1, T = 80ºC

6711.69

Btu lbmol

279

hF

CPL,c6 z c6

CPL,c7 z c7

80 25 1.8

hF

51.7 .5

50.8 .5

55 1.8

QR

500 4094.64

500 6711.69

A

3.423 10 7 Btu h

Q U

1000 5073.75

T

Btu 110 98.4 C h ft 2 F where U is average from Table 11-2. Condenser:

50

U

TAvg

32, 800 ft 2

1.8 F C

3.39 10 7

Qc

A

3.39 10 7

3.423 107 Btu h

QR Reboiler:

5073.75 Btu lbmol

2850 ft 2

110 70 2 Note these areas are very approximate. For detailed design need a much better estimate of U.

A

Costs: Condenser

180 156.2

2850 ft

2

1m

2

3.2808 ft

264.8 m 2

2

Fixed Tube Sheet S&T Fig. 11-3, Cost = $125/m2 Reboiler A

32,800 ft

2

1m

2

3.2808 ft

2

3047 m 2

large because of low U.

Extrapolate $70/m2 1 atm, Fp

Condenser Reboiler 11.D3.

1.0 . Eq. (11-9) CBM

1.0, Fm

C BM

3.29

$125

264.8 m 2

m2

3.29 $70 m 2

C BM

Cp 1.63 1.66 Fm FQ

3.29 C p

$109, 000

$702, 000

304 m 2

$811, 000

Very sensitive to U.

Note: This solution requires the solution to Problem 11-D2. lb Q R 3.423 10 7 lb Steam rate, 35, 704.6 h 958.7 h where Q R is from Solution to problem 11-D2. Steam Cost

$

35, 704.6

h

Cooling water,

lb

$20.00

h

1000 lb

lb

Qc

h

C p w Tw

3.39 10 7 1.0

40

$714 h. 847, 500

lb h

where Q R is from Solution to problem 11-D2. Water Cost

$ h

847, 500

lb

$3.00

1

h

1000 gal

8.3 lb gal

$306 h

280

11.D4.

From Example 11-1 needed 21.09 eq stages + P.R. h pack 21.09 1.1 ft stage 23.2 ft 7.07 m

D2

Vol

4

2

h pack

4

15 23.2

4099.6 ft

3

4099.6 ft

1m

3

3

3.2808 ft

3

116.1 m 3

Cp ~ $250 m3 packing Tower 23.2 ft + 2 ft between sections + 2 ft top + 2 ft top = 29.2 ft 1m h = 29.2 ft 8.9 m , Vessel Vol. 146.1 m 3 3.2808 ft Fig. 11.1 Cp $700 m3 for tower From Fig. 11-2

Fp

1.0 1 atm , Tower FM

1 carbon steel , C BM

C p 2.25 1.82

$700 146.1 4.07 $416, 312

Packing, Fm

4.1. C BM

C p 1.63 1.66 4.1

250 116.1 8.436

244, 855

Total

$661,000

Does not include cost distributors, supports, hold down plates, etc. 11.D.5. New Problem for 3rd edition. n = [log (cost A) – log (cost B)]/[ log (size A) – log (size B)] Let size A = 10 m2 and size B = 1.0 m2. The cost A = $400/m2 = ($400/m2)(10 m2) = 4000, and cost B = $2100. n = [log (4000) – log (2100)]/[log (10) – log (1)] = [3.602 – 3.322]/[1 - 0] = 0.28 11.D6. See residue curves in Figure. The recycle is pure MB. Mixing point is determined in same way as in Fig. 11-11. Now mixing point splits into light (L) component methanol on B1 . Thus line LM is extended to 0.0 mole fraction methanol to find location of B1 (0.73333 MB and 0.26667 toluene). We can use mass balance to find point B1 accurately. If D1 is pure methanol, D1 = 50 (all methanol in feed) and B1 = 150. Then from toluene balance 0.2 × 200 = 150 x tol,B1 , which gives x tol,B1 0.266667 . B1 which is toluene product. F Re cycle

F2 which is then split into I (Some of which is recycled) and B 2 100 100

M

Since D1 contains no toluene B1 0.26667

B1

F .4

D1

Re cycle 0

.4

100 150 kmo h. F2 , D1 M B1 0.26667 For Column 2: B 2 essentially pure toluene, B2 0.266667 F2 B1

OK – Satisfies overall external M.B. D 2 Re cycle F2

D2

B2

50.0 kmol h

0.26667 150

40

150 40 110

10.0, which also satisfies external M.B.

281

11-D7. a.)

Proposed Split: Bottoms – Essentially pure toluene Distillate ~ .83 methanol. (See figure for Solution to 11.D7) F 100, z M 0.5, z MB .1, z T .4 Assume all toluene in bottoms & bottoms is pure. B 100 .4 40, D 60

60 x M,dist b.)

50 , x M,dist

0.8333 & xMB,dist

0.166667

Proposed Split. Distillate pure M Bottoms 0.2 MB, 0.8 T (See figure). D 100 .5 50, B 50 Note – Doubtful this will work.

282

11.D.8. New Problem for 3rd edition. Part a. Cost in June 2010 $947, 000 556 397

Part b.

F

$1,326, 000

2 x F of Example 11-2.

Since Dia

Tower

F , Dia 2F11

12 ft

2

1m

Diameter

17

Volume

488.2m3

CP,tower

$550 m3

3.2808 ft

2

16.97 or 17 feet.

5.182m, Tray Area

Fig. 11-1, C0p 488.2 m3

21.08m 2 .

Cost vol ~ $550 m3

$268,500

Fig. 11-2. Tray cos t area ~ $750 m3 extrapolate , Cp,tray CBM,tower

$750 m 2

21.08m2

268,500 2.25 1.82 1.0 1.0

$15,800 tray

$1, 093, 000 283

CBM,trays

$15,800 36 1.0 1.0

$569, 000

Total bare module cost September 2001 = $1,662,000 In June 2010, Cost $1, 662, 000 556 397 $2,328, 000 Part c.

Original feed rate 1000 lbmol h . At Foriginal , cost lbmol $1326 lbmol At

11.G1.

2 x Foriginal , cost lbmol

$1164 lbmol

- Use Fig. 11-10b as flowsheet. Use NRTL. Feed : 1 atm, sat'd Liq, 100 kmol/h, 0.5 MeOH, 0.4 T, 0.1 MB Fed to Stage 30 Int. Recycle sat’d Liq, 100 kmol/h, 100% MB, fed to Stage 20 Block B1 : N = 46, total condenser, partial reboiler, D = 50, L/D = 3

Dist : 0.999759 MeOH, 0.000241 tol, 1.537 E 0.8 MB Bot : 8.02 E 0.5 MeOH, 0.266586 tol, 0.73333 MB Block B2: N = 85, L/D = 9, B = 40, feed = 41 Dist: 0.0001094 MeOH, 0.0008684 Tol, 0.999022 MB Bot : 2.855 E 35 MeOH, 0.99731 tol, 0.0026888 MB Thus, this is feasible. 11.G2.

a)

For Fig. 11-10A.

Col 1. N

90, N f

41 ,

L D

8 D

60

Bottom 0.999186 tol 0.000814 MB

Dist. 0.99938 tol, 0.000605 MB, Col 2. N

20, N f

10 , L D

2, B

10 Bot .996679 MB, 0.003088 MeOH, 0.000233 toluene

If increase L/D in column 1, Col 1. L D

Col 2. L D

2, N

24, N f

9, N

90, N f

41, Bot. 0.99941 toluene.

12, Dist. 0.99954 MeOH, Bot. 0.9976 MB

Which now meets specifications. Thus Figure 11-10a without recycle is feasible. b.)

For Fig. 11-10B – converged N = 30, L/D = 6 Dist. Col 1. 88.7% M & 11.3% T – No MB (azeotrope) Would not converge higher N. Does not appear to work; thus, not feasible.

11.G.3. New Problem for 3rd edition. F 100, 10% Ethanol, 5.0 atm, Sat’d liquid feed N = 10 includes partial reboiler, total condenser, D = 10, L D

2

284

P=1

NF = 5 NF = 6 NF = 7 NF = 8 NF = 9 NF = 10

Pcol = 3 atm

Pcol = 5 atm

L D 2 NF = 8 NF = 9 NF = 10 L D 2 NF = 8 NF = 9 NF = 10

QR = 52,948 QR = 53,029 53,148 53,175 52896.9 cal/s

QR = 67995.5 QR = 68022.1 67794

QC = -78614.5

Qc = - 74274

QR = 76435.3

7646 Qc

76264

71692.8

xD,E = .72033 .74564 .76332 .77539 .78037 .69964

.75453 0.76027 .69695

.74276 .74840 .69411

xB,E = .031085 .028263 .026298 .024957 .024363← .033373

0.027275 0.026637← .033672

.028582 .027955 ← 0.033988

Size optimal feed columns. Sieve tray 1 uses 18 inch tray spacing at 85% approach flooding, Fair calculation method for flooding. Pcol 5 Max diameter tray 2 0.34867m

Pcol

3

Max diameter tray 2 0.38070m

Pcol

1

Max diameter tray 2 0.46429m

Part d. D1. 1.0 atm gives the best separation because the relative volatility is highest. D2. The lowest Qreboiler is 1.0 atm. The effect of pressure on Qreboiler in this problem occurs because the feed is always a saturated liquid at 5.0 atm. For the 5 atm column this feed remains a saturated liquid and the feed line is vertical. At lower column pressures the feed flashes and is a twophase feed in the column. These feed lines have a negative slope. For the feed lines at lower pressures the slopes of the bottom operating lines are steeper, which means lower boilup ratios, Vreboiler/B. This means lower Qreboiler at the lower pressures. Another way to think about this is the flashing feed produces vapor and thus less vapor is required from the reboiler. D3. The lowest absolute value of Qcondenser is 5 atm. All columns have the same D and L/D. Thus, V entering the condenser is the same. At higher pressures the latent heat of vaporization λ is lower. Since Qcondenser = Vλ, the result is a lower absolute value of Qcondenser at the higher pressures. D4. The smallest diameter column is at 5 atm. Vapor density is highest. Part e. Increasing pressure above 1 atm for the same purity requires more stages, but smaller column diameter. Thus capital cost initially goes down. Above 8 atm the column must be designed for high pressure operation, which makes it more expensive. Operating cost may go up if a higher L/D is required to achieve the desired purity. 11.G.4. New Problem for 3rd edition. a. (L/D)min = 1.3962 → L/D = 1.5358. Obtain N (Aspen Notation) = 19 with Nfeed = 9 (on stage). Distillate is 0.75056 ethanol and bottoms is 0.00005987 mole fraction ethanol. Q R = 569,172 cal/s, 285

Qc = - 443187 cal/s, Dia = 1.0755 m, A = 0.90843 m2, H = 11.5824 m, Vol = 10.52 m3 Part b. Note that balancing the flow rates to achieve the desired separation in each column is trial and error. The easiest was to do this is to first find a flow rate that works for the high pressure column (note that D changes every time the flow rate is changed). This gives a value for Q C,high pressure = - QR,low pressure. Then design the low pressure column with this QR. Check that both columns work. The correct flow rate into the high pressure column is between 570.6 and 577.275. The results are reported below with the latter value: High pressure column: F = 577.275 kmol/h, QR = 0.7555(569172) = 430,000 cal/s. N = 20, Nfeed = 9, distillate is 0.75004 mole fraction ethanol and bottoms is 0.00009733 mole fraction ethanol. Q c = 243,900 cal/s, L/D = 1.5618, T cond = 382.16 K, Dia = 0.79443 m, A = 0.49568, H = 12.192 m, Vol = 6.043 m3. Low pressure column: F = 422.725 kmol/h, QR = 243,900 cal/s, N = 19, Nfeed = 9, distillate is 0.75087 mole fraction ethanol and bottoms is 0.00003829 mole fraction ethanol. Qc = -190,627 cal/s, L/D = 1.5803, Tcond = 351.56 K, Treb =373.16 K, Dia = 0.7014 m, A = 0.3864, H = 11.5824 m, Vol = 4.475 m3. Part c. Energy requirement of multieffect distillation system was set at 75.55 % of the single column. This is not optimized, but was set because it was known to work. Cooling is only in the low pressure column of the multieffect system, and is significantly less than with only one column. The total volume of columns is the same; however, this is misleading because volume in the single column would have been less if it was designed at 3.0 atm. The capital cost will be higher for building two columns than one larger one, but energy costs are less.

286

SPE 3rd Edition Solution Manual Chapter 12 New Problems and new solutions are listed as new immediately after the solution number. These new problems are: 12.A5, 12.A6, 12.C2, 12.D1, 12.D3, 12D8, 12D13, 12D19, 12D21, 12D22, 12.G3. 12.A1.

By raising T or dropping p can make gas desorb. The direction of transfer of solute controls whether a column is a stripper or absorber. If operating line (on Y vs X) is below equilibrium have a stripper.

12.A5.

New Problem in 3rd edition. c. AspenPlus

12.A6.

New Problem in 3rd edition. A. a. B. d. C. e. D. h. E. i

12.B1.

Calculate: N or y out , or N for A and y out for B, or m, or N for A and m for B, or L/V, or feed composition, or b, or m and b from 2 experiments, or Overall Efficiency.

12.B2.

Two feeds, Sidestream, Reboiled absorber, Coupled absorber and stripper (see Figure 12-2), Interstage coolers (absorption) or heaters (strippers), Packed columns, Two different solvents, Two different stripping gases, Add solid adsorbed to solvent (see Chapt. 14). Cross-flow, Co-current flow, Combinations of flow patterns, etc.

12.B3.

See Isom and Rogers (1994).

12.C4.

Eo

Eq 12 22

N equil N actual n 1 E mV

Eo

(10-1)

Eq 12 34 mV L

1

n ( L mV)

n L mV

n 1 E mV Eo

mV 1 L

n 1 E mV

mV 1 L

(10-4)

n mV L QED

Ref. Lacks, D.J., Chem. Engr. Educ., 302 (Fall 1998).

y

12.C5.

12.C6.

x 1

1 x

When

x

0,

When

x

1,

Apply Kremser as

x

Graphically,

L V

dx

dy

, b

dx dy

1

dx 0 or x

yN MIN

dy

, m

xN

1

, b

1

1 x

2

0

1

1.0 y1

x0

where y N

1

& x N are in equilibrium

287

yN L

m

For absorbers

mV

min

N

Thus as Eq. (12-23) becomes

yN yN

or

mV 1

y1

1

* 1

m

y

p tot

x

0.395

y1

L

x 0 .0001

H

0 min

L

y

3 b 0

x

yN

V

1

xN

min

b

y1

L

x0

V

min

0.395 x

0.0002 unknown

unknown

1

L x0 y1*

mx 0

min

L x0

1.186

where y1*

mV

y1

1

yN 1 b m m which agrees with graphical analysis.

12.D1. New Problem 3rd Edition.

min

L

,

yN

b for linear.

L

&1

V

mx N

1

Gy N

m x0

1

Gy1

L xN

.395 0.0001 0.0000395

N

xN

L

yN

G

1

0.004 Since x N unknown,

unknown

5

1

y N 1 y1* y1 y1*

mV L n

n 5

1

39.5 L

1

unknown.

Eq. (12-30) or (12-31) easiest to use

n Eq. 12-30

y*N

0.004 0.0000395 0.0002 0.0000395 L n 39.5

mV L

mV L 39.5 L

n

1

39.5 L n

24.67

39.5 L

L 39.5

288

T & E → be sure L mV

L 50 75 65 63 62 xN

G

yN

L

1

mV/L 0.79 0.52668 0.60769 0.62698 0.637097 y1

1

L/mV RHS 1.2658 7.58 1.8987 3.90 1.64557 4.67 1.59494 4.895 1.56962 5.015 100 0.004 0.0002 62

x0

OK 0.0001 0.006229

There are alternative solution paths, but L = 39.5 is not valid, it becomes Alternative: Trial-and-error McCabe-Thiele solution. 12.D2.

a)

n1

0

n1

0

.

L & V constant. 97% rec. H2S 3% left in gas. y out 0.03 0.0012 0.000036 Equil.:

y

p H 2S

423x

p tot

2.5

169.2x

M.B. with Const. L & V: 0.0012 V

x out b) See Figure. y c)

L V

0.000036 V

0.0012 0.000036 x

L V

y in

V

x out L

0.001164

L

10

5.82E 6

2000

x out where y in , x out & y out , x in are on op. line .

Eq

yin

169.2

0.000012

LG

x*

7.09E 6

Note that L G min can be calculated from a crude sketch.

min

y

L G y out

L

0.0000036 0

200

G act 0

0.0012 0.000036 7.09 E 6

min

M

164.124

L G

min

M = 1.2186

x

289

d) L/G too high – L too high – Liquid too dilute. Need much better solvent (e.g., MEA solution).

12.D3. New Problem 3rd Edition. Mass Balance:

Vy IN

Lx IN

y out V Lx , Equil., y

Substitute equil. into M.B. HVx HVx Lx IN Lx p tot p tot Solve p tot b.

y out

H p tot

x out

x 10.96 0.274

100 2 10 0.4 10

6

p tot

x

x 6

10.96 10 0.4 10

HVx L x IN

L x IN

H

6

0.4 10

16.0 10

6

0.274 atm

6

Can also do graphically, with Kremser equation (trial and error) or by solving mass balance first, L y out x in x out 16 10 6 Then p tot H x out yout 0.274 atm. V 12.D4.

Mass bal.

yin Vin

Lin x in

y out V Lx out

290

y out

y in

L V

x in

Op. Eq.,

point yin , x out line. See graph.

L V

x out

L V

x in

y

x out

L

x

12 0.0002 0.00001 y in

0.00228

L

x out V V 0.0, 0.00001 and point y out , x in

0.00228, 0.0002 are on op.

Can also use Kremser equation for this problem.

291

12.D5.

Since equilibrium is linear in weight ratio units can do either McCabe-Thiele or Kremser solutions. For McCabe-Thiele solution know that points Yout , Xin and Yin , X out are on operating line. McCabe-Thiele diagram is shown in Figure. N = 5.9 stages. HETP = 10/.59 = 1.69 ft.

Kremser: Several different forms can be used. We will illustrate with Eq. (12-26) written in ratio units. * YN 1 Yin 0.02, YN+1 mX out 1.5 0.06 0.09

Y1 N

Yout

0.50, Y1*

mX in

n

.02 .09

.5 .6

n

.02 .5

.09 .6

HETP

1.5 .40

0.60

5.88

h N 10 5.88 1.70 feet

12.D6. First, assume Nitrogen is an ideal gas: 1 lbmol = 359 cu ft at 0˚C and 1 atm 333.16 359 437 cu ft/lbmol at 60˚C 273.16 N 2 flow rate 2500 437 5.72 lbmol/h Water flow rate: Ignore CO 2 in water. MWw

100, 000 18

18

5560 lbmol/h

292

Equilibrium at 60˚C: H = 3410, y =

L

H PTOT

x

3410x

5560

972.0 . External mass balance is: 5.72 L L L y out x in y in x out x in x out 972. 9 10 6 0.00875 V V V Can solve with either McCabe-Thiele diagram or with Kremser Eq. (12-34). The McCabe-Thiele solution is shown in the figure. Note different scales on axes. Need 5 real stages. V

Kremser: y N

1

0, y1

0.00875, y1* n

mx 0

3410 9.2 10

0.03137 , m V L =3.508

0 0.03137 0.00875 0.03137

1 3.508

N

6

n 1

3.508 5.07

.4 3.508 1

Probably use 6 real stages.

G Wt

12.D7.

0.828 wt frac air 1050

G mole

Yin

Inlet

869.4 29

y in

kg gas h

29.98 kmol air/h

0.172

0.172

1 y in

1 0.172 .828 29 Yin ,molar 0.2077 0.3543 17 NH3 0.172 1050 180.6 kg NH 3 2% remains in gas

869.4 kg air/h

0.2077

kg NH 3 kg air

3.612 kg NH 3

293

3.612 17

Yout mole

Equilibrium data Table 12-2 - y mole

X wt

kgNH 3 kgW

0.05 0.075 0.100 0.15 0.20 0.25 0.30 0.40 0.50 0.60 M.B.

Ymole

0.007087

869.4 29

p NH 3

p, mmHg

p, mmHg

PTOT

1.30 760

988

p

y mole frac

11.2 17.7 25.1 42.7 64 89.5 119 190 275 380

0.01134 0.017915 0.0254 0.04322 0.06478 0.090587 0.12044 0.19231 0.27834 0.3846

X in L

G mol Ymol

17

L wt 17 G mole

X wt

y

Ymole

1 y

0.011466 0.01824 0.02607 0.04517 0.06926 0.09961 0.13694 0.2381 0.3857 0.6250

LX

wt

GYout ,

kg NH 3 17 kmol NH 3

Yout

Note: Units, although mixed, work in Mass Balance & in Operating equation. See graph. Minimum Solvent

Slope

L wt,min

0.3543 0.007087 0.477 0

= 0.7279=

.7279 17 29.98

Actual Solvent, L wt Op Line Slope

1.5 L min

L wt ,min 17 G mole

370.99 kg W h

556.48

L wt

556.48

17 G mole

17 29.98

kg W h

1.092

294

295

12.D8. New Problem 3rd Edition. X IN

y out

0

0.002

L

Yout

p tot

0.002004

2 atm

760. 2 mmHg

1520 mmHg

L G

G

L

F2

0.5 .475 air 0.05 mole frac HCl

.05

y

0.05263

.95

mol HCl mol air

X out

F1 1.0, G=.8 y1IN

0.20 mole frac HCl

YIN

.2 .8

mol HCl

=0.25

mol air

Assume Water (not total liquid) flow rate is constant in both sections. Assume air flow rate (not total gas) constant is each section. In bottom section G 0.8 mol air h . In top section

G

0.8 0.5 .95

1.275 mol air h. Keep X as kg HCl kg water (from equil. data).

Convert p to y (mole fraction) to Y (mole ratios). kg HCl y X p kg water

p p tot

Y

kg HCl

y

kmol air

1 y

0

0

0

0

0.0870

0.000583

3.8355 E-7

3.8355E-7

0.1905

0.016

0.00001053

0.00001053

0.316

0.43

0.0002829

0.0002830

0.47

11.8

0.007763

0.0078240.

0.563

56.4

0.037105

0.038535

0.667

233

.15329

.18104

0.786

840

.55263

1.2353

296

Y vs X equilibrium data is curved. Using these units, we need L in kg water hr and G and G in kmol air hr , and we need to convert the X terms from kg HCl/h to kmol HCl/h. Top Operating Line

Y

L G(MWHCl )

(L / MWHCl )X GYout , X IN

GY (L / MWHCl )X IN

L

X Yout

0.002004 Goes through (0, 0.002004)

1.275(MWHCl )

Bottom Operating Line GY (L / MWHCl )X out

L

0

GYin

(L / MWHCl )X

L

L 0.8MWHCl 0.25 L 0.8MWHCl X out X out G(MWHCl ) G(MWHCl ) a) Feed line saturated gas at Y 0.05263. Two operating lines intersect at feed line. Y

X YIN

Y1,IN

0.25

X *out

Sketch for Min L determination

Y

0.002 *

0.69 (see figure)

L (G(MW)) L MIN

(G(MWHCl ))

X *OUT

YFIN

0.8(MW) 0.69

L

Yout

(G(MWHCl ))

G(MWHCl ) YF2

YF2

YF2

L

G G

YF2

.25 0.05263

X *OUT

YOUT

G(MWHCl )

X *OUT

X int er sec t

(G(MWHCl )) L

Bottom operating line: YF2

L

0.05263

X

Top operating line:

YF2

YF2

X int er sec t L /(G(MWHCl ))

Yout

From plot X OUT

Solve for X int er sec t

L (G (MWHCl ))

YOUT

X int er sec t

L.

YF2

YFIN

Then

YOUT

L MIN

0

YFIN

G(MW) X

* OUT

YFIN

YF2

G(MW) X *OUT

YF2

YOUT

1.275(MW)

L MIN / MWHCl

0.05263 0.00200 0.69 0.22883 0.09355 .3224 kg water h

Since MWmin = 36.461, Lmin = 11.755 kg/h, L = 1.2407 Lmin = 14.584 kg/h. L/(MW)HCl = 0.40

297

298

b. M.B.

F2 YF2 X out

X out Top Bottom

GYIN

(L / MWHCl )X IN

F2 YF2

GYIN

GYout

(L(MWHCl ))

0.475 0.05263

GYout

(L / MWHCl )X out

X IN

0.8 0.25

1.275 0.002

14.584 / 36.461

(L / MW)

0.4

G (L / MW)

1.275 0.4

G

Top goes through

0.8

0

0.5561 kg HCl kg water

0.3137 0.5

X IN

0, Yout

0.002, Slope

(L/MWHCl )

G Arbitrary point for plotting: X .4, Y .3137 .4 0.1255 Y .1255 .002 0.1275 Bottom from X out 0.5561, YIN 0.25 To intersection Top Operating and Feed Line. Need ~ 1.6 stages. Opt. Feed for F2 is Stage 1 (Feed 1 is at bottom.) Check Slope bottom

.25 0 .5561 .06

0.3137

0.504 0K.

299

Figure for 12.D8.

300

12.D9.

Repeat 12.D2 with Kremser.

y

y1*

mx

m x0 n

Eq. (12-22)

m

b

169.2, b

0, x 0

mV

0,

169.2

L 0 , y1

x in

0.0012 0 0.000036 0

1 0.846

N

2000

0.846 , L V

200 y out 0.000036, y N

1

200 10 0.0012

0.846 10.69

n 1 0.846

Graphical solution was 10.4. Pretty close! 12.D10.

Use Kremser equation such as y A out y A* 1 L mV out

y A in

N

y A*

out

4, m 1.414, L V

L mV

.65, y Ain

L mV Equation becomes:

1

y Aout

0, y A*

12.D11.

V L

y out

V L

m x Ain

out

1.414 .02

.02828

.65 1.414 .459 .02828

Overall mass balance: yin V L x in

x out

N 1

y in

x in

.02828 .552

.01267

y out V L x out 1 .65

.01267

0 .02

4.93 10

4

Any of the vapor forms of Kremser equation can be used but problem is trial and error. For example, use Eq. (12-21) inverted for L m V 1

L mV becomes, .27 N 1 y N 1 y1* L 1 mV Set up table and try values of m. y1

y1*

1

1 1

1.2 m 1.2 m

5

m 1.0 1.2 1.3 1.4 1.41 1.415 RHS .1344 0/0 .233 .2658 .2691 .2706 By linear interpolation m = 1.414. Note that m = 1.2 is a trap for the unsuspecting student since L/(mV) = 1.0 and special form of Kremser is required. 12.D12.

Note this requires information in Section 13.4.

301

L

X in

.796 1000 .204

796 kg solvent/h ,

.256, X out

.796 x .05 .10 .15 .20 .25

0.025

L G

796 All stages

25,190

0.02564, Y1,in

0.975

0.0316

.0012 1 .0012

.001201

Equilibrium, y = 0.04 x X 0.0526 .11111 .1765 .25 .3333

y .002 .004 .006 .008 .01

Y .002004 .00402 0.00604 .00806 .0101

Plot weight ratios.

Yj

L

Xj

Y0

L

Xj 1 G G Slope = - L/G = - 0.0316. Step off stages backwards (start w. stage N) since it is different than other stages and we wouldn’t be sure when to switch if stepping off forwards. Need 4 equilibrium stages. Note: Can also plot y vs X, since y ~ Y and G ~ V Op. Line:

302

12.D13. New Problem 3rd Edition. Strip Vinyl chloride from water at 25ºC and 850 mmHg. H 1243.84x y x 1147.904x p tot 850 760 Want 0.1 ppm water leaving. Entering air is pure, L 1 kmole hr.

y

x0

* out

x IN

y out

y1

y IN

0

1 y

N 1147.904

1

x out

yN

5.0 xN

x out

y1* G b and c. Want

y*out

1147.904 x IN 5739.52 0

x0

5739.52 ppm (mole)

1171.33

5.0 0.1 G 2 G MIN

mix

L G

G MIN

0.00085373

L F

x *N n

0

585.665 1147.904

1

y out

Lx IN

For

L 1 kmol h , G

x IN

h

0.00170746 kmol h air

5.0 0 0.1 0

585.665 1147.904

n 1147.904 585.665 d.

kmol

585.665 (See figure for part b – labeled HW5 Prob 1b). m = 1147.904.

c. Eq. 12-28

N

x IN

1 ppm

L

y out

yN

x out

Gy IN

y IN

Lx out

n 25.0 0.672944

4.78327

G 0.00170746 kmol h

585.665 5.0 0.1

0

2869.76 ppm 0.00286976 mole frac. Probably send waste gas to incinerator. Will require additional fuel to burn. e. All concentrations are dilute enough that L G and equilibrium are straight and operation is very close to isothermal.

303

304

12.D14. a)

95% removal CH4, 5% remains – Constant V: Yout b.

yin

CH 4 Eq.

0.00129

L V

y CH 4

min

x*

y CH 4 y out

0.05 0.00129

0.0000645

CH 4

p CH 4

3600 x CH 4

p TOT

175

y CH 4 in

0.00129

20.5714

20.5714

20.5714 x CH 4

0.00006271

0.000645

CH 4

x CH 4

x in Slope

0.00129 0.0000645

L V

min

L V actual c) Ext. bal.

x out

CH 4

x CH 4 ,out

V

V L

0.00006271 0 1.4 L V min 27.360; L

y in

x CH 4 ,in

V

L 100

CH 4

19.5429

27.360 V

y CH 4 ,out

y CH 4 ,in y CH 4 ,out 0.00129 0.0000645 L 2736 d) Use methane values in Kremser eqn. (12-22) to find N 20.57 100 mV m 3600 / 175 20.57, 0.75183; y1* L 2736 0.00129 0 0.0000645 0

n 1 0.75183 N

2736.0

0.00004479

mx 0

b

0

0.75183 6.11 stages

n 1 0.75183

e) Now use Argon values with N = 6.11 to find y Ar,out & x Ar,out .

m Ar

7700 175

Eq. (12-23)

x Ar ,out

44.00,

y Ar ,in

mV

44 100

L

2736

0.00024

y Ar ,1

0.00024 0

V L

yN

y Ar,out

y Ar,1

y Ar ,in

y Ar ,out

1,Ar

0.00024, x Ar ,in

1.6082, 1 1.6082 1

y1*

mx 0

1 1.6082 1 1.6082

7.11

x Ar ,0

b

Ar

2736

0

7.11

0.60846

0.00024 0.00024 0.60846 100

0.0

0.00024 0.00009397

0.00009397 0.00000534

305

0.00000534 2736

% Argon recovery in liquid 12D15.

100 0.00024

Need equilibrium data. From DePriester chart: K C3 y C3 x C3 1.23 m C3 , K C4

L

Butane is a design problem:

100

y C4 x C4

N

y N 1 y1* y1 y1*

mV L

mV L

n

Propane is simulation: y N

m C4

0.17 C4

m C4 x 0,C4

.0006 0 .0000072 0

.83

L n mV

0.34

mV

5.882,

mV C 4 L 1.2% of the butane leaves as a gas. Thus, * y1,C4 0.006 0.012 0.0000072, y1,C4 1

60.85%

.17 2.39

n 5.882 * 0.0017, y1,C3

1,C3

0,

L

1.626,

mV

C3

0

mV L

0.615 C3

N 1

yN y1

1

y1* y

yN

* 1

1

y1

L 1 mV , y1 1 L mV

12.D16. Was 12.D19 in 2nd edition. .

a.)

Equil.

y CO 2

H CO2 PTOT

yN

0.000298

1

5.7034

x CO 2 , H CO 2 25 C

1640

50 mmHg 50 760 0.06579 atm PTOT 1.0 x CO 2 y CO 2 .00035 Feed H CO 2 1640 Equilibrium in column: y CO 2

H CO 2 PTOT

x CO 2

1640 0.06579

x CO 2

atm mol frac 2.134 10

7

24982 x CO 2

Basis: L = 1 kmol total/h. Assume L & G constant. Input = 2.1341 × 10-7 kmol CO2/h. 95% removal = (.95) (2.1341 × 10-7) = 2.027395 × 10-7 kmol CO 2 h in outlet gas. 5% CO2 remains in liquid

x in

x out

.05 2.1341 10

7

outlet liquid mole frac

0.106705 10 7 kmol CO 2 h 0.106705 10 1 kmol h

7

0.106705 10

7

b.)

306

Slope

y max out

24982

y

L V

Slope

yin

24982 2.1341 10

7

5.3314 10

3

y max

y in

5.3314 10

x in

x out

2.1341 0.10605

max

3

0 10

7

L Vmax

0

x out

x

x in

2.62968 10

V

2.62968 10

4

max

1

Vmin

Since L = 1,

L

7

2.1341 10

3.803 10 5 kmol h

4

c.) V 1.5 Vmin 1.5 3.803 105 5.704 10 5 kmol h Conditions for Kremser eq. are satisfied. CO2 Mass Bal: 2.1341 10 7 in 0.106705 10 5 out w. water 5.704 105 yCO2,out

y1

y CO 2 ,out

Eq. 12-29

2.027395 10 5.704 10 N

n

7

3.5543 10

5

x *N

xN

n

yN

x *N

x0

N

0,

1

y1*

y1

x *0

3.5543 10

m

24982

0.106705 10 2.1341 10

7

7

1.4227 10

24982 5.704 10

.0002 1.414

3

1.4227 10

7

0

1.0

n 12D.17.

x *0

L mV

m n

3

.0002828,

7

5.3569

5

L

L

mV 14.14 Can use variety of forms of Kremser equation, but cannot easily use forms with y*N 1 since

y*N

1

mx N and x N is unknown and hard to calculate. Try Eq. (12-21).

yN y1 Do by trial-and-error

L/mV RHS

1

y1*

.0083 .0002828

* 1

.0005 .0002828

y

2 15

3 40

2.9 36.699

L 1 mV 36.91 L 1 mV

N 1

2.91 37.02

307

Linearly interpolate L/mV = 2.907. Then, L = (2.907) (14.14) = 41.10 kmol/h. 12.D18. a.

L, x 0

y

p

11.5

p total

1520

0.00757 , x

y1

0 1

x1

0.0004,

L

y1

VT

x

L

yi

V

xN

Bot. of Column:

yN

1

y

1

y x

0.00757 0.0127

0.596

.00596 L

slope

VN

1

100

VT 150 100 , VF

Ext. MB: Lx 0 150 0.0004

xN

VN

K

.01, y

x0

VF y F 0.003

N

0.0127,

23

50, L 100

VF y F 50 .003

VN 1 y N

1

Lx N

100 .0058

100

VT y1

0.0067

0.0058

L V

x

yN

L 1

V

xN,

Slope

L VN

100 1

100

1.0

308

c. Minimum L.

xN yN

1

Eq.

y

L min

yF

VN

y1 L min

L min

VT

150

x0

L min 100 slope

1

slope x

Pinch is at y F . x F

y F 0.596

L min VT L min

0.003 0.596

Slope min

0.51653 VT

0.00503

0.003 0.0004 0.51653 150

0.00503 0

0.51653

77.48 kmol h .

309

12.D19*. New Problem 3rd Edition. Found m and L/V in Example 12-1. L/(mV) = (L/V)/m = 133/105.6 = 1.259, (mV)/L = 0.794, y1* = 105.6×0 = 0 If use Eq. (12-22), N = {ln[(1 - .794)((100-0)/(10-0)) + .794]}/ln[1.259] = 4.5

.000024

12.D20. a) 99% removal H 2 O , 1% left,

1000 .01

0.00024 moles out in L

0.00024

x H 2Sout

0.00000024 1000 Moles H 2S out in gas = (.000024) (1000) (.99) = 0.02376 0.02376

y H 2Sout b)

H H 2S

Equil. H 2S. y H 2S

p tot

V 26800

x H 2S

15.5

H CO2

y CO2

p tot

0.02376

x H 2S

1729.03 x H 2S , m H 2S

728

x CO 2

0.00691

3.44

15.5

x CO 2

1729.03

46.9677 x CO 2 , m CO 2

46.9677

Can use Kremser eq. for H 2S design [dilute linear system]. For example, Eq. (12-28)

n N

1000

mV

x*N

1729.03 3.44

H 2S

yN

1

n

m

0, x 0

* N * N CO 2

xN

x

x0

x

x N ,CO 2

mV L

1

mV L

1

1

x0 1

0.000024, x N

H 2S

CO 2 N 1

yN

1,CO2

, with N

mCO2

H 2S

0.00000024

H 2S

0.16813 2.4807

0, x 0

2.4807 and

5.9479

L

n 5.9479

unknown, x*N

For CO 2 . x N

mV

0.16813,

.000024 .00000024

1 0.16813

N c)

H 2S

L mV

mV L

n

L

x *N x *N

x0 xN

L mV

1

0.000038 Kremser (12-31) 46.9677 3.44

mV L

CO 2

1000

0.16157

CO 2

mV L N mV L

1

0.000038

1 0.16157 1

.16157

3.4807

0.000031916 CO 2

Little amount of CO 2

310

23.78

12.D21. New Problem 3rd Edition. Abs. y Stripper

Abs

y out

23.78

y

y IN y

Slope

0.2

5 x

x

4.756x

118.90x

0.00098

4.756

x EQ

L MIN G

L MIN

x IN

735.302 Abs

100

0.00001

x

L act

m

y IN

y1* mx IN Eq. (12-22).

N

0.00098, y1

4.756 .00001 y N 1 y1* y1 y1*

mV L n

N abs

459.56 kmol h

735.302 kmol h

V y abs,IN

y abs,out

100 0.00098 0.000079 735.302

Kremser Eq.

1

100 4.5956

4.5956

0.00001

L abs

x abs,out

n

0.000079

0.00020605

1.6 L abs,MIN

L abs x abs,IN

735.302 0.00001

x abs,out

1

0.00098

0.00020605

7.353 x abs,out

External M.B.

yN

0.00098 4.756

Operating line Slope

0.000079

L V

slope equilibrium

x strip,IN

4.756, b

y out

0.00013253 0,

L mV

0.000079

7.353 4.756

1.5460,

mV L

0.64681

0.00004756 mV L

n

L mV

.35319

0.00098 0.00004766 0.000079 .00004756

0.64681

n 1.5460

n 11.1216

2.40889

n 1.5460

0.4357

5.53

311

Stripper 118.90 = Slope Equilibrium

y EQ

y

L

x out

0.015758

= Slope Operating line

VMIN

y IN 0 118.90x N

118.90 0.00013253

0.00001

x IN

x out,abs

0.00013253

x0

118.90

L

0.015758 0

V

VMIN ,Strip

VStrip y Strip out

128.6037

0.00013253 0.00001

MIN Strip

735.302

L Strip 128.6037

1.5 5.718

VStrip y Strip IN

5.718 kmol h.

128.6037 8.576 kmol h.

L Strip x Strip IN

x Strip out

0

735.302 0.00013253 0.00001

VStrip

yStrip,IN

yStrip,out

8.576

x *N

0.010506,

0

yN

1

118.9

Kremser (12-28)

n

1

N

x *N x *N

x0 xN

L mV

L mV

n mV L n

0.00013253 0 0.00001 0

0.27889

N

,

L

735.302

mV

118.90 8.576

0.721106 4.54

n 1.38676

12.D22. New Problem 3rd Edition. K

0.22. y

L 75, V 150, L V External M.B.: Lx N xN

Vy1 V L

Lx 0 yN

1

Vy N y1

0.721106

Kx

0.22x. Plot e.g., at x

0.006, y

0.00132

0.5

1

2 0.003 0.0004

0.0052

Points x 0 , y1

0, 0.0004 , and x N , y N 1 0.0052, 0.003 are on op. line. Plot Op. Line. See graph (labeled 12.D.b). 2 stages more than sufficient.

312

313

12D23.

Apparatus similar to Figure 12-2, except part of treated gas is heated and used as stripping gas. Absorber: Work in terms of mole ratios. Yi n .15 .85 0.1765, Yout 0.005 .995

G

1400 .85

Equilibrium: y

0.00502, X in

1190 mol carrier gas/day, L

800 .995

0.005 .995

0.00502

796 mol solvent/day

.5x . Convert to mole ratios

.05 .1 .15 .2

x 0

X 0 .0526 .1111 .1765 .25

Plot ratios on McCabe-Thiele diagram L Absorber: Op. Line: Y X Yout G External balance Absorber: 796 .00502

y 0 .025 .05 .075 .1 L

X in ,

Y 0 .0256 .0526 .0811 .1111 L

0.669 G G 1190 .1765 1190 00502

796 X out

X out Abs 0.2614 Step off stages as shown in figure. Need 8 equilibrium stages. Stripper:

Yin

.00502 same as Yout abs , Xin

0.2614 same as Xout abs

X out 0.00502 same as X in abs , L 796 mol solvent day, have 4 stages Stripper equilibrium: y = 3 x. Convert to mole ratios.

x 0 .025 .05 .075 .1 .15

X 0 .0256 .0526 .0811 .111 .1765

y 0 .075 .15 .225 .3 .45

Y 0 .0811 .1765 .290 .4286 .818

Problem is trial-and-error. Select Yout . Draw operating line from Yout , X in to

Yin , X out .

See if need 4 stages. When need exactly 4 stages, L/G = slope. From Figure, final result shown: G strip

L slope

796 1.0998

723.7 mol carrier gas/day.

Yout strip ~ 0.287 mol ratio

314

315

12.E1.

Convert ppm(wt) to mole fraction: ppm 10 6 wt frac Basis 1000 g. of steam: Feed Liquid 1000 ppm = 0.001 wt frac mole frac

Liquid Water

lg Nitrobenzene

0.00812 gmole

999 g water =

55.4507

F L F C where C

28.1 10

L mV

S Vout

99 61.8

18.0314

37.2g water , C

2.0648 mol

95.7219 2.0648 97.7867 mol h

Outlet: Basis 1000 g 28.1 ppm 28.1 10

S

0.9998535

Total 55.4588 mol 1000g Avg mol. wt 55.4588 1726 g h 95.7219 mol h

L

V

0.0001465

6

wt. frac.

moles

3

Mol frac. 4

g nitrobenzene 2.2825 10 999.9997 18.016 55.5062 1000 28.1 10 3 g water 55.5064 mol 99 18.016 5.495 mol h .

4.11 10

6

~ 1.0

Equil. y mx b, b 0, m H p tot 28.0 Kremser Eq. – Several forms can be used. → Use Eq. 12-28. 97.7867 mV yN 1 b 0 0.63555, 1.5734 , x *N m 28.0 5.495 L

n

1

N

Effic

x0 xN

L mV

x *N x *N

L mV

mV n L

N Eq

5.763

5.763

0.524 N act 11 Ref. Hwang et al, IEC Research, 31 (7) 1992, 1753 & 1759. 12-F1.

H = 59.3 (Perry’s 4th ed., p 14-4).

y

H

59.3

x 11.86x PTOT 5 To be absolutely correct should convert this to mole ratios, although at these low concentrations could use mole fractions with small error. y x Y , X 1 y 1 x

316

x in

x y X Y 0 0 0 0 .001 .01186 .001001 .012 .0015 .01779 .001501 .01810 .002 .02372 .002003 .0243 Change specified conditions to mole ratios. 0, X in 0; yin .02, Yin .02041; y out .002; Yout .002003; x out .001, X out See Figure for plot of operating line, equilibrium and stages. N = 3.3 Height 5 ft HETP 1.515 N 3.3 equil stage Yin Yout L .02041 .002003 18.4 G X out X in .001001 0 If use mole fractions find L/V = 18.0

12.F2.

K E 26.0, K p 0.6, K After one pass of mass balance obtain:

.001001

0.019

317

xi,1

ethane 0.032

pentane 0.005

octane 0.963

xi,2 xi,3

0.035 0.033

0.031 0.162

0.934 0.805

yi,1

0.975

0.003

0.021

yi,2 yi,3

0.962 0.885

0.019 0.099

0.019 0.016

For new temperature used multi-variant Newtonian convergence. T1,New 73.90 F, T2,New 80.93 F, T3,New 2.G1.

a) N

4, L

570, y A,out

0.00317

b) N

8, L

500, y A,out

0.00315 while with L

c) N 16, L 490, y A,out d) Have a pinch point.

0.002978 while with L

490, y A 480, y A

99.15 F.

0.00358 0.00337

12.G2. Used Peng-Robinson. a. Total number of stages required 8 b. Feed stage location for the solvent 1 c. Feed stage location for stream A 8 d. Feed stage location for stream B 6 e. Outlet mole fractions of gas stream leaving absorber

0.9991009, 0.00024691, 0.00019067, 0.0004617 f. Outlet mole fractions of liquid leaving absorber 0.012878, 0.0310992, 0.0198928, 0.93612992 g. Outlet gas flow rate 161.6478 kmol/h h. Outlet liquid flow rate 213.352 kmol/h i. Highest temperature in column 19.1287 ˚C and stage it occurs on 8

12.G.3. New Problem 3rd Edition. Used NRTL. Column pressure = 1.0 atm. Feed gas flow rate = 752 kmol/h. Feed gas temperature = 100oC. Liquid feed temperature = 75oC. Recovery of isopropyl alcohol = 0.98000. T1 = 302.5K, T2 = 298.6K, T3 = 299.9K, T4 = 301.9K, T5 = 303.3K. Leaving gas: G = 802.2 kmol/h, Mole fractions: IPA = 0.02443, W = 0.033836, N2 = 0.93721 Leaving liquid: L = 149.8 kmol/h, Mole fractions: IPA = 0.002671, W = 0.99638, N2 = 0.000950 Column diameter = 1.5455 m.

318

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

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