SOLUTION MANUAL

for

SEPARATION PROCESS ENGINEERING.

Includes Mass Transfer Analysis

3rd Edition

(Formerly published as Equilibrium-Staged Separations)

by

Phillip C. Wankat

17

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

10.8 kg

.6 kmol Water .6 MW 18total 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.

18

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. high LIQF,P p F ref

h T c T T . 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 TF 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.

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 increasesb. K decreases as P increasesc. 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.

1.0

.5 0

1.0

.5

0 xw

Flash

operating

line

yw Equilibrium

(pure water)

2.A4

zw = 0.965

19

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. The answer is 3.5 to 3.6 b. The answer is 36C

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

2.B1. Must be sure you dont violate Gibbs phase rule for intensive variables in equilibrium. Examples:

drum drumF,z,T ,P FF,T , z, p FF, h , z, p

drumF, z, y, P FF,T , z, y FF, h , z, y

drumF, z, x, p FF,T , z, x etc.

drumF, z, y, p F drum drumF,T ,z,T , p

drumF, z, x,T FF,T , y, p

Drum dimensions, drum drumz,F , p F drumF,T , y,T

Drum dimensions, drumz, y, p FF,T , x, p

etc. F drumF,T , x,T

FF,T , 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.

If lb mole

F 1000 , D .98 and L 2.95 ft from Problem 2-D1ehr

.

Since D V and for constant V/F, V F, we have D F . With F = 25,000:

new old new old new newF F = 5, D = 5 D = 4.90, and L = 3 D = 14.7 .

Existing drum is too small.

Feed rate drum can handle: F D2. 2 2

existing existF D 4

1000 .98 .98gives

existingF 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

20

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

v

drum L v v

V MWD

3K 3600

Bypass reduces V

c1) Kdrum is already 0.35. Perhaps small improvements can be made with a better demister Talk to the manufacturers.

c2) v 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. A B

B A

z zV

F K 1 K 1

2.C5. a. Start with iii

Fzx and let V F L

L VK

i ii i

ii

Fz zx or x

L LL F L K1 K

F F

Then i ii i i

i

K zy K x

L L1 K

F F

From i ii i

i

K 1 zy x 0 we obtain 0

L L1 K

F F

V = .25 (16660) = 4150

LTotal x

y = .58,

8340

16,660 25,000

21

2.C7. i

i

z V1 f

V F1 K 1F

From data in Example 2-2 obtain:

V/F 0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1.0 f 0 -.09 -.1 -.09 -.06 -.007 .07 .16 .3 .49 .77

2.C8. New Problem.

drump

x y

z

drumT

F = L + V

zF Lx Vy

Solve for L & V

Or use lever arm-rule

22

2.C9. New Problem. Derivation of Eqs. (2-62) and (2-63). Overall and component mass balances are,

1 2F V L L

i 1 i,L1 2 i,L2 iFz L x L x Vy

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

i 1 i,L1 L2 i,L2 2 i,L2 iV L2 i,L2Fz L K x L x VK x

Solving,

i ii,L2

1 i,L2 2 i,V L2 1 i,L1 L2 1 i,V L2

Fz Fzx

L K L VK L K F V L VK

Dividing numerator and denominator by F and collecting terms.

ii,liq 2

1i,L1 L 2 i,V L 2

zx

L V1 K 1 K 1

F F

Since i i,V L2 i,L2y K x , i,V L2 i

i1

i,L1 L2 i,V L2

K zy

L V1 K 1 K 1

F F

Stoichiometric equations, C C C C

i,L2 i i i,L2i 1 i 1 i 1 i 1

x 1 , y 1 , thus, y x 0

which becomes C i,V L 2 i

i 1 1i,L1 L 2 i,V L 2

K 1 z0

L V1 K 1 K 1

F F

(2-62)

Since i,L1 L2 ii,liq1 i,L1 L2 i,liq 2 i,liq11

i,L1 L2 i,V L2

K zx K x , we have x

L V1 K 1 K 1

F F

In addition, C i,L1 L 2 i

i,liq1 i,liq 2i 1 1

i,L1 L 2 i,V L 2

K 1 zx x 0

L V1 K 1 K 1

F F

(2-63)

2.D1. a. V 0.4 100 40 and L F V 60 kmol/h

Slope op. line L V 3 2, y x z 0.6See graph. y 0.77 and x 0.48

b. V 0.4 1500 600 and L 900 . Rest same as part a.

c. Plot x 0.2 on equil. Diagram and int ercepty x z 0.3. y zF V 1.2

V F z 1.2 0.25 . From equil y 0.58 .

d. Plot x 0.45 on equilibrium curve.

23

L F V 1 V F .8Slope 4

V V V F .2

Plot operating line, y x z at z 0.51 . From mass balance F 37.5 kmol/h. e. Find Liquid Density.

L m m w wMW x MW x MW .2 32.04 .8 18.01 20.82

Then, wmL m wm w

MWMW 32.04 18.01V x x .2 .8 22.51 ml/mol

.7914 1.00

LL LMW V 20.82 22.51 0.925 g/ml

Find Vapor Density. vvP MW

RT (Need temperature of the drum)

v m wm wMW y MW y MW .58 32.04 .42 18.01 26.15 g/mol

Find Temperature of the Drum T: From Table 3-3 find T corresponding to

y .58, x 20, T=81.7 C 354.7K

4v

ml atm1 atm 26.15 g/mol 82.0575 354.7 K 8.98 10 g/ml

mol K

Find Permissible velocity:

24

perm drum L v v

2 3 4

drum lv lv lv lv

u K

K exp A B nF C nF D nF E nF 2-60

Since V

V F 0.25 1000 250 lbmol/h,F

vvlb

W V MW 250 26.15 6537.5 lb / hlbmol

LLL F V 1000 250 750 lbmol/h, and W L MW 750 20.82 15, 615 lb/h,

4

VLlv lv

V L

W 15615 8.89 10F 0.0744, and n F 2.598

W 6537.5 .925

Then 4

drum perm 4

.925 8.98 10K .442, and u .442 14.19 ft/s

8.98 10

v

2cs 4 3

perm v

V MW 250 26.15 454 g/lbA 2.28 ft .

u 3600 14.19 3600 8.98 10 g/ml 28316.85 ml/ft

Thus, csD 4A 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 0.34Fz 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 x2, find y2 = 0.62 from equilibrium. From equilibrium point

plot op. line of slope 2 2

2

VL V 1 V F 3 7.

F This gives

2 1z 0.51 x (see Figure). Then from equilibrium, 1y 0.78 .

For stage 1, 1 1

1 1

z xV 0.55 0.510.148

F y x 0.78 0.51.

2.D3. a. z 0.4 V F 0.6 V 6.0 k mol h, L 4.0

Op. eq.

Ly x z V F

V2

y x 2 33

See graph: My 0.55 x 0.18 T ~ 82.8 C linear interpretation on Table 2-7 .

25

b. Product 78.0 C x 0.30, y 0.665,

Mass Bal: Fz Lx Vy F V x Vy

or 4.0 10 V 0.3 0.665V

V 2.985 and V F 0.2985Can also calculate V/F from slope.

c. V

F 10, 0.3 V 3 & L 7F

L z 7 zy x x

V V F z 0.3

If y 0.8, x 0.545 @ equil

Then 7

z 0.3 0.8 0.545 0.6215.3

Can also draw line of slope 7

3 through equil point.

26

2.D4. New problem in 3rd edition. Highest temperature is dew point V F 0

Set i i i i iz y . K y x

Want i i ix y K 1.0

ref New ref Old i iK T K T y K

If pick C4 as reference: First guess bu tan eK 1.0, T 41 C : C3 C6K 3.1, K 0.125

i

i

y .2 .35 .454.0145

K 3.1 1.0 .125 T too low

Guess for reference

C4K 4.014 T 118 C : C3 C6K 8.8, K .9

i

i

y .2 .35 .450.6099

K 8.8 4.0145 .9

C4,NEWK 4.0145 .6099 2.45, T 85 : C2 C6K 6.0, K 0.44

i

i

y .2 .35 .451.20

K 6 2.45 .44

C4,NEWK 2.45 1.2 2.94, T 96 C : C3 C6K 6.9, K 0.56

i

i

y .2 .35 .450.804 Gives 84 C

K 6.9 2.94 .56

27

Use 90.5 Avg last two T C4 C3 C6K 2.7, K 6.5, K 0.49

i i.2 .35 .45

y K 1.0796.5 2.7 .49

T ~ 87 88 C Note: hexane probably better choice as reference.

2.D5. a)

b) 1 11 1

L Fy x z

V V Plot 1st Op line.

y = x = z = 0.55 to x1 = 0.3 on eq. curve (see graph)

Slope L 0.55 0.80 .25

0.454545V .55 0 .55

1 1 1 11

L0.45454 & L V F 1000

V

c) Stage 2 2V L 0.75F

0.25 , 3 , y x z 0.66.F V 0.25F

Plot op line

At 20.66 F z 0.66

x 0, y z V F 2.64. At y 0, x z 0.880.25 L L F 0.75

From graph 2 2y 0.82, x 0.63 .

2 22

VV F 0.25 687.5

F 171.875 kmol/h

F1 = 1000

z1 = 0.55 p1,2 = 1 atm

x2

x1 = 0.30

v2

y2

v1 = F2

y1 = z2

2 1

2

V0.25

F

y1 = 0.66 = z2

V1 = 687.5 kmol/h = F2 1

V 687.50.6875

F 1000

28

2.D6.

RR eq., i i

i

K 1 z0

1 K 1 V F, First guess V/F = 0.6

1

1.4 .45 0.2 .35 0.7 .2f 0.0215

1 1.4 .6 1 0.2 0.6 1 0.7 0.6

Use Newtonian Convergence 2

ci ik

2i 1

i

K 1 zdf

d V F 1 K 1 V F

k

k 1 k

fV VdfF F

d V F

T = 50C

P = 200 kPa

Kc4 = 2.4

zc4 = 0.45

F = 1.0 kmol/min

zc5 = 0.35

Zc6 = 0.20 L

V

Kc5 = 0.80

Kc6 = 0.30

29

2 22

12 2 2

1.4 0.45 0.2 0.35 0.7 0.20df0.570

V 1 1.4 .6 1 0.2 0.6 1 0.7 0.6dF

2

V 02150.6 0.6377

F 0.570

2

1.4 .45 0.2 .35 0.7 0.2f 0.00028

1 1.4 0.6377 1 0.2 0.6377 1 0.7 0.6377

Which is close enough.

i i iic4

c4i

y K xz 0.45x 0.2377,

V y 2.4 0.2377 0.57051 1.4 .63771 K 1F

c5 c5

0.35x 0.4012, y 0.8 0.4012 0.3210

1 0.2 0.6377

c6 8

i i

0.20 0.3613 0.1084x , y 0.30 0.4012

1 0.7 0.6377 x 1.0002 y 0.9998

2.D7. A B

B A

z zV

F K 1 K 1

M PK 5.6 and K 0.21

V 0.3 0.70.2276

F 0.21 1 5.6 1

Eq. (2-38) MMM

z 0.3x 0.1466

V 1 4.6 0.22761 K 1F

P Mx 1 x 0.8534 , M M My K x 5.6 0.1466 0.8208

P My 1 y 0.1792

2.D8. Use Rachford-Rice eqn: i i

i

K 1 zVf 0

F 1 K 1 V / F . Note that 2 atm = 203 kPa.

Find iK from DePriester Chart: 1 2 3K 73, K 4.1 K .115

Converge on V F .076, V F V F 152 kmol/h, L F V 1848 kmol/h .

From iii

zx

V1 K 1

F

we obtain 1 2 3x .0077, x .0809, x .9113

From i i i 1 2 3y K x , we obtain y .5621, y .3649, y .1048

2.D9. Need hF to plot on diagram. Since pressure is high, feed remains a liquid

LF P F ref refh C T T , T 0 from chart

L EtOH wP P EtOH P wC C x C x

30

Where EtOH wx and x are mole fractions. Convert weight to mole fractions.

Basis: 100 kg mixture 30

30 kg EtOH 0.651 kmol46.07

70 kg water 70 18.016 3.885 Total = 4.536 kmol

Avg. 100

MW 22.0464.536

Mole fracs: E w0.6512

x 0.1435, x 0.85654.536

.

Use LEtOH

P PC at 100 C as an average C value.

LP

kcalC 37.96 .1435 18.0 .8565 20.86

kmol C

Per kg this is LP

avg

C 20.86 kcal0.946

MW 22.046 kg C

Fh 0.946 2000 189.2 kcal/kg

which can now be plotted on the enthalpy composition diagram.

Obtain drum E ET 88.2 C, x 0.146, and y 0.617 .

For F 1000 find L and V from F = L + V and Fz Lx Vywhich gives V = 326.9, and L = 673.1

Note: If use wt. fracs. L LP P avg

C 23.99 & C MW 1.088 Fand h 217.6 . All wrong.

31

2.D.10 New Problem. Solution 400 kPa, 70C C4z 35 Mole % n-butane C6x 0.7

From DePriester chart C3 C4 C6K 5, K 1.9, K 0.3

Know ii i i i i i ii

zy K x , x , x y 1 z

V1 K 1

F

R.R. i i C3 C6 C4 C6i

K 1 z0 z 1 z z .65 z

V1 K 1

F

For C6 C6 C6 C6C6

z z V0.7 z 0.7 1 0.7

V V F1 K 1 1 0.7F F

C6

Vz 0.7 0.49

F

RR Eq: C6 C64 .65 z 0.9 .35 0.7z

0V V V

1 4 1 0.9 1 0.7F F F

2 equations & 2 unknowns. Substitute in for C6z . Do in Spreadsheet.

Use Goal Seek to find V F.

V0.594

F when R.R. equation 0.000881 .

C6

Vz 0.7 0.49 0.7 (0.49)(0.594) 0.40894

F2.D11. L F 0.6 V F 0.4 & L V 1.5

Operating line: Slope 1.5, through y x z 0.4

32

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.

L m m w wMW x MW x MW .2 32.04 .8 18.01 20.82

Then, wmL m wm w

MWMW 32.04 18.01V x x .2 .8 22.51 ml/mol

.7914 1.00

LL LMW V 20.82 22.51 0.925 g/ml

Find Vapor Density. vvp MW

RT (Need temperature of the drum)

v m wm wMW y MW y MW .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

33

4v

ml atm1 atm 26.15 g/mol 82.0575 354.7 K 8.98 10 g/ml

mol KFind Permissible velocity:

perm drum L v v

2 3 4

drum,horizontal drum,vertical lv lv lv lv

u K

K 1.25 K exp A B nF C nF D nF E nF 1.25

Since V V F 0.25 1000 250 lbmol/h,

vvW V MW 250 26.15 lb lbmol 6537.5 lb / h

LLL F V 1000 250 750 lbmol/h, and W L MW 750 20.82 15, 615 lb/h,

4VL

lv lv

V L

W 15615 8.98 10F 0.0744, and n F 2.598

W 6537.5 .925

drum,vertical drum,horiz

4

perm 4

K 0.442, and K 0.5525

0.925 8.98 10u 0.5525 17.74 ft/s

8.98 10

v

cs 4 3perm v

V MW 250 26.15 454 g/lbmA

u 3600 17.74 3600 8.98 10 g/ml 28316.85 ml/ft

2CsA 1.824 ft ,

2T CsA A 0.2 9.12 ft

With L/D = 4, TD 4A 3.41 ft and L 13.6 ft

2.D13. New Problem. The answer is ycresol = 0.19582

Since pc p pp

x 1.76 .7x 0.3, x 0.7, y 0.80418

1 1 x 1 .76 .7

c py 1 y 0.19582

Or cp c

c

cp c

xy

1 1 x, cp

pc

10.5682

2.D14. Raoults Law: C4C4Tot

VPK

P

410 C C 4

10 C6 C6

log VP 4.04615 , VP 11121 mm Hg

log VP 3.2658 , VP 1844.36 mm Hg

ii

i

zx 1.0 1.0

1 K 1 V F

34

0.3 0.71

11121 1844.361 1 0.4 1 1 0.4

P P

Solve for Pdrum = 3260 mmHg

ii

i

zx

V1 K 1

F

C4 C4 C4 C4

.3 11121x 0.1527, y K x 0.1527 0.52091

11121 32601 1 .4

3260

C6 C4 C6

1844.36x 1 x 0.84715, y 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 isnt. We can still write R-R eqn. Will have three variables: zC2, ziC4, znC4. Need two other eqns: iC4 nC4 C2 iC4 nC4z z constant, and z z z 1.0

Thus, solve three equations and three unknowns simultaneously.

Do It. Rachford-Rice equation is,

C2 C2 iC4 iC4 nC4 nC2

C2 iC4 nC4

K 1 z K 1 z K 1 z0

V V V1 K 1 1 K 1 1 K 1

F F FCan solve for zC2 = 1 ziC4 and ziC4 = (.8) znC4. Thus zC2 = 1 1.8 znC4 Substitute for ziC4 and zC2 into R-R eqn.

C2 iC4 nC4nC4 nC4 nC4

C2 iC4 nC4

K 1 .8 K 1 K 11 1.8 z z z 0

V V V1 K 1 1 K 1 1 K 1

F F F

Thus,

C2

C2

nC4C2 iC4 nC4

C2 iC4 nC4

K 1

V1 K 1

FzK 1 .8 K 1 K 1

1.8V V V

1 K 1 1 K 1 1 K 1F 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. C1 C4 C5 C6 C1 C4 C5 C6z 0.5, z 0.1, z 0.15, z 0.25, K 50, K .6, K .17, K 0.05

1st guess. Can assume all C1 in vapor, ~ 1/3 C4 in vapor, C5 & C6 in bottom

1V / F .5 .1 / 3 .53 This first guess is not critical.

35

R.R. eq. i i

i

K 1 zVf 0

F 1 K 1 V F

49 .5 .4 .1 .83 .15 .95 .25

0.1571 49 .53 1 .4 .53 1 .83 .53 1 .95 .53

Eq. 3.33 12

2 1 i i2

i

f V FV V

F F z K 1

V1 K 1

F

where 1 1

V / F 0.53 and f V / F 0.157 .

calculate 2

V / F .53 0.157 2.92 0.584

V .584 150 87.6 kmol/h and L 150 87.6 62.4

C1C1

C1

z .5x 0.016883

1 K 1 (V / F) 1 49 .584

C1 C1 C1y K x 50 0.016883 0.844

Similar for other components. 2-D17. a. V 0.4F 400, L 600 Slope L F 1.5 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

36

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

2.D18. From iii

zx

V1 K 1

F

, we obtain

h

h

h

z1

xV

F K 1

Guess drumT , calculate h b pK , K and K , and then determine V F .

Check: 1 i

1

K 1 z0

1 K 1 V F ?

Initial guess: If hx .85 then drumT must be less than temperature to boil pure hexane

hK 1.0, T 94 C . On this basis 85 to 90 would be reasonable. Try 85C.

h b pK =0.8, K 4.8, K =11.7 .

0.61

V 0.85 1.471F 0.8 1

. Not possible. Must have h0.6

K 0.7060.85

Try T 73 C where hK 0.6 . Then b pK 3.8, K 9.9 .

0.61

V .85 0.735F .6 1

Check:

i i

i

K 1 z 8.9 .1 2.8 .3 .4 .60.05276

1 K 1 V F 1 8.9 .735 1 2.8 .735 1 .4 735

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

2.D19. 90% recovery n-hexane means C6 C60.9 Fz L x

Substitute in L F V to obtain C6 C6z .9 1 V F x

8C balance: C6 C6 C6 C6 C6 C6z F Lx Vy F V x K Vx

or C6 C6 C6 C6z 1 V F x x K V F

Two equations and two unknowns. Remove C6x and solve

C6C6 6.9 z KV F

z .93C1 V F

Solve for C6

V .1V F.

F .9K .1. Trial and error scheme.

Pick T, Calc C6K , Calc V F, and Check f V F 0 ?

If not new

ref oldref

K TK

1 d f T

37

Try C4 C5 C6 refT 70 C. K 3.1, K .93, K .37 K

V .10.231

F .9 .37 .1.

Rachford Rice equation

2.1 .4 .08 .25 .63 .35f .28719

1 2.1 .231 1 .08 .231 1 .63 .231

ref new

.37K T 0.28745 use .28

1 0.28719Converge on New C4 C8T ~ 57 C. Then K 2.50, K .67, and V F 0.293 .

2.D20. New Problem. The K values are: EK 8.7 , BK 0.54 , PK 0.14

Can use Eq. (2-40), (2-41) or (2-42). If we use (2-42) the R R eqn

i iF B E B

i

K 1 zVf 0 Use z 1 z z .8 z

VF 1 K 1F

Then B B7.7 .2 .46 z .86 .8 z

RR eq = 1+7.7 25 1 .46 .25 1 0.86 .25

B B0 0.5265 0.51977 z 0.8764 1.0955z

B B0 .5757z 0.3499 z 0.6078

2.D21. a.) C2 C5K 4.8 K 0.153

Soln to Binary R.R. eq. A B

B A

z zV

F K 1 K 1

V 0.55 0.450.5309

F .153 1 4.8 1

C2C2 C2

C2

z 0.55x 0.1823, y 0.8749

V 1 3.8 .53091 K 1F

C5 C5x 0.8177 , y 0.1251

Need to convert F to kmol.

Avg MW 0.55 30.07 0.45 72.15 49.17

kg kmolF 100,000 2033.7 kmol/h

hr 49.17 kg

V V F F 1079.7, L F V 954.0 kmol/h

b.) L vPerm drumv

u K

To find LMW 0.1823 30.07 0.8177 72.15 64.48

38

VMW 0.8749 30.07 0.1251 72.15 35.33

For liquid assume ideal mixture:

C2 C5

1 C2 C2,liq C5 C5,liq C2 C5

C2,liq C5,liq

MW MWV x V x V x x

L

30.07 72.15V 0.1823 0.8177 103.797 ml/mol

0.54 0.63

L

L

L

MW 64.480.621 g/ml

V 103.797

For vapor: ideal gas: v

v

MW

RT

v

atm g700 kPa 35.33

101.3 kPa mol0.009814 g / ml

ml atm82.0575 303.16K

mol K

drumK : Use Eq. (2-60) with vL

lV

V L

WF

W

Vkmol 64.48 kg

W 997.7 6, 4331.7 kg/hh kmol

VW 881.5 35.33 31,143.4 kg/h

lV64331.7 0.009814

F 0.259731,143.3 0.621

2

drum

3 4

K exp 1.877478 0.81458 n .2597 0.18707 n 0.2597

0.0145229 n 0.2597 0.0010149 n 0.2597 0.3372

Perm

0.621 0.009814u 0.3372 2.6612 ft/s

0.009814

ft 1.0 m 2.6612 0.8111 m/s

s 3.2808 ft

2VC 6 3

Perm v3 3

kmol kg1079.7 35.33

V MW h kmolA 1.392 m

u 3600 m s g kg 10 cm0.8111 3600 0.009814

s h cm 1000g m

CD 4A 1.33 m

Arbitrarily L D 4, L 5.32 m

39

2.D22. i i iP iP NP NP

iiP NP

K 1 z K 1 z K 1 zVf 0

V VF 1 K 1 V F 1 K 1 1 K 1F F

Solve for V/F.

NP iP iP iP NP NP

V V1 K 1 K 1 z 1 K 1 K 1 z 0

F F

iP iP NP NP

NP iP

K 1 z K 1 zV

F K 1 K 1 where iP NPz z 1.0

drum totp p 760 mm Hg, T 90 C

10 NP

1499.2log VP 7.84767 2.75943

90 204.64

NPVP 574.68 mm Hg , NP totK 574.68 p 0.75616

10 iP

1580.9log VP 8.11778 3.011679

20 219.61

iPVP 1027.256 mm Hg , iPK 1027.256 760 1.35165

Note: iP NP iPMW MW . z 0.5 in both wt & mol frac., as does NPz .

0.35165 0.5 0.24384 0.5V0.629

F 0.24384 0.35165

iPiP

iP

zx 0.4095

V1 K 1

F

NP iP ip iP iP NPx 1 x 0.5905; y K x 0.55347 y 0.44653

2.D23. 5. 0C, 2500 kPaFig 2.12: M ethylene Ethane C6K 5.7, K 1.43, K 0.98, K 0.007

First, try 1

V0.6

F (equal split ethylene and ethane)

1

.47 .4 0.43 .05 .02 0.35 0.993 0.2Vf 0.0108

F 1 4.7 .6 1 .43 .6 1 .02 .6 1 .993 6

Eq. (2-46) 1 22 1 i i

2

i

Vf

V V F0.6059

F F z K 1

V1 K 1

F

Then Eq. (2-38), ii M ethylenei

zx . x 0.104, x 0.040

1 K 1 V F

ethane C6x 0.354, x 0.502, 1.0001 OK

Find i i iy K x

40

2.D24. New Problem. p = 300 kPaAt any T. C3 C3 C3K y x

Ks are known. C6 C6 C6 C3 C3K y x 1 y 1 x

Substitute 1st equation into 2nd C6 C3 C3 C3K 1 K x 1 x

Solve for xC3, C3 C6 C3 C31 x K 1 K x

C3 C3 C6 C6x K K 1 K

C3 C6C6C3 C3

C3 C6 C3 C6

K 1 K1 Kx & y

K K K K

At 300 kPa pure propane C3K 1.0 boils at -14C (Fig. 2-11)

At 300 kPa pure n-hexane C6K 1.0 boils at 110C

Check: at -14C C6C3C6

1 Kx 1,

1 K C6C3

C6

1 1 Ky 1.0

1 K

at 110C C3C3

0x 0,

KC3

C3

C3

K 0y 0

K

Pick intermediate temperatures, find C3K & C6K , calculate C3x & C3y .

T

0C 1.45 0.0270.9915

10C 2.1 0.044 0.465 0.976 See20C 2.6 0.069 0.368 0.956 Graph30C 3.3 0.105 0.280 0.92440C 3.9 0.15 0.227 0.88450C 4.7 0.21 0.176 0.82760C 5.5 0.29 0.136 0.7570C 6.4 0.38 0.103 0.659

C6KC3K C3x

1- 0.027= 0.684

1.45 - 0.027

C3 C3 C3y K x

41

42

b. C3x 0.3 , V F 0.4, L V 0.6 0.4 1.5

Operating line intersects y x 0.3, Slope 1.5

L Fy x z

V V

at F 0.3

x 0, y z 0.75V 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: C3 C3 C3K y x 0.63 0.062 10.2 , DePriester Chart T = 109C

d. y .8, x ~ .16 L y .8 .6

Slope 0.45V x .16 .6

1 f.45

f

V 1f 0.69

F 1.45

2.D25. New Problem. 20% Methane and 80% n-butane

drumT .50 C , V

0.40F

, Find drump

A A B B

BA

K 1 z K 1 zV0 f

VVF 1 K 11 K 1FF

Pick drum C4 nC4p 1500 kPa: K 13 K 0.4

(Any pressure with C1 C4K 1 and K 1.0 is OK)

Trial 1 112 .2 .6 .8

f 0.21781 12 .4 1 .6 .4

Need lower drump

C 4 oldC 4 new

old

K P 0.4K P 0.511

1 d f P 1 .2138

1.0

new C1P 1160 K 16.5

2

15.5 .2 .489 .8f 0.4305 .4863 0.055769

1 15.5 .4 1 .489 .4

C4 new

0.511K P 0.541

1 0.055769

new C1P 1100 K 17.4

3

16.4 .2 .459 .8f 0.0159

1 16.4 .4 1 .459 .4, OK. Drum pressure = 1100 kPa

43

b.) ii C1i

z 0.2x , x 0.02645

V 1 16.4 .41 K 1F

C1 C1 C1y K x 17.4 0.02645 0.4603

2.D26. New Problem. a) 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

b) Stage is equil. C3C3C3

y 0.8K 3.700

x 0.2162

C50.2

K .25520.7838

These K values are at same T, P. Find these 2 K values on DePriester chart. Draw straight line between them. Extend to drum drumT , p . Find 10C, 160 kPa.

2.D27. New Problem. a.) C5 101064.8

VP : log VP 6.853 2.28320 233.01

,

VP 191.97 mmHg

b.) VP 3 760 2280 mmHg ,

10log VP 6.853 1064.8 / T 233.01

Solve for T = 71.65C c.) totP 191.97 mm Hg [at boiling for pure component totP VP ]

d.) C5: 101064.8

log VP 6.853 2.804530 233.01

VP 637.51 mm Hg

C5 C5 totK VP P 637.51 500 1.2750

C6: 10 C61171.17

log VP 6.876 2.272530 224.41

C6VP 187.29 mm Hg

C6K 187.29 500 0.3746

e.) A A A B B B A AK y x K y x (1 y ) / (1 x )

If A BK & K are known, two eqns. with 2 unknowns A AK & y Solve.

C6C5C5 C6

1 K 1 0.3746x 0.6946

K K 1.2750 0.3746

C5 C5 C5y K x 1.2750 0.6946 0.8856

f.) Overall, M.B., F = L + V or 1 = L + V FC5: Fx 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.

44

2.D28. New Problem. From example 2-4, H drum H Hx 0.19, T 378K, V F 0.51, y 0.6, z 0.40

With v

perm v

V MWh D C, D

u 3600 C

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

3v 3 31 28316.85ml lbm

3.14 10 g mol 0.198454g lbm ft ft

Example 2.4

L vperm drum horiz verticalV

u K , K 1.25 K

From Example 2-4, verticalK 0.4433 , horizK 1.25 0.4433 0.5541

1 2

perm

0.6960 0.00314u 0.5541 8.231 ft s

0.00314 [densities from Example 2-4]

V lbmol

V F 0.51 3000 1530 lbmol hrF hr

3

lbmol lbm1530 97.39

h lbmolD 5.067 ft

ft s lbm8.231 3600 0.1958

s h ft

h 4D 20.27 ft

Use 1

5 20 or 5 222

ft drum.

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 Peng-Robinson 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.

h

F

D

L

V

45

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 = 4D2. Then the equation for the drum diameter is D = {[(MWV) V]/[V 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 uperm (in ft/s) can be determined by combining Eqs. (2-59) and (2-60) for vertical drums with Eq. (2-64a). Flv = (WL/WV)[V/ L]

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 xA = 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.

46

Graph for problem 2.D30.

47

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 uperm in ft/s) is D = {[4(MWV) V]/[3600 V 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)[V/ L]

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 V 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)[V/ L]

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 V 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)[V/ L]

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 V 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)[V/ L]

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. Tdrum = 115.42

oC, Q = 6630.39 kW,

48

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 C6 C8 C6 C8 C6 C6 C8 C8x x 1.0, and y y K x K x 1.0

Substitute for xC6 C8C6C6 C8

1 Kx

K K

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

TC KC6 KC8 xC6 yC6 = KC6 xC6

125 4 1.0 0 0 120 3.7 .90 .0357 .321 110 3.0 .68 .1379 .141 100 2.37 .52 .2595 .615 90 1.8 .37 .4406 .793 80 1.4 .26 .650 .909 66.5 1.0 .17 1.0 1.0

Op Line Slope L 1 V F .6

1.5V V F .4

, 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 = 86C = 359K

b. Follows Example 2-4.

49

L C8 C8C6 C8MW x MW x MW .52 86.17 .48 114.22 99.63

C6 C8L C6 C8

C6 C8

MW MW 86.17 114.22V x x .52 .48 145.98 ml/mol

.659 .703

3

L

L 3L

MW 99.63 28316 ml/ft lbm.682 g/ml 42.57

V 145.98 454 g/lbm ft

v C6 C6 C8 C8MW y MW y MW .85 86.17 .15 114.22 90.38

v 3

v

1.0 90.38 g/molpMW0.00307 g/ml 0.19135 lbm/ft

ml atmRT 82.0575 359Kmol K

Now we can determine flow rates

V

V F .4 10,000 4000 lbmol/hF

vvW V MW 4000 90.38 361,520 lb/h

LLL F V 6000 lbmol/h, W L MW 6000 99.63 597, 780 lb/h

vLlv lvv L

W 597, 780 0.19135F 0.111, nF 2.1995

W 361,520 42.57

2

drum

3 4

K exp 1.87748 .81458 2.1995 .18707 2.1995

0.01452 2.1995 0.00101 2.1995 0.423

Perm drum L v vu K 0.423 42.57 19135 .19135 6.30 ft/s

v

2Cs

Perm v

V MW 4000 90.38A 83.33 ft

u 3600 6.3 3600 0.19135

CsD 4A 4 83.33 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

75% 85%Perm Perm

u 0.75 0.85 u 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 .7 .8 .9 1

yB 0 .22 .38 .52 .62 .71 .79 .85 .91 .96 1 Benzene-toluene equilibrium is plotted in Figure 13-8 of Perrys Chemical Engineers

Handbook, 6th ed. 2.F2. See Graph. Data is from Perrys Chemical Engineers Handbook, 6th ed., p. 13-12.

50

Stage 1) 1F

2 3z .4 f 1 3 Slope 2,

1 3

Intercept 1 1 2.4

1.2 y .872 x .164 z1 3

Stage 2) 2F

1 3z .164 f 2 3 Slope 1 2

2 3

Intercept 2 2 3.164

.246 x .01 y .240 z2 3

Stage 3) 3F

z .240 f 1 2 Slope 1

Intercept 3 3.240

.480 x .022 y .4611 2

2.F3. Bubble Pt. At P = 250 kPa. Want 1 1K z 1

Guess 1 2 3T 18 C, K 1, K .043, K .00095, .52

Converge to T 0 C

Dew Pt. Calc. Want 1

1

z1.0

K

Try 1 2 3T 0 C, K 1.93, K 0.11, K 0.0033, 120.26

51

Converge to T 124 C . This is a wide boiling feed. Tdrum must be lower than 95C since that is feed temperature.

First Trial: Guess d,1 1 2 3T 70 C : K 7.8, K 1.07, K .083

Guess V F 0.5 . Rachford Rice Eq.

7.8 1 .517 .07 .091 .083 1 .392

f V F .141 6.8 .5 1 .07 .5 1 .083 1 .5

V F .6 gives f .6 .101

By linear interpolation V F .56. f 0.56 .0016 which is close enough for first

trial.

V V F F 56, L 44

ii i i ii

zx and y K x

1 K 1 V F

1 2 3x .1075 x .088 x .806 x 1.001

1 2 3y .839 y .094 y .067 y .9999

Data: Pick refT 25 C . (Perrys 6th ed; p. 3-127), and (Perrys 6th ed; p. 3-138)

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 pL1T 0 C, C 0.576 cal / (g C) 44 25.34 kcal/(kmol C) .

For pL3T 20 to 123 C, C 65.89 kcal/(kmol C)

at pL2T 75 C, C 39.66 kcal/(kmol C) . (Himmelblau/Appendix E-7)

2pv C a bT cT

propane a = 16.26 b = 5.398 10-2 c = -3.134 10-5 n-pentane a = 27.45 b = 8.148 10-2 c = -4.538 10-5 **n-octane a = 8.163 b = 140.217 10-3 c = -44.127 10-6 ** Smith & Van Ness p. 106 Energy Balance: E(Td) = VHv + LhL FhF = 0

FFh 100 .577 25.34 .091 39.66 .392 65.89 95.25 297,773 kcal/h

LLh 44 .1075 25.34 .088 39.66 .806 65.89 70.25 117,450

2v

2

3

VH 56 .839 3597.4 16.26 5.398 10 45

0.94 6302.88 27.45 8.148 10 45

0.67 9895.3 8.163 140.217 10 45 240, 423

drumE T 60,101 Thus, Tdrum is too high.

Converge on drum 1 2 3T 57.2 C : K 6.4, K .8, K .054

For V F 0.513, f 0.513 0.0027. V 51.3, L 48.7

52

1 2 3 1x .137, x .101, x .762, x 1.0000

1 2 3 1y .878, y .081, y .041, y 1.0000

F L v drumFh 297,773; Lh 90,459; VH 209,999; E T 2685

Thus Tdrum must be very close to 57.3C. 1 2 3x .136, x .101, x .762

1 2 3y .328, y .081, y .041

V 51.3 kmol/h, L 48.7 kmol/h Note: With different data Tdrum may vary significantly.

2.F4. New Problem. yV Lx Fz or L F

y x zV V

L

V F 0.4, V 4kmol h , L 6, 1.5 slopeV

F

x 0, y z 2.5 .25 0.625V

Find: V = 4 kmol/h, L = 6 kmol/h. From the graph, x = 0.19 y = 0.34 Equilibrium is from NRTL on Aspen Plus.

53

FIGURE 2.F.4.

2.G1. Used Peng-Robinson for hydrocarbons.

Find drumT 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. Used Peng-Robinson. Find drumT 30.11 C, L 31.348, V 68.66 kmol/h.

In same order as 2.G1,

i ix y are: 0.0189 0.1123 , 0.0906 0.3023 , 0.3255 0.3495 , 0.0402 0.0501 , 0.5248 0.1858 .

54

2.G3. Used NRTL-2. drumT 79.97 C , M Mx 0.2475, y 0.6287 . Compares to graph with

M Mx 0.18 and y 0.55 . Different equilibrium data.

2.G4. New Problem.

COMP x(I) y(I) METHANE 0.12053E-01 0.84824 BUTANE 0.12978 0.78744E-01 PENTANE 0.29304 0.47918E-01 HEXANE 0.56513 0.25101E-01 V/F = 0.58354

2.G5. New Problem. Used NRTL. T = 368.07, Q = 14889 kW, 1st liquid/total liquid = 0.4221,

Comp Liquid 1, x1 Liquid 2, x2 Vapor, y Furfural 0.630 0.0226 0.0815 Water 0.346 0.965 0.820 Ethanol 0.0241 0.0125 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 x(I) y(I) METHANE 0.000273 0.04959 BUTANE 0.18015 0.47976 PENTANE 0.51681 0.39979 HEXANE 0.30276 0.07086 V/F = 0.40001

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 i-butane

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

T deg R 509.688 p psia 36.258 F 150

zM 0.5 z iB 0.1 z np 0.15 znhex V/F 0.602698586

0.25

55

guess KM 51.86751896

KiB 0.926804057 KnPen 0.175621816 KnHex 0.05400053

Use goal seek for cell B24 to = 1.0 change B9

xM 0.015793905 xib 0.104615105 xnPen 0.29812276 xnHex 0.581601672 Sum 1.000133443 RR M 0.803396766 RR nB -0.007657401 RRnP -0.2457659 RRnHex -0.550194874 sum RR -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

56

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. Simultaneous solution is likely when one of the key variables can be found only from the energy balances. For example, if only 1 of D Bx , x , D, B, A distFR 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.

3.B1. a. D Bx , x , opt feed, RebQ

D Bx , x , opt feed, CQ

D Bx , x , opt feed, S (open steam), satd vapor steam

All of above with fractional recoveries set instead of D Bx , x

D, Bx , opt feed, L/D

b. N, FN , col diameter, frac. recoveries both comp.

N, FN , col diameter, A oFR dist, L D

N, FN , col diameter, A RFR dist, Q

N, FN , col diameter, A CFR dist, Q

N, FN , col diameter, D C Bx , Q or x

N, FN , col diameter, S (satd steam), satd vapor steam, D Bx or x

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. mix 1 2F F F D B

mix mix 1 1 2 2 D BF z Fz F z Dx Bx (Mole frac. MVC)

1 1 2 2mix

mix

Fz F zz

F

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

mix Bmix

D B

z xD F

x x

kmol/h.

mixB F D kmol/h.

3.C4. New Problem in 3rd Edition. See solution to 3D4, Part b.

57

3.D1.

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

a) o C 0 D 1L

3, Eq 3-14 Q 1 L / D D h HD

Dh is a saturated liquid at Dx 0.85 wt. frac. From Fig. 2-4, Dh ~ 45 kcal/kg

1H is saturated vapor at D 1 1x y 0.85, H ~ 310 kcal/kg

CQ 1 3 764.62 45 310 810,497 kcal/hour EB around column.

1 21 F 2 F col C R D BFh F h Q Q Q Dh Bh

1 2F F

h 81 C, 60 wt% ethanol ~ 190kcal / kg; h 20 C, 10 wt% ethanol ~ 10kcal / kg

Bh (satd liquid leaves equil contact, ~ 0 wt% ethanol) ~ 100 kcal/kg, Qcol = 0 (adiabatic)

RQ 764.62 45 735.38 100 1000 190 500 10 810,497 657,259 kcal/kg

(b) V B 2.5 mass.

1F

2F

DD, x

L V RQ

BB, x

oL

1V

1

CQ 1 2 BF F B D x 0.0001 0.01%

1 1 2 2 B D DFz F z Bx D x x 0.85

avg Btotal

D B

z xSolve D F

x x

total 1 2F F F 1500 kg/h

0.43333 0.0001

D 1500 764.62 kg/h0.85 0.0001

total

kgB F D 1500 764.62 735.38

h

1 1 2 2avg

total

1000 .60 500 0.10Fz F zz 0.43333

F 1500

58

Approximately B V Lx ~ y ~ x . Thus BLh h 100 . VH 640kcal kg

RQ 1838.45 640 735.38 100 2573.83 100 992,763 kcal / h EB.

1 2C D B col 1 F 2 F RQ Dh Bh Q Fh F h Q

CQ 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 + S D BSy Dx Bx (2)

Note: Sy 0

energy bal: f s C D BFh SH Q Dh Bh (3)

Condenser: mass bal. : 1 oV L D (4)

energy bal.: 1 1 C o DV H Q L D h (5) Solve Eqs. (1) and (2) to get:

B B

D B

100 .3 100 .05 100 .05Fz Fx SxD 36.4

x x .6 .05

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

C D 1Q D 1 L D h H

Substitute CQ into Eq. (3):

D B F

D 1

Dh F S D h Fh SHL1

D D h H

S

From Figure 2-4: F D B 1h 8, h 65, h 92, H 638, H 608 kcal/kg. S

36.4 65 163.6 92 100 8 100 638L D 1 2.77

36.4 65 408

3.D3. External balances: F + C = B + D (1) C B DFz Cx Bx Dy (2)

R F C B DQ Fh Ch Bh DH (3)

R BL VLh Q VH Bh

L

Reboiler

Satd liqd

B, Bx =0.0001

RQ

(Satd vapor) V

Satd liqd

L V B L 1838.45 735.38 2573.83

V B 2.5 or V 2.5B 1838.45

59

F = 2000, C = 1000, z = .4, C B Dx 1.0, x .05, y .80, Fh 20 C 30.7, Ch sat 'd liquid 50, B Dh sat 'd liquid 92, H sat 'd vapor 327 kcal/kg

Around reboiler: L V B N Reb BLx Vy Bx

N R reb BLh Q VH Bh

For a total reboiler: N B N N B N Bx x , y x x , h h 92

M.B.: N R reb BV B h Q VH Bh

or R B Nreb N

QV since h h

H h

RebH 617 (saturated vapor at N 1y 0.05 )

Solve Eqs. (1) and (2) for B: C D D

B D

Fz Cx Fy CyB

x y

Thus

800 1000 1600 800B 800 and D 2200

.05 .8

From Eq. (3), R B D F CQ Bh DH Fh Ch

RQ 800 92 2200 327 2000 30.75 1000 50 804,500 cal/h

R

reb N

Q 804500V 1532.4 kg/h

H h 617 92

3.D4. New Problem in 3rd Edition. F B D

MVC B DFz Bx Dy But given recoveries. Thus, use:

F,M D,MFz (Frac Rec Methanol in distillate) Dy

and F,W B,WFz (Frac Rec water in bottom) Bx

F,M F,W F,Mz 0.3, z 1 z 0.7

D,M B,Wy unknown, x unknown.

D,MMethanol 29.7 100 0.3 .99 Dy

If 99% methanol recovered in distillate, 1% is in bottoms B,M0.3 100 0.3 0.01 Bx Water B,W68.6 100 0.7 0.98 Bx 2% water in distillate

D,W1.4 100 0.7 0.02 Dy

Since ix 1 and i D,i B,iy 1, Dy D, and Bx B Thus, D,M D,WD Dy Dy 29.7 1.4 31.1 kmol h

B,M B,WB Bx Bx 0.3 68.6 68.9 kmol h

Check: B D 100 F OK

60

D,Md,MDy

y 29.7 31.1 0.955D

B,M B,Mx Bx B 0.3 68.9 0.00435

a) 0D 31.1 & L D 2. Thus 0L 62.2

Reflux liquid is in equilibrium with vapor D,My 0.955

From equilibrium data (Table 2-7) M,0x ~ 0.893 (linear interpolation)

b) E.B. Partial condenser: 1 1 c 0 0 0V H Q DH L h

1 0V D L 93.3

1 1 0 0 0 1 0 0 0 1V y Dy L x y Dy L x V

1,My 29.7 62.2 0.893 93.3 0.914; 1,W 1,My 1 y 0.086 M 35,270J mol@64.5 C 35,270 kJ / kmol choose MeOH reference 64.5C.

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

1 M 1,M W 1,W 1,M P,V,M 1 1,W P,V,W 1H y y y C T 64.5 y C T 100

1V is at 1T in equilibrium with 1,My 0.914. From Table 2-7 1T ~ 67.6 C

Assuming only constant & linear T term are important in P,VC eqs., P,V P avgC C T . For

methanol avg67.6 64.5

T 66.65 C2

. For water, avg

67.6 100T 83.8 C

2

.

PV,M

J 1000 mol kJ kJC 42.93 0.08301 66.05 48.41 48.41

mol kmol 1000J kmol C

P,V,W oJ kJ

C 33.46 0.00688 83.8 34.04 34.04mol C kmol C

MT 67.6 64.5 3.1 ; WT 67.6 100 32.4 Then

1H 35270 0.914 40656 0.086 48.41 3.1 0.914 34.04 32.4 0.086

1H 35775.5 kJ kmol Note terms dominate. DH is at Dy 0.955@66.1 C

D M D,M W D,W D,M P,V,M D D,W P,V,W DH y y y C T 64.5 y C T 100

P,V,M P,V,M M,avgC C T . avg,M66.1 64.5

T 65.32

P,V,MC 42.93 0.08301 65.3 48.35

P,V,W P,V,W W,avgC C T . W,avg100 66.1

T 83.052

P,V,WC 33.46 0.00688 83.05 34.03

WT 66.1 100 33.9

61

DH 35270 .955 40656 0.045 48.35 0.955 1.6 34.03 0.045 33.9 DH 33682.85 1829.5 73.88 51.91 35534.3

Reflux liquid at 66.1C and M,0 W,0x 0.893, x 0.107

Reference MeOH 64.5C, water reference 100 C 0 PL,M M,0 M PL,W W,0 Wh C x T C x T

PL,M PL,M avgkJ

C C T 75.86 0.1683 65.3 86.85kmol C

0h 86.85 0.893 1.6 75.4 0.107 33.9 149 kJ kmol C D 0 0 1 1Q DH L h V H 31.1 35534.3 62.2 149 93.3 35775.5 2,242,030 kJ h

Overall EB F C R D BFh Q Q DH Bh or R D B C FQ DH Bh Q Fh

Bh is saturated liquid with B,Mx 0.00435 and B,Wx 0.99565

Interpolating in Table 2.7 BotT 99.2 C

B PL,M PL,Wh C 0.00435 99.2 64.5 C 0.99565 99.2 100

avg,M99.2 64.5

T 81.862

and PL,MC 75.86 0.1683 81.86 89.64

Bh 13.53 60.06 46.5 kJ kmol Feed is saturated liquid at M Wz 0.3, z 0.7.

From Table 2-7, FT 78 C

F P,LM PL,Wh C 0.03 78 64.5 C 0.7 78 100

avg,MT 78 64.5 2 71.25 and P,L,MC 75.86 0.1683 71.25 87.85

Fh 805.4 kJ kmol Then

RQ 31.1 35534.3 68.9 46.5 2,242,030 100 805.4

1,105,116 3204 2, 242,030 80,536 3,424,479 kJ h

3.D5. Mass Balances: F = D + S + B, D S BFz Dx Sx Bx Solving simultaneously, B = 76.4 kg/min, D = 13.6 kg/min.

Condenser: C 1 0 1Q V h H

1 0V L D L D 1 D 4 13.6 54.4 kg/min

From Figure 2-4, 0h 7.7 kcal/kg (x = .9, T = 20C),

1H 290 kcal/kg (y = .9, satd vapor).

Thus, CQ = 54.4 (7.7 290) = -15,357 kcal/min

Overall Energy Balance: F R C D S BFh Q Q Dh Sh Bh

R D S B F CQ Dh Sh Bh Fh Q From Figure 2-4,

S S Fh 61 x .7, sat'd Liq'd ; h 200 z .2, 93 C ,

62

B B D oh 99 x .01, sat'd Liq'd , h h 7.7 Thus,

RQ 13.6 7.7 10 61 76.4 99 100 200 15357 3635.3 kcal/min

3.D6. From Eq. (3-3), D = F B

D B

z x .4 .0022500 998

x x .999 .002

lbmol/h.

Then B = F = 1502.

Condenser: o oV L D L D D D

C D V oQ h H D L D 1

With 99.9% 5nC have essentially pure 5nC . Thus, it is at its boiling point.

D V C5h H 11,369 Btu/lbmol.

CQ 11,369 998 4 45,385,048 Btu/h

Overall: R D B F CQ Dh Bh Fh Q

Distillate is at boiling point of pure 5 C5nC K 1.0 on DePriester Chart) = 35C. Bottoms

is at boiling point of 6 C6nC K 1.0 67C. Converting to F: 35C = 95F, 67C = 152.6F, 30C = 86F. Note feed is obviously a subcooled liquid. Arbitrarily, pick a liquid at 0F as reference.

(This will not affect the result and other values can be used.)

F LC5P C5 P C6 PLC6C x C z C

FP

C .4 39.7 .6 51.7 46.9 Btu/lbmol F

FF P F

h C T 0 46.9 86 4033.4 Btu/lbmol

Distillate is almost pure 5nC . Liquid at 95F

LC5D P Dist

h C T 0 39.7 95 3771.5 Btu/lbmol

Bottoms is almost pure liquid 6nC at 152.6F.

pLC6 bot

hC T 0 51.7 152.6 7889.4 Btu/lbmol

RQ 998 3771.5 1502 7889.4 2500 4033.4 45,385,048 50,861,491 Btu/h

3.D7. Eq. (3-3), B

D B

z x 0.7 0.001D F 1000 700.4

x x 0.999 0.001

kmol/h

B F D 299.6 kmol/h Condenser: o oL L D D 2.8 700.4 1961.1 kmol/h Only this reflux is condensed since product is a vapor.

C oQ L where is for essentially pure n-pentane.

Ckmol Btu 2.20462 lbmol

Q 1966.1 11,369h lbmol 1 kmol

10C -4Btu 1 J J

Q 49,154,204.85 5.18176 10 h 9.486 10 Btu h

63

From overall balance R D B F CQ DH Bh Fh Q

Distillate is vapor at b.p. of pure n-pentane (35C from DePriester chart, C5K 1.0 ) Bottoms is boiling n-hexane (67C) Conversions: 35C = 95F - distillate & Feed and 67C = 152.6F - bottoms As reference, arbitrarily choose liquid at 0F. Feed is subcooled liquid.

oPF C5 PLC5 C6 PLC6C z C z C 0.7 39.7 0.3 51.7 43.3Btu lbmol F

F PF Fh C T 0 43.3 95 0 4113.5Btu lbmol

Distillate D C5 PLC5 distH C T 0

DH 11,369 39.7 95 0 15,140.5 Btu lbmol

Bottoms is pure 6C @152.6 F

B PLC6 both C T 0 51.7 152.6 0 7889.4 Btu lbmol

R

kmol Btu kmol BtuQ 700.4 15,140.5 299.6 7889.4

h lbmol h lbmol

kmol Btu 2.20462 lbmol Btu1000 4113.5 49,154,204.85

h lbmol kmol h

10R -4Btu 1 J

Q 68,675,167.9 7.240 10 J hh 9.486 10 Btu

3.D8. New Problem in 3rd Edition.

E wF 300, z .3, z .7

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

D Dist. .98 90 1 .81 300 .7 128.1 kmol/h

DE

.98 90y 0.6885

128.1

B Bottoms .02 90 .81 210 171.9kmol h

b. Partial Condenser.

0 0L

2, L 2D 2 128.1 256.2 kmol h.D

0x in equilibrium with 0y , thus from equation data 0x 0.575.

Entering vapor 1y (from graph) 0.61

D Dy ,D,H

cQ

0 0 0x ,L ,h

1 1 1H y V Vapor

64

1 0V L D 256.2 128.1 384.3 kmol h.

c. E.B. on PC. 1 1 c dist 0 0V H Q DH L h . Can use Figure 2-4 by converting mole fracs to mass fracs. Basis 1 kmole.

Distillate .6885 mol E MW 46 31.671 kgE

5.607 kgW

.3115 mole W MW 1837.28 kg total

Mass frac. E = 31.671 37.28 0.8496

Vapor 1V 0.61 mole E 46 28.06 kgE

7.02 kgW

.39 mole W 1835.08 kg total

Mass frac E = 28.06 35.08 0.7999

Liquid reflux 0L 0.575 mole E 46 26.45 kgE

7.65 kgE

0.425 mole W 1834.1 total

Mass frac E = 26.45 34.1 0.7757

From Figure 2-4, dist 1 0H ~ 310 kcal kg, H ~ 330kcal kg, h ~ 65 kcal kg

c dist 0 0 1 1

kmol 37.28 kg kcalQ DH L h V H 128.1 310 256.2 34.1 65

hr kg kg

35.08 kg 364.3 330 2, 400,517 kcal hr

kmol

Overall EB. F R c dist BFh Q Q DH Bh

Know cQ 2,400,517 kcal h

and distDH 1,480,426 kcal h.

To find F BFh and Bh , need to convert mole frac to wt frac. Basis 1 kmol

Feed 30 mole % E: .3 mole 46 13.8

12.6

70% W : .7 18total 26.4 kg kmol

Mass frac E 13.8 26.4 0.5227

Bottoms 0.01047 mole 46 .48162

17.811

0.98953 mole 18total 18.293 kg kmol

Mass frac E 0.48162 18.28 0.0263

From Figure 2-4 Fh satd liqd 70kcal kg

Bh satd liqd 97

65

Then R dist B F cQ DH Bh Fh Q

R

kmol 18.29316 kg 97 kcal 26.4 kg kcalQ 1,480,426 171.9 300 70

h kmol kg kmol kg

2,400,517 1,480,426 305,525 554,408 2,400,517 3,632,069kcal h 3D9. New Problem 3rd Edition. B = (xD z)/(xD xB)F = [(0.9999 - 0.76)/(0.9999 0.00002)](500) = 120

R BQ Lh VH Bh and L V B

Assume Bh h . RQ L B h Vh V H h V

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

6RQ 180 kmol/h 9720 1.750 10 kcal h 3.D10. 2 atm 101.3 kPa/atm = 202.6 kPa. Pentane Recovery: P D0.995 Fz Dx

0.995 1000 0.55D 547.6333

0.9993 kmol/h

B = 1000 547.6333 = 452.3667

Since B pBx Pentane Recovery Bot F z ,

B

1 .995 1000 0.55x 0.006079

452.3667

mol frac pentane

Distillate is essentially pure Pentane. Bottoms Pure in Hexane. From DePriester Chart

P distK 1@p 202.6 kPa when T 59.5 C

n H botK 1@p 202.6 kPa when T 94 C

For Total Condenser, Eq. (3-14) oC D 1L

Q 1 D h HD

D PLC5 dist ref refh pure pentane C T T choose T 25 C

Dkcal kcal

h 39.7 59.5 25 1369.65kmol C kmol

1 DH h assuming is independent of temperature

1Btu 1 lbmol 0.252 kcal kcal

H 1369.65 11369 7680.196lbmol 0.454 kmol Btu kmol

Eq. (3-14) is Ckmol kcal

Q 1 2.8 547.6333 6310.5 13,132,288h h

B PLC6 bot refkcal kcal

h pure hexane C T T 51.7 94 25 3567.3kmol C kmol

Feed is a liquid at 65C

F PC5 C5 PC6 C6 F refh C z C z T T

Fh 39.7 0.55 51.7 0.45 65.25 1804kcal / kmol

66

R D B F CQ Dh Bh Fh Q

RQ 547.6333 1369.65 452.3667 3567.3 1000 1804 13,132,288

RQ 13,692,081 kcal/h.

Note that C RQ and Q are relatively close.

3.E1. Was 3.D8 in 2nd Edition. Pick as basis liquid at 0C, W Mh 0 & h 0 (essentially steam table choice) Assume ideal mixtures.

avg avg W,L M,LF P P ref P P P

h C T T where C 0.4 C 0.6 C

Felder & Rouseau p. 637 PWC 0.0754 kJ/mol

5PMC 0.07586 16.83 10 T kJ/mol

PM PM avg PM PMC C T C 0 40 2 C 20 C 5PMC 0.07586 16.83 10 20 0.079226

avgP

C 0.4 0.0754 0.6 0.079226 0.077696

Fh 0.077696 40 0 3.1078 kJ/mol feed = 3107.8 kJ/kmol Can also use steam table for water

SH is satd vapor steam 1 atm, SkJ 18.0 kg

H 2676.0kg kmol

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

Dh is satd liquid at Dx 0.99 . From Table 2-7, T = 64.6C

avg avgD P P PW PM

h C 64.6 0 where C 0.01 C 0.99 C

5PM PM avg PMC C T C 32.3 C 0.07586 16.83 10 32.3 0.081296

avgPC 0.01 0.0754 0.99 0.081296 0.08124

Bx 0.02 B

CQ

p = 1.0 atm

Mz 0.6

100 kmol h

F

DD, x 0.990

M.B. F + S = D + B

M S,M D BFz Sy Dx Bx

Since steam is pure, S,My 0

Know M D BF, z , x , x

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

F C S D BFh Q SH Dh Bh

S

67

DH 0.08124 64.6 5.2479 kJ mol 5247.9 kJ kmol

Bh : Since leaving an equilibrium stage, satd liqd. 2% MeOH Table 2-7, T = 96.4C

avg avgB P P PW PM

h C 96.4 0 where C 0.98 C 0.02 C

PM PM PM C C 96.4 0 2 C 48.2 C 5PM C 0.07586 16.83 10 48.2 0.08397

avgP

C 0.98 0.0754 0.02 0.08397 0.07557 kJ mol

Bh 0.07557 96.4 7.28509 kJ mol 7285.09 kJ mol CQ do EB around condenser

oC dist 1 dist o distL

Q V L D 1 DD

dist MeOH W0.99 0.01

Felder & Rousseau: M W35.27 kJ mol & 40.656 kJ mol

distkJ

0.99 35.27 0.01 40.656 35.324 35,323,86 kJ kmolmol

CQ 35,323.86 2.3 1 D 116,568.7D kJ h Plug CQ & 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: D = 56.33 kmol/h, B = 211.71 kmol/h CS 168.04 kmol/h, Q 6,566,000 kJ/h.

E2. Was 3.D9 in 2nd Edition. kmol mol

F 500 500,000h h

F + S = D + B S D BFz Sy Dx Bx 2 eq. 3 unknowns

Condenser: C 0 o 1Q 1 L / D D h H Note Eq (3-14) not valid. For enthalpy pick reference pure liquid water 0C and pure liquid methanol 0C. Felder &

Rouseau: MeOHP

C 75.86 0.01683T at MeOHavg P

64.5 0T 3225, C 76.4 J mol C.

2

68

Assuming distillate pure methanol, boils at 64.5C

Meoh ,liqD P ref

h C T T 76.4 J/mol 64.5 0 4928.0 J mol

1 D 1H h at 64.5 F 4928 35270 J/mol 40,198 J/mol

CQ 4 4928 40198 D 141,080D J/h where D is mol/h

Overall Energy balance: F S C D BF h SH Q Dh Bh

Bottoms is essentially pure 2H O at 100C

W,liqB P ref

J Jh C T T 75.4 100 0 7540

mol C mol

S B WH h at 100 C 7540 40656 J/mol 48196 For feed. 60 mole % Methanol boils at 71.2C (Table 2-7).

iF P i ref

h C z T T 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) D BFz Dx Bx or (500,000) (.6) = 0.998D + 0.0013B

(3) oC D 1 CL

Q 1 D h H or Q 141,080DD

(4) F S C D BFh Sh Q Dh Bh or (500,000) (54137) + S (48196)

+ CQ = D(4928) + B (7540) Solve simultaneously: D = 298.98, B = 1245.5, S = 1044.2 kmol/h CQ = - 4.218 10

+7 kJ/h

3.F1. An enthalpy composition diagram is available on p. 272 of Perrys Chemical Engineers

Handbook, 3rd ed., 1950.

Eq. (3-3) BD B

z x 0.79 0.004D F 25,000 19,788.5

x x 0.997 0.004

kmol/h

Note that 2N mole fractions were used since 2N is more volatile. B = F D = 5211.5

From enthalpy comp. diag. D 1h 0, H 1350 kcal/kmol, B Fh 160, h 1575 . Then,

C o D 1Q 1 L D D h H 5 19788.5 0 1350 133,572,000 kcal/h

R D B F CQ Dh Bh Fh Q

R CQ 0 5211.5 160 25,000 1575 Q 95,030,000 kcal/h 3.F2. We will use the enthalpy composition diagram on p. 3-171 of Perrys 6th edition or p. 3-158 of

Perrys 5th ed.Do for 1 kmol of feed: Conversion of feed from kg to moles. Basis 100 kg 30 kg 3NH = 1.765 kmol

70 kg 2H O 3.888 Total 5.653 kmol Thus 1 kmol is 100/5.653 = 17.69 kg

69

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

B Dx Fz Dx B 17.69 .3 5.15 98 12.54 0.021

From diagram: D 1 B Fh 55, H 415, h 150, h 5 kcal/kg

Eq. (3-14), C o D 1Q 1 L D D h H 3 5.15 55 415 5562 kcal/kmol feed

and R D B F CQ Dh Bh Fh Q

R CQ 5.15 55 12.54 150 17.69 5 Q 7815 kcal/kmol of feed G1. a.) Using NRTL. C RQ 778,863 kcal/h, Q 709,520 kcal/h

b.) C RQ 1,064,820 kcal/h, Q 995,478 kcal/h G2. New Problem in 3rd Edition. ASPENPlus. D = 988, L/D = 3, Peng Robinson, FN 40, N 20 (arbitrary values in Radfrac)

7D DC6x 1.000 x 1.211 10

Bx 0.0013316

7

C7

R

Q 4.4426 10 Btu h,

Q 4.9852 10 Btu h

70

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 2 2K y / x , 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. New Problem in 3rd Edition. Answer is a.

4A7. See Table 11-3 and 11-4 for a partial list.

4A13. 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

4A14. 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 sufficientfor 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. Fewerstages, 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 Dz, y satisfies operating equation.

Solution: Op. Eq. By L V x L V 1 x

Substitute in Dy y , x z

D By V Lz L V x

But q 1.0, V D, L F, L V B

D By D Fz Bx

Which is external mass balance. QED.

71

Can do similar for enriching column for a saturated vapor feed. 4.C10. New Problem in 3rd Edition. If we consider , the latent heat per mole to be a positive quantity,

then RQ V . With CMO and a saturated liquid feed (1 / )V V L D D , and then

/ (1 / )RQ D L D .

4.C16. New Problem in 3rd Edition. Define a fictitious total feed T T TF , z , h

T 1 2F F F , 1 1 2 2

T

T

F z F zz

F, 1 21 F 2 FT

T

F h F hh

F

Intersection of top & bottom operating lines must occur at feed line for fictitious feed FT. (Draw a column with a single mixed feed to prove this.)

This feed line goes through Ty x z

b.) Does 1 1 2 2TT

q F q Fq

F

x Bx

B

Tz

y

0z A

2z

1z

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

opt feed locations, 1 2 1 2 F1 F2z , z , F , F , h , h

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

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

bot. op. line. Connect pts B & C to get

middle op. line.

TF

with slope T Tq q 1

where mix TTmix mix

H hq

H h and mix, mixH h are saturated

vapor and liquid enthalpies at feed stage of column with

mixed feed.

72

check

1 2

1 2

1 F 2 F

mix1 2 mix 1 F 2 FTmix T

T

mix mix mix mix mix mix T

F h F hH

F F H F H F hFH hq

H h H h H h F

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

1 21 mix F mix F

2

mix mix mix mix

T

T

F H h H hF

H h H hq

F

Usual CMO assumption is >> latent heat effects in either vapor or liquid.

Then 1 2mix F mix F1 2mix mix mix mix

H h H hq and q

H h H h

Thus 1 1 2 2TT

F q F qq

F if CMO is valid.

4D1. a. Top op line: DL L

y x 1 xV V

and L L D 1.25

0.5555V 1 L D 2.25

Intersects Dy x x 0.9

When DL

x 0, y 1 x 0.4V

Plot See diagram

b. Bottom op line: BL L

y x 1 xV V

, and L V B V B 1 3

V V V B 2

Intersects y = x = xB = 0.05

1 0.5 / 2

@ y 1 x 0.683 this is convenient point to plot3 2

c. See diagram for stages. Optimum feed stage is #2 above partial reboiler. 5 equilibrium stages + PR is more than sufficient.

73

d. Feed line goes from y = x = z = 0.55 to intersection of two operating lines.

q

Slope 1.0 or q 0.5q 1

.

This is a 2 phase feed which is liquid & vapor. 4D2. New Problem in 3rd Edition. Part a.

E

L F V 1 V / F .63y x z .6 Slope 1.703

V V V F .37

b. From Table 2-1, at 84.1 C y .5089

c. liquid at 20C FH h

qH h

and 40 mole % ethanol.

The pressure in Figure 2-4 is very close to 1.0 atm, thus it can be used, but must convert to wt frac.

74

Basis 1kmol feed.

.4 kmole E .4 MW 46 18.4 kg 0.63 wt frac.

10.8 kg

.6 kmol Water .6 MW 18total 29.2 kg

From Figure 2-4 FH 398 kcal kg ,h 75,h 20 C 10

398 10

q 1.20398 75

q 1.2

Slope 6q 1 .2

Alternate Solution: 40 mole % ethanol boils at 84.1C (Table 2-1). Then if pick reference as saturated liquid at 40 mole %

F p,40%liqh C 20 84.1

40%E

h 0, H

d. vaporF p

kcal40 mole %E 63 wt%, H 398 kcal kg , h 65, h 398 C 120 84.1

kg

vapor Evapor w ,vaporP E P w P

C y C y C

Assume only 1st and 2nd terms in PC equations are significant. From Problem 2.D9

vaporPC .4 14.66 0.03758T .6 7.88 .0032T

kcal/kmol T is C which simplifies to

vaporPC 10.592 0.16952T

For linear 120

p

84.1

C dT is equal to vaporP avg

C @T

avgT 84.1 120 2 102.05 . Then v ,avgPkcal

C 10.592 0.16952 102.05 12.32kmol

Fkcal kcal 1 kmol

h 398 12.32 120 84.1kg kmol 27.2 kg

Fh 398 15.149 413.15kcal kg

398 413.15 15.147

q 0.045.398 65 333

e. F 13

q L L f , L L F L F12 12

13 12

q 13 12, slope q q 1 131 12

f. Flash V L 1 V F .3 3

.7, F V V F .7 7

See graph for feed lines.

75

Graph for 4.D2

76

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

200C. Assume Cp,liq is not a function of T. Estimate,

L

P,liq

h 60 C h 20 46.1 23h kcalC .63 wt frac ~ 0.864

T 60 20 80 kg C

Then LF L P L

h h 200 C 200 60 h 60 C .864 200 60 46.1 167.1

v F

v L

H h 395 167.1 q 0.691q 0.691, 2.24

H h 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

F P,liq F ref P,liqh C T T C 250 0

wP,liq P,EtOH EtOH P w

C C z C z 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

P,liqC 37.96 .281 18.0 .719 23.61

PP,liqAVG

C 23.61C in kcal kg C 0.911

MW 25.91. Then,

Fkcal

h 0.911 250 C 228kg C

v F

v L

H h 446 228q 0.58

H h 446 70

4.D4*. a. F Pvh H C 350 50 H 25 300

F25 300H h

q 1.5H h

Slope q q 1 0.6. y x z 0.6 is intersection.

b. q L L F where L L 0.6F. Then q L 0.6F L / F 0.6, and

slope q q 1 1.5

c. q L L F where L L F 5. q L F 5 L F 1 5 , slope q q 1 1 6

z

y = x

feed line

.5 .6 .7

4.D4a

77

4.D5*. liq refluxL

vap liq

h h 3100 1500f 0.1111

H h 17500 3100

0 0

1 0

L L D 1.10.524

V L D 1 2.1

c 0 11

2 c 0 1

1 f L V 1.1111 .524L0.55

V 1 f L V 1 .111 .524

Alternate Solution

For subcooled reflux, 01

0 1

H hLq

L H h

17500 15001.111

17500 3100

Then, 1 0 0L qL 1.1111 L

1 1 1

2 1 1

L L L D

V L D L D 1, 01

1.111 LL1.111 1.1 1.2222

D D

1

2

L 1.2220.55

V 2.222

4D6. New Problem in 3rd Edition. a) 1 2175 F F 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. 1q 1, vertical at 1y x z 0.6

Feed 2. 60% vapor = 40% liquid 2q 0.4

Slope feed line 2

2

q 0.42 3

q 1 .06 through 2y x z 0.4

Bottom Op. Line By L V x L V 1 x . Through By x x

V B 1

Slope L V 3 2V B

Middle 2L F B V

2 2 B2 2 BF z BxL

L x F z Bx V y y xV V

When 2 2 BF z Bx

x 0, y , Slope L VV

Also intersects bot. op. line and Feed line 2.

Do External Balances and Find D & B. Then V V / B B 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 0.126241.25

Plot Middle Op Line.

78

DL L

y x 1 xV V

Know that Dy x x and gives through interaction Middle and Feed line 1.

Also, 1L L F 231.875 75 156.875 and V V 241.25 ; thus,

L V 156.875 241.25 0.65 c) See graph.

Graph for 4D6.

79

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

.8 , x y x .9.V

Step off stages as shown on Figure.

b. Plot bottom op. line: slope BL V

1 1 2 , x y x 0.13.BV

Step off stages

(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.

slope 9 q

4 q 1 gives q = 0.692.

4.D8*. The equilibrium data is plotted and shown in the figure. From the Solution to 4.D7c,

q 0.692 and q q 1 9 4

a. total reflux. Need 5 2/3 stages (from large graph) 5.9 from small diagram shown.

b. min

.9 .462L V 0.660

.9 .236 (see figure)

minmin

min

L VL D 1.941

1 L V

c. In 4.D7, actL V .8

L D 41 L V .2

act minL D Multiplier L D

Multiplier = 4/1.941 = 2.06

80

d. Operating lines are same as in Problem 4.D7. Start at bottom of column. Reboiler is an equilibrium contact. Then use MVE 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)

1 2 1 1F F D B 100 F 80 B F B 20

1 1 2 2 B B 1Fz F z Dx Bx F .42 18 .66 80 0.04 B

Solve simultaneously, 1B 113.68, F 93.68

b) 1 L L D 1 2 1

L D , 2 V 1 L D 3 2 3

L

L D 40, V L D 120D

Saturated Liquid Feed V V 120 1L L F 40 93.68 133.68, L V 1.114

c) Top Op. Line Normal: Dy L V x 1 L V x

Through D2

y x x , Slope 1 3, y intercept .66 .443

Bottom Normal: B By L V x L V 1 x , through y x x

Also through intersection, 2F feed line and middle op. line. 2

2

F

2

F

L .7Feed line F slope

V .3

81

Middle 1 1DF zD

y L V x xV V

(or do around bottom)

Slope L V . Through intersection feed line 1F and top op. line.

Also, D 1 180 .66 93.68 .42Dx F z

x 0, y 0.11212V 120

d)Opt. Feed 2F stage 1 from bottom, Opt feed 1F , Stage 2. 4 stages + PR more than sufficient.

Graph for 4D9.

82

4.D10*. Operating Line DL L L D 4

y L V x 1 x , where .8V V 1 L D 5

Thus, operating line is y = .8x + .192

a. Equilibrium is 111

yyx or x

1 y 1.79 .76y

Start with y1 = .96 = Dx

Equilibrium: 111

y .96x 0.9317

1.76 .76y 1.76 .76 .96

Operating: 2y .8x .192 .8 .9317 .192 0.93736

Equilibrium: 222

y .93736x 0.89476

1.76 .76y 1.76 .76 .93736

Operating: 3 2y .8x .192 .8 .89476 .192 0.9078

b. Generate equilibrium data from: 1.76x

y1 .76x

x 1.0 .9 .8 .7 .6 .5 .4y 1.0 .9406 .8756 .8042 .7253 .6377 .5399

Plot equilibrium curve and operating line. (See Figure). Slope = L/V = .8, y intercept (x = 0)

= 0.192, y = x = Dx = 0.96. Find 6x = 0.660.

4.D11. a) Same as 4.D2 part g. q = 1.0668, slope feed line = 15.97.

b) Top Dy L V x 1 L V x goes through y = x = Dx = 0.99

L D

L V 0.69691 L D

@ x = 0 y = (1-L/V) Dx = (1-0.6969) 0.99 = 0.300

Feed line: Slope q q 1 , y x z 0.6

83

Bottom Op line:

y= 0, x = Bx . 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. min

0.99 0.57L V Slope 0.4242

0.99 0

min min

L L V 0.42420.73684

D 1 L V 1 0.4242

Actual L/D is 3.12 this value.

V L

B S

M,S ByV Lx Sy Bx But M,Sy 0 (Pure steam)

With CMO B L

B

L Ly x x

V V

84

4.D12. L D

L V 3 41 L D

slope. Top op line goes throug Dy x x 0.998

DL L

y x 1 x @ x 0, y .25 .998 0.2495V V

Bottom slope From Soln to 3.D9 orB 1245167

L V 1.19 from graph. 1.169S 1044168

Feed line is vertical at z = 0.6. Can also plot top and feed lines, and then find bottom from 2

points By 0, x x & 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.

85

86

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

b.) See figure. MIN

L 0.665 0.950.4385

V 0.30 0.95

MIN

L L L V 0.43850.7808

D V L 1 L V 0.5615

c.) MIN

L2.0 L D 1.5616

D,

L L D 1.56160.6096

V 1 L D 2.5616

y intersect DL

1 y 0.3709V

. Top operating line DL L

y x 1 yV V

Goes through Dy x y 0.95

Bottom BL L

y x 1 xV V

Goes through By x x & 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.

d.) Slope bottom: See figure for parts c & d. 0.85 0.025

L V 1.9410.45 0.025

1 1

V B V L V 1.0625L V 1 0.941

.

87

Graph for problem 4.D13.

88

4.D14a.

89

4.D14b. Two approaches to answer. Common sense is all methanol leaks out and MAx 0 .

McCabe-Thiele diagram: This is enriching column with sz y 0 . Intersection top op. line and

horizontal feed line is at M,bx x 0 , which is also a pinch point. Thus M,dx 0 also.

4.D15. New Problem in 3rd Edition. Saturated liquid. q

q 1, ,q 1

feed line vertical @ z .3 .

Top operating line DL L

y x 1 yV V

, L L D 2

SlopeV 1 L D 3

Dy intercept 1 L V y 1 3 0.6885 0.2295 and Dy x y

Bottom operating line By L V x L V 1 x goes through By x x

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.

S B

External M.B. S = B

Sys = BxB . Since yS = 0 (pure water)

xB = 0

New S.S.

90

Graph for problem 4.D15.

91

4.D16*. v F

v L

H hL-Lq

F H h

Using 32F = 0C as reference T, Fh 4033.4 Btu/lbmole.

v LH h . Approx. at feed conditions.

.4 11369 .6 13572 12691 Btu/lbmole

For approx. temperature of feed stage, do bubble pt. calc.

1 1 1 1 1y 1 K x K z

Pick T = 48C (~ 40% of way between boiling pts.)

C5 C6 1 1K 1.5, K .54, K x 1.5 .4 .54 .6 .92

C5 new new.54

K T = =.594, T 50 C .92

C5 1 1K 50 C 1.6, K x 1.6 .4 .584 .6 .99 Close enough.

v L feedH h 50 C , 50 C 122 F

feed ,Liqv P

H C 122 46.9 122 5721.8

P feed,liqNote : C 46.9 is from Prob. 3-D6.

vH 5721.8 12691 18412.8 Btu/lbmole

V F

V L

H h 18412.8 4033.4q 1.133

H h 12691

Note: Fh is from Prob. 3.D6.

4.D17. Top Op. Line: D DL L

y x 1 x , goes through y x x 0.9V V

L L D 7 2

7 9V 1 L D 9 2

DL 2

x 0, y 1 x .9 =0.2 V 9

Plot Top. Step off 2 stages. Find Sx ~ 0.81

The vertical line at Sx x 0.81 is the withdrawal line.

Bot. Op. Line intersects Top at Sx x .

Also know it intersects feed line at Bx x (unknown)

External Balances F D B S Dont know BD,B, or x .

D B SFz Dx Bx Sx

Feed enters as saturated vapor. Thus q 0 & V F

Bottoms leaves an equilibrium contact, it is saturated liquid L B Do flow balances

V F 100

V V 100 since S is removed as saturated liquid.

92

L

L V 7 9 100 77.777. D V L 100 77.777 22.222V

L L S 77.7777 15 62.7777. L V 62.7777 100 0.6278

B L 62.777

D SB60 22.222 0.9 15 0.81Fz Dx Sx

x 0.444B 62.7777

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

4.D18. New Problem in 3rd Edition. Feed 1 1F : z 0.6, saturated liquid, q 1,q / (q 1)

93

Feed 2 2F : z 0.4, 80% vapor hence 20% liquid Fq L / F 0.2F / F .2

q .2

1 4q 1 .8

Part a.) Bottom operating line goes through point, By x x 0.04

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

Slope max

L 1.0 .042.2326

V .47 .04

min

V V 1 10.8113

B L V 1.2326L V 1

Part b. 1L F 100

2F

L L L 100 .2 80 116

L V B and V B 1.5

116

116 L 1.5B B B 46.42.5

116

V L B 116 46.4 69.6 L V 1.6666769.6

2F

D V V V 69.6 .8 80 133.6

L 100

0.7485V 133.6

Check overall balance 1 2F F D B 180 133.6 46.4 180.0 OK

To find Dy use MVC mass balance

1 1 2 2 D BFz F z Dy Bx

or 1 1 2 2 BD100 .6 80 .4 46.4 0.4F z F z Bx

y 0.675D 133.6

Actual bottom op. line: BL L

y x 1 xV V

L V B V B 1 2.5 5

V V V B 1.5 3

Goes through By x x 0.04 , Slope 5 3

2nd point y = 1, x = 0.616 (this was arbitrarily found by setting y = 1.) Plot bottom op. line

Top Op. line: 1 1 DyV Fz Lx Dy . D 1 1Dy F zLy x

V V

Goes through intersection feed line for F2 and bot. op. line. Does NOT go through

Dy x y .

Since D & F, passing streams, Point 1 Dz , y is on op. line.

94

Figure for 4D18

4.D19*. B = 0. Then from external balance F = D + B must have D = F = 1000. Acetone balance

becomes D DFz Dx or x z 0.75 .

To predict Bx need operating lines. Top: DL L

y x 1 xV V

DL L V 2

and y x x .75V 1 L D 3

95

Bottom: L V 1.0 . Thus y = x is operating line. From Figure Bx 0.01 to 0.02

Feed line can be calculated but is not needed.

4.D20*. To use enthalpy composition diagram change to wt. fractions. Basis = 1 kg mole Distillate: 0.8 ETOH = (.8)(46.07) = 36,856 0.2 Water = (.2)(18.016) = 3.6032 Total = 40.459 Weight Fractions: EtOH = .911, Water = .089 Feed: 0.32 (EtOH) = (.32)(46.07) = 14.7424 0.68 (W) = (.68)(18.016) = 12.25088 Total = 2