Capitulo 4 del wankat. solucionario

8/10/2019 Capitulo 4 del wankat. solucionario

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SPE 3rdEd. 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: know2 2

K 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 3rdEdition. Answer is a.

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

4A13. New Problem in 3rdEdition.

A. Answer is bB. Answer is aC. Answer is a

D. Answer is a

E. Answer is b

F. Answer is aG. Answer is b

4A14. If feed stage is non-optimum, the feed conditions can be changed to have an optimumfeed 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 pointD

z, y satisfies operating equation.

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

Substitute inD

y 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.


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Can do similar for enriching column for a saturated vapor feed.

4.C10. New Problem in 3rdEdition. If we consider , the latent heat per mole to be a positive quantity,

thenR

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

/ (1 / )R

Q D L D .

4.C16. New Problem in 3rdEdition. Define a fictitious total feedT T TF , z , h

T 1 2F F F , 1 1 2 2

T

T

F z F zz

F, 1 2

1 F 2 F

T

T

F h F hh

F

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 throughT

y x z

b.) Does 1 1 2 2T

T

q F q Fq

F

xB

x

B

Tz

y

0z

A

2z

1z

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

x , x

opt feed locations,1 2 1 2 F1 F2

z , z , F , F , h , h

TF

with slopeT T

q q 1

where mix TT

mix mix

H hq

H h and

mix, mixH h are saturated


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check

1 2

1 2

1 F 2 F

mix

1 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

wheremix 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 h

qF

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

Then 1 2mix F mix F

1 2

mix mix mix mix

H h H hq and q

H h H h

Thus 1 1 2 2T

T

Fq F qq

Fif CMO is valid.

4D1. a. Top op line:D

L Ly x 1 x

V Vand

L L D 1.250.5555

V 1 L D 2.25

IntersectsD

y x x 0.9

WhenD

Lx 0, y 1 x 0.4

V Plot See diagram

b.Bottom op line:B

L Ly x 1 x

V 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.


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

qSlope 1.0 or q 0.5

q 1.

This is a 2 phase feed which is liquid & vapor.

4D2.New Problem in 3rdEdition. 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.


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Basis 1kmol feed.

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

10.8 kg.6 kmol Water .6 MW 18

total 29.2 kg

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

q 1.20398 75

q 1.2Slope 6

q 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%Eh 0, H

d. vaporF pkcal

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

vapor Evapor w,vapor P E P w PC y C y C

Assume only 1stand 2ndterms inP

C equations are significant.

From Problem 2.D9

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

kcal/kmol T is C

which simplifies tovaporP

C 10.592 0.16952T

For linear

120

p

84.1

C dT is equal tovaporP avgC @T

avgT 84.1 120 2 102.05 . Then

v,avgP

kcalC 10.592 0.16952 102.05 12.32

kmol

F

kcal kcal 1 kmolh 398 12.32 120 84.1

kg kmol 27.2 kg

Fh 398 15.149 413.15kcal kg

398 413.15 15.147q 0.045.

398 65 333

e.F 13

q L L f , L L F L F12 12

13 12q 13 12 , slope q q 1 13

1 12

f. FlashV L 1 V F .3 3

.7,F V V F .7 7

See graph for feed lines.


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Graph for 4.D2


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4.D3*. a. Basis 1 mole feed.

0.4 moles EtOH 46 = 18.4 kg EtOH0.6 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). hFis liquid at

200C. Assume Cp,liqis 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

ThenLF 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 wC 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

P

P,liqAVG

C 23.61

C in kcal kg C 0.911MW 25.91 . Then,

F

kcalh 0.911 250 C 228

kg C

v F

v L

H h 446 228q 0.58

H h 446 70

4.D4*. a.F Pv

h 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

feedline

.5 .6 .7

4.D4a


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4.D5*.liq reflux

L

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 .524L 0.55V 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 0

L 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 3rdEdition. 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 1.1

q 1, vertical at1y x z 0.6

Feed 2. 60% vapor = 40% liquid2

q 0.4

Slope feed line 2

2

q 0.42 3

q 1 .06through

2y x z 0.4

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

V B 1Slope L V 3 2

V B

Middle2L F B V

2 2 B

2 2 B

F z BxLL x F z Bx V y y x

V 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 9.625x 0, y 0.126

241.25 Plot Middle Op Line.


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D

L Ly x 1 x

V V

Know thatDy x x and gives through interaction Middle and Feed line 1.

Also,1

L 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.


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

L.8 , x y x .9.

V Step off stages as shown on Figure.

b. Plot bottom op. line: slopeB

L V1 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, andintersection of two operating lines.

slope9 q

4 q 1gives 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)

min

min

min

L VL D 1.941

1 L V

c. In 4.D7,act

L V .8L D 4

1 L V .2

act minL D Multiplier L D

Multiplier = 4/1.941 = 2.06


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d. Operating lines are same as in Problem 4.D7. Start at bottom of column. Reboiler is an

equilibrium contact. Then useMV

E 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 3rdEdition. 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

LL D 40, V L D 120

D

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

ThroughD

2y x x , Slope 1 3, y intercept .66 .44

3

Bottom Normal:B B

y L V x L V 1 x , through y x x

Also through intersection,2

F feed line and middle op. line. 2

2

F

2

F

L .7Feed line F slope

V .3


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Middle 1 1D

F zDy L V x x

V V(or do around bottom)

Slope L V . Through intersection feed line1

F and top op. line.

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

x 0, y 0.11212

V 120

d)Opt. Feed2

F stage 1 from bottom, Opt feed1

F , Stage 2. 4 stages + PR more than sufficient.

Graph for 4D9.


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4.D10*. Operating LineD

L L L D 4y L V x 1 x , where .8

V V 1 L D 5

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

a.

Equilibrium is 11

1

yyx or x

1 y 1.79 .76y

Start with y1= .96 = Dx

Equilibrium: 11

1

y .96x 0.9317

1.76 .76y 1.76 .76 .96

Operating:2

y .8x .192 .8 .9317 .192 0.93736

Equilibrium: 22

2

y .93736x 0.89476

1.76 .76y 1.76 .76 .93736

Operating:3 2

y .8x .192 .8 .89476 .192 0.9078

b. Generate equilibrium data from:1.76x

y1 .76x

x 1.0 .9 .8 .7 .6 .5 .4

y 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 =D

x = 0.96. Find6

x = 0.660.

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

b) TopD

y L V x 1 L V x goes through y = x =D

x = 0.99

L DL V 0.6969

1 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


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Bottom Op line:

y= 0, x =B

x . 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

ButM,S

y 0 (Pure steam)

With CMO B L

B

L Ly x x

V V


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4.D12.L D

L V 3 41 L D

slope. Top op line goes througDy x x 0.998

D

L Ly x 1 x @ x 0, y .25 .998 0.2495

V V

Bottom slope From Soln to 3.D9 orB 1245167L V 1.19from 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

pointsB

y 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.


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4.D13.New Problem in 3rdEdition. a.) See Figure

b.) See figure.

MIN

L 0.665 0.950.4385

V 0.30 0.95

MIN

L L L V 0.4385

0.7808D 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 intersectD

L1 y 0.3709

V. Top operating line

D

L Ly x 1 y

V V

Goes throughDy x y 0.95

BottomB

L Ly x 1 x

V V

Goes throughB

y 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 1V B V L V 1.0625

L V 1 0.941.


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Graph for problem 4.D13.


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4.D14a.


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4.D14b. Two approaches to answer. Common sense is all methanol leaks out andMA

x 0 .

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

horizontal feed line is atM,b

x x 0 , which is also a pinch point. ThusM,d

x 0 also.

4.D15. New Problem in 3rdEdition. 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 lineB

y 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 + 3equilibrium stages are more than enough to obtain separation.

S B

External M.B. S = B

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

New S.S.


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Graph for problem 4.D15.

Capitulo 4 del wankat. solucionario

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