### 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 K y / x , can get T from temperature-composition graph or _{2} _{2}
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 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 L V x L V 1 x_{B}

Substitute in y y , x zD D B y V Lz L V x But q 1.0, V D, L F, L V B D B y D Fz Bx

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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 *Q _{R}*

*V . With CMO and a saturated liquid feed*

*V V*(1

*L D D , and then*/ ) / (1 / )

*R*

*Q D* *L D . *

4.C16. *New Problem in 3rd _{ Edition. Define a fictitious total feed }*

T T T F , z , h T 1 2 F F F , 1 1 2 2 T T F z F z z F , 1 2 1 F 2 F T T F h F h h 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 y x z T

b.) Does 1 1 2 2 T T q F q F q F x B x B T z y 0 z A 2 z 1 z

Given p, L/D, saturated liquid reflux,x , x _{D} _{B}
opt feed locations, z , z , F , F , h , h _{1} _{2} _{1} _{2} _{F1} _{F2}

Plot top op line. Plot all 3 feed lines. Draw
line from point A to y = x = x to obtain _{B}
bot. op. line. Connect pts B & C to get
middle op. line.

T
F
with slope q_{T} q_{T} 1
where _{T} mix T
mix mix
H h
q

H h and Hmix, mixh are saturated vapor and liquid enthalpies at feed stage of column with mixed feed.

### 72

check 1 2 1 2 1 F 2 F mix 1 2 mix 1 F 2 F T mix T Tmix mix mix mix mix mix T

F h F h
H _{F F H} _{F H} _{F h}
F
H h
q
H h H h H h F

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

1 2

1 mix F mix F

2

mix mix mix mix

T T F H h H h F H h H h q F

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

Then mix F1 mix F2

1 2

mix mix mix mix

H h H h
q and q
H h H h
Thus 1 1 2 2
T
T
Fq F q
q
F if CMO is valid.
4D1. a. Top op line: y Lx 1 L x_{D}
V V and L
L D 1.25 _{0.5555}
V 1 L D 2.25
Intersects y x x_{D} 0.9
When D
L
x 0, y 1 x 0.4

V Plot – See diagram
b. Bottom op line: y Lx L 1 x_{B}
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 plot 3 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.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. }*

E
L F V 1 V / F .63
y 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 20°C _{q} H hF

H 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.

.6 kmol Water .6 MW 18 10.8 kg
total 29.2 kg
From Figure 2-4 H 398 kcal kg,h 75,h 20 C_{F} 10

q 398 10 1.20

398 75

Slope q 1.2 6

q 1 .2

Alternate Solution: 40 mole % ethanol boils at 84.1°C (Table 2-1).

Then if pick reference as saturated liquid at 40 mole %

h_{F} C_{p,40%liq} 20 84.1

h 0, H _{40%E}

d. 40 mole %E 63 wt%, H 398 kcal kg , h 65, h_{F} 398 kcal C_{p}_{vapor} 120 84.1
kg

C_{P}_{vapor} y C_{E} _{P}_{Evapor} y C_{w} _{P}_{w ,vapor}

Assume only 1st_{ and 2}nd_{ terms in }
P

C equations are significant. From Problem 2.D9

CP_{vapor} .4 14.66 0.03758T .6 7.88 .0032T

kcal/kmol T is C
which simplifies to CP_{vapor} 10.592 0.16952T
For linear 120 _{p}

84.1

C dT is equal to CP_{vapor} @Tavg

T_{avg} 84.1 120 2 102.05. Then C_{P}_{v ,avg} 10.592 0.16952 102.05 12.32 kcal

kmol
h_{F} 398kcal 12.32 120 84.1 kcal 1 kmol

kg kmol 27.2 kg hF 398 15.149 413.15kcal kg q 398 413.15 15.147 0.045. 398 65 333 e. q L L f , L L F F L 13F 12 12 q 13 12, slope q q 1 13 12 13 1 12 f. Flash V .7, L 1 V F .3 3 F V V F .7 7

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

L
P,liq
h 60 C h 20 46.1 23
h kcal
C .63 wt frac ~ 0.864
T 60 20 80 kg C
Then h_{F} h 200_{L} C_{P}_{L} 200 60 h 60 C_{L} .864 200 60 46.1 167.1
v F
v L
H h 395 167.1 q 0.691
q 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,liq

h C T T C 250 0

w
P,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,liq C 37.96 .281 18.0 .719 23.61 P P,liq AVG C 23.61 C in kcal kg C 0.911 MW 25.91 . Then, F kcal h 0.911 250 C 228 kg C v F v L H h 446 228 q 0.58 H h 446 70 4.D4*. a. hF H C 350 50Pv H 25 300 F 25 300 H h q 1.5 H 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 reflux L vap liq h h 3100 1500 f 0.1111 H h 17500 3100 0 0 1 0 L L D 1.1_{0.524}V L D 1 2.1 c 0 1 1 2 c 0 1 1 f L V 1.1111 .524 L 0.55 V 1 f L V 1 .111 .524

*Alternate Solution*

For subcooled reflux, 1 0

0 1
H h
L
q
L H h
17500 1500 _{1.111}
17500 3100
Then, L_{1} qL_{0} 1.1111 L _{0}
1 1 1
2 1 1
L L L D
V L D L D 1,
0
1 1.111 L
L _{1.111 1.1} _{1.2222}
D D
1
2
L 1.222 _{0.55}
V 2.222
*4D6. New Problem in 3rd _{ Edition. a) }*

1 2 175 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. q1 1, vertical at y x z1 0.6 Feed 2. 60% vapor = 40% liquid q2 0.4 Slope feed line 2

2

q 0.4

2 3

q 1 .06 through y x z2 0.4
Bottom Op. Line y L V x L V 1 x . Through _{B} y x x _{B}

V B 1
Slope L V 3 2
V B
Middle L F2 B V
2 2 B
2 2 B
F z Bx
L
L x F z Bx V y y x
V V
When _{x 0, y} F z2 2 BxB_{, 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 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

### 78

DL L

y x 1 x

V V

Know that y x x and gives through interaction Middle and Feed line 1. _{D}
Also, L L F 231.875 75 156.875 and V V_{1} 241.25 ; thus,

L V 156.875 241.25 0.65

c) See graph.

### 79

4.D7*. a. Plot top op. line: slope L .8 , x y xD .9.V Step off stages as shown on Figure. b. Plot bottom op. line: slope B

L V

1 1 2 , x y x 0.13. B

V 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. L V _{min} .9 .462 0.660
.9 .236 (see figure)
min
min
min
L V
L D 1.941
1 L V
c. In 4.D7, act
L V .8
L D 4
1 L V .2
act min
L 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 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) }*

1 2 1 1 F F D B 100 F 80 B F B 20 1 1 2 2 B B 1 Fz F z Dx Bx F .42 18 .66 80 0.04 B Solve simultaneously, B 113.68, F 93.68 1 b) L D 1 L, L D 1 2 1 2 V 1 L D 3 2 3 L L D 40, V L D 120 D

Saturated Liquid Feed V V 120 1

L L F 40 93.68 133.68, L V 1.114 c) Top Op. Line – Normal: y L V x 1 L V xD

Through D

2

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

Bottom – Normal: y L V x L V 1 x , through y x x _{B} _{B}

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

F

L _{.7}

Feed line F slope

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Middle 1 1 D F z D y L V x xV V (or do around bottom)

Slope L V . Through intersection feed line F and top op. line. _{1}
Also, _{x 0, y} DxD F z1 1 80 .66 93.68 .42 _{0.11212}

V 120

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

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4.D10*. Operating Line y L V x 1 L x , where_{D}L L D 4 .8

V V 1 L D 5

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

a. Equilibrium is 1
1
1
y
y
x or x
1 y 1.79 .76y
Start with y1 = .96 = x _{D}
Equilibrium: 1
1
1
y .96
x 0.9317
1.76 .76y 1.76 .76 .96
Operating: y2 .8x .192 .8 .9317 .192 0.93736
Equilibrium: 2
2
2
y .93736
x 0.89476
1.76 .76y 1.76 .76 .93736
Operating: y_{3} .8x_{2} .192 .8 .89476 .192 0.9078

b. Generate equilibrium data from: y 1.76x 1 .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 = x = 0.96. Find _{D} x = 0.660. _{6}

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 = 0.99 _{D}

L D

L V 0.6969

1 L D @ x = 0 y = (1-L/V) x = (1-0.6969) 0.99 = 0.300 D Feed line: Slope q q 1 , y x z 0.6

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Bottom Op line:y= 0, x = x . Also goes through intersection of feed line and top op.line. B 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 0.99 0.57 0.4242
0.99 0
min min
L L V 0.4242
0.73684
D 1 L V 1 0.4242

Actual L/D is 3.12 × this value. V L

B S

M,S B

yV Lx Sy Bx

But y_{M,S} 0 (Pure steam)
With CMO B L

B

L L

y x x

### 84

4.D12. L V L D 3 41 L D slope. Top op line goes througy x xD 0.998 D

L L

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

V V

Bottom slope L V B 1245167 1.19 From Soln to 3.D9 or from graph. 1.169 S 1044168

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 & intersect top & feed . _{B}

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.

### 86

*4.D13. New Problem in 3rd*

_{ Edition. a.) See Figure }b.) See figure.
MIN
L 0.665 0.95 _{0.4385}
V 0.30 0.95
MIN
L L L V 0.4385 _{0.7808}
D V L 1 L V 0.5615
c.) L 2.0 L D _{MIN} 1.5616
D , L
L D 1.5616 _{0.6096}
V 1 L D 2.5616
y intersect 1 L y_{D} 0.3709

V . Top operating line D

L L
y x 1 y
V V
Goes through y x y_{D} 0.95
Bottom y Lx L 1 x_{B}
V V

Goes through y x x & intersection top operating line & feed line. _{B}
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. L V 0.85 0.025 1.941

0.45 0.025

1 1

V B V L V 1.0625

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Graph for problem 4.D13.### 88

4.D14a.### 89

4.D14b. Two approaches to answer. Common sense is all methanol leaks out and xMA 0 .McCabe-Thiele diagram: This is enriching column withz ys 0 . Intersection top op. line and
horizontal feed line is at x x_{M,b} 0 , which is also a pinch point. Thus x_{M,d} 0 also.

*4.D15. New Problem in 3rd _{ Edition. Saturated liquid. }*

_{q 1,}q

_{,}

q 1 feed line vertical @ z .3 .

Top operating line D

L L y x 1 y V V , L L D 2 Slope V 1 L D 3 y intercept 1 L V yD 1 3 0.6885 0.2295 andy x yD

Bottom operating line y L V x L V 1 x goes through_{B} y 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.

S B

External M.B. S = B

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

New S.S.