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Solutions Manualto accompany

Unit Operationsof

Chemical EngineeringSeventh Edition

Prepared byJulian C. SmithPeter Harriott

PROPRIETARY AND CONFIDENTIAL

This Manual is the proprietary property of The McGraw-Hill Companies, Inc.(McGraw-Hill) and protected by copyright and other state and federal laws. Byopening and using this Manual the user agrees to the following restrictions, and ifthe recipient does not agree to these restrictions, the Manual should be promptlyreturned unopened to McGraw-Hill: This Manual is being provided only toauthorized professors and instructors for use in preparing for the classesusing the affiliated textbook. No other use or distribution of this Manual ispermitted. This Manual may not be sold and may not be distributed to orused by any student or other third party. No part of this Manual may bereproduced, displayed or distributed in any form or by any means, electronicor otherwise, without the prior written permission of McGraw-Hill.

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I

= =

r

Alternative solution. Mass of 0.389 m:s: 0.389 x 2291 = 891.2 kg

Mass fraction: 891.2/ 1500 = 0.594.

2. 11. Temperature: 1 0C = 50FSpecific volume: V:. = 0.016024 f P/lb (Appendix 7)Density: p = 1 I~= 63.41 lb/f P or

62.41 x 16.018 = 999. 7 kg/m:sFrom Eqs. ( 1. 15) and ( 1 .6),

1 atm = 1.01325 x 1 o~ Pa = 1.01325 x 1 o~ kg/m s2

From Eq. ( 1.13), g = 9.80665 m/s2

Substituting in Eq. (2.4),Za - Z,, = 1.013:1.:5 x 105

9.8066:5 x 999.7= 10.34 or, f ram Append.X 1 ,

10.34 x 3.281 = 33.91 ft

2. 1 2. From Eq. (2.4 ),9.8066:5 x 78:5 x 10 . kPP, - p,, 76.98 a

a 1000

2.13. Use Eq. (2.7). Assume p6= 1 atm. Height Z6 - Z~ = - 8848 mM= 29 T= 273.15 K R = 8314.4 7 kg m2/kg mol s2 (See

Eq. ( 1.8)). Then the exponent .s - 9 .80665 x :1.9 x -88438314.47 x 273.15

= 1.108

From Eq. (2. 7), P~ = 1 I e1.1oa = 1 /3.0283 = 0.330 atm2. 14. (a) Radius of small tank: r; = 0.3 m Length L1 = 3 m

Volume V, = 3(1t x 0.32) = 0.848 ml

Large tank: Vz = 12 V, = 12 x 0.848 = 10.12 mlIf r2/ L2 = 0.3/3 = 0.1 as before,

Volume V2 = xr/ x 1 Or2 = 31.42r2:s = 10.12rz:s = 0.322 r2 = 0.685 m L2 = 6.85 m

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....

f..)0

9v{u

C.1 70

c ;

cs

30

. -~

o ,'l.

i...C - ...

--

.

. '~ .

PROPRIETARY MATERIAL. The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission ofthe publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.

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=

At duldy= 1000 s 1, f.,= 20.02 x 100(11!.'s = 1062!-'~pp= 1 062/1 000 = 1.062 Pa s or 1 062 cP.

(b) For the clay suspension, from Table 3.3, n' = 0.185,K' = 1590 g/m s2-n or 1.59 kg/m s2-nAt du/dy= 10 s 1, from Eq. (3.9),

fy= 1.59 X 1 Q O.lBS = 2.43

~i~pp= 2.43/1 O = 0.243 Pa so r 243 cP

At duldY= 1 ooo s 1, ty= 1.59 x 1 oooo-11s = 3.59

~pp= 3.59/1000 = 0.00359 Pa so r 3.59 cP

3. 11. For the conditions of the problem, the Reynalds number can be

calculated to be 2029 ( not necessary for the solution).

(a) If Tis increased to 200 C, the density falls slightly, but the

velocity increases in the same proportion, so there is no effect.

From Eq. (3.6), the viscosity increases by (473/373)06s = 1.17.

This reduces Re, so flow would be laminar.

(b) If Dis increased to 2.0 cm, the velocity falls by ( 1.5/2)2 or0.5625, and Re would be reduced to 0.75 X 2029 = 1521. Flow

would still be laminar.

(c) If pis increased to 5 atm, the density increases but thevelocity decreases in the same proportion. There is no net

effect, and flow is still laminar.

3.12. Diameters: D,= 0.15 m 00= 0.158 mGap width: x = (0.158 - 0.15)/2 = 0.004 m

Shear rat.e . l!:iu = .1r D,.n =.1r x O. 15 x O .:S .._8

9 s_ 1. -

x ., 0.004du

.J

Shear stress: rv= u- = O.'.!l x 58.9 = l'.!.4 N/m2 d).1Page 3 - 9

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8" 26

4. :l . --:- -i O,~ ,Sds 2p 2 x 0.05

Half~.fa}' bet,,-reen the plates I x = 0. OOOS re,and dv I dx - 8 8 .. 2 E / 2 - 44 .. 13 s-L

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may be displayed, reproduced or distributed in any form or by any means, without the prior written permission ofthe publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.

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Rate of heat ,generation:0/t - tllt = 3028 \ol = 3028 J/o0: 3028 J Prom Appendu 1,c,.. = 0.5 x 1000 x 4.184 ;;; 2092 J/kg-0c

Tetnpe:rat.-ur,e rise: AT e Q/crl;, = 307.8/(2092 x 22. 78}c O. 064'"5.6. Sv~';jt1 vte V 4m/TlJ)~r t n Eq. (S.8) .>nd solvf~ir- o,S' in terns 01 f-. Cuunt1t1 needed ar,e,:r. 2.0 x 02.s = 162 lb/ft,

_...(,\:.. 2 :it: 2.42::.. 4-.94 lb/1t-h~ L 1 7Z >t 5280 9134 f tg~ "~.17 X 108 (t-lb /lb-h2

~P~ 0.10: ~ 144 ~b.3S b lit~

~2 ~ 91~4 ~ 10002f0S.

rr"J.x 162 x 4. l 7 >C 1 QD )( o.915.f2b. SSFrom Eq. < 4.62) , Re = 'ia11ILpfo 4m/n D~ 'Cl

(4 x 7000)/(Tfx 4.9'11H 1941/D.

8),1 tP-1:il from, FLQ. 5.10 D o .s7:, ft or 'L5 110.

Check: o.!:> S

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Bypass Bypassed sectionV1 = 2.02 ft/sRei = 31, 310

f1 = 0.0063

f1V/ = 0.0257

V4= 5.87 ft/sRe4 = 363,000

f4 = 0.0041

0.0182 f4V42 = 0.0257

Mass flow rate through bypass:

mi= 2.02 x 3600 x (1t/4) * 0.16672 x 62.3= 9,888 lb/h

Total flow rate:

mo = 6 x 3600 x (n/4)x 0.66672 x 62.3

= 469,800 lb/h

Fraction of total flow through bypass:

. .mi/mo= 9,888 /469,800 = 0.021 or 2.1%

5.8 V = 15ft/s D = 2ft

Assume= 1 cP, p = 52.3 lb/ft3

Re= 2 x 15 x 62.3 = 2.78 x 106

6. 72 x 10 -4

Let subscript 1 refer to the original pipe, 2 to

lined pipe.

For k/D = 0.0003/2 = 0.00015, from Fig. 5.10:

f1 = 0. 0032

For smooth pipe, f2 = 0.0023. From Eq. (5.7),since 02= 1.9ft:

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S.17. ~p5/p = !!. Z = 12m. From Eq. (5.8),

f = l!.ZD/2~LV2

D 0.12m 600m

= 1 x 10-3 Pa-s p = 998 kg/m3

f = (12 x o.12)/(2 x 6oov2i = o.00121v2

Re= (0.12 x 998V)/(l x 10-3) = 1.2 x 105VFrom Fig 5.10, by trial for smooth tubing,

V = 0.49 mis, Re= 5.88 x 104, f = 0.0050

If V is increased by a factor of 1.25, to0.613m/s, then

Re = 5.88 x 1.25 x 104 = 7.35 x 104

f= 0.0012/0.6132 = 0.0032From Fig. 5.12, this will require about 12 ppm of

polyethylene oxide.

5.18. Basis: A 1-foot cube of packing

Many assumptions and approximations are needed.

Assume the surfaces of the rings are inclined at

an average angle of 45. Then cos~= 0.071 and

the average length of the path from top to bottom

of the cube is 1/0.7071 = 1.41 ft.

The total surface area of 1-inch rings is 58

ft2/ft3 (Table 18.1). The wetted area is 58/2 = 29ft2/ft3 The width of a single plate with this

area and a length of 1.41 ft is 29/1.41 = 20.6

ft. Also,

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