Solucionario Bird 2da Edición, Ejercicios de Mecánica de Fluidos

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SOLUTIONS TO THE PROBLEMS

in

TRANSPORT PHEOMENA

Second Edition

b y

R. Byron Bird

Warren E. Stewart

Edwin N. Lightfoot

Department of Chemical Engineering

University of Wisconsin

Madison, Wisconsin 53706 USA

This solutions manual has been prepared by the

authors of the textbook for use by professors teaching

courses in transport phenomena. It contains the

solutions to 500 of the unsolved problems in the

textbook. No part of this material may be reproduced

in any form, electronic, mechanical, photocopying,

recording, scanning, or otherwise.

JOHN WILEY & SONS, Inc.

New York, New York

a. Table E .l gives Tc = 126.2 K, pc = 33.5 atm, and p c = 180 x 10~6 g/cm-s for N2. The reduced conditions for the viscosity estimation are then:

pr - P/ Pc = (1000 + 14.7)/33.5 x 14.7 = 2.

Tr = T / T c = (273.15 + (68 - 32)/1.8)/126.2 = 2. At this reduced state, Fig. 1.3-1 gives pr = 1.15. Hence, the predicted viscosity

is p = fir / p c — 1.15 x 180 xlO-6 = 2.07 xlO-4 g/cm-s. This result is then converted

into the requested units by use of Table F.3-4:

p = 2.07 x 10"4 x 6.7197 x 10~2 = 1.4 x 1 0 "4 lbm/ft-s

1A.1 Estimation of dense-gas viscosity.

1A.3 Computation of the viscosities of gases at low density.

Equation 1.4-14, with molecular parameters from Table E .l and collision integrals from Table E.2, gives the following results:

For O2: 293.15/113 =

M = 32.00, <r = 3.433A , e/K = 113 K. Then at 20°C, k T / e =

- 2.594 and = 1.086. Equation 1.4-14 then gives

0 ^ 0 i ^ - 5 V32.00 x 293. /x = 2.6693 x IO 77— — — — r (3.433)2 x 1. = 2.02 x 10-4 g/cm-s = 2.02 x 10~5 Pa-s = 2.02 x 10-2 mPa-s.

The reported value in Table 1.1-3 is 2.04 x 10 2 mPa-s. o For N2: 293.15/99.8 =

M = 28.01, a = 3.667A , s / k = 99.8 K. Then at 20°C, k T / e = = 2.937 and = 1.0447. Equation 1.4-14 then gives

, , V28.01 x 293. a = 2.6693 x 10 5 *, ** (3.6672 x 1. = 1.72 x 10-4 g/cm-s = 1.72 x 10~5 Pa-s = 1.72 x 10-2 mPa-s.

The reported value in Table 1.1-3 is 1.75 x 10 2 mPa-s.

For CH4, M = 16.04, a = 3.780A , s / k = 154 K. Then at 20°C, k T / e 293.15/154 == 1.904 and = 1.197. Equation 1.4-14 then gives

^ - 5 ^ 1 6 .0 4 x 293. r (3.780)2 x 1. = 1.07 x 10-4 g/cm-s = 1.07 x 1 0 "5 Pa-s = 1.07 x 10-2 mPa-s.

The reported value in Table 1.1-3 is 1.09 x 10 2 mPa-s.

1A.4 Gas-mixture viscosities at low density.

The data for this problem are as follows:

Component M 1(H2) 2. 2(CC12F 2) 120.

/i, poise x 106

Insertion of these data into Eq. 1.4-16 gives the foloowing coefficients for mixtures of H2 and Freon-12 at this temperature:

$ n = $ 22 = 1.

$ 2 1

1 / 120.92~1/2 L / 124.0\ 1/2 / 2.016 \ 1/

V8 V + 2.016 ) + V 88.4 ) V120.92 J

Equation 1.4-15 then gives the predicted mixture viscosities: 1 = (^) £ i = £ 2 = A : =^ B :=^ A + B = 1 — x 2 (^) £ ®y3$l/8 (^) £/»$2/> ^1/^ 1/£1 x 2 ^ / £ 2 /^mix 19 ^obs,poise X 10 0.00 3.934 1.000 0.0 124.0 (124.0)^ 124. 0.25 3.200 0.773 6.9 120.3 127.2 128. 0.50 2.467 0.546 18.1 113.6 131.7 131. 0.75 1.734 0.319 38.2 97.2 135.4 135. 1.00 1.000 0.092^ 88.4 0.0 (88.4)^ 88.

1A.6 Estimation of liquid viscosity.

a. The calculated values for Eq. 1.5-9 at 0°C and 100° C are as follows: T,K (^) 273.15 373. p, g/cm 3 0.9998 0. V = M / p, cm3/g-mole 18.01 18. At7vap,Tt, cal/g-mole= 897.5 x 18.016 x 252.16/453.59 8989. 8989. A U ^ tJ R T ^ 8989/1.98721 / T 16.560 12. exp 0.408z*7 vap,T6/ R T 859.6 140. N h / V , g/cm-s 2.22 x 10“ 4 2.12 x 10 Predicted liquid viscosity, g/cm-s 0.19 0. b. The predicted values for Eq. 1.5-11 at 0°C and 100°C (^) are:

~ TJC 273.15^ 373. N h / V , g/cm-s 2.22 x 1 0 "4 2.12 x 10 exp(3.8Tft/T) 179.7 (^) 44. Predicted liquid viscosity, g/cm-s 0.0398 0.

Summary of results: Temperature, °C (^0 ) Observed viscosity, centipoise[=]g/cm-sxl00 1.787 0. Prediction of Eq. 1.5-9 19. 2. Prediction of Eq. 1.5-11 3.98 0.

Both equations give poor predictions. This is not surprising, since the empirical formulas in Eqs. 1.5-8 et stq. are inaccurate for water and for other associated liquids.

1 A .7 M olecular velocity and m ean free p ath.

From eq. 1.4-1, the mean molecular velocity in O2 at 273.2 K is

>8RT 1 8 x 8.31451 x 107 x 273. u = (^) = 4.25 x 104 cm /s 7 tM V 7r x 32.

From eq. 1.4-3, the mean free path in O2 at 1 atm and 273.2 K is

R T 82.0578 x 273. A = V2 tt d2pN /2?r(3 x 10~8 )2 x 1 x 6.02214 x 1023

= 9.3 x 10 6 cm

Hence, the ratio of the mean free path to the molecular diameter is (9.3 x 10- 4 /3 x 10- 8 ) = 3.1 x 104 under these conditions. At liquid states, on the other hand, the corresponding ratio would be on the order of unity or even less.

1B.2 A fluid in a state of rigid rotation

a. A particle within a rigid body rotating with an angular

velocity vector w has a velocity given by v = [w xr], If the angular

velocity vector is in the +z-direction, then there are two nonzero

velocity components given by vx = - w zy and vy = +wzx. Hence the

magnitude of the angular velocity vector is b in Problem lB.l(c).

b. For the velocity components of Problem IB. 1(c),

dx dy

= 0 and = 2b

c. In Eq. 1.2-4, we selected only the linear sy m m etric

combinations of derivatives of the velocity, so that in pure rotation

there would be no viscous forces present. In (b) we see that the

antisymmetric combination is nonzero in a purely rotational motion.

1B.3 Viscosity of suspensions

Expanding the Mooney expression, we get (with e = 0/00)

/£eff _ 1 + _J_*

1 1 0^5 a^ ^ +•

J - e ) 2\~ e )

= l + f 0 ( l + e + £ 2 +•••) + -^ 0 2 (1 + 2 e + - ••)+ ^ f 0 3 (!+•••)+•

= l + | 0 + 0 2

125 25 J _ 5 J _

48 + 4 0„ + 2 0 o 2

The first two terms match exactly with the first two terms in Eq.

1B.3-1. We can make the third term match exactly, by setting

^ = 7.17 whence 0O= 0.

8 2 0O Vo

and the coefficient of 0 3 becomes

If we try 0O= 0.70, the coefficients of 0 2 and 0 3 become 6.70 and 17.

respectively. This gives a somewhat better find of Vand's data.

1C.2 The wall collision frequency

When we change to dimensionless variables in the second

line of Eq. 1C.2-1, we get

Z = r l ^ - T

{ I tikT J

= n

____ m V/2( 2 k T 1

2 ttkT J \ m 2

= n-y *L 2 irni

1C.3 The pressure in an ideal gas

a. The dimensions of the quantities in Eq. 1C.3-1 are

s H L

«* H L/t

At H t

m H M

f [=] (l/L3)(L/f)

duxduyduz [=] m 3

Using these units, one finds that the expression on the right of Eq.

1C.3-1 has units of M /Lt2 (which are the same as the units of force

per area).

b. Combining Eqs. 1C. 1-1 and 1C.3-1 we get

1D.2 Force on a surface of arbitrary orientation.

a. We can specify the surface area and the orientation of the

surface of A OBC as ndS. To project this surface onto the yz-plane,

we take the dot product with d x/ so that the area of A OBC is

("• 8 x)dS.

b. The force per unit area on three triangles perpendicular

to the three coordinate axes are

Force on A OBC = $ xnxx + § ynxy + § znxz

Force on A OCA = &xnyx + &y 7 iyy + 8 znyz

Force on A OAB = d xnzx + h n + 8 z 7 T 22

c. Force balance on the volume OABC is then

TtndS = {8x7txx + $ y7txy + Bznxz )(n '8x}dS

+ ( s * v +8 yttm + 8 27ry. )(n -8y)dS

+ (8I» ZI + 8 , ^ + 5 r/r22)(n-S 2)dS

or

^n “ [** ’^x^x^xx] ’^x^y^xy ] ’^x^z^xz]

+[n •8 , 8 , ^ ] + [n •5 ,5 , nm ] + [n •8,,52^ ]

+[n *8 2§ xtczx ] + [n •8zSy7Tz^ J + [n • 8 z8 z7tzz ]

= S 2 [ n 8(8y®,y ] = [n •It]

2A.1 Thickness of a falling film.

a. The volume flow rate w/p per unit wall width W is obtained from Eq. 2.2-25: w i/Re (1.0037 x 1 0 "2)(10) = 2.509 x 10-2 cm2/s pW 4 Here the kinematic viscosity v for liquid water at 20°C was obtained from Table 1.1-

2. Since 1 ft= 12x2.54 cm, 1 hr =3600 s, and 1 gal=231.00 in3x(2.54cm /in)3=3785. cm3 (see Appendix F ), the result in the requested units is

w = 0.02509 cm2/s x pW ' 3785. = 0.727 U.S. gal/hr-ft

gal/cm 2 x 30.48 cm /ft x 3600 s/hr

b. The film thickness is calculated from Eqs. 2.2-25 and 2.2-22 as

3 u w g cos 0 pW Zv i/Re

1/

1/

g cos /3 4 J 3 x 1.0037 x 10“ 2 (980.665)(1.0)

(2.509 x 10“ 2)

= 0.00361 in.

1/ = 0.009167 cm

Assuming the flow to be laminar, we use Eq. 2.4-17 to calculate the volume flow rate w/p, with the specifications

2A .3 Volume rate of flow thrugh an annulus.

k = 0.495/1.1 = 0. H = 136.8 (lbm/ft-hr)(l hr/3600s) = 3.80 x 10-2 lbm/ft-s (Vo —V l ) = (5.39 psi)(4.6330 x 103 poundals/ft2/psi) = 2.497 x 104 lbm/ft- •-s R = 1.1 in. = 1.1/12 ft

Here Appendix F has been used for the conversions of units. With these specifica tions, Eq. 2.4-17 gives

w P

(tt)( 2.497 x 104)(1.1/12) (8)(3.80 x 10~2)(27)

ln (l/0.45)

= (0.49242) (1 - 0.04101)

ln (l/0.495) = (0.6748)[0.1625] = 0.110 ft3/s

As a check on our assumption of laminar flow, we calculate the Reynolds num ber: 2R(1 — k ) ( v z) p 2 w Re = --------------------- = ————------ r p, TrRp(l + K) = __________ 2(0.110X80.3) __________ = (3.1416)(1.1/12)(3.80 x 1 0 -2)(1.45) This value is well within the laminar range, so our assumption of laminar flow is confirmed.

2 A .4 Loss o f cataly st particles in stack gas.

a. Rearrangement of Eq. 2.6-17 gives the terminal velocity

vt = _D(p3_* - p)g/lSfi

in which D is the sphere diameter. Particles settling at vt greater than the centerline gas velocity will not go up the stack. Hence, the value of D that corresponds to Vt = 1.0 ft/s will be the maximum diameter of particles that can be lost in the stack gas of the present system.

Conversions of data to cgs units give

vt = (1 ft/s)(12 x 2.54 cm /ft) = 30.48 cm /s p = (0.045 lbOT/ft3)(453.59 g/lbm)((12 x 2.54)~3 ft3/cm 3) = 7.2 x 10~4 g/cm 3

Hence,

Dr,

lSpvt (p* ~ P)

(1.2 - 7.2 x 10_4)(980.7)

= /l.21 x 10-4 = 1.1 x 10 2 cm = 110 microns

b. Equation 2.6-17 was derived for R e< < 1, but holds approximately up to R e = l. For the system at hand,

Re =

Dvtp _ (1.1 x 10- 2 )(30.5)(7.2 x 10“ 4) fi ~ (0.00026)

Hence, the result in a. is approximately correct. Methods are given in Chapter 6 for solving problems of this type without the creeping-flow assumption.

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