Transport Phenomena (Bird, Stewart & Lightfoot) – Complete Solutions Guide & Study Notes | Revised 2nd Edition | Chemical Engineering Exam Prep

All Chapters Covered
BH BH




SOLUTION MANUAL
BH

,1A.1 Estimation of dense-gas viscosity.
B H B H BH B H




a. Table E.1 gives T, = 126.2 K, p = 33.5 atm, and = 180x 10°
B H BH B H BH BH BH BH BH BH BH BH B H BH BH B H g/cm·s
for N. The reduced conditions for the viscosity estimation are then:
BH BH BH BH BH BH BH BH BH BH BH




P+ =P/Pe = (1000 + 14.7)/33.5 x 14.7 = 2.06
B H B H B H BH BH B H B H B H B H




T, =T/T. =(273.15+ (68 -32)/1.8)/126.2 = 2.32
BH BH B
H BH BH BH BH




At this reduced state, Fig. 1.3-1 gives , = 1.15. Hence, the predicted viscosity
BH BH BH BH BH BH BH B
H BH BH BH BH BH




is = ,/, = 1.15180x10° = 2.0710 g/cm·s. This result is then converted
BH B
H BH B
H BH BH BH B H B H BH BH BH BH




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




=2.07 10' 6.7197 x 10 BH BH B H B H B H B H = B H 1.4 x B H B H 10 1b,~/fts




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,1A.2 Estimation of the viscosity of methyl fluoride.
BH BH BH BH BH BH BH




a. CH%F has M = 16.04--1.008+19.00 = 34.03 g/g-mole, T,
B H BH BH BH BH BH BH BH BH BH = 4.55+273.15 =
BH BH




277.70 K, p = 58.0 atm, and BH B
H BH B H B H B H BH V, = BH B H 34.03/0.300 = 113.4 cm/g-mole. BH B H B H B H The
BH critical
viscosity is then estimated as
B H B H B H B H




, = 61.6(34.03 x 277.70)/(113.4)2/ 255.6 micropoise
B
H BH BH BH B
H BH BH




from Eq. 1.3-1a, and
B H B H B H




, = 7.70(34.03)/(58.0)/(277.7)
BH BH -/° 263.5 micropoise
B
H BH BH




from Eq. 1.3-1b.
B H B H




The reduced conditions for the viscosity estimate are T, = (370 4273.15)/277.70 =
BH BH BH BH BH BH BH BH BH BH BH BH




2.32, p, = 120/58.0 = 2.07, and the predicted , from Fig. 1.3-1 is 1.1. The
BH BH BH BH BH BH BH BH BH BH B
H BH BH BH BH B H




resulting predicted viscosity is
BH BH BH BH




=r=1.1 B H x B H 255.6 B H x B H 10° B H =2.8 B H x 10 g/cm·s via Eq.1.3-1a, or
B H BH BH BH




1.1 263.5 x 10° = 2.9 10g/cm·s via E.1.3-1b.
B H B H B H B H B H B H B H




J-2

, lA.3 Computation of the viscosities of gases at low density.
B H B H BH B H B H BH B H B H B H




Equation 1.4-14, with molecular parameters from Table E.1 and collision integrals
B H BH BH BH BH BH BH B H BH BH




from Table E.2, gives the following results:
BH BH BH BH BH BH BH




For O: M = 32.00, o = 3.433A,
0
e/K = 113 K. Then at
20°C, Te
B H B H B H B H B H B H B H B H B H B H B H B H



B H B H


=


293.15/113 BH = 2.594 BH BH and 9, BH BH = 1.086. BH B H Equation 1.4-14 then gives B H BH BH




= 2.6693 x 10-s V32.00 293.15
B H B H B H BH




(3.433) 1.086
=2.02 x 10 BH BH g/cm·s
=2.02 10 Pas
= 2.02 x 10
BH BH BH mPas.

The reported value in Table 1.1-3 is 2.04 x 10
BH BH BH BH B H B H B H B H B H mPa.s.

For N: M B H B H B H B H = B H 28.01, B H o B H = B H 3.667~, B H e/K B H = B H 99.8 B H K. B H Then B H at
B 20°C, T/e
H B H B H =
293.15/99.8 = 2.937 and 9, = 1.0447. Equation 1.4-14 then gives
BH BH BH BH BH BH B H BH BH BH




= 2.6693
BH 10-s V28.01 B H 293.15
(3.667 1.0447
= 1.72x 10 g/cm·s
BH BH




= 1.72BH 10 Pas
= 1.72 10
B H mPas.

The reported value in Table 1.1-3 is 1.75 x 10
BH BH BH BH B H BH B H B H BH mPass.
e/K
0

For B M = H CH,, B H B H B H 16.04, B H o B H = B H 3.780A, B H B H = B H 154 B H K. B H Then B H at
B H 20°C, T/e = B H B H




293.15/154 BH = 1.904 BH BH and 9, = 1.197. Equation 1.4-14 then gives
BH BH B H B H B H B H BH




= 2.6693
BH 10-s VI6.04 293.15 B H B H




(3.780) x 1.197 B H B H




= 1.07 BH 10 g/cm·s
= 1.07 x 10
BH BH BH Pa.s
= 1.07 x 10 mPass.
BH BH BH




The reported value in Table 1.1-3 is 1.09 x 10
BH BH BH BH B H B H B H B H B H mPass.




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