All Chapters Covered
SOLUTION MANUAL
,1A.1 Estimation of dense-gas viscosity.
a. Table E.1 gives T, = 126.2 K, p = 33.5 atm, and = 180x 10° g/cm·s
for N. The reduced conditions for the viscosity estimation are then:
+ 14.7)/33.5 x 14.7 = 2.06
P+ =P/Pe = (1000
T, =T/T. =(273.15 + (68 -32)/1.8)/126.2 = 2.32
At this reduced state, Fig. 1.3-1 gives , = 1.15. Hence, the predicted viscosity
is = ,/, = 1.15180x10° = 2.0710 g/cm·s. This result is then converted
into the requested units by use of Table F.3-4:
=2.07 10' 6.7197 x 10 = 1.4 x 10 1b,~/fts
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,1A.2 Estimation of the viscosity of methyl fluoride.
a. CH%F has M = 16.04--1.008+19.00 = 34.03 g/g-mole, T, = 4.55+273.15 =
277.70 K, p = 58.0 atm, and V, = 34.03/0.300 = 113.4 cm/g-mole. The critical
viscosity is then estimated as
, = 61.6(34.03 x 277.70)/(113.4) 2/ 255.6 micropoise
from Eq. 1.3-1a, and
, = 7.70(34.03)/(58.0)/(277.7) -/° 263.5 micropoise
from Eq. 1.3-1b.
The reduced conditions for the viscosity estimate are T, = (370 4273.15)/277.70 =
2.32, p, = 120/58.0 = 2.07, and the predicted , from Fig. 1.3-1 is 1.1. The
resulting predicted viscosity is
=r=1.1 x 255.6 x 10° =2.8 x 10 g/cm·s via Eq.1.3-1a, or
1.1 263.5 x 10° = 2.9 10g/cm·s via E.1.3-1b.
J-2
, lA.3 Computation of the viscosities
p h p h ph p h p h of gases at
ph p h p h low p h density.
Equation 1.4-14, with molecular parameters from Table E.1
p h ph ph ph ph ph ph p h and collision ph
integrals from Table E.2, gives the following results:
ph ph ph ph ph ph ph ph
For O: M p h p h = p h 32.00, p h o p h = 0
p h 3.433A, p h e/K = p h p h 113 p h K. p h Then p h at
20°C, Te
p h p h
=
293.15/113 ph = 2.594
ph ph and 9, ph ph = 1.086.
ph p h Equation 1.4-14 then gives ph ph ph
= 2.6693 x 10-s V32.00 293.15
ph ph ph ph
(3.433) 1.086
=2.02 x 10 g/cm·s ph ph
=2.02 10 Pas
= 2.02 x 10 mPas.
ph ph ph
The reported value in Table 1.1-3 is 2.04 x 10 mPa.s.
ph ph ph ph ph ph ph ph ph
For N: M = 28.01,
p h o ph ph p h p h p h p h = p h 3.667~, p h e/K p h = p h 99.8 p h K.
Then at 20°C, T/e =
p h p h p h p h p h
293.15/99.8 = 2.937 and 9, = 1.0447. Equation 1.4-14 then gives
ph ph ph ph ph ph ph ph ph ph
= 2.6693
ph 10-s V28.01 293.15
ph
(3.667 1.0447
= 1.72x 10 g/cm·s
ph ph
= 1.72
ph 10 Pas
= p h 1.72 10 mPas.
The reported value in Table 1.1-3 is 1.75 x 10mPass.
ph ph ph ph ph ph ph ph ph
0
For p h CH,, M = p h p h p h 16.04, p h o p h = p h 3.780A, p h e/K p h = p h 154 p h K. p h Then
p h at p h 20°C, T/e p h p h =
293.15/154 ph = 1.904 ph ph and 9, = 1.197.
ph ph ph p h Equation 1.4-14 then gives ph ph ph
= 2.6693
ph 10-s VI6.04 ph p h 293.15
(3.780) p h x 1.197
p h
= 1.07ph 10 g/cm·s
= 1.07ph ph x 10 Pa.s
ph
= 1.07 x 10 mPass.
ph ph ph
The reported value in Table 1.1-3 is 1.09 x 10
ph ph ph ph ph ph ph ph ph mPass.
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SOLUTION MANUAL
,1A.1 Estimation of dense-gas viscosity.
a. Table E.1 gives T, = 126.2 K, p = 33.5 atm, and = 180x 10° g/cm·s
for N. The reduced conditions for the viscosity estimation are then:
+ 14.7)/33.5 x 14.7 = 2.06
P+ =P/Pe = (1000
T, =T/T. =(273.15 + (68 -32)/1.8)/126.2 = 2.32
At this reduced state, Fig. 1.3-1 gives , = 1.15. Hence, the predicted viscosity
is = ,/, = 1.15180x10° = 2.0710 g/cm·s. This result is then converted
into the requested units by use of Table F.3-4:
=2.07 10' 6.7197 x 10 = 1.4 x 10 1b,~/fts
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,1A.2 Estimation of the viscosity of methyl fluoride.
a. CH%F has M = 16.04--1.008+19.00 = 34.03 g/g-mole, T, = 4.55+273.15 =
277.70 K, p = 58.0 atm, and V, = 34.03/0.300 = 113.4 cm/g-mole. The critical
viscosity is then estimated as
, = 61.6(34.03 x 277.70)/(113.4) 2/ 255.6 micropoise
from Eq. 1.3-1a, and
, = 7.70(34.03)/(58.0)/(277.7) -/° 263.5 micropoise
from Eq. 1.3-1b.
The reduced conditions for the viscosity estimate are T, = (370 4273.15)/277.70 =
2.32, p, = 120/58.0 = 2.07, and the predicted , from Fig. 1.3-1 is 1.1. The
resulting predicted viscosity is
=r=1.1 x 255.6 x 10° =2.8 x 10 g/cm·s via Eq.1.3-1a, or
1.1 263.5 x 10° = 2.9 10g/cm·s via E.1.3-1b.
J-2
, lA.3 Computation of the viscosities
p h p h ph p h p h of gases at
ph p h p h low p h density.
Equation 1.4-14, with molecular parameters from Table E.1
p h ph ph ph ph ph ph p h and collision ph
integrals from Table E.2, gives the following results:
ph ph ph ph ph ph ph ph
For O: M p h p h = p h 32.00, p h o p h = 0
p h 3.433A, p h e/K = p h p h 113 p h K. p h Then p h at
20°C, Te
p h p h
=
293.15/113 ph = 2.594
ph ph and 9, ph ph = 1.086.
ph p h Equation 1.4-14 then gives ph ph ph
= 2.6693 x 10-s V32.00 293.15
ph ph ph ph
(3.433) 1.086
=2.02 x 10 g/cm·s ph ph
=2.02 10 Pas
= 2.02 x 10 mPas.
ph ph ph
The reported value in Table 1.1-3 is 2.04 x 10 mPa.s.
ph ph ph ph ph ph ph ph ph
For N: M = 28.01,
p h o ph ph p h p h p h p h = p h 3.667~, p h e/K p h = p h 99.8 p h K.
Then at 20°C, T/e =
p h p h p h p h p h
293.15/99.8 = 2.937 and 9, = 1.0447. Equation 1.4-14 then gives
ph ph ph ph ph ph ph ph ph ph
= 2.6693
ph 10-s V28.01 293.15
ph
(3.667 1.0447
= 1.72x 10 g/cm·s
ph ph
= 1.72
ph 10 Pas
= p h 1.72 10 mPas.
The reported value in Table 1.1-3 is 1.75 x 10mPass.
ph ph ph ph ph ph ph ph ph
0
For p h CH,, M = p h p h p h 16.04, p h o p h = p h 3.780A, p h e/K p h = p h 154 p h K. p h Then
p h at p h 20°C, T/e p h p h =
293.15/154 ph = 1.904 ph ph and 9, = 1.197.
ph ph ph p h Equation 1.4-14 then gives ph ph ph
= 2.6693
ph 10-s VI6.04 ph p h 293.15
(3.780) p h x 1.197
p h
= 1.07ph 10 g/cm·s
= 1.07ph ph x 10 Pa.s
ph
= 1.07 x 10 mPass.
ph ph ph
The reported value in Table 1.1-3 is 1.09 x 10
ph ph ph ph ph ph ph ph ph mPass.
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