Page 294 - Bird R.B. Transport phenomena

278 Chapter 9 Thermal Conductivity and the Mechanisms of Energy Transport

EXAMPLE 9.3-2 Estimate the thermal conductivity of molecular oxygen at 300K and low pressure.
Estimation of the SOLUTION
Thermal Conductivity
m
of a Polyatomic Gas ^he ° l e c u l a r weight of O is 32.0000; its molar heat capacity C at 300°K and low pressure is
2 p
p
at Low Density 7.019 cal/g-mole • K. From Table E.I we find the Lennard-Jones parameters for molecular
oxygen to be a = 3.433 A and е/к = 113K. At 300K, then, кТ/s = 300/113 = 2.655. From
Table E.2, we find П = 1.074. The viscosity, from Eq. 1.4-18, is
м
/LL = (2.6693 X 10" ) - MT
5
1 0 1
2
(3.433) (1074)
(9.3-19)
5
= 2.065 X 10~ g/cm • s
Then, from Eq. 9.3-15, the Eucken approximation to the thermal conductivity is
k = (C, + fKXjt/M)
4
= (7.019 + 2.484Ж2.065 X 10~ )/(32.000)) (9.3-20)
5
= 6.14 X 10~ cal/cm • s • К
This compares favorably with the experimental value of 6.35 X 10 5 cal/cm • s • К in Table
9.1-1.

EXAMPLE 9.3-3 Predict the thermal conductivity of the following gas mixture at 1 atm and 293K from the
given data on the pure components at the same pressure and temperature:
Prediction of the
Thermal Conductivity
of a Gas Mixture at Mole Molecular 7
Low Density fraction weight fl a X 1 0 7 k a x 10
Species a (g/cm • s) (cal/cm • s • K)
CO 2 1 0.133 44.010 1462 383
o 2 2 0.039 32.000 2031 612
N 2 3 0.828 28.016 1754 627

SOLUTION
Use Eqs. 9.3-17 and 18. We note that the Ф р for this gas mixture at these conditions have al-
а
ready been computed in the viscosity calculation in Example 1.5-2. In that example we evalu-
ated the following summations, which also appear in Eq. 9.3-17:
a-* 1 2 3
х Ф 0.763 1.057 1.049
р ф
Substitution in Eq. 9.3-17 gives

itmiv
(0.133)(383)(10" ) (0.039)(612)(10 ) (0.828)(627)(10~ )
7
7
7
+
0.763 1.057 1.049
7
= 584 X (10~) cal/cm • s • К (9.3-22)
No data are available for comparison at these conditions.