TRANSPORT PHENOMENA DO NUMBER 1 Calculate the viscosity of carbon dioxide at 450

Question

TRANSPORT PHENOMENA
DO NUMBER 1

Calculate the viscosity of carbon dioxide at 450K using equation 7-12 in the textbook and compare with the value given in Appendix I by determining the percent difference.
Note. Answer from Appendix I is 2.1332*10^-5 Pa.s

(7-11) expresses the viscosity entirely in terilg Ul huld plu pioperties ion id alistic molecular model utilizing a force field rather than the rigid-sph yield a viscosity-temperature relationship much more consistent than the T result. The most acceptable expression for sed upon the Lennard-Jones potential energy function. This function ment leading to the viscosity expression will not be included here. The ata onpola may refer to Hirschfelder, Curtiss, and Bird for the details of thi spression for viscosity of a pure gas that results is = 2.6693 × 10-6 VMT (7-12) sity, in pascal-seconds; T is absolute temperature, in K; M is the molecular llision diameter," a Lennard-Jones parameter, in (Angstroms); s the a Lennard-Jones parameter that varies in a relatively slow manner with mperture KTE ; is the Boltzmann constant, 1.38 . 10-16 ergs/K ; and e is of interaction between molecules. Values of and for various gases rgy x K, and a table of versus kTIe is also included in Appendix K.

Explanation / Answer

given
viscosity mu = 2.6693810^-6*sqrt(MT)/sigma^2*omegau
where T is abnsolute temperature
M is molecular mass
sigma is collision diameter ( leonard joines parameter) in angstrons
omegau = kT/epsilon
k is boltzmanns constant = 1.38*10^-16 ergs/K

T = 450 K
gas is CO2
from taBLE k2
for CO2, ea/k = 190 K
sigma = 3.996 A

now for T = 450 K
T(k/e) = 450/190 = 2.386
omegau = 1.107
also
M = 0.004 kg
hence
mu = 2.02604*10^-7 Pa s

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