Question 16.4: Employing rigorous transport theory, determine the following...
Employing rigorous transport theory, determine the following transport properties for gaseous argon:
(a) its dynamic viscosity at 300 K;
(b) its thermal conductivity at 300 K;
(c) its self-diffusion coefficient at 300 K and 1 atm.
(a) From Appendix O, ε/k = 124 K and σ = 3.42 Å for argon. Hence, from Eq. (16.48), the reduced temperature is
T^{\ast} =\frac{kT}{\varepsilon} = \frac{300}{124} = 2.42.
Consequently, from Appendix P, the normalized collision integral is Ω^{(2,2)∗} = 1.104. On this basis, from Eq. (16.50), the dynamic viscosity becomes
\eta=2.6696 \times 10^{-6} \frac{\sqrt{M T}}{\sigma^{2} \Omega^{(2,2) *}} N · s/m², (16.50)
=2.6696 \times 10^{-6} \frac{\sqrt{(39.948)(300)}}{(3.42)^{2}(1.104)}
= 2.263 × 10^{−5} N · s/m.
(b) For any atomic species, the specific heat ratio is γ = 5/3. Moreover, the gravimetric specific heat at constant volume for argon is
\tilde{c}_{v}=\frac{3}{2}\left(\frac{R}{M}\right)=\frac{3}{2}\left(\frac{8.3145 kJ / kmol \cdot K }{39.948 kg / kmol }\right) = 0.3122 kJ/kg · K.
Hence, from Eq. (16.51), the thermal conductivity for argon at 300 K is
λ = 250(9γ − 5)\tilde{c}_{v}\eta W/m · K , (16.51)
= 2500(0.3122)(2.263 × 10^{−5}) = 1.766 × 10^{−2} W/m · K.
(c) From the reduced temperature of part (a) and Appendix P, the normalized collision integral \Omega^{(1,1) *} = 1.010. Therefore, from Eq.(16.52), the self-diffusion coefficient for argon at 300 K and P = 1 atm becomes
D=2.6295 \times 10^{-7} \frac{T^{3 / 2}}{\sqrt{M} P \sigma^{2} \Omega^{(1,1) *}} m²/s, (16.52)
=2.6295 \times 10^{-7} \frac{(300)^{3 / 2}}{\sqrt{39.948}(3.42)^{2}(1.010)}
= 1.830 × 10^{−5} m²/s.
While the dynamic viscosity evaluated when using the L–J potential is only 33% higher than that obtained in Example 16.3 when using the rigid-sphere model, the thermal conductivity is a whopping 232% greater and the self-diffusion coefficient is 76% greater than the associated rigid-sphere calculations.