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Exercise 5.2. In the treatment of energy transport in this section we in effect assumed (by considering only the energy of the random and internal motions) that there was no macroscopic motion of the gas. Consider now a gas with not only a temperature gradient $dT/dx_2$ but also a gradient of flow velocity $du_1/dx_2$ as in the treatment of momentum transport. By the mean-free-path methods of this section, show that the flux of molecular energy in the $x_2$-direction is now given by $\Lambda = -K \frac{dT}{dx_2} - \mu_1 \mu \frac{du_1}{dx_2} = q - \mu_1 \tau$. The second term on the right is interpreted at the macroscopic level as the work done by the shear stress.

          Exercise 5.2. In the treatment of energy transport in this section we in effect assumed (by considering only the energy of the random and internal motions) that there was no macroscopic motion of the gas. Consider now a gas with not only a temperature gradient $dT/dx_2$ but also a gradient of flow velocity $du_1/dx_2$ as in the treatment of momentum transport. By the mean-free-path methods of this section, show that the flux of molecular energy in the $x_2$-direction is now given by
$\Lambda = -K \frac{dT}{dx_2} - \mu_1 \mu \frac{du_1}{dx_2} = q - \mu_1 \tau$.
The second term on the right is interpreted at the macroscopic level as the work done by the shear stress.

Exercise 5.2. In the treatment of energy transport in this section we in effect assumed (by considering only the energy of the random and internal motions) that there was no macroscopic motion of the gas. Consider now a gas with not only a temperature gradient dT/dx2 but also a gradient of flow velocity du1/dx2 as in the treatment of momentum transport. By the mean-free-path methods of this section, show that the flux of molecular energy in the x2-direction is now given by
Λ = -K (dT)/(dx2) - μ1 μ(du1)/(dx2) = q - μ1 τ.
The second term on the right is interpreted at the macroscopic level as the work done by the shear stress.

Added by Miguel -Ngel W.

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