Determine the heat flux at the wall of a microchannel of width \(1 \mu \mathrm{m}\) if the wall temperature is \(50^{\circ} \mathrm{C}\) and the average gas temperature near the wall is \(100^{\circ} \mathrm{C}\) for the cases of (a) \(\sigma_{T}=1.0, \gamma=1.667, k=0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \lambda / \operatorname{Pr}=0.5\) (b) \(\sigma_{T}=0.8, \gamma=2, k=0.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \lambda / \operatorname{Pr}=5\)

First, calculate the temperature difference between the wall and the gas temperature, which is given by: $$ \Delta T = T_w - T_g $$ For both cases, this will be: $$ \Delta T = 50^{\circ} \mathrm{C} - 100^{\circ} \mathrm{C} = -50^{\circ} \mathrm{C} $$ Note that \(\Delta T\) is negative since we expect the heat to flow from the gas to the wall.

In case (a), we have values for \(k_a = 0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and the microchannel width \(\Delta x_a = 1 \mu\mathrm{m}\). Let's calculate the heat flux in case (a): $$ q_a = k_a * \frac{\Delta T}{\Delta x_a} = 0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} * \frac{-50^{\circ} \mathrm{C}}{1 \mu\mathrm{m}} $$ Converting the width from \(\mu\mathrm{m}\) to meters, we get \(1 \mu\mathrm{m} = 1 * 10^{-6} \mathrm{m}\). Thus, the heat flux for case (a) becomes: $$ q_a = 0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} * \frac{-50^{\circ} \mathrm{C}}{1 * 10^{-6} \mathrm{m}} = -75,\!000 \,\mathrm{W}/\mathrm{m}^2 $$

In case (b), we have values for \(k_b = 0.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and the microchannel width \(\Delta x_b = 1 \mu\mathrm{m}\). Let's calculate the heat flux in case (b): $$ q_b = k_b * \frac{\Delta T}{\Delta x_b} = 0.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} * \frac{-50^{\circ} \mathrm{C}}{1 \mu\mathrm{m}} $$ Using the same width conversion as before, the heat flux for case (b) becomes: $$ q_b = 0.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} * \frac{-50^{\circ} \mathrm{C}}{1 * 10^{-6} \mathrm{m}} = -50,\!000 \,\mathrm{W}/\mathrm{m}^2 $$ In conclusion, the heat flux at the wall of the microchannel is \(-75,\!000 \,\mathrm{W}/\mathrm{m}^2\) for case (a), and \(-50,\!000 \,\mathrm{W}/\mathrm{m}^2\) for case (b).