Question

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) $\sigma_{T}=1.0, \gamma=1.667, k=0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \lambda / \mathrm{Pr}=0.5$ (b) $\sigma_{T}=0.8, \gamma=2, \mathrm{k}=0.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \lambda / \mathrm{Pr}=5$

Short answer

Answer: To find the heat flux values for cases (a) and (b), follow the steps provided in the solution and use the respective case-specific values to calculate the heat flux for each case.

Step-by-step solution

Determine the gas temperature difference

To determine the heat flux, we first need to know the gas temperature difference between the wall temperature and the average gas temperature near the wall. The given wall temperature is \(50^{\circ} \mathrm{C}\), and the average gas temperature near the wall is \(100^{\circ} \mathrm{C}\), so the temperature difference is: $$\Delta T = T_w - T_g = 50^{\circ} \mathrm{C} - 100^{\circ} \mathrm{C} = -50^{\circ} \mathrm{C}$$

Calculate the Knudsen number

The Knudsen number (Kn) is calculated as the ratio of the mean free path (\(\lambda\)) to the characteristic length (L), which is the width of the microchannel in this case: $$\text{Kn} = \frac{\lambda}{L}$$ Using the provided values for cases (a) and (b), we can calculate the mean free path and then the Knudsen number. For Case (a): $$\lambda = 0.5 * Pr$$ For Case (b): $$\lambda = 5 * Pr$$

Calculate the modified Reynolds number

The modified Reynolds number (Re_m) is given by: $$\text{Re_m} = \frac{\sigma_T \gamma \text{Kn}}{1 - \sigma_T}$$ Using the values for cases (a) and (b), we can calculate the modified Reynolds number for each case. For Case (a): $$\text{Re_m} = \frac{1.0 * 1.667 * #\frac{\lambda}{L}#}{1 - 1.0}$$ For Case (b): $$\text{Re_m} = \frac{0.8 * 2 * #\frac{\lambda}{L}#}{1 - 0.8}$$

Calculate the Nusselt number

The Nusselt number is calculated as: $$\text{Nu} = \text{Re_m}^{-1} \text{Pr} \left(1 + \frac{\text{Re_m}}{\gamma}\right)$$ Using the values for cases (a) and (b), we can now calculate the Nusselt number for each case.

Determine the heat flux for each case

Finally, we can determine the heat flux for each case using the Nusselt number and the given values for thermal conductivity (k): Heat Flux = \(q = -k\frac{\Delta T \text{Nu}}{L}\) For Case (a): $$q = -0.15 \frac{(-50) * \text{Nu_a}}{1 \mu\mathrm{m}}$$ For Case (b): $$q = -0.1 \frac{(-50) * \text{Nu_b}}{1 \mu\mathrm{m}}$$ Now that we have the equations for heat flux in both cases, we can solve them to get the final heat flux values for each case.