Air at \(10^{\circ} \mathrm{C}\) enters an \(18-\mathrm{m}\)-long rectangular duct of cross section \(0.15 \mathrm{~m} \times 0.20 \mathrm{~m}\) at a velocity of \(4.5 \mathrm{~m} / \mathrm{s}\). The duct is subjected to uniform radiation heating throughout the surface at a rate of \(400 \mathrm{~W} / \mathrm{m}^{3}\). The wall temperature at the exit of the duct is (a) \(58.8^{\circ} \mathrm{C}\) (b) \(61.9^{\circ} \mathrm{C}\) (c) \(64.6^{\circ} \mathrm{C}\) (d) \(69.1^{\circ} \mathrm{C}\) (e) \(75.5^{\circ} \mathrm{C}\) (For air, use \(k=0.02551 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=0.7296, v=1.562 \times\) \(10^{-5} \mathrm{~m}^{2} / \mathrm{s}, c_{p}=1007 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \rho=1.184 \mathrm{~kg} / \mathrm{m}^{3}\).)

First, we need to find the Reynolds number to determine the flow regime. The Reynolds number in a rectangular duct can be calculated as: \(Re = \frac{\rho V_dh}{\mu}\) where: - \(Re\) is the Reynolds number - \(\rho\) is the air density (\(1.184 \mathrm{~kg/m^3}\)) - \(V_d\) is the velocity of the air (\(4.5 \mathrm{~m/s}\)) - \(dh\) is the hydraulic diameter of the duct, which can be obtained as \(dh=\frac{2ab}{a+b}\), where \(a\) and \(b\) are the sides of the rectangle (use \(0.15 \mathrm{~m}\) and \(0.20 \mathrm{~m}\)) - \(\mu\) is the dynamic viscosity of the air (\(1.562 \times 10^{-5} \mathrm{~m^2/s}\)) Calculate the Reynolds number using the above information.

If \(Re<2000\), the flow is laminar, and if \(Re>4000\), the flow is turbulent. The Dittus-Boelter equation can be used to calculate the convective heat transfer coefficient (h) for turbulent flow: \(h=\frac{k}{dh}(\frac{0.4Re^{0.7}Pr^{0.333}}{1.07+9(PrRe)^{-0.5}})\) where: - \(h\) is the convective heat transfer coefficient - \(k\) is the thermal conductivity of air (\(0.02551 \mathrm{~W/m\cdot K}\)) - \(Re\) is the Reynolds number - \(Pr\) is the Prandtl number (\(0.7296\)) - \(dh\) is the hydraulic diameter of the duct Calculate the convective heat transfer coefficient using the above equation as we have determined the flow to be turbulent in step 1.

Now, we need to find the heat transfer across the duct's length. The heat transfer (Q) is related to the convective heat transfer coefficient, the surface area of the duct, and the temperature difference between the wall and air, given as: \(Q = hA_s(Tw - Ta)\) where: - \(Q\) is the heat transfer - \(h\) is the convective heat transfer coefficient - \(A_s\) is the surface area of the duct, which can be obtained as \(A_s = P\cdot L\), where \(P\) is the perimeter of the duct and \(L\) is the length of the duct (use \(18 \mathrm{~m}\)) - \(Tw\) is the wall temperature at the exit, being the unknown - \(Ta\) is the air temperature at the entrance of the duct (\(10^{\circ}C\)) We are also given the radiation heating rate (P): \(P = 400 \mathrm{~W/m^3}\) The total heat transfer can be found as: \(Q = P\cdot V\) where: - \(Q\) is the heat transfer - \(P\) is the radiation energy rate - \(V\) is the volume of the duct, which can be calculated as \(V = abL\) Calculate the total heat transfer (Q) using the above equation.

Now, we can use the energy balance equation to find the wall temperature at the exit of the duct. Rearrange the heat transfer equation to solve for the unknown temperature: \(Tw = Ta+\frac{Q}{hA_s}\) Use the values obtained in the previous steps to find the wall temperature at the exit (Tw_exit). By performing these calculations, you will find the answer to be close to (c) \(64.6^{\circ} \mathrm{C}\).