Question

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 $400 \mathrm{~W} / \mathrm{m}^{2}$. 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, \nu=1.562 \times\( \)\left.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} .\right)$

Short answer

Question: Calculate the wall temperature at the exit of the duct given the initial temperature, length of the duct, cross-sectional dimensions, velocity of air entering, and uniform radiation heating rate. Also, use the provided air properties such as thermal conductivity, Prandtl number, kinematic viscosity, specific heat, and density.

Step-by-step solution

Calculate the Reynolds number

Reynolds number is given by: $$Re = \frac{uL}{\nu}$$ where,\(u\) is the velocity, \(L\) is the characteristic length, \(\nu\) is the kinematic viscosity. Here, we will use the hydraulic diameter (\(D_h\)) as the characteristic length, which is given by: $$D_h=\frac{2ab}{a+b}$$ where, a and b are the cross-sectional dimensions of the duct. Let's calculate the Reynolds number.

Calculate the Nusselt number

Using Dittus-Boelter correlation formula, we can calculate the Nusselt number(\(Nu\)) for a smooth rectangular duct with fully developed flow as follows: $$Nu=0.023Re^{0.8}Pr^{0.4}$$ where, \(Re\) is the Reynolds number and \(Pr\) is the Prandtl number. Let's calculate the Nusselt number.

Calculate the heat transfer coefficient

Using the Nusselt number, we can calculate the heat transfer coefficient (\(h\)) as follows: $$h = \frac{Nu \times k}{D_h}$$ where, \(Nu\) is the Nusselt number, \(k\) is the thermal conductivity, and \(D_h\) is the hydraulic diameter. Let's calculate the heat transfer coefficient.

Calculate the wall temperature at the exit of the duct

The energy balance for the entire duct gives: $$Q_{in} = \dot{m} \times c_p \times (T_{w,exit} - T_{in})$$ The mass flow rate \(\dot{m}\) can be calculated as: $$\dot{m} = \rho \times A \times u$$ where, \(\rho\) is the density, \(A\) is the cross-sectional area, and \(u\) is the velocity. Heat transfer into the duct due to radiation, \(Q_{in}\) is given by: $$Q_{in} = h \times A_s \times \Delta T_{avg}$$ where, \(A_s\) is the surface area of the duct and \(\Delta T_{avg}\) is the average temperature difference between the wall and the air during the heating process (which we assume to be linear along the duct). The surface area of the duct is given by: $$A_s = P \times L$$ where, \(P\) is the perimeter of the duct and \(L\) is the length of the duct. Calculating \(\Delta T_{avg}\), we get: $$\Delta T_{avg} = \frac{(T_{w,exit} - T_{in})}{2}$$ Now, we can equate the heat transfer equations and solve for \(T_{w,exit}\). After calculating all relevant values, we can now find the wall temperature at the exit of the duct.