Liquid mercury at \(250^{\circ} \mathrm{C}\) is flowing in parallel over a flat plate at a velocity of \(0.3 \mathrm{~m} / \mathrm{s}\). Surface temperature of the \(0.1-\mathrm{m}\)-long flat plate is constant at \(50^{\circ} \mathrm{C}\). Determine \((a)\) the local convection heat transfer coefficient at \(5 \mathrm{~cm}\) from the leading edge and \((b)\) the average convection heat transfer coefficient over the entire plate.

We are given the following information: Temperature of liquid mercury, \(T_L = 250^\circ C\) Velocity of liquid mercury, \(u = 0.3 m/s\) Surface temperature of the flat plate, \(T_P = 50^\circ C\) Length of the flat plate, \(L = 0.1 m\) Position we would like to compute local (convection) heat transfer coefficient, \(x = 0.05 m\)

To make use of the Nusselt number and heat transfer coefficient formulas, first, we need to calculate the Reynolds number at the position \(x = 5 cm\). Reynolds number, \(Re_x = \frac{u x}{\nu}\) where, \(\nu\) is the kinematic viscosity of liquid mercury. Based on the provided information, you can look up the kinematic viscosity of liquid mercury at \(250^\circ C\) in a reference table. For instance, \(\nu \approx 1.125 \times 10^{-7} m^2/s\). Now we can calculate the Reynolds number at \(x = 0.05 m\): \(Re_x = \frac{0.3 \times 0.05}{1.125 \times 10^{-7}} \approx 133333\)

Next, we need to calculate the Prandtl number, which characterizes the relative thickness of the momentum and thermal boundary layers. Prandtl number, \(Pr = \frac{\nu}{\alpha}\) where, \(\alpha\) is the thermal diffusivity of liquid mercury. You can look up the thermal diffusivity of liquid mercury at \(250^\circ C\) in a reference table. For instance, \(\alpha \approx 1.0 \times 10^{-7} m^2/s\). Now we can calculate the Prandtl number: \(Pr = \frac{1.125 \times 10^{-7}}{1.0 \times 10^{-7}} \approx 1.125\)

Now that we have the Reynolds and Prandtl numbers, we can determine the local Nusselt number using the following formula: Local Nusselt number, \(Nu_x = C \cdot Re_x^m \cdot Pr^n\) Here, \(C\), \(m\), and \(n\) are empirical constants, typical values are \(C = 0.332\), \(m = 0.5\), and \(n = 1/3\) for laminar flow over a flat plate. Therefore, \(Nu_x = 0.332 \cdot (133333)^{0.5} \cdot (1.125)^{1/3} \approx 4598\) Now, we can determine the local convection heat transfer coefficient at \(x = 0.05 m\) by using the following formula: Local convection heat transfer coefficient, \(h_x = \frac{k \cdot Nu_x}{x}\) where, \(k\) is the thermal conductivity of liquid mercury. You can look up the thermal conductivity of liquid mercury at \(250^\circ C\) in a reference table. For instance, \(k \approx 26.3 W/(m \cdot K)\). Now we can calculate the local convection heat transfer coefficient: \(h_x = \frac{26.3 \times 4598}{0.05} \approx 241048 W/(m^2 \cdot K)\) The local convection heat transfer coefficient at \(5 cm\) from the leading edge is approximately \(241048 W/(m^2 \cdot K)\)

Next, we will determine the average convection heat transfer coefficient over the entire plate. First, we need to calculate the average Nusselt number with the following formula: Average Nusselt number, \( \overline{Nu}_L = C \cdot Re_L^m \cdot Pr^n\) Here, \(Re_L\) is the Reynolds number at the end of the plate, \(L = 0.1 m\): \(Re_L = \frac{0.3 \times 0.1}{1.125 \times 10^{-7}} \approx 266667\) Now, we can calculate the average Nusselt number: \( \overline{Nu}_L = 0.332 \cdot (266667)^{0.5} \cdot (1.125)^{1/3} \approx 6515\) Finally, we can determine the average convection heat transfer coefficient over the entire plate using the following formula: Average convection heat transfer coefficient, \(\overline{h}_L = \frac{k \cdot \overline{Nu}_L}{L}\) Now we can calculate the average convection heat transfer coefficient: \(\overline{h}_L = \frac{26.3 \times 6515}{0.1} \approx 1712950 W/(m^2 \cdot K)\) The average convection heat transfer coefficient over the entire plate is approximately \(1712950 W/(m^2 \cdot K)\). Answer: (a) The local convection heat transfer coefficient at \(5 cm\) from the leading edge is approximately \(241048 W/(m^2 \cdot K)\). (b) The average convection heat transfer coefficient over the entire plate is approximately \(1712950 W/(m^2 \cdot K)\).