A \(5-\mathrm{m} \times 5-\mathrm{m}\) flat plate maintained at a constant (?) temperature of \(80^{\circ} \mathrm{C}\) is subjected to parallel flow of air at \(1 \mathrm{~atm}, 20^{\circ} \mathrm{C}\), and \(10 \mathrm{~m} / \mathrm{s}\). The total drag force acting on the upper surface of the plate is measured to be \(2.4 \mathrm{~N}\). Using momentum-heat transfer analogy, determine the average convection heat transfer coefficient, and the rate of heat transfer between the upper surface of the plate and the air.

To begin, we will calculate the Reynolds number for the flow over the flat plate. The Reynolds number is given by the formula: $$ Re = \frac{\rho u L}{\mu} $$ where \(Re\) is the Reynolds number, \(\rho\) is the density of the fluid, \(u\) is the free-stream velocity, \(L\) is the characteristic length (in this case, the length of the plate), and \(\mu\) is the dynamic viscosity of the fluid. For air at \(20^{\circ}\text{C}\) and \(1~\text{atm}\), we can take the density \(\rho~=~1.204~\text{kg/m}^3\) and dynamic viscosity \(\mu~=~1.825 \times 10^{-5}~\text{Pa} \cdot \text{s}\). The length of the plate is \(L=5~\text{m}\), and the free-stream velocity is \(u=10~\text{m/s}\). Calculating the Reynolds number: $$ Re = \frac{(1.204~\text{kg} / \text{m}^3)(10~\text{m/s})(5~\text{m})}{1.825 \times 10^{-5}~\text{Pa} \cdot \text{s}} \approx 3.3 \times 10^5 $$

We are given the total drag force on the upper surface of the plate, which is \(2.4~\text{N}\). The drag coefficient is calculated using the formula: $$ C_D = \frac{2 \cdot F_D}{\rho u^2 A} $$ where \(F_D\) is the drag force, and \(A\) is the area of the flat plate. In this case, the area of the plate is \((5 \cdot 5)~\text{m}^2 = 25~\text{m}^2\). Now, we can find the drag coefficient: $$ C_D = \frac{2 \cdot 2.4~\text{N}}{(1.204~\text{kg} / \text{m}^3)(10^2~\text{m}^2 / \text{s}^2)(25~\text{m}^2)} \approx 0.0032 $$

The Reynolds-Colburn analogy relates the drag coefficient (\(C_D\)) and the Nusselt number (\(Nu\)) as follows: $$ C_D \approx \frac{2}{Nu} $$ Using the calculated drag coefficient, we can now determine the Nusselt number: $$ Nu \approx \frac{2}{0.0032} \approx 625 $$

The Nusselt number is related to the average convection heat transfer coefficient (\(h\)) through the equation: $$ Nu = \frac{hL}{k} $$ where \(k\) is the thermal conductivity of the fluid. For air at \(20^{\circ}\)C, we can take \(k = 0.0262~\text{W/m} \cdot \text{K}\). We can now solve for the average heat transfer coefficient: $$ h = \frac{625 \cdot (0.0262~\text{W/m} \cdot \text{K})}{5~\text{m}} \approx 32.7~\text{W/m}^2\cdot \text{K} $$

Finally, we can calculate the rate of heat transfer between the upper surface of the plate and the air using the formula: $$ Q = h \cdot A \cdot \Delta T $$ where \(\Delta T\) is the temperature difference between the plate and the air. In this case, \(\Delta T = (80 - 20)~\text{K} = 60~\text{K}\). So, the rate of heat transfer is: $$ Q = (32.7~\text{W/m}^2 \cdot \text{K})(25~\text{m}^2)(60~\text{K}) \approx 49,050~\text{W} $$ In conclusion, the average convection heat transfer coefficient is approximately \(32.7~\text{W/m}^2\cdot \text{K}\), and the rate of heat transfer between the upper surface of the plate and the air is approximately \(49,050~\text{W}\).