A thin, square flat plate has \(1.2 \mathrm{~m}\) on each side. Air at \(10^{\circ} \mathrm{C}\) flows over the top and bottom surfaces of a very rough plate in a direction parallel to one edge, with a velocity of \(48 \mathrm{~m} / \mathrm{s}\). The surface of the plate is maintained at a constant temperature of \(54^{\circ} \mathrm{C}\). The plate is mounted on a scale that measures a drag force of \(1.5 \mathrm{~N}\). Determine the total heat transfer rate from the plate to the air.

In order to determine if the flow is laminar or turbulent, we should first calculate the Reynolds number using the following equation: $$Re = \frac{\rho V L}{\mu}$$ Where: - \(Re\) is the Reynolds number, - \(\rho\) is the density of the fluid (air), - \(V\) is the velocity of the fluid, - \(L\) is the characteristic length of the object (in our case, the side length of the square plate), - \(\mu\) is the dynamic viscosity of the fluid. According to the problem statement, we have \(L = 1.2 \mathrm{~m}\) and \(V = 48 \mathrm{~m / s}\). At \(10^{\circ} \mathrm{C}\), we can approximate the density and dynamic viscosity of air as \(\rho = 1.25 \mathrm{~kg / m^3}\) and \(\mu = 1.8 \times 10^{-5} \mathrm{~Pa \cdot s}\). With these values, we can calculate the Reynolds number: $$Re = \frac{(1.25 \mathrm{~kg / m^3})(48 \mathrm{~m / s})(1.2 \mathrm{~m})}{1.8 \times 10^{-5} \mathrm{~Pa \cdot s}} \approx 4 \times 10^5$$ The Reynolds number \(4 \times 10^5\) indicates that the flow is turbulent.

Now we will find the Nusselt number for the flat plate, which relates the convective heat transfer coefficient and the thermal conductivity of the fluid. Since the flow is turbulent, we will use a correlation for a turbulent boundary layer, given by: $$Nu = 0.0296 Re^{\frac{4}{5}} Pr^{\frac{1}{3}}$$ Where: - \(Nu\) is the Nusselt number, - \(Re\) is the Reynolds number, - \(Pr\) is the Prandtl number. The Prandtl number for air at \(10^{\circ} \mathrm{C}\) is approximately \(Pr = 0.71\). Thus, we can calculate the Nusselt number as: $$Nu = 0.0296 (4 \times 10^5)^{\frac{4}{5}} (0.71)^{\frac{1}{3}} \approx 363.4$$

The convective heat transfer coefficient (\(h\)) can be found using the Nusselt number, given as: $$h = \frac{Nu \cdot k}{L}$$ Where: - \(h\) is the convective heat transfer coefficient, - \(Nu\) is the Nusselt number, - \(k\) is the thermal conductivity of the fluid, - \(L\) is the characteristic length. For air at \(10^{\circ} \mathrm{C}\), the thermal conductivity is approximately \(k = 0.0261 \mathrm{~W / (m \cdot K)}\). Using this value, we can find the convective heat transfer coefficient: $$h = \frac{(363.4)(0.0261 \mathrm{~W / (m \cdot K)})}{1.2 \mathrm{~m}} \approx 7.92 \mathrm{~W / (m^2 \cdot K)}$$

Finally, we can find the total heat transfer rate (\(\dot{Q}\)) from the plate to the air using the equation: $$\dot{Q} = h A \Delta T$$ Where: - \(\dot{Q}\) is the total heat transfer rate, - \(h\) is the convective heat transfer coefficient, - \(A\) is the surface area of the plate, - \(\Delta T\) is the temperature difference between the plate and the air. The surface area of the plate is \(A = (1.2 \mathrm{~m})^2\), and the temperature difference is \(\Delta T = T_{plate} - T_{air} = (54^{\circ} \mathrm{C} - 10^{\circ} \mathrm{C}) = 44^{\circ} \mathrm{C}\). Now we can find the total heat transfer rate: $$\dot{Q} = (7.92 \mathrm{~W / (m^2 \cdot K)})(1.44 \mathrm{~m^2})(44^{\circ} \mathrm{C}) \approx 503 \mathrm{~W}$$ Thus, the total heat transfer rate from the plate to the air is approximately \(503 \mathrm{~W}\).