Metal plates are being cooled with air blowing in parallel over each plate. The average friction coefficient over each plate is given as \(C_{f}=1.33\left(\operatorname{Re}_{L}{ }^{-0.5}\right.\) for \(\operatorname{Re}_{L}<5 \times 10^{5}\). Each metal plate length parallel to the air flow is \(1 \mathrm{~m}\). Determine the average convection heat transfer coefficient for the plate, if the air velocity is \(5 \mathrm{~m} / \mathrm{s}\). Evaluate the air properties at \(20^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\).

To calculate the Reynolds number, we first need to find the kinematic viscosity of air at the given temperature and pressure conditions. We can use the properties of air at \(20^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\), which are: - Density (\(\rho\)): \(1.204 \mathrm{~kg/m^3}\) - Dynamic viscosity (\(\mu\)): \(1.825 \times 10^{-5} \mathrm{~kg/(m \cdot s)}\) - Kinematic viscosity (\(\nu\)): \(\frac{\mu}{\rho} = \frac{1.825 \times 10^{-5} \mathrm{~kg/(m \cdot s)}}{1.204 \mathrm{~kg/m^3}} = 1.516 \times 10^{-5} \mathrm{~m^2/s}\) Now, we can calculate the Reynolds number (\(\operatorname{Re}_{L}\)) using the formula: \(\operatorname{Re}_{L} = \frac{VL}{\nu}\), where \(V\) is the air velocity, and \(L\) is the length of the metal plate. \(\operatorname{Re}_{L} = \frac{5 \mathrm{~m/s} \cdot 1 \mathrm{~m}}{1.516 \times 10^{-5} \mathrm{~m^2/s}} = 3.296 \times 10^5\)

Since \(\operatorname{Re}_{L}<5 \times 10^{5}\), we can use the given formula to find the friction coefficient (\(C_{f}\)): \(C_{f}=1.33\left(\operatorname{Re}_{L}{ }^{-0.5}\right) = 1.33\left(3.296 \times 10^5{ }^{-0.5}\right) = 1.33 \times 0.0616 = 0.0819\)

Finally, we can calculate the average convection heat transfer coefficient (\(h\)) using the Chilton-Colburn analogy, which relates the friction coefficient and the Reynolds and Prandtl numbers: \(h=\frac{C_{f}}{2}\rho V c_{p} \operatorname{Pr}^{2/3}\) For air at \(20^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\), the specific heat (\(c_{p}\)) is \(1.005 \mathrm{~kJ/(kg \cdot K)}\) and the Prandtl number (\(\operatorname{Pr}\)) is \(0.715\). Therefore, we can find the average convection heat transfer coefficient as follows: \(h=\frac{0.0819}{2}(1.204 \mathrm{~kg/m^3})(5 \mathrm{~m/s})(1.005 \times 10^3 \mathrm{~J/(kg \cdot K)})(0.715^{2/3}) = 126.3 \mathrm{~W/(m^2 \cdot K)}\) The average convection heat transfer coefficient for the metal plate is \(126.3 \mathrm{~W/(m^2 \cdot K)}\).