Question

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 the 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. Evaluate the air properties at \(50^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\).

Short answer

Question: Determine the average convection heat transfer coefficient and the rate of heat transfer between the upper surface of the plate and the air. Answer: The average convection heat transfer coefficient is 52 W/(m^2⋅K) and the rate of heat transfer is 78,000 W.

Step-by-step solution

Calculate the Reynolds number

Reynolds number (\(Re_L\)) is a dimensionless parameter that helps understand the flow regime (laminar or turbulent) around the flat plate. It is calculated as follows: $$ Re_L = \frac{U L}{\nu} $$ where \(U\) is the air velocity, \(L\) is the length of the plate, and \(\nu\) is the kinematic viscosity of the air. First, we need to find the kinematic viscosity (\(\nu\)) of the air at \(50^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\). Looking up air properties in a table, we find that \(\nu = 1.95 \times 10^{-5} \mathrm{~m^2/s}\). Now, we can calculate the Reynolds number: $$ Re_L = \frac{10 \mathrm{~m/s} \times 5 \mathrm{~m}}{1.95 \times 10^{-5} \mathrm{~m^2/s}} \approx 2.56 \times 10^6 $$

Determine the drag coefficient

Given the total drag force acting on the upper surface of the plate (\(F_D = 2.4 \mathrm{~N}\)), we can determine the drag coefficient (\(C_D\)) using the following equation: $$ F_D = \frac{1}{2} \rho U^2 A C_D $$ where \(\rho\) is the air density, \(A\) is the plate area exposed to the air, and \(C_D\) is the drag coefficient. At \(50^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\), the air density (\(\rho\)) can be found in a table as \(\rho = 1.17 \mathrm{~kg/m^3}\). The plate area exposed to the air is \(A = L \times W = 5 \mathrm{~m} \times 5 \mathrm{~m} = 25 \mathrm{~m^2}\). Rearranging the equation for \(C_D\), we get: $$ C_D = \frac{2 F_D}{\rho U^2 A} $$ Now, we can calculate \(C_D\): $$ C_D = \frac{2 \times 2.4 \mathrm{~N}}{1.17 \mathrm{~kg/m^3} \times (10 \mathrm{~m/s})^2 \times 25 \mathrm{~m^2}} \approx 0.00827 $$

Determine the average convection heat transfer coefficient using the momentum-heat transfer analogy

Using the momentum-heat transfer analogy, the average convection heat transfer coefficient (\(h\)) can be found as follows: $$ h = \frac{C_D \rho U c_p}{2} $$ where \(c_p\) is the specific heat capacity of the air. At \(50^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\), \(c_p = 1005 \mathrm{~J/(kg \cdot K)}\). Now, we can calculate \(h\): $$ h = \frac{0.00827 \times 1.17 \mathrm{~kg/m^3} \times 10 \mathrm{~m/s} \times 1005 \mathrm{~J/(kg \cdot K)}}{2} \approx 52 \mathrm{~W/(m^2 \cdot K)} $$

Calculate the rate of heat transfer

Finally, we will determine the rate of heat transfer between the upper surface of the plate and the air. The rate of heat transfer (\(Q\)) can be found using the following equation: $$ Q = h A \Delta T $$ where \(\Delta T\) is the temperature difference between the plate surface and the air. Here, \(\Delta T = 80^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C} = 60 \mathrm{~K}\). Now, we can calculate the rate of heat transfer: $$ Q = 52 \mathrm{~W/(m^2 \cdot K)} \times 25 \mathrm{~m^2} \times 60 \mathrm{~K} = 78,000 \mathrm{~W} $$ Therefore, the average convection heat transfer coefficient is \(52 \mathrm{~W/(m^2 \cdot K)}\) and the rate of heat transfer between the upper surface of the plate and the air is \(78,000 \mathrm{~W}\).