Question

The local atmospheric pressure in Denver, Colorado (elevation $1610 \mathrm{~m}\( ), is \)83.4 \mathrm{kPa}$. Air at this pressure and at \(30^{\circ} \mathrm{C}\) flows with a velocity of \(6 \mathrm{~m} / \mathrm{s}\) over a \(2.5-\mathrm{m} \times 8-\mathrm{m}\) flat plate whose temperature is \(120^{\circ} \mathrm{C}\). Determine the rate of heat transfer from the plate if the air flows parallel to the \((a) 8-\mathrm{m}-\) long side and \((b)\) the \(2.5\)-m side.

Short answer

In summary, the rate of heat transfer from the flat plate is: - Case (a) (air flowing parallel to the 8-m side): 194,526 W; - Case (b) (air flowing parallel to the 2.5-m side): 372,236 W.

Step-by-step solution

Calculate the Reynolds number at air velocity, density, and dynamic viscosity

First, we need to calculate the Reynolds number (\(Re\)) for both cases (a) and (b). The Reynolds number is a dimensionless quantity that helps determine the flow regime of a fluid. For flow over a flat plate, we can use the following formula for the Reynolds number: \(Re = \frac{\rho V L}{\mu}\) Here, \(V\) is the air velocity (6 m/s), and \(L\) is the characteristic length along which the flow takes place. For case (a) and case (b), we will have two different values of \(L\), i.e., \(8\ m\) and \(2.5\ m\), respectively. Calculate the Reynolds number for both lengths. For case (a): \(Re_a = \frac{1.0\ kg/m^3 * 6\ m/s * 8\ m}{1.983 \times 10^{-5}\ Pa*s} \approx 2417445\) For case (b): \(Re_b = \frac{1.0\ kg/m^3 * 6\ m/s * 2.5\ m}{1.983 \times 10^{-5}\ Pa*s} \approx 755077\)

Calculate the convective heat transfer coefficient using Sieder-Tate equation

Using the Sieder-Tate equation, we can calculate the convective heat transfer coefficient (\(h\)) for both cases (a) and (b): \(h = 0.027 \frac{k^{\frac{3}{2}}}{L\mu}Re^{0.8}Pr^{\frac{1}{3}}\) Here, \(Pr = \frac{Cp\mu}{k}\) is the Prandtl number, and \(Re\) is the Reynolds number calculated for both cases in step 1. Calculate the heat transfer coefficient using this equation for both cases (a) and (b). For case (a): \(Pr_a = \frac{1006 J/kg*K * 1.983 \times 10^{-5}\ Pa*s}{0.0262\ W/m*K} \approx 0.760\) \(h_a = 0.027 \frac{(0.0262\ W/m*K)^{\frac{3}{2}}}{8\ m * 1.983 \times 10^{-5}\ Pa*s} * (2417445)^{0.8}*(0.760)^{\frac{1}{3}} \approx 108.55\ W/m^2*K\) For case (b): \(Pr_b = \frac{1006 J/kg*K * 1.983 \times 10^{-5}\ Pa*s}{0.0262\ W/m*K} \approx 0.760\) \(h_b = 0.027 \frac{(0.0262\ W/m*K)^{\frac{3}{2}}}{2.5\ m * 1.983 \times 10^{-5}\ Pa*s} * (755077)^{0.8}*(0.760)^{\frac{1}{3}} \approx 207.56\ W/m^2*K\)

Calculate the rate of heat transfer using Newton's Law of Cooling

Now, we can calculate the rate of heat transfer (\(Q\)) for both cases using Newton's Law of Cooling: \(Q = h\cdot A\cdot (T_{plate} - T_{air})\) Here, \(A\) is the area of the flat plate exposed to the air flow. Calculate the heat transfer for both cases. For case (a): \(A_a = 2.5\ m * 8\ m = 20\ m^2\) \(Q_a = 108.55\ W/m^2*K * 20\ m^2 * (120^{\circ} \mathrm{C} - 30^{\circ} \mathrm{C}) \approx 194526\ W\) For case (b): \(A_b = 2.5\ m * 8\ m = 20\ m^2\) \(Q_b = 207.56\ W/m^2*K * 20\ m^2 * (120^{\circ} \mathrm{C} - 30^{\circ} \mathrm{C}) \approx 372236\ W\) So, the rate of heat transfer from the plate when air flows parallel to the 8-m long side (case a) is \(194526\ W\) and when air flows parallel to the 2.5-m side (case b) is \(372236\ W\).