Question

An airfoil with a characteristic length of \(0.2 \mathrm{ft}\) is placed in airflow at 1 atm and \(60^{\circ} \mathrm{F}\) with free stream velocity of $150 \mathrm{ft} / \mathrm{s}\( and convection heat transfer coefficient of \)21 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}$. If a second airfoil with a characteristic length of \(0.4 \mathrm{ft}\) is placed in the airflow at \(1 \mathrm{~atm}\) and \(60^{\circ} \mathrm{F}\) with free stream velocity of \(75 \mathrm{ft} / \mathrm{s}\), determine the heat flux from the second airfoil. Both airfoils are maintained at a constant surface temperature of \(180^{\circ} \mathrm{F}\).

Short answer

Answer: The heat flux from the second airfoil is 1260 Btu/h·ft².

Step-by-step solution

Given Data for Airfoil 1

Characteristic length: \(L_1 = 0.2 \ \mathrm{ft}\). Free stream velocity: \(V_1 = 150 \ \mathrm{ft/s}\). Convection heat transfer coefficient: \(h_1 = 21 \ \mathrm{Btu/h \cdot ft^2 \cdot ^\circ F}\).

Given Data for Airfoil 2

Characteristic length: \(L_2 = 0.4 \ \mathrm{ft}\). Free stream velocity: \(V_2 = 75 \ \mathrm{ft/s}\).

Common Data for Both Airfoils

Air pressure: \(1 \ \mathrm{atm}\). Air temperature: \(T_\infty = 60 \ ^\circ\mathrm{F}\). Surface temperature: \(T_s = 180 \ ^\circ\mathrm{F}\).

Heat Transfer Rate Equation

The convection heat transfer equation we'll use is: \(q = h * A * (T_s - T_\infty)\).

Heat Flux from Airfoil 1

First, we need to find the heat transfer rate from airfoil 1: \(q_1 = h_1 * A_1 * (T_s - T_\infty)\). Since we don't have the surface area \(A_1\), we can rewrite the equation in terms of heat flux, \(\phi_1 = q_1/A_1\): \(\phi_1 = h_1 * (T_s - T_\infty)\). Now, substitute the known values: \(\phi_1 = 21 \ \mathrm{Btu/h \cdot ft^2 \cdot ^\circ F} * (180 \ ^\circ\mathrm{F} - 60 \ ^\circ\mathrm{F}) = 2520 \ \mathrm{Btu/h \cdot ft^2}\).

Finding the Convection Heat Transfer Coefficient for Airfoil 2

Since the airfoils have the same geometry and properties, the Reynolds number and Prandtl number will remain constant. Since we know the convection heat transfer coefficient \(h_1\), we can use the following relation to find \(h_2\): \(\frac{h_2}{h_1} = \frac{L_2}{L_1} * \frac{V_2}{V_1}\). Substitute the given values: \(\frac{h_2}{21 \ \mathrm{Btu/h \cdot ft^2 \cdot ^\circ F}} = \frac{0.4 \ \mathrm{ft}}{0.2 \ \mathrm{ft}} * \frac{75 \ \mathrm{ft/s}}{150 \ \mathrm{ft/s}}\). Solving for \(h_2\): \(h_2 = 10.5 \ \mathrm{Btu/h \cdot ft^2 \cdot ^\circ F}\).

Heat Flux from Airfoil 2

Now, we can find the heat flux from airfoil 2 using the convection heat transfer equation in terms of heat flux, \(\phi_2 = q_2/A_2\): \(\phi_2 = h_2 * (T_s - T_\infty)\). Since we have the values of \(h_2\), \(T_s\), and \(T_\infty\), we can substitute them into the equation: \(\phi_2 = 10.5 \ \mathrm{Btu/h \cdot ft^2 \cdot ^\circ F} * (180 \ ^\circ\mathrm{F} - 60 \ ^\circ\mathrm{F}) = 1260 \ \mathrm{Btu/h \cdot ft^2}\). So, the heat flux from the second airfoil is \(1260 \ \mathrm{Btu/h \cdot ft^2}\).