Question

Object 1 with a characteristic length of \(0.5 \mathrm{~m}\) is placed in airflow at \(1 \mathrm{~atm}\) and \(20^{\circ} \mathrm{C}\) with free stream velocity of \(50 \mathrm{~m} / \mathrm{s}\). The heat flux transfer from object 1 when placed in the airflow is measured to be $12,000 \mathrm{~W} / \mathrm{m}^{2}$. If object 2 has the same shape and geometry as object 1 (but with a characteristic length of \(5 \mathrm{~m}\) ) and it is placed in the airflow at \(1 \mathrm{~atm}\) and \(20^{\circ} \mathrm{C}\) with free stream velocity of \(5 \mathrm{~m} / \mathrm{s}\), determine the average convection heat transfer coefficient for object 2 . Both objects are maintained at a constant surface temperature of \(120^{\circ} \mathrm{C}\).

Short answer

Based on the given data and the calculations performed, the average convection heat transfer coefficient for the second object is \(12 \mathrm{~W}/ \mathrm{m}^2 \mathrm{K}\).

Step-by-step solution

Calculate the Reynolds numbers for both objects

The Reynolds number (Re) is a dimensionless quantity that helps determine whether the flow around an object is laminar or turbulent. We can calculate it using the following formula: $$ Re = \frac{V * L}{\nu} $$ where \(V\) is the velocity of the flow, \(L\) is the characteristic length of the object, and \(\nu\) is the kinematic viscosity of the fluid (air). Given that both objects are in airflow at \(1 \mathrm{~atm}\) and \(20^{\circ} \mathrm{C}\), we can find the kinematic viscosity of the air at these conditions, which is approximately \(1.51 * 10^{-5} \mathrm{~m}^2/ \mathrm{s}\). Now, we can calculate the Reynolds number for each object: $$ Re_1 = \frac{50 * 0.5}{1.51 * 10^{-5}} = 1655629 $$ $$ Re_2 = \frac{5 * 5}{1.51 * 10^{-5}} = 1655629 $$

Calculate the Nusselt numbers for both objects

The Nusselt number (Nu) is a dimensionless quantity that represents the ratio of convective to conductive heat transfer. For a given object with the same shape and geometry, the Nusselt number will be the same under different flow conditions. We can relate the Nusselt number to the heat transfer coefficient (h) and the thermal conductivity (k) of the fluid using the following equation: $$ Nu = \frac{h * L}{k} $$ Given the heat flux transfer for object 1, which is \(12,000 \mathrm{~W}/ \mathrm{m}^2\), we can find the heat transfer coefficient for object 1. Since both objects are maintained at a constant surface temperature of \(120 ^{\circ}\mathrm{C}\), the temperature difference between the surface and the free stream is the same (\(120 - 20 = 100 ^{\circ}\mathrm{C}\)). The heat transfer equation is: $$ q = h * A * \Delta T $$ Solving for \(h_1\): $$ h_1 = \frac{q}{A * \Delta T} = \frac{12,000}{1 * 100} = 120 \mathrm{~W}/ \mathrm{m}^2 \mathrm{K} $$ Now, we can find the Nusselt number for object 1 using the thermal conductivity of air at the given conditions, approximately \(0.0262 \mathrm{~W}/ \mathrm{m} \mathrm{K}\). $$ Nu_1 = \frac{h_1 * L_1}{k} = \frac{120 * 0.5}{0.0262} \approx 2293 $$ Since the Nusselt number remains the same for objects with the same shape and geometry, we can assume \(Nu_1 = Nu_2\).

Calculate the average convection heat transfer coefficient for object 2

Now that we have the Nusselt number for object 2, we can find the average convection heat transfer coefficient for object 2 (h_2) using the following equation: $$ h_2 = \frac{Nu_2 * k}{L_2} = \frac{2293 * 0.0262}{5} \approx 12 \mathrm{~W}/ \mathrm{m}^2 \mathrm{K} $$ So, the average convection heat transfer coefficient for object 2 is \(12 \mathrm{~W}/ \mathrm{m}^2 \mathrm{K}\).