Question

Consider two irregularly shaped objects with different characteristic lengths. The characteristic length of the first object is \(L_{1}=0.5 \mathrm{~m}\), and it is maintained at a uniform surface temperature of $T_{s, 1}=350 \mathrm{~K}$. The first object is placed in atmospheric air at a temperature of \(T_{\infty, 1}=250 \mathrm{~K}\) and an air velocity of $V_{1}=20 \mathrm{~m} / \mathrm{s}$. The average heat flux from the first object under these conditions is \(8000 \mathrm{~W} / \mathrm{m}^{2}\). The second object has a characteristic length of \(L_{2}=2.5 \mathrm{~m}\), is maintained at a uniform surface temperature of \(T_{s, 2}=350 \mathrm{~K}\), and is placed in atmospheric air at a temperature of \(T_{\infty, 2}=250 \mathrm{~K}\) and an air velocity of \(V_{2}=4 \mathrm{~m} / \mathrm{s}\). Determine the average convection heat transfer coefficient for the second object.

Short answer

Question: Calculate the heat transfer coefficient for the second object using the given conditions and the Reynolds Analogy. Solution: To find the heat transfer coefficient for the second object, use the formula: \(h_2 = h_1 \times \left(\frac{L_2}{L_1}\right)^{2}\left(\frac{V_2}{V_1}\right) \times \frac{\phi_2}{\phi_1}\) However, we need to determine the values of \(h_1\) and \(\phi_2\) through experimentation or other sources before we can plug them into the equation and find the heat transfer coefficient \(h_2\).

Step-by-step solution

Write the heat transfer equation for both objects.

According to the convection heat transfer equation, we have: \(q = h \times A \times (T_s - T_\infty)\) Where: - \(q\) is the heat transfer rate (W) - \(h\) is the convection heat transfer coefficient (W/m²·K) - \(A\) is the surface area of the object (m²) - \(T_s\) is the surface temperature of the object (K) - \(T_\infty\) is the ambient temperature (K) We can rewrite the equation for both objects as: \(q_1 = h_1 \times A_1 \times (T_{s,1} - T_{\infty,1})\) \(q_2 = h_2 \times A_2 \times (T_{s,2} - T_{\infty,2})\)

Determine the relationship between heat flux and heat transfer.

The heat flux, \(\phi\), is given as the heat transfer per unit area, so we have: \(\phi = \frac{q}{A}\) Thus, the heat transfer rates for both objects can be written in terms of heat flux as: \(q_1 = \phi_1 \times A_1\) \(q_2 = \phi_2 \times A_2\)

Combine equations and solve for the heat transfer coefficient for the second object.

First, we can rewrite the heat transfer equations for both objects using the heat flux: \(h_1 \times A_1 \times (T_{s,1} - T_{\infty,1}) = \phi_1 \times A_1\) \(h_2 \times A_2 \times (T_{s,2} - T_{\infty,2}) = \phi_2 \times A_2\) Now divide the first equation by \(A_1\) and the second equation by \(A_2\): \(h_1 \times (T_{s,1} - T_{\infty,1}) = \phi_1 \quad (1)\) \(h_2 \times (T_{s,2} - T_{\infty,2}) = \phi_2 \quad (2)\) From the problem statement, we have \(\phi_1 = 8000\) W/m², \(T_{s,1} = T_{s,2} = 350\) K, and \(T_{\infty,1} = T_{\infty,2} = 250\) K. We need to find \(h_2\). We can divide equation (1) by equation (2): \(\frac{h_1 \times (T_{s,1} - T_{\infty,1})}{h_2 \times (T_{s,2} - T_{\infty,2})} = \frac{\phi_1}{\phi_2}\) Given that \(T_{s,1} = T_{s,2}\) and \(T_{\infty,1} = T_{\infty,2}\), we can rewrite this as: \(\frac{h_1}{h_2} = \frac{\phi_1}{\phi_2}\) Rearrange to find \(h_2\): \(h_2 = h_1 \times \frac{\phi_2}{\phi_1}\)

Evaluate the given conditions to solve for the heat transfer coefficient for the second object.

From the problem statement, we have: \(L_1 = 0.5\) m, \(V_1 = 20\) m/s, \(\phi_1 = 8000\) W/m² \(L_2 = 2.5\) m, \(V_2 = 4\) m/s We will use the following ratio to relate the heat transfer coefficients: \(h_2 = h_1 \times \frac{\phi_2}{\phi_1}\) Notice that we haven't been given the value for \(h_1\) or \(\phi_2\). However, we can apply the Reynolds Analogy, which states that: \(\frac{h_2}{h_1}=\left(\frac{L_2}{L_1}\right)^{2}\left(\frac{V_2}{V_1}\right)\) Now we can find \(h_2\): \(h_2 = h_1 \times \left(\frac{L_2}{L_1}\right)^{2}\left(\frac{V_2}{V_1}\right) \times \frac{\phi_2}{\phi_1}\) However, since we don't have the values for \(h_1\) or \(\phi_2\), we must rely on the given conditions and the Reynolds Analogy to determine the relationship between these quantities for both objects. So, the heat transfer coefficient for the second object can be determined by: \(h_2 = h_1 \times \left(\frac{L_2}{L_1}\right)^{2}\left(\frac{V_2}{V_1}\right) \times \frac{\phi_2}{\phi_1}\) \(h_2 = h_1 \times \left(\frac{2.5}{0.5}\right)^{2}\left(\frac{4}{20}\right) \times \frac{\phi_2}{8000}\) We now simply need to determine the \(h_1\) and \(\phi_2\) values through experimentation or other sources and plug them into the equation to determine the heat transfer coefficient for the second object, \(h_2\).