Question

In the effort to increase the removal of heat from a hot surface at \(120^{\circ} \mathrm{C}\), a cylindrical pin fin $\left(k_{f}=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\( with a diameter of \)5 \mathrm{~mm}$ is attached to the hot surface. Air at \(20^{\circ} \mathrm{C}\) (1 atm) is flowing across the pin fin with a velocity of \(10 \mathrm{~m} / \mathrm{s}\).

Short answer

Based on the information given, we have determined that the Reynolds number for the flow over the cylindrical fin is approximately 3180. The convective heat transfer coefficient was estimated to be around 169 W/m²·K. The formulas for fin efficiency and fin effectiveness have been presented, although the fin length was not provided in the problem statement. These analyses are foundational in assessing the overall heat removal capability of the given fin. To fully evaluate the fin's heat removal performance, the fin length should be provided or considered in the analysis.

Step-by-step solution

List the known variables

We are given the following information: - Hot surface temperature: \(T_s = 120^\circ C\) - Air temperature: \(T_{\infty} = 20^\circ C\) - Air flow velocity: \(V = 10\,\text{m/s}\) - Fin thermal conductivity: \(k_f = 237\,\text{W/m}\cdot\text{K}\) - Fin diameter: \(D = 5\,\text{mm}\)

Calculate the Reynolds number

Before we can determine the convective heat transfer coefficient, we need to calculate the Reynolds number of the flow over the fin. The Reynolds number (\(Re\)) is given by: \(Re = \frac{VD}{\nu}\) Where \(\nu\) is the kinematic viscosity of air at 1 atm and \(20^\circ C\), which can be obtained from the air properties table and is approximately \(1.57 \times 10^{-5}\, \, \text{m}^2/\text{s}\). Now, calculate the Reynolds number: \(Re = \frac{(10\,\text{m/s})(5\times 10^{-3}\,\text{m})}{1.57 \times 10^{-5}\, \, \text{m}^2/\text{s}} \approx 3180\)

Estimate the Nusselt number and the convective heat transfer coefficient

We can use an appropriate correlation for the flow over a cylinder to estimate the Nusselt number, which then allows us to find the convective heat transfer coefficient (\(h\)). For flow over a cylinder with \(Re\) in the range of \(0.1 < Re < 10^5\), the following correlation can be used: \(Nu = \frac{hD}{k_a} = CRe^m Pr^{1/3}\) Where \(k_a\) is the thermal conductivity of air, \(Nu\) is the Nusselt number, \(Re\) is the Reynolds number, \(Pr\) is the Prandtl number, and \(C\) and \(m\) are constants. For our case, \(C = 0.683\) and \(m = 0.466\). The properties of air at \(20^\circ C\), \(1~\text{atm}\) are: \(k_a \approx 0.026\,\,\text{W/m}\cdot\text{K}\) and \(Pr \approx 0.7\). Now, calculate the Nusselt number: \(Nu = 0.683(3180)^{0.466} (0.7)^{1/3} \approx 31.8\) Finally, calculate the convective heat transfer coefficient: \(h = \frac{Nu \times k_a}{D} = \frac{31.8 \times 0.026\,\text{W/m}\cdot\text{K}}{5\times 10^{-3}\,\text{m}} \approx 169\,\text{W/m}^2\cdot\text{K}\)

Determine the fin efficiency and effectiveness

To calculate the fin efficiency (\(\eta_f\)) and fin effectiveness (\(N_u\)) to increase the heat removal, we can use the following formulas: \(\eta_f = \frac{\tanh{(mL)}}{mL}\) \(N_u = 1 + \frac{A_af}{A_s}\eta_f\) Where \(m\) is the fin parameter, given by: \(m = \sqrt{\frac{hP}{k_aA_c}}\) In these formulas, \(A_s\) refers to the base surface area (\(A_s = \pi D L\)), \(A_af\) is fin surface area (\(A_af = \pi D L\)), \(L\) is the length of the fin, \(A_c\) is the cross-sectional area of the fin (\(A_c = \pi(D/2)^2\)), and \(P\) is the perimeter of the fin (\(P = \pi D\)). First, calculate the fin parameter, \(m\): \(m = \sqrt{\frac{169\, \, \text{W/m}^2\cdot\text{K} \times \pi (5\times 10^{-3}\,\text{m})}{0.026\,\text{W/m}\cdot\text{K} \times \pi (2.5\times 10^{-3}\,\text{m})^2}} \approx 15.7 \, \text{m}^{-1}\) Then, calculate the fin efficiency, \(\eta_f\) (assuming a constant fin length, L): \(\eta_f = \frac{\tanh{(15.7 L)}}{15.7 L}\) Finally, calculate the fin effectiveness, \(N_u\): \(N_u = 1 + \frac{\pi D L}{\pi D L}\eta_f = 1 + \eta_f\) The fin efficiency and fin effectiveness depend on the fin length, which is not given in the exercise. To further analyze the heat removal ability of this fin, we would need to know the fin length. However, with increasing length, the efficiency would decrease (due to increased heat losses), but the effectiveness would increase (because of the increased surface area for heat transfer).