Question

A plane wall surface at \(200^{\circ} \mathrm{C}\) is to be cooled with aluminum pin fins of parabolic profile with blunt tips. Each fin has a length of $25 \mathrm{~mm}\( and a base diameter of \)4 \mathrm{~mm}$. The fins are exposed to ambient air at \(25^{\circ} \mathrm{C}\), and the heat transfer coefficient is \(45 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the thermal conductivity of the fins is 230 \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\), determine the heat transfer rate from a single fin and the increase in the rate of heat transfer per \(\mathrm{m}^{2}\) surface area as a result of attaching fins. Assume there are 100 fins per \(\mathrm{m}^{2}\) surface area.

Short answer

Answer: The heat transfer rate from a single fin is approximately 8.247 W, and the increase in the rate of heat transfer per square meter surface area as a result of attaching the fins is approximately 653384.1 W/m².

Step-by-step solution

Define the given parameters

We are given the following parameters: - Wall surface temperature: \(T_s = 200^{\circ} \mathrm{C}\) - Length of the fin: \(L = 25 \mathrm{~mm}\) - Base diameter of the fin: \(D_b = 4 \mathrm{~mm}\) - Ambient air temperature: \(T_{\infty} = 25^{\circ} \mathrm{C}\) - Heat transfer coefficient: \(h = 45 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) - Thermal conductivity of the fins: \(k = 230 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) - Number of fins per \(\mathrm{m}^{2}\) surface area: \(N = 100\)

Calculate cross-sectional area of the fin

We first need the cross-sectional area of the fin for calculating the heat transfer rate through it. For a parabolic profile fin with a base diameter \(D_b\) and length \(L\), the cross-sectional area can be obtained as: \(A_{c} = \frac{\pi}{4} D_b^{2} = \frac{\pi}{4} \cdot (0.004\,\mathrm{m})^{2} = \mathrm{1.257 \times 10^{-5} \, m^2}\)

Calculate the fin efficiency

The efficiency of a fin can be calculated using the formula: \(\eta_\mathrm{f} = \frac{\tanh(mL)}{mL}\) Where \(m\) is given by the formula: \(m = \sqrt{\frac{2h}{kA_{c}}}\) First, we will calculate the value of \(m\): \(m = \sqrt{\frac{2(45)}{230(1.257 \times 10^{-5})}} = 27.73 \mathrm{~m}^{-1}\) Now, we can find the fin efficiency: \(\eta_\mathrm{f} = \frac{\tanh(27.73 \cdot 0.025\,\mathrm{m})}{27.73 \cdot 0.025\,\mathrm{m}} = 0.923\)

Calculate the heat transfer rate from a single fin

The heat transfer rate from a single fin, \(q_f\), can be calculated using the following formula: \(q_\mathrm{f} = \eta_\mathrm{f} \cdot (hA_{t})(T_s - T_{\infty})\) Where \(A_t\) is the total surface area of the fin. In this case, we have a parabolic profile, so the formula for \(A_t\) is: \(A_\mathrm{t} = \pi DL = \pi \cdot (0.004\,\mathrm{m}) \cdot (0.025\,\mathrm{m}) = \mathrm{3.1416 \times 10^{-4}\, m^2}\) Now, we can calculate the heat transfer rate: \(q_\mathrm{f} = (0.923) \cdot (45\,\mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K})(3.1416 \times 10^{-4}\,\mathrm{m^2})(200 - 25) = \mathrm{8.247\, W}\)

Calculate the increase in the rate of heat transfer per \(\mathrm{m}^2\) surface area

To find the increase in heat transfer rate per \(\mathrm{m}^2\) of surface area, we will first calculate the total heat transfer rate from \(N = 100\) fins: \(Q_\mathrm{total} = N \cdot q_\mathrm{f} = 100 \cdot 8.247\,\mathrm{W} = 824.7\,\mathrm{W}\) The increase in the rate of heat transfer per \(\mathrm{m}^2\) surface area is given by: \(\Delta Q = \frac{Q_\mathrm{total}}{A_\mathrm{s}}\) The surface area of the base exposed to the ambient air can be calculated as: \(A_\mathrm{s} = N \cdot A_\mathrm{c} = 100 \cdot (1.257 \times 10^{-5}\,\mathrm{m^2}) = 0.00126\,\mathrm{m^2}\) Finally, we can find the increase in the rate of heat transfer per \(\mathrm{m}^2\) surface area: \(\Delta Q = \frac{824.7\,\mathrm{W}}{0.00126\,\mathrm{m^2}} = 653384.1\,\mathrm{W/m^2}\) So, the heat transfer rate from a single fin is approximately \(8.247\,\mathrm{W}\), and the increase in the rate of heat transfer per \(\mathrm{m}^2\) surface area as a result of attaching the fins is approximately \(653384.1\,\mathrm{W/m^2}\).