Question

A hot surface at \(100^{\circ} \mathrm{C}\) is to be cooled by attaching 3 -cm- long, \(0.25\)-cm-diameter aluminum pin fins \((k=237\) $\mathrm{W} / \mathrm{m} \cdot \mathrm{K})\( to it, with a center-to-center distance of \)0.6 \mathrm{~cm}\(. The temperature of the surrounding medium is \)30^{\circ} \mathrm{C}\(, and the heat transfer coefficient on the surfaces is \)35 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Determine the rate of heat transfer from the surface for a \(1-\mathrm{m} \times 1-\mathrm{m}\) section of the plate. Also determine the overall effectiveness of the fins.

Short answer

Answer: The rate of heat transfer from the 1 m x 1 m finned surface is 83,141.10 W and the overall effectiveness of the pin fins is 33.79.

Step-by-step solution

Calculate the number of fins on the plate

To determine the number of fins on the 1 m x 1 m plate, we will first find the number of fins in one row and column and then find the total number of fins. The center-to-center distance between the fins is 0.6 cm, so we will first convert it to meters. Number of fins per row = Length of the plate (in meters)/ Center-to-center distance (in meters) = 1/0.006 = 166.67 ~ 167 fins Number of fins per column = Width of the plate (in meters)/ Center-to-center distance (in meters) = 1/0.006 = 166.67 ~ 167 fins Total number of fins on the plate = Number of fins per row × Number of fins per column = 167 × 167 = 27889 fins

Calculate the fin's efficiency

To calculate the fin's efficiency, we will first need to calculate the fin parameter \(m\), length of the fin \(L\), and the circumference of the fin \(P\). Fin parameter \(m = \sqrt{\frac{hP}{kA_c}}\), where \(h\) is the heat transfer coefficient, \(P\) is the circumference of fin, \(k\) is the thermal conductivity of the material, and \(A_c\) is the cross-sectional area of the fin. \(L = 3 \times 10^{-2} m\), Diameter \(D = 0.25 \times 10^{-2} m\), \(A_c = \pi\frac{D^2}{4} = \pi\frac{(0.25 \times 10^{-2})^2}{4}\), \(P = \pi D = \pi(0.25 \times 10^{-2})\) \(h=35 \mathrm{~W} / \mathrm{m}^2 \cdot \mathrm{K}\) and \(k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) Fin parameter \(m = \sqrt{\frac{35\pi(0.25 \times 10^{-2})}{237\pi\frac{(0.25 \times 10^{-2})^2}{4}}} = 24.34\) \(m^{-1}\) Fin efficiency \(\eta_{fin} = \frac{\tanh{(\frac{mLhP}{kA_c})}}{\frac{mLhP}{kA_c}}= \frac{\tanh{(24.34 \times 0.03)}}{24.34 \times 0.03} = 0.952\)

Calculate the rate of heat transfer from the finned surface

The heat transfer rate of one fin, \(q_f\) is given by the formula, \(q_f = \eta_{fin} h A_f (T_b - T_{ambient})\). \(A_f\) is the fin's heat-transfer area, \(A_f = PL\). Using the fin efficiency \(\eta_{fin}=0.952\), heat transfer coefficient \(h = 35\) W/m²K, \(A_f = \pi(0.25 \times 10^{-2}) \times 0.03\), base temperature \(T_b = 100 ^\circ \mathrm{C}\) and ambient temperature, \(T_{ambient} = 30 ^\circ \mathrm{C}\), we can find the rate of heat transfer from one fin: \(q_f = 0.952\times 35 \times \pi(0.25 \times 10^{-2}) \times 0.03 \times (100-30) = 2.980\) \(\mathrm{W}\) Total heat transfer rate from the surface with fins, \(q_t = q_f \times N\), where \(N\) is the total number of fins. \(q_t = 2.980 \times 27889 = 83141.10\) \(\mathrm{W}\)

Calculate the overall effectiveness of the fins

The overall effectiveness of the fins can be calculated using the formula, \(\eta_o = \frac{q_t}{q_{max}}\), where \(q_{max}\) is the maximum heat transfer rate, which would occur without fins. \(q_{max} = h A_{unfinned} (T_b - T_{ambient})\), \(A_{unfinned}\) is the unfinned area of the plate \(A_{unfinned} = A_{plate} - N \times A_c = 1 \times 1 - 27889 \times \pi\frac{(0.25 \times 10^{-2})^2}{4} = 0.942\) \(m^2\) \(q_{max} = 35 \times 0.942 \times (100 - 30) = 2459.10\) \(\mathrm{W}\) Overall effectiveness of the fins, \(\eta_o = \frac{83141.10}{2459.10} = 33.79\) So the rate of heat transfer from the surface for a 1 m x 1 m section of the plate is \(83141.10\) \(\mathrm{W}\) and the overall effectiveness of the fins is \(33.79\).