Question

A total of 10 rectangular aluminum fins $(k=203 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})$ are placed on the outside flat surface of an electronic device. Each fin is \(100 \mathrm{~mm}\) wide, \(20 \mathrm{~mm}\) high, and $4 \mathrm{~mm}$ thick. The fins are located parallel to each other at a center- to-center distance of \(8 \mathrm{~mm}\). The temperature at the outside surface of the electronic device is \(72^{\circ} \mathrm{C}\). The air is at $20^{\circ} \mathrm{C}\(, and the heat transfer coefficient is \)80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\(. Determine \)(a)$ the rate of heat loss from the electronic device to the surrounding air and \((b)\) the fin effectiveness.

Short answer

b) What is the fin effectiveness of the rectangular aluminum fins?

Step-by-step solution

Compute the parameters

First, we need to calculate surface area of a single fin and the total fin surface area. We also need to find the effective conductance and overall fin resistance. The surface area of a single fin is given by: \(A_{f} = 2LWH\) Where \(L\) is the height, \(W\) is the width, and \(H\) is the thickness of the fin. The total surface area of all the fins: \(A_{total} = 10A_{f}\) The effective conductance is given by: \(Q_{effective} = \frac{k}{H}A_{total}\) The overall resistance is calculated using: \(R = \frac{1}{hA_{total}} + \frac{1}{Q_{effective}}\)

Calculate the rate of heat loss

We can find the rate of heat loss using the equation: \(Q = \frac{T_{device}-T_{air}}{R}\)

Calculate the fin effectiveness

The fin effectiveness can be found using the equation: \(\epsilon = \frac{Q}{Q_{max}}\) Where \(Q_{max}\) is the maximum possible heat transfer rate from the electronic device to the surrounding air without using fins, and can be calculated using: \(Q_{max} = hA_{base}(T_{device}-T_{air})\) Where \(A_{base}\) is the fin base area. Now we can plug in the values and calculate the required quantities.

Plug in the values and find the solution

Given: \(k = 203\mathrm{~W/ m\cdot K}\) \(L = 0.02\mathrm{~m}\) \(W = 0.1\mathrm{~m}\) \(H = 0.004\mathrm{~m}\) \(T_{device} = 72^{\circ} \mathrm{C}\) \(T_{air} = 20^{\circ} \mathrm{C}\) \(h = 80\mathrm{~W/m^{2}\cdot K}\) Calculating the surface area of a single fin: \(A_{f} = 2(0.02 \mathrm{~m})(0.1 \mathrm{~m})(0.004 \mathrm{~m}) = 0.0016\mathrm{~m^{2}}\) The total surface area of all fins: \(A_{total} = 10(0.0016\mathrm{~m^{2}}) = 0.016\mathrm{~m^{2}}\) Now, we can calculate the effective conductance and overall resistance: \(Q_{effective} = \frac{203\mathrm{~W/m\cdot K}}{0.004\mathrm{~m}}(0.016\mathrm{~m^{2}}) = 812\mathrm{~W/ K}\) \(R = \frac{1}{80\mathrm{~W/m^{2}\cdot K} \cdot 0.016\mathrm{~m^{2}}} + \frac{1}{812\mathrm{~W/ K}} = 0.0109375\mathrm{~K/ W}\) Calculation the rate of heat loss: \(Q = \frac{72^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C}}{0.0109375\mathrm{~K/ W}} = 4757.3\mathrm{~W}\) Calculating the base area of fins: \(A_{base} = 10(0.004\mathrm{~m})(0.1\mathrm{~m}) = 0.004\mathrm{~m^{2}}\) The maximum possible heat transfer rate without the use of fins: \(Q_{max} = 80\mathrm{~W/m^{2}\cdot K}\cdot0.004\mathrm{~m^{2}}(72^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C}) = 166.4\mathrm{~W}\) Finally, we can find the fin effectiveness: \(\epsilon = \frac{4757.3\mathrm{~W}}{166.4\mathrm{~W}} = 28.58\) Thus, \((a)\) the rate of heat loss from the electronic device to the surrounding air is approximately \(4757.3\mathrm{~W}\), and \((b)\) the fin effectiveness is approximately \(28.58\).