Question

A triangular-shaped fin on a motorcycle engine is \(0.5 \mathrm{~cm}\) thick at its base and \(3 \mathrm{~cm}\) long (normal distance between the base and the tip of the triangle), and is made of aluminum $(k=150 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})$. This fin is exposed to air with a convective heat transfer coefficient of \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) acting on its surfaces. The efficiency of the fin is 75 percent. If the fin base temperature is \(130^{\circ} \mathrm{C}\) and the air temperature is $25^{\circ} \mathrm{C}$, the heat transfer from this fin per unit width is (a) \(32 \mathrm{~W} / \mathrm{m}\) (b) \(57 \mathrm{~W} / \mathrm{m}\) (c) \(102 \mathrm{~W} / \mathrm{m}\) (d) \(124 \mathrm{~W} / \mathrm{m}\) (e) \(142 \mathrm{~W} / \mathrm{m}\)

Short answer

Answer: (e) 142 W/m

Step-by-step solution

Compute the heat transfer area

In order to compute the heat transfer area of the fin, we need to take into account its triangular shape. We can calculate the area by multiplying the length by the base and dividing by 2. Since the thickness is given in cm and the length in cm, we should convert them into meters before proceeding: \(0.5 \mathrm{~cm} \times 10^{-2} = 5 \times 10^{-3} \mathrm{~m}\) \(3 \mathrm{~cm} \times 10^{-2} = 3 \times 10^{-2} \mathrm{~m}\) Area \(A = 0.5 \times 3 \times 10^{-2} \times 5 \times 10^{-3} = 7.5 \times 10^{-5} \mathrm{~m}^{2}\) Step 2: Calculate the temperature difference

Compute the temperature difference

The temperature difference between the base of the fin and the air can be calculated as: \(\Delta T = T_{base} - T_{air} = 130^{\circ}\mathrm{C} - 25^{\circ}\mathrm{C} = 105^{\circ}\mathrm{C}\) Step 3: Compute the heat transfer without considering efficiency

Calculate the heat transfer without considering efficiency

To compute the heat transfer without considering efficiency, we can use the formula: \(Q = hA\Delta T\) , where \(h\) is the convective heat transfer coefficient, \(A\) is the heat transfer area, and \(\Delta T\) is the temperature difference. Plugging in the given values: \(Q = 30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \times 7.5 \times 10^{-5} \mathrm{~m}^{2} \times 105^{\circ}\mathrm{C} = 236.25 \mathrm{~W} / \mathrm{m}\) Step 4: Compute the heat transfer with considering efficiency

Calculate the heat transfer with considering efficiency

Now, we can calculate the heat transfer per unit width considering the efficiency of the fin, which is 75 percent, or 0.75. So, \(Q_{eff} = Efficiency \times Q = 0.75 \times 236.25= 177.1875 \mathrm{~W} / \mathrm{m}\) Step 5: Compare the calculated value to the options provided

Compare and find the closest option

Now, we will compare the calculated value of \(Q_{eff}\) to the given options: (a) 32 W/m (b) 57 W/m (c) 102 W/m (d) 124 W/m (e) 142 W/m The closest option to our calculated value of \(177.1875 \mathrm{~W} / \mathrm{m}\) is (e) \(142 \mathrm{~W} / \mathrm{m}\). This will be our final answer for the heat transfer per unit width from this fin, considering its efficiency.