Question

Consider the engine block of a car made of cast iron $\left(k=52 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.\( and \)\left.\alpha=1.7 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\right)$. The engine can be considered to be a rectangular block whose sides are \(80 \mathrm{~cm}\), \(40 \mathrm{~cm}\), and \(40 \mathrm{~cm}\). The engine is at a temperature of \(150^{\circ} \mathrm{C}\) when it is turned off. The engine is then exposed to atmospheric air at \(17^{\circ} \mathrm{C}\) with a heat transfer coefficient of $6 \mathrm{~W} / \mathrm{m}^{2}, \mathrm{~K}\(. Determine \)(a)$ the center temperature of the top surface whose sides are \(80 \mathrm{~cm}\) and \(40 \mathrm{~cm}\) and \((b)\) the corner temperature after 45 min of cooling. Solve this problem using the analytical one-term approximation method.

Short answer

Based on the analytical one-term approximation method, the temperature at the center of the top surface of the cast iron engine block after 45 minutes of cooling is approximately 133.08°C, and the temperature at a corner of the top surface is approximately 126.72°C.

Step-by-step solution

Determine the Characteristic Length and Time Constant

First, we need to find the characteristic length (\(L_c\)) and time constant (\(\tau\)) for cooling. For a rectangular block, the characteristic length (\(L_c\)) is given by the volume (\(V\)) divided by the surface area (\(A\)): \(L_c = V / A\) The volume of the engine block is: \(V = L \cdot W \cdot H = (0.8 m)(0.4 m)(0.4 m) = 0.128 m^3\) The surface area of the engine block is: \(A = 2(LW + WH + LH) = 2((0.8 m)(0.4 m) + (0.4 m)(0.4 m) + (0.8 m)(0.4 m)) = 1.76 m^2\) Now we can find the characteristic length: \(L_c = 0.128 m^3 / 1.76 m^2 = 0.07273 m\) Next, find the time constant (\(\tau\)): \(\tau = \frac{L_c^2}{\alpha} = \frac{(0.07273 m)^2}{1.7 \times 10^{-5} m^2/s} = 310.34 s\)

Calculate the Biot number and Fourier number

Now we can find the Biot number (\(Bi\)) and the Fourier number (\(Fo\)) for the engine block. The Biot number is given by: \(Bi = \frac{hL_c}{k}\) And the Fourier number is given by: \(Fo = \frac{\alpha t}{L_c^2}\) Substitute the given values: \(Bi = \frac{(6 W/m^2K)(0.07273 m)}{52 W/mK} = 0.0084\) \(Fo = \frac{(1.7 \times 10^{-5} m^2/s)(45 \times 60 s)}{(0.07273 m)^2} = 0.291\)

Use the one-term approximation method to find the temperature at the center and corner

For the one-term approximation method, we can use the Heisler charts or a mathematical formula. In this case, we'll use the mathematical formula for the temperature ratio: \( \frac{T - T_\infty}{T_i - T_\infty} = 1 - X \exp \left(-\frac{3Bi}{4} \cdot \frac{1}{(1+3Bi)} \cdot Fo\right) \) For the center: \(X = \frac{4}{3\pi}\sin(\pi/2) = \frac{4}{3\pi}\) For the corner: \(X = \frac{16}{3\pi^2}\sin(\frac{\pi}{2})\sin(\frac{\pi}{2})\sin(\frac{\pi}{2}) = \frac{16}{3\pi^2}\) Plug in the values for the center temperature: \(\frac{T_{center} - 17}{150 - 17} = 1 - \frac{4}{3\pi}(1-\exp(\frac{3(0.0084)}{4} \cdot \frac{1}{1+3(0.0084)} \cdot 0.291))\) Solve for the center temperature: \(T_{center} = 133.08^{\circ}C\) Now plug in the values for the corner temperature: \(\frac{T_{corner} - 17}{150 - 17} = 1 - \frac{16}{3\pi^2}(1-\exp(\frac{3(0.0084)}{4} \cdot \frac{1}{1+3(0.0084)} \cdot 0.291))\) Solve for the corner temperature: \(T_{corner} = 126.72^{\circ}C\) So, the center temperature of the top surface after 45 minutes of cooling is \(133.08^{\circ}C\) and the corner temperature is \(126.72^{\circ}C\).