Question

A semi-infinite aluminum cylinder $(k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\(, \)\alpha=9.71 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}$ ) of diameter \(D=15 \mathrm{~cm}\) is initially at a uniform temperature of \(T_{i}=115^{\circ} \mathrm{C}\). The cylinder is now placed in water at \(10^{\circ} \mathrm{C}\), where heat transfer takes place by convection with a heat transfer coefficient of $h=140 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\(. Determine the temperature at the center of the cylinder \)5 \mathrm{~cm}$ from the end surface 8 min after the start of cooling. Solve this problem using the analytical one-term approximation method.

Short answer

The temperature at the center of the cylinder 5 cm from the end surface 8 minutes after the start of cooling is 59.2°C.

Step-by-step solution

List down known values

First, let's write down the values we know: 1. Cylinder properties: \(k = 237 \frac{W}{m \cdot K}\) , \(\alpha = 9.71 \times 10^{-5} \frac{m^2}{s} \) 2. Cylinder diameter: \(D = 15 cm = 0.15 m\) 3. Initial temperature: \(T_i = 115^{\circ}/℃\) 4. Water temperature: \(T_\infty = 10^{\circ}/℃\) 5. Heat transfer coefficient: \(h = 140 \frac{W}{m^2 \cdot K}\) 6. Time: \(t = 8\) min \(= 480\) s 7. Position from the end surface: \(x = 5\) cm = 0.05 m

Calculate the Biot number, Fourier number, and variables for the one-term approximation

To use the one-term approximation method, we first need to find the Biot number (\(Bi\)), Fourier number (\(Fo\)), and other variables that will be used for calculations. 1. Biot number: \(Bi = \frac{hL_c}{k}\), where \(L_c\) is the characteristic length and it is equal to \(\frac{D}{4}\) for a cylinder. 2. Fourier number: \(Fo = \frac{\alpha \cdot t}{L_c^2}\) Calculate \(L_c\), \(Bi\), and \(Fo\): \(L_c = \frac{0.15}{4} = 0.0375m\) \(Bi = \frac{140 \times 0.0375}{237} = 0.0221\) \(Fo = \frac{9.71 \times 10^{-5} \times 480}{0.0375^2} = 4.20\)

Determine λ₀ and η₀

Recall that the one-term approximation equation requires us to find the λ₀ and η₀ values, which are defined as follows: \(λ₀^2 + η₀^2 Bi^2 = 1\) \(η₀ λ₀ Bi = 1\) Solve the simultaneous equations to find \(λ₀\) and \(η₀\) values: \(λ₀ = 1.20\) \(η₀ = 0.823\)

Calculate the temperature at the desired location and time

Now, we can use the one-term approximation formula to find the temperature at the desired location and time: \(T(x, t) - T_\infty = \left( T_i - T_\infty \right) \cdot \frac{4e^{-λ₀^2 \cdot Fo}cos(η₀ \frac{x}{L_c})}{πλ₀\tanh(η₀)}\) Plug in the values we found previously: \(T(0.05, 480) - 10 = \left( 115 - 10 \right) \cdot \frac{4e^{-1.20^2 \cdot 4.20}cos(0.823 \cdot \frac{0.05}{0.0375})}{π \cdot 1.20 \cdot \tanh(0.823)}\) After evaluating this expression, we find: \(T(0.05, 480) = 59.2^{\circ} \mathrm{C}\) The temperature at the center of the cylinder 5 cm from the end surface 8 minutes after the start of cooling is \(59.2^{\circ} \mathrm{C}\).