Question

A \(10-\mathrm{cm}\)-thick aluminum plate $\left(\alpha=97.1 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)$ is being heated in liquid with temperature of \(500^{\circ} \mathrm{C}\). The aluminum plate has a uniform initial temperature of \(25^{\circ} \mathrm{C}\). If the surface temperature of the aluminum plate is approximately the liquid temperature, determine the temperature at the center plane of the aluminum plate after 15 s of heating. Solve this problem using the analytical one-term approximation method.

Short answer

Answer: After 15 seconds of heating, the temperature at the center plane of the aluminum plate is approximately \(463.9^{\circ}\mathrm{C}\).

Step-by-step solution

Define the heat conduction equation

We need to start with the analytical one-term approximation of the transient heat conduction equation for a plate with temperature-dependent boundary conditions. The equation for temperature as a function of time and thickness is given by: $$ T(x, t) = T_\infty + (T_0 - T_\infty) \sum_{n=0}^\infty \frac{4}{(2n+1)\pi} \cos \left( \frac{(2n+1) \pi x}{2L} \right) e^{- \frac{(2n+1)^2 \pi^2 \alpha t}{4L^2}} $$ Where \(T(x, t)\) is the temperature at location \(x\) and time \(t\), \(T_\infty\) is the liquid temperature, \(T_0\) is the initial temperature of the plate, \(L\) is half the thickness of the plate, \(n\) is the number of terms used in the approximation (we will consider only the first term, i.e. \(n=0\)), and `alpha` is the thermal diffusivity of the plate.

Convert the thickness to meters and calculate L

We need to convert the plate thickness from cm to m, and then calculate the value of L, which is half the thickness of the plate. Plate thickness = \(10\ \mathrm{cm} = 0.1\ \mathrm{m}\) L = \(0.1\ \mathrm{m} / 2 = 0.05\ \mathrm{m}\)

Calculate the temperature at the center plane (x=L) after 15 seconds of heating using the first term of the equation (n=0)

Now we can substitute the values into the equation and calculate the temperature at the center plane of the aluminum plate after 15 seconds of heating using only the first term (\(n=0\)) of the equation. $$ T(0.05, 15) = 500 + (25 - 500) \frac{4}{(2 \times 0+1)\pi} \cos \left( \frac{(2 \times 0+1) \pi (0.05)}{2 \times 0.05} \right) e^{- \frac{(2 \times 0+1)^2 \pi^2 (97.1 \times 10^{-6} \times 15)}{4 \times (0.05)^2}} $$

Calculate the temperature

After inserting the given values, we can calculate the temperature at the center plane of the aluminum plate after 15 seconds of heating. $$ T(0.05, 15) = 500 + (25 - 500) \frac{4}{\pi} \cos(\pi) e^{- \frac{\pi^2 (97.1 \times 10^{-6} \times 15)}{(0.05)^2}} $$ $$ T(0.05, 15) \approx 463.9\ ^{\circ}\mathrm{C} $$ After 15 seconds of heating, the temperature at the center plane of the aluminum plate is approximately \(463.9^{\circ}\mathrm{C}\).