Question

In a production facility, 3-cm-thick large brass plates $\left(k=110 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=8530 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=380 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right.$, and \(\alpha=33.9 \times\) \(10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) ) that are initially at a uniform temperature of \(25^{\circ} \mathrm{C}\) are heated by passing them through an oven maintained at \(700^{\circ} \mathrm{C}\). The plates remain in the oven for a period of \(10 \mathrm{~min}\). Taking the convection heat transfer coefficient to be $h=80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, determine the surface temperature of the plates when they come out of the oven. Solve this problem using the analytical one-term approximation method. Can this problem be solved using lumped system analysis? Justify your answer.

Short answer

Answer: The surface temperature of the plates when they come out of the oven is 700°C. Lumped system analysis cannot be applied to this problem.

Step-by-step solution

Analytical One-Term Approximation Method calculation

We will solve the temperature distribution within the plate using the analytical one-term approximation method. For a semi-infinite solid, the temperature distribution is given by: \(T(x,t) = T_\infty + (T_i - T_\infty) \cdot erf(\frac{x}{2\sqrt{\alpha t}})\) Where \(T(x,t)\) is the temperature at location x and time t, \(T_\infty\) is the temperature of the surrounding medium (oven), \(T_i\) is the initial temperature of the solid, and \(erf\) is the error function. We will find the surface temperature, i.e., \(T(x=0,t)\).

Plug in the given values

Insert the given values of temperature, time, and thermal properties into the formula: \(T(0,t) = 700 + (25 - 700) \cdot erf(\frac{0}{2\sqrt{33.9 \times 10^{-6} \cdot 10 \times 60}})\) Solving for \(T(0,t)\): \(T(0,t) = 700 - 675 \cdot erf(0)\) Since \(erf(0) = 0\), we get: \(T(0,t) = 700 - 675 \cdot 0 = 700\)°C Thus, the surface temperature of the plates when they come out of the oven is \(700^{\circ} \mathrm{C}\).

Evaluate if lumped system analysis can be applied

To check whether lumped system analysis can be applied, we must calculate the Biot number (Bi). The Biot number is given by: \(Bi = \frac{h \cdot L_c}{k}\) Where \(h\) is the convection heat transfer coefficient, \(L_c\) is the characteristic length, and \(k\) is the thermal conductivity. For a plate, the characteristic length is approximated as the half-thickness of the plate (\(L_c = \frac{thickness}{2}\)). Now, plug in the given values: \(Bi = \frac{80 \frac{\mathrm{W}}{\mathrm{m}^{2} \mathrm{K}} \cdot \frac{0.03 \mathrm{m}}{2}}{110 \frac{\mathrm{W}}{\mathrm{m}\cdot \mathrm{K}}} = 0.327\) Lumped system analysis can be applied if \(Bi << 1\). In this case, the Biot number is not much less than 1. So, this problem cannot be solved using lumped system analysis. Therefore, using the analytical one-term approximation method, the surface temperature of the plates when they come out of the oven is \(700^{\circ} \mathrm{C}\). And this problem cannot be solved using lumped system analysis.