Question

Air at \(5^{\circ} \mathrm{C}\), with a convection heat transfer coefficient of \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), is used for cooling metal plates coming out of a heat treatment oven at an initial temperature of \(300^{\circ} \mathrm{C}\). The plates $\left(k=180 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=2800 \mathrm{~kg} / \mathrm{m}^{3}\right.$, and \(\left.c_{p}=880 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) have a thickness of \(10 \mathrm{~mm}\). Using appropriate software, determine the effect of cooling time on the temperature gradient in the metal plates at the surface. By varying the cooling time from 0 to \(3000 \mathrm{~s}\), plot the temperature gradient in the plates at the surface as a function of cooling time. Hint: Use the lumped system analysis to calculate the plate surface temperature. Make sure to verify the application of this method to this problem.

Short answer

The key objective in this exercise is to determine the effect of cooling time on the temperature gradient at the metal surface and plot the temperature gradient as a function of time.

Step-by-step solution

Verify the applicability of the lumped system analysis

To verify the application of the lumped system analysis to this problem, we need to check if the Biot number (Bi) is much less than 1. The Biot number is defined as the ratio of the convective resistance to the conductive resistance. It can be calculated using the following formula: \(\text{Bi} = \frac{hL_c}{k}\) Where: \(h\) - convection heat transfer coefficient (\(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)) \(L_c\) - characteristic length, which is the thickness of the plate for this problem (\(10 \mathrm{~mm}\) or \(0.01 \mathrm{~m}\)) \(k\) - thermal conductivity of the metal plate (\(180 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\)) Now, plug in the values to calculate the Biot number.

Calculate the plate surface temperature

We will use the lumped system analysis to calculate the surface temperature of the metal plate during the cooling process: \(T(t) = T_\infty + (T_i - T_\infty)e^{-\dfrac{ht}{\rho V c_p}}\) Where: \(T(t)\) - temperature of the plate at time \(t\) \(T_\infty\) - air temperature (\(5^\circ \mathrm{C}\)) \(T_i\) - initial temperature of the metal plate (\(300^\circ \mathrm{C}\)) \(h\) - convection heat transfer coefficient (\(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)) \(t\) - cooling time \(\rho\) - density of the metal plate (\(2800 \mathrm{~kg} / \mathrm{m}^{3}\)) \(V\) - volume of the metal plate (depends on the length and width, not given in the problem, but it will be canceled out when calculating the temperature gradient) \(c_p\) - specific heat capacity of the metal plate (\(880 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\))

Determine the temperature gradient

To determine the temperature gradient in the metal plate at the surface, we will find the derivative of the temperature equation concerning the thickness of the plate: \(\dfrac{dT(t)}{dx} = -\dfrac{hk}{\rho c_p}\left(T(t) - T_\infty\right)\) Here, \(x\) is the distance from the surface, and we need to find the gradient on the surface (\(x = 0\)).

Plot the temperature gradient as a function of cooling time

Now, the temperature gradient equation can be used to calculate the temperature gradient on the plate's surface for different cooling times (0s to 3000s). Then it can be plotted as a function of cooling time using appropriate software. Keep in mind: Since the volume of the metal plate was not given in the problem, the plotted temperature gradient values will have an unknown constant factor that depends on the volume. Students can easily find the correct factor using the dimensions and shape of the specific metal plate under investigation.