Question

Consider a large 5 -cm-thick brass plate \((k=111\) $\mathrm{W} / \mathrm{m} \cdot \mathrm{K})\( in which heat is generated uniformly at a rate of \)2 \times 10^{5} \mathrm{~W} / \mathrm{m}^{3}$. One side of the plate is insulated, while the other side is exposed to an environment at \(25^{\circ} \mathrm{C}\) with a heat transfer coefficient of $44 \mathrm{~W} / \mathrm{m}^{2}, \mathrm{~K}$. Explain where in the plate the highest and the lowest temperatures will occur, and determine their values.

Short answer

Answer: The highest temperature occurs at the insulated side of the plate and is equal to \(25^{\circ}C\), while the lowest temperature occurs at the exposed side of the plate, which can be calculated using the formula \(T(L) = \frac{kC_{1} - q'L^2}{hL} + T_{\infty}\).

Step-by-step solution

Convert Units

First, let's convert the thickness of the brass plate from cm to m: Thickness, \(L = \frac{5\,\text{cm}}{100} = 0.05\,\text{m}\)

Heat Transfer Differential Equation

The temperature distribution in the plate can be determined by the one-dimensional heat transfer equation with heat generation: $$ k\frac{d^2T}{dx^2} + q' = 0 $$ Here, we are given: Thermal conductivity, \(k = 111\) W/m.K Heat generation rate, \(q' = 2\times10^5\) W/m³

Integrate the Heat Transfer Equation Twice

First, we will integrate the heat transfer equation once to get: $$ \frac{dT}{dx} = -\frac{q'x}{k} + C_{1} $$ Then, we integrate it once more to find the temperature distribution: $$ T(x) = -\frac{q'x^2}{2k} + C_{1} x + C_{2} $$

Apply Boundary Conditions

For the insulated side (x=0), heat transfer is zero: $$ 0 = -h[T(0) - T_{\infty}] \Rightarrow T(0) = T_{\infty} = 25^{\circ}C $$ For the exposed side (x=L) with heat transfer coefficient h=44 W/m².K: $$ -k\frac{dT}{dx} |_{x=L} = -h[T(L) - T_{\infty}] $$

Determine Constants C1 and C2

Using the boundary conditions, we can determine the constants C1 and C2: From \(T(0) = T_{\infty}\), we get \(C_{2} = T_{\infty} = 25^{\circ}C\). Let's plug the boundary condition for x=L into the equation: $$ -k(\frac{-q'L}{k} + C_{1}) = -h[T(L) - T_{\infty}] $$ Solving for \(C_{1}\): $$ C_{1} = \frac{hL}{k}(T(L) - T_{\infty}) + q'L $$

Determine the Temperature Distribution

Substitute C1 and C2 back into the temperature equation: $$ T(x) = -\frac{q'x^2}{2k} + (\frac{hL}{k}(T(L) - T_{\infty}) + q'L) \cdot x + 25 $$

Identify the highest and the lowest temperature locations

Since the brass plate has one side insulated, the highest temperature occurs at this boundary (x=0) whereas the lowest temperature is exposed to the environment on the other boundary (x=L). Therefore, the highest temperature is \(T_{max}=T(0)=25^{\circ}C\), and the lowest temperature can be calculated at the exposed side (x=L).

Calculate the lowed temperature

To find \(T_{min}\), we need to calculate T(L) which we can get from the following relation from Step 5: $$ C_{1} = \frac{hL}{k}(T(L) - T_{\infty}) + q'L $$ Plug the known values and solve for \(T(L)\): $$ T(L) = \frac{kC_{1} - q'L^2}{hL} + T_{\infty} $$ Now, we can find \(T_{min}=T(L)\). Thus, the highest temperature occurs at the insulated side of the plate and is equal to \(25^{\circ}C\) while the lowest temperature occurs at the exposed side of the plate, which can be calculated using the formula above.