Question

Consider a large 3 -cm-thick stainless steel plate \((k=\) \(15.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) in which heat is generated uniformly at a rate of \(5 \times 10^{5} \mathrm{~W} / \mathrm{m}^{3}\). Both sides of the plate are exposed to an environment at \(30^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(60 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Explain where in the plate the highest and the lowest temperatures will occur, and determine their values.

Short answer

Answer: The highest temperature within the stainless steel plate is approximately \(50.981^{\circ}C\), and the lowest temperature is \(30^{\circ}C\).

Step-by-step solution

Write down the heat equation for the plate

The one-dimensional heat conduction equation with heat generation is given by: \(\frac{d}{dx}\left(k \frac{dT}{dx}\right) + \dot{q}_{gen} =0 \) Where \(x\) is the direction perpendicular to the plate, \(k\) is the thermal conductivity of the material, \(T\) is the temperature as a function of \(x\), and \(\dot{q}_{gen}\) is the heat generation rate per unit volume.

Apply boundary conditions

Since both sides of the plate are exposed to the same environment, we can assume the temperature distribution is symmetric across the plate. Due to this symmetry, we only need to analyze half of the plate (from \(x=0\) to \(x=L/2\), where \(L\) is the plate's thickness). The boundary conditions are: 1. At the center of the thickness, there is symmetry, so \(\frac{dT}{dx}(0) = 0\). 2. At \(x=L/2\), we have a convective boundary condition, and can use Newton's law of cooling: \(-k \frac{dT}{dx}(L/2) = h[T(L/2) - T_\infty]\), where \(h\) is the heat transfer coefficient, and \(T_\infty\) is the ambient temperature.

Solve the heat equation

Integrate the heat equation once, considering the first boundary condition: \(\int_{0}^x \frac{d}{dx'}(k \frac{dT}{dx'})dx' = -\int_{0}^x \dot{q}_{gen} dx'\) \(k \frac{dT}{dx} = -\dot{q}_{gen}x + C_1 \) Using the first boundary condition, we find \(C_1 = 0\): \(k \frac{dT}{dx}(0) = 0 \Rightarrow C_1 = 0\) Then the temperature distribution is given by: \(\frac{dT}{dx} = -\frac{\dot{q}_{gen}}{k}x \) Integrating this expression, we obtain the temperature distribution as a function of \(x\): \(T(x) = -\frac{\dot{q}_{gen}}{2k}x^2 + C_2\)

Solve for the integration constant C_2

To determine the integration constant \(C_2\), we use the second boundary condition: \(-k \frac{dT}{dx}(\frac{L}{2}) = h[T(\frac{L}{2}) - T_\infty]\) Substitute the temperature expression: \(-k\left(-\frac{\dot{q}_{gen}}{k}\frac{L}{2}\right) = h\left[-\frac{\dot{q}_{gen}}{2k}\left(\frac{L}{2}\right)^2 + C_2 - T_\infty\right]\) Solve for \(C_2\): \(C_2 = T_\infty - \frac{\dot{q}_{gen}L^2}{16k} + \frac{\dot{q}_{gen}L}{8h}\) Now, we have the complete temperature distribution within the plate: \(T(x) = -\frac{\dot{q}_{gen}}{2k}x^2 + T_\infty - \frac{\dot{q}_{gen}L^2}{16k} + \frac{\dot{q}_{gen}L}{8h}\)

Determine the highest and the lowest temperatures

The highest temperature occurs at the center of the plate, which is at \(x=0\): \(T_{max} = T(0) = T_\infty - \frac{\dot{q}_{gen}L^2}{16k} + \frac{\dot{q}_{gen}L}{8h}\) The lowest temperature occurs at the surface of the plate, which is at \(x=L/2\): \(T_{min} = T(\frac{L}{2}) = T_\infty\) Plug in the given values and calculate the highest and lowest temperatures: \(T_{max} = 30 - \frac{5\times10^5 \times (0.03)^2}{16\times 15.1} + \frac{5\times 10^5\times 0.03}{8\times 60} \approx 50.981 ^{\circ}C\) \(T_{min} = 30^{\circ}C\) Therefore, the highest temperature occurs at the center of the plate and is approximately \(50.981^{\circ}C\), while the lowest temperature occurs at the surface of the plate and is \(30^{\circ}C\).

Key concepts

Thermal Conductivity

Thermal conductivity is a property of materials that describes their ability to conduct heat. In this exercise, we consider a stainless steel plate with a thermal conductivity of \(15.1 \, \mathrm{W/m \cdot K}\). This value tells us how quickly or slowly the material can transfer heat across its surface.

Higher thermal conductivity means better heat conduction. Metals like stainless steel are known for decent thermal conductivity compared to insulators like wood or plastic, which have low values. The formula for thermal conductivity is often used in heat equations to find out how heat distributes in a material.

In our case, the heat conduction in the steel plate is affected by its thermal conductivity, influencing how temperature changes from one part of the plate to another. This property is essential in determining where the hottest and coolest points will be on the plate when heat is generated.

Heat Generation

Heat generation refers to the creation of heat due to internal sources within an object. In this problem, the stainless steel plate generates heat uniformly at a rate of \(5 \times 10^{5} \, \mathrm{W/m^3}\). This means that throughout every unit volume of the material, a substantial amount of heat energy is produced.

Understanding heat generation helps predict how internal temperatures change over time. The heat generated inside the plate adds to the heat flowing through it, which influences the overall temperature distribution.

This internal heat production must be considered along with thermal conductivity to figure out how heat spreads in the plate. The generated heat contributes to increasing temperatures towards the center of the plate, which is a crucial aspect in solving such problems.

Boundary Conditions

Boundary conditions are rules that provide constraints at the borders of a material's surface. They are essential in determining the behavior of heat flow within an object.

In this example, the boundary conditions are set by the plate's environment. Both sides of the plate are at \(30^{\circ}\mathrm{C}\), and a heat transfer coefficient of \(60 \, \mathrm{W/m^2 \cdot K}\) applies, which describes how easily heat passes between the plate and its surroundings.

There is symmetry at the center of the plate, simplifying the problem. At \(x=0\), there's no heat flux, while at the surface \(x=L/2\), Newton's law of cooling applies. These boundary conditions greatly aid in finding the temperature profile across the plate. They impact where and how heat convection and conduction occur.

Temperature Distribution

The temperature distribution is the variation of temperature within the material. Determining how heat spreads along the plate provides insight into the warmest and coldest areas.

Using the thermal conductivity, heat generation, and boundary conditions, the temperature distribution can be derived. In our scenario, the temperature equation is \( T(x) = -\frac{\dot{q}_{gen}}{2k}x^2 + T_\infty - \frac{\dot{q}_{gen}L^2}{16k} + \frac{\dot{q}_{gen}L}{8h}\), where each part contributes to shaping the temperature profile.

With this formula, we calculated that the highest temperature is at the center of the plate, occurring because of the symmetry and high internal heat generation. Meanwhile, the surface closest to the environment remains at the lowest temperature, primarily dictated by external boundary conditions. This distribution is crucial in understanding where to observe the extremes in temperature changes across the material.