Question

Consider a large plate of thickness \(L\) and thermal conductivity \(k\) in which heat is generated uniformly at a rate of \(\dot{e}_{\text {gen. }}\) One side of the plate is insulated while the other side is exposed to an environment at \(T_{\infty}\) with a heat transfer coefficient of \(h\). \((a)\) Express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the plate, \((b)\) determine the variation of temperature in the plate, and \((c)\) obtain relations for the temperatures on both surfaces and the maximum temperature rise in the plate in terms of given parameters.

Short answer

Answer: The temperature distribution equation within the large plate is given by: $$ T(x) = \frac{-\dot{e}_{\text{gen}}}{2k}x^{2} + \frac{(\dot{e}_{\text{gen}}L+2LhT_{\infty})}{2h}. $$

Step-by-step solution

Steady-State Heat Diffusion Equation

The heat diffusion equation for a large plate with a constant heat generation rate per unit volume, \(\dot{e}_{\text{gen}}\), is given by: $$ k\frac{d^{2}T}{dx^{2}}=-\dot{e}_{\text{gen}} $$ where \(T\) is the temperature within the plate, \(x\) is the coordinate perpendicular to the surfaces of the plate along the thickness, and \(k\) is the thermal conductivity.

Boundary Conditions

Since one side of the plate is insulated, there is no heat transfer across that surface. Therefore, the heat flux is zero at x = 0, which leads to the first boundary condition: $$ \frac{dT}{dx}\Bigg|_{x=0} =0 $$ The other side of the plate is exposed to an environment at temperature \(T_\infty\) with a heat transfer coefficient of \(h\). The second boundary condition is given by Newton's law of cooling at x = L: $$ k\frac{dT}{dx}\Bigg|_{x=L} = h(T(L) - T_{\infty}) $$

Solve the Differential Equation

To solve the steady-state heat diffusion equation \(k\frac{d^{2}T}{dx^{2}}=-\dot{e}_{\text{gen}}\), we first need to integrate it with respect to x. The integration gives us: $$ k\frac{dT}{dx} = -\dot{e}_{\text{gen}} x + C_1, $$ where \(C_1\) is the integration constant. Then, we apply the first boundary condition: \(\frac{dT}{dx}\Bigg|_{x=0} =0\). This concludes that \(C_1=0\), so the equation becomes: $$ k\frac{dT}{dx} = -\dot{e}_{\text{gen}} x. $$ Now, to get the temperature distribution equation, \(T(x)\), we integrate the equation again with respect to \(x\): $$ T(x) = \frac{-\dot{e}_{\text{gen}}}{2k}x^{2} + C_2, $$ where \(C_2\) is another integration constant. To find \(C_2\), we can use the second boundary condition: \(k\frac{dT}{dx}\Bigg|_{x=L} = h(T(L) - T_{\infty})\). After solving for \(C_2\), the temperature distribution equation becomes: $$ T(x) = \frac{-\dot{e}_{\text{gen}}}{2k}x^{2} + \frac{(\dot{e}_{\text{gen}}L+2LhT_{\infty})}{2h} . $$

Surface Temperatures and Maximum Temperature Rise

The temperature at the insulated surface (x = 0) is: $$ T_s = T(0) = \frac{\dot{e}_{\text{gen}}L+2LhT_{\infty}}{2h} . $$ The temperature at the exposed surface (x = L) is: $$ T_e = T(L) = \frac{hT_{\infty}}{h} = T_{\infty}. $$ The maximum temperature rise in the plate is the temperature difference between the insulated surface and the exposed surface: $$ \Delta T_{max} = T_s - T_e = \frac{\dot{e}_{\text{gen}}L+2LhT_{\infty}}{2h} - T_{\infty}. $$

Key concepts

Heat Diffusion Equation

The heat diffusion equation forms the heart of understanding heat transfer in materials. Let's start by looking at the scenario where a plate with certain thickness is experiencing heat generation inside it uniformly. The role of the heat diffusion equation is to mathematically represent how temperature changes within the plate.

The heat diffusion equation for steady-state one-dimensional conduction with a constant heat generation is expressed as \(k\frac{d^{2}T}{dx^{2}}=-\dot{e}_{\text{gen}}\), where \(k\) is the thermal conductivity, \(\frac{d^{2}T}{dx^{2}}\) is the second derivative of temperature with respect to \(x\), the direction of heat transfer, and \(\dot{e}_{\text{gen}}\) is the uniform heat generation rate per unit volume. What this equation essentially states is that the rate of heat conduction is proportional to the negative of the rate at which the temperature changes.

To solve it, you integrate it in steps leading to an equation representing the temperature distribution. This process incorporates constants of integration that will later be defined by boundary conditions, allowing for a solution tailored to our specific setup.

Boundary Conditions In Heat Transfer

In heat transfer problems, boundary conditions are essential to solve the heat diffusion equation. They define the behavior or state of a system at the boundaries, and for our plate scenario, we have two boundary conditions.

The first boundary condition arises because one side of the plate is insulated. An insulated surface implies that there is no heat flux across it, so the first derivative of the temperature with respect to \(x\) is zero at that point: \(\frac{dT}{dx}\bigg|_{x=0} =0\).

The second boundary condition comes from the other side of the plate, which is exposed to an environment at temperature \(T_{\infty}\) with a known heat transfer coefficient \(h\). This is expressed using Newton’s law of cooling: \(k\frac{dT}{dx}\bigg|_{x=L} = h(T(L) - T_{\infty})\). These two conditions are crucial as they allow us to solve for the integration constants in the heat diffusion equation and ultimately find the temperature profile within the plate.

Temperature Distribution

Once the heat diffusion equation has been integrated and the boundary conditions applied, we gain insight into the temperature distribution within the plate. The temperature distribution, \(T(x)\), shows how temperature varies with position within the solid.

For the plate, the temperature distribution is \(T(x) = \frac{-\dot{e}_{\text{gen}}}{2k}x^{2} + \frac{(\dot{e}_{\text{gen}}L+2LhT_{\infty})}{2h}\). This quadratic relationship indicates that temperature variation within the plate is not linear but follows a parabolic profile due to the constant heat generation.

The temperature distribution is also used to determine surface temperatures and the maximum temperature within the plate. The exercise additionally asks for the temperature at both surfaces and the maximum temperature rise which are critical for understanding the thermal behavior of the plate when in operation. This information can influence material selection, thermal insulation considerations, or safety protocols linked to the plate's use in various applications.