Question

Consider a large plane wall of thickness \(I_{\text {, }}\) thermal conductivity \(k\), and surface area \(A\). The left surface of the wall is exposed to the ambient air at \(T_{\infty}\) with a heat transfer coefficient of \(h\) while the right surface is insulated. The variation of temperature in the wall for steady one-dimensional heat conduction with no heat generation is (a) \(T(x)=\frac{h(L-x)}{k} T_{\infty}\) (b) \(T(x)=\frac{k}{h(x+0.5 L)} T_{\infty}\) (c) \(T(x)=\left(1-\frac{x h}{k}\right) T_{\infty}\) (d) \(T(x)=(L-x) T_{\infty}\) (e) \(T(x)=T_{\infty}\)

Short answer

Answer: The correct equation for the temperature distribution in the wall, under the given conditions, is $$T(x) = T_{\infty}$$

Step-by-step solution

Calculate heat flux through the wall

Since the system is in the steady state, there is no change in temperature with time and no internal heat generation. So, the heat flux \(q''\) through the wall must be constant and determined by the surface heat transfer at the left surface. This can be calculated using convective heat transfer equation at the left surface (x=0): $$ q'' = h(T(0) - T_{\infty}) $$

Calculate temperature gradient in the wall

Since the heat flux is constant due to steady-state condition and no heat generation, the temperature gradient in the wall should also be constant. Therefore, the change in temperature with respect to position in the wall should be constant: $$ \frac{dT}{dx}=C $$ Where \(C\) is a constant value.

Check the boundary conditions

We should check the given options against the boundary conditions: 1. At x = 0, the left surface should have a temperature \(T_{0}\) which can be calculated from the heat flux equation mentioned in step 1: $$ T(0)=T_{0} = T_{\infty} +\frac{q''}{h} $$ 2. At x = L, the right surface is insulated, so there should be no heat flux through the surface. This means that the derivative of the temperature with respect to x should be zero at the right surface: $$ \frac{dT}{dx}(L) = 0 $$

Find the correct option by analyzing the given choices with respect to boundary conditions

- Option (a): At x=0, \(T(0)=0\), which does not meet the boundary condition 1. At x=L, \(\frac{dT}{dx}(L) = -\frac{h}{k}\neq0\), which also does not meet the boundary condition 2. So, this option is not correct. - Option (b): At x=0, \(T(0)=\frac{k}{0.5hL}T_{\infty}\), which does not meet the boundary condition 1. At x=L, \(\frac{dT}{dx}(L) \) is not zero (derivative calculation needed), which does not meet the boundary condition 2. So, this option is also not correct. - Option (c): At x=0, \(T(0)= T_{\infty}\), which meets the boundary condition 1. At x=L, \(\frac{dT}{dx}(L) = -\frac{h}{k}\neq0\), which does not meet the boundary condition 2. So, this option is not correct. - Option (d): At x=0, \(T(0)=(L) T_{\infty}\), which does not meet the boundary condition 1. Derivative calculation is not needed, as it already doesn't follow boundary condition 1. - Option (e): Since we've eliminated the other options, this must be the correct choice. However, let's verify: At x=0, \(T(0)=T_{\infty}\), which meets the boundary condition 1. At x=L, \(\frac{dT}{dx}(L) = 0\), which meets the boundary condition 2. Therefore, the correct equation for the temperature distribution in the wall, under given conditions, is: $$T(x) = T_{\infty}$$

Key concepts

Steady-State Condition

The steady-state condition describes a situation in heat conduction where the temperature does not change over time. This means that heat flowing into a material is equal to the heat flowing out, maintaining a constant temperature throughout the process.
In a steady-state, there is no accumulation of heat within the system, implying energy conservation. For a plane wall, this indicates a constant temperature gradient across the wall's thickness—no matter where you measure the temperature, the gradient remains unchanged.
This simplifies calculations as it allows engineers to make sure that the rate of heat transfer does not vary over time, enabling the use of simpler, straightforward equations.

Heat Flux

Heat flux refers to the rate of heat energy transfer through a given surface per unit area, typically measured in watts per square meter \( \left(\mathrm{W/m^2} \right) \). Understanding heat flux is crucial in thermal engineering, as it helps determine how much energy is moving through a system.
When analyzing a wall in contact with air, the heat flux can be calculated using the convective heat transfer equation. At the wall's surface, the heat transferred is related to the difference between the wall temperature and the surrounding air temperature, modulated by the heat transfer coefficient \(h\). The formula used is:

• \(q'' = h(T(0) - T_{\infty})\)

In the context of the problem, this equation allows for calculating how energy leaks via conduction, providing a foundational understanding of how well materials insulate.

Thermal Conductivity

Thermal conductivity is a material property that measures its ability to conduct heat. It is denoted by the symbol \(k\) and is measured in watts per meter-kelvin \( \left( \mathrm{W/(m \cdot K)} \right) \).
In the context of the plane wall, thermal conductivity helps determine how easily heat can pass through the material. A high thermal conductivity value means heat moves through the material quickly, while a low value indicates a good insulator.
When heat conduction is in a steady-state condition, thermal conductivity is crucial in establishing the temperature gradient. With a known thermal conductivity, the temperature difference across the wall's thickness under steady conditions is directly proportional to the heat flux.

Convective Heat Transfer

Convective heat transfer is the process of heat transfer between a solid surface and a fluid (air, water, etc.) in motion. The efficiency of this transfer is largely determined by the convective heat transfer coefficient, symbolized as \(h\).
Convective heat transfer occurs at the boundary between the wall's surface and ambient air. The coefficient \(h\) reflects how effectively heat is exchanged; a higher value means better conduction from the wall to the air or vice versa.
For steady one-dimensional heat conduction, like in this exercise, calculating the convective heat transfer helps determine the heat flux from the wall to the environment. Therefore, understanding this concept supports efficient engineering designs, ensuring materials' surfaces do not lose too much or too little heat, maintaining desired temperature conditions.