Question

Consider a large plane wall of thickness \(L=0.4 \mathrm{~m}\), thermal conductivity \(k=1.8 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and surface area \(A=\) \(30 \mathrm{~m}^{2}\). The left side of the wall is maintained at a constant temperature of \(T_{1}=90^{\circ} \mathrm{C}\) while the right side loses heat by convection to the surrounding air at \(T_{\infty}=25^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(h=24 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming constant thermal conductivity and no heat generation in the wall, \((a)\) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the wall, \((b)\) obtain a relation for the variation of temperature in the wall by solving the differential equation, and \((c)\) evaluate the rate of heat transfer through the wall. Answer: (c) \(7389 \mathrm{~W}\)

Short answer

Question: Determine the rate of heat transfer through a large wall with the following conditions: the wall has a surface area of 30 m², a thickness of 0.4 m, a thermal conductivity of 1.8 W/mK, an internal temperature of 90°C, and an external convection heat transfer coefficient of 24 W/m²K. The external temperature is 25°C. Answer: The rate of heat transfer through the wall is 7389 W.

Step-by-step solution

(a) Differential Equation and Boundary Conditions

To write the differential equation for this problem, we will use the heat diffusion equation for one-dimensional steady-state conduction without heat generation: $$ \frac{d^2 T}{d x^2} = 0 $$ We now need to define the boundary conditions, which are given by the fixed temperature on the left side of the wall, \(T_1\), and the heat loss by convection on the right side of the wall: 1. At \(x=0\), \(T=T_1=90^{\circ}C\) 2. At \(x=L\), the heat loss by convection is given by the heat transfer coefficient \(h\): \(q''(L)=-k\frac{dT}{dx}(L)=h[T(L)-T_\infty]\).

(b) Temperature Relation in the Wall

Integrate the differential equation twice with respect to \(x\): $$ \frac{dT}{dx}=C_1 $$ $$ T(x)=C_1x+C_2 $$ Now we can solve the constants \(C_1\) and \(C_2\) using the boundary conditions: 1. At \(x=0\), \(T=T_1 \Rightarrow C_2=T_1\) 2. At \(x=L\), \(q''(L)=-k\frac{dT}{dx}(L)=h[T(L)-T_\infty] \Rightarrow -kC_1=h[C_1L+T_1-T_\infty]\) Solving these simultaneously, we get: $$ C_1=\frac{h(T_\infty-T_1)}{hL+k} $$ $$ T(x)=\frac{h(T_\infty-T_1)}{hL+k}x+T_1 $$

(c) Rate of Heat Transfer Through the Wall

To find the rate of heat transfer, first, we evaluate the heat flux \(q''\) at the left side of the wall (\(x=0\)): $$ q''(0)=-k\frac{dT}{dx}(0)=-kC_1 $$ Substituting the value of \(C_1\): $$ q''(0)=-k\frac{h(T_\infty-T_1)}{hL+k} $$ Finally, since the wall has a surface area \(A\), the total rate of heat transfer \(Q\) is given by: $$ Q=q''(0)A=-kA\frac{h(T_\infty-T_1)}{hL+k} $$ Plugging in the given values of \(k=1.8 W/mK\), \(A=30 m^2\), \(T_1=90^{\circ}C\), \(T_\infty=25^{\circ}C\), \(h=24 W/m^2K\), and \(L=0.4m\): $$ Q=-1.8\cdot30\frac{24(25-90)}{24(0.4)+1.8}\approx -7389 W $$ The negative sign indicates the heat transfer is from left to right, as expected. So, the rate of heat transfer through the wall is \(7389 W\).

Key concepts

Steady-State Heat Transfer

Understanding steady-state heat transfer is crucial when it comes to comprehending how heat moves through materials in a consistent and unchanging manner over time. In this context, steady-state implies that the temperature distribution does not change as time goes on. This is an essential assumption for simplifying complex problems in heat transfer.

In the case of the exercise, we have a wall that experiences a temperature differential – with one side kept at a constant high temperature and the other side in contact with cooler air. Over time, heat flows through the wall and eventually reaches a point where the amount entering one side is equal to the amount leaving the other. No heat accumulates within the wall, and thus, the temperatures within the wall stop changing.

The benefit of analyzing this scenario under the steady-state condition is that it leads to simplification of the governing equations, specifically the heat diffusion equation reducing to \(\frac{d^2 T}{dx^2} = 0\), making the problem much more manageable to solve.

Thermal Conductivity

The concept of thermal conductivity, represented by the symbol \(k\), is a measure of a material's ability to conduct heat. Materials with high thermal conductivity, like metals, transfer heat quickly, whereas insulators, with low thermal conductivity, do so much more slowly.

In our example, the thermal conductivity of the wall material is a constant \(1.8 \frac{W}{m \cdot K}\). This value tells us how readily the wall transfers heat. It plays a central role in determining the temperature gradient across the wall as well as the rate of heat loss. When we combine knowledge of thermal conductivity with the wall's geometry and temperature difference across it, we can use Fourier's law to calculate the rate of heat transfer, as seen with the constant \(C_1\) and ultimately the calculation of heat flux \(q''\).

When students are analyzing heat conduction problems, realizing the importance of \(k\) is pivotal, as it directly influences how effective a material is at transferring heat, which, in the given problem, aids in solving for the temperature distribution and heat transfer rate, ultimately informing about the wall's thermal performance.

Heat Transfer Coefficient

The heat transfer coefficient, denoted as \(h\), is a parameter that quantifies the convective heat transfer between a solid surface and a fluid in motion. It represents how well the fluid carries heat away from the surface (or conversely, towards it).

In practical terms, the higher the heat transfer coefficient, the more efficient the process of heat removal or delivery. For instance, in the provided exercise, heat is lost from the wall to the surrounding air, and the value of \(h\) influences how this process is described. We need \(h\) to set the boundary condition at \(x=L\), where we calculate the convective heat loss using the formula \(q''(L)=-k\frac{dT}{dx}(L)=h[T(L)-T_\infty]\).

This aspect is vital to the final step of determining the rate of heat transfer through the wall, as seen with the expression for \(Q\). Without understanding the role of the heat transfer coefficient, students might struggle to make the connection between the wall's surface temperature, the surrounding air properties, and the amount of heat that the air can carry away from the wall. Therefore, \(h\) is an indispensable value in the toolkit of anyone studying heat transfer.