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

Expert verified
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

01

(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 \(h\): \(q''(L)=-k\frac{dT}{dx}(L)=h[T(L)-T_\infty]\).
02

(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: $$ C_1=\frac{h(T_\infty-T_1)}{hL+k} $$ $$ T(x)=\frac{h(T_\infty-T_1)}{hL+k}x+T_1 $$
03

(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 \(7389 W\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider a spherical shell of inner radius \(r_{1}\) and outer radius \(r_{2}\) whose thermal conductivity varies linearly in a specified temperature range as \(k(T)=k_{0}(1+\beta T)\) where \(k_{0}\) and \(\beta\) are two specified constants. The inner surface of the shell is maintained at a constant temperature of \(T_{1}\) while the outer surface is maintained at \(T_{2}\). Assuming steady one- dimensional heat transfer, obtain a relation for \((a)\) the heat transfer rate through the shell and ( \(b\) ) the temperature distribution \(T(r)\) in the shell.

How do differential equations with constant coefficients differ from those with variable coefficients? Give an example for each type.

Consider a homogeneous spherical piece of radioactive material of radius \(r_{o}=0.04 \mathrm{~m}\) that is generating heat at a constant rate of \(\dot{e}_{\text {gen }}=5 \times 10^{7} \mathrm{~W} / \mathrm{m}^{3}\). The heat generated is dissipated to the environment steadily. The outer surface of the sphere is maintained at a uniform temperature of \(110^{\circ} \mathrm{C}\) and the thermal conductivity of the sphere is \(k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Assuming steady one-dimensional heat transfer, \((a)\) express the differential equation and the boundary conditions for heat conduction through the sphere, \((b)\) obtain a relation for the variation of temperature in the sphere by solving the differential equation, and \((c)\) determine the temperature at the center of the sphere.

The outer surface of an engine is situated in a place where oil leakage can occur. Some oils have autoignition temperatures of approximately above \(250^{\circ} \mathrm{C}\). When oil comes in contact with a hot engine surface that has a higher temperature than its autoignition temperature, the oil can ignite spontaneously. Treating the engine housing as a plane wall, the inner surface \((x=0)\) is subjected to \(6 \mathrm{~kW} / \mathrm{m}^{2}\) of heat. The engine housing \((k=13.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) has a thickness of \(1 \mathrm{~cm}\), and the outer surface \((x=L)\) is exposed to an environment where the ambient air is \(35^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). To prevent fire hazard in the event the leaked oil comes in contact with the hot engine surface, the temperature of the engine surface should be kept below \(200^{\circ} \mathrm{C}\). Determine the variation of temperature in the engine housing and the temperatures of the inner and outer surfaces. Is the outer surface temperature of the engine below the safe temperature?

Consider a large plane wall of thickness \(L=0.05 \mathrm{~m}\). The wall surface at \(x=0\) is insulated, while the surface at \(x=L\) is maintained at a temperature of \(30^{\circ} \mathrm{C}\). The thermal conductivity of the wall is \(k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and heat is generated in the wall at a rate of \(\dot{e}_{\text {gen }}=\dot{e}_{0} e^{-0.5 x / L} \mathrm{~W} / \mathrm{m}^{3}\) where \(\dot{e}_{0}=8 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\). Assuming steady one-dimensional heat transfer, \((a)\) express the differential equation and the boundary conditions for heat conduction through the wall, \((b)\) obtain a relation for the variation of temperature in the wall by solving the differential equation, and (c) determine the temperature of the insulated surface of the wall.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free