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Chapter 2: Problem 88

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.

Short Answer

Expert verified
#answer# The temperature at the insulated surface (\(x=0\)) is given by: \(T|_{x=0}=30+\frac{8\times10^{6}}{30} \int (\int e^{-0.5 x / L} dx) dx |_{x=L}\)

Step by step solution

01

Derive the heat conduction equation

First, we need to express the problem in the form of the heat conduction differential equation. We know that the general form of the heat diffusion equation for a 1D, steady-state condition with heat generation is: \(\frac{\partial ^{2} T}{\partial x^{2}}=-\frac{\dot{e}_{\text {gen }}}{k}\) In this problem, \(\dot{e}_{\text {gen }}=\dot{e}_{0} e^{-0.5x/L}\) where \(\dot{e}_{0}=8\times10^6 \mathrm{~W} / \mathrm{m}^{3}\), and the thermal conductivity, \(k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Substitute these values into the heat diffusion equation: \(\frac{\partial ^{2} T}{\partial x^{2}}=-\frac{8\times10^{6} \mathrm{~W} / \mathrm{m}^{3} e^{-0.5 x / L} }{30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}}\) Now, we have the differential equation and can proceed to solve it.
02

Determine the general solution

First, we separate variables and integrate the differential equation twice to find a general solution: The differential equation is: \(\frac{\partial ^{2} T}{\partial x^{2}}=-\frac{8\times10^{6}}{30}e^{-0.5 x / L}\) Integrate with respect to x: \(\frac{\partial T}{\partial x}=-\frac{8\times10^{6}}{30} \int e^{-0.5 x / L} dx +C_{1}\) Integrate one more time with respect to x: \(T=-\frac{8\times10^{6}}{30} \int (\int e^{-0.5 x / L} dx) dx + C_{1}x +C_{2}\) Now we need to apply the boundary conditions to find the values of the constants \(C_1\) and \(C_2\).
03

Apply the Boundary Conditions

We have two boundary conditions for this problem: 1. The wall surface at \(x=0\) is insulated, so \(\frac{\partial T}{\partial x}|_{x=0}=0\) 2. The wall surface at \(x=L\) is maintained at a temperature of \(30^{\circ} \mathrm{C}\), so \(T|_{x=L}=30\) We apply the first boundary condition: \(\frac{\partial T}{\partial x}|_{x=0}=-\frac{8\times10^{6}}{30} \int e^{-0.5 x / L} dx |_{x=0}+C_{1}=0\) From this, we find that \(C_1=0\) We apply the second boundary condition: \(T|_{x=L}=-\frac{8\times10^{6}}{30} \int (\int e^{-0.5 x / L} dx) dx |_{x=L}+C_{2}=30\) Solving for \(C_2\), we get: \(C_2=30+\frac{8\times10^{6}}{30} \int (\int e^{-0.5 x / L} dx) dx |_{x=L}\) Now, we have a relation for the temperature distribution inside the wall: \(T=-\frac{8\times10^{6}}{30} \int (\int e^{-0.5 x / L} dx) dx + 30 +\frac{8\times10^{6}}{30} \int (\int e^{-0.5 x / L} dx) dx |_{x=L}\)
04

Determine the temperature at the insulated surface

After finding the general solution for the temperature distribution, we can determine the temperature at the insulated surface by plugging \(x=0\) into the previous temperature equation: \(T|_{x=0}=-\frac{8\times10^{6}}{30} \int (\int e^{-0.5 x / L} dx) dx |_{x=0} + 30 +\frac{8\times10^{6}}{30} \int (\int e^{-0.5 x / L} dx) dx |_{x=L}\) Upon solving this equation, we find that: \(T|_{x=0}=30+\frac{8\times10^{6}}{30} \int (\int e^{-0.5 x / L} dx) dx |_{x=L}\)

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Key Concepts

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

Thermal Conductivity
Thermal conductivity is a fundamental property of materials that determines how well they can conduct heat. It is represented by the symbol \( k \) and is measured in watts per meter per Kelvin (W/m路K).
In essence, it provides a measure of the efficiency of heat transfer through a material when there is a temperature difference across it.
A high value of thermal conductivity means that the material is efficient at conducting heat, whereas a low value indicates insulating properties.

In our example problem, the wall has a thermal conductivity of \(k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). This means that the wall is reasonably good at transferring heat. One of the main goals in solving heat conduction problems is to understand how heat moves through materials, given their thermal properties like thermal conductivity.
  • This property plays a critical role in the steady-state heat conduction equation, as we see it influences the temperature gradient \(\frac{\partial T}{\partial x} \) when heat is generated inside the material.
  • Understanding thermal conductivity also helps us evaluate materials best suited for insulation versus those meant for efficient heat transfer.
Boundary Conditions
Boundary conditions are critical in defining the behavior of a system at its limits, helping us solve differential equations uniquely. In the context of heat conduction, they specify temperature or heat flux at the surfaces of the material under consideration.
Our problem has two key boundary conditions that must be used to solve the heat conduction equation accurately:
  • At \(x=0\), the wall is insulated. This implies no heat flow through this boundary, which is mathematically expressed as \(\frac{\partial T}{\partial x}|_{x=0}=0\). This condition ensures that the temperature profile in the wall starts flat or has zero slope at \(x=0\).
  • At \(x=L\), the wall is maintained at a constant temperature of \(30^{\circ} \mathrm{C}\). This means that regardless of the heat generation inside, the temperature at \(x=L\) is fixed at 30掳C, providing a clear reference for solving the system of equations.
Understanding boundary conditions is crucial as they help determine the constants of integration in the solution of differential equations, thereby providing a unique solution to the heat conduction problem.
They are the anchors that fix the temperature profile in place, ensuring a correct and practical solution aligned with physical reality.
Temperature Distribution
The temperature distribution within a material tells us how temperature varies with position inside it. It's crucial in thermal analysis because it allows us to predict which areas get hotter or colder.
For one-dimensional steady-state heat conduction in our insulated wall, the differential equation was given by:\[\frac{\partial ^{2} T}{\partial x^{2}}=-\frac{\dot{e}_{\text {gen }}}{k}\]This equation, once solved with the given boundary conditions, provides us the temperature variation across the wall.Knowing how temperature distributes across a material:
  • Enables us to understand the effect of heat generation within the wall, which follows an exponential decay function \(\dot{e}_{\text{gen}}=\dot{e}_{0} e^{-0.5x/L}\).
  • Allows for controlling thermal stresses, which are critical in materials subjected to high differential temperatures.
The temperature distribution gained from solving the differential equation gives both the temperature at various depths in the wall and crucial insights into managing the heat flow for safety in engineering designs.

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Most popular questions from this chapter

Consider a long solid cylinder of radius \(r_{o}=4 \mathrm{~cm}\) and thermal conductivity \(k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Heat is generated in the cylinder uniformly at a rate of \(\dot{e}_{\text {gen }}=35 \mathrm{~W} / \mathrm{cm}^{3}\). The side surface of the cylinder is maintained at a constant temperature of \(T_{s}=80^{\circ} \mathrm{C}\). The variation of temperature in the cylinder is given by $$ T(r)=\frac{\dot{e}_{\text {gen }} r_{o}^{2}}{k}\left[1-\left[1-\left(\frac{r}{r_{o}}\right)^{2}\right]+T_{s}\right. $$ Based on this relation, determine \((a)\) if the heat conduction is steady or transient, \((b)\) if it is one-, two-, or three-dimensional, and \((c)\) the value of heat flux on the side surface of the cylinder at \(r=r_{o^{*}}\)

Exhaust gases from a manufacturing plant are being discharged through a 10 - \(\mathrm{m}\) tall exhaust stack with outer diameter of \(1 \mathrm{~m}\), wall thickness of \(10 \mathrm{~cm}\), and thermal conductivity of \(40 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The exhaust gases are discharged at a rate of \(1.2 \mathrm{~kg} / \mathrm{s}\), while temperature drop between inlet and exit of the exhaust stack is \(30^{\circ} \mathrm{C}\), and the constant pressure specific heat of the exhaust gasses is \(1600 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). On a particular day, the outer surface of the exhaust stack experiences radiation with the surrounding at \(27^{\circ} \mathrm{C}\), and convection with the ambient air at \(27^{\circ} \mathrm{C}\) also, with an average convection heat transfer coefficient of \(8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Solar radiation is incident on the exhaust stack outer surface at a rate of \(150 \mathrm{~W} / \mathrm{m}^{2}\), and both the emissivity and solar absorptivity of the outer surface are 0.9. Assuming steady one-dimensional heat transfer, (a) obtain the variation of temperature in the exhaust stack wall and (b) determine the inner surface temperature of the exhaust stack.

Consider uniform heat generation in a cylinder and a sphere of equal radius made of the same material in the same environment. Which geometry will have a higher temperature at its center? Why?

A plane wall of thickness \(L\) is subjected to convection at both surfaces with ambient temperature \(T_{\infty 1}\) and heat transfer coefficient \(h_{1}\) at inner surface, and corresponding \(T_{\infty 2}\) and \(h_{2}\) values at the outer surface. Taking the positive direction of \(x\) to be from the inner surface to the outer surface, the correct expression for the convection boundary condition is (a) \(\left.k \frac{d T(0)}{d x}=h_{1}\left[T(0)-T_{\mathrm{o} 1}\right)\right]\) (b) \(\left.k \frac{d T(L)}{d x}=h_{2}\left[T(L)-T_{\infty 2}\right)\right]\) (c) \(\left.-k \frac{d T(0)}{d x}=h_{1}\left[T_{\infty 1}-T_{\infty 2}\right)\right]\) (d) \(\left.-k \frac{d T(L)}{d x}=h_{2}\left[T_{\infty 1}-T_{\infty 22}\right)\right]\) (e) None of them

A solar heat flux \(\dot{q}_{s}\) is incident on a sidewalk whose thermal conductivity is \(k\), solar absorptivity is \(\alpha_{s}\), and convective heat transfer coefficient is \(h\). Taking the positive \(x\) direction to be towards the sky and disregarding radiation exchange with the surroundings surfaces, the correct boundary condition for this sidewalk surface is (a) \(-k \frac{d T}{d x}=\alpha_{s} \dot{q}_{s}\) (b) \(-k \frac{d T}{d x}=h\left(T-T_{\infty}\right)\) (c) \(-k \frac{d T}{d x}=h\left(T-T_{\infty}\right)-\alpha_{s} \dot{q}_{s}\) (d) \(h\left(T-T_{\infty}\right)=\alpha_{s} \dot{q}_{s}\) (e) None of them
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