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Consider steady one-dimensional heat conduction through a plane wall, a cylindrical shell, and a spherical shell of uniform thickness with constant thermophysical properties and no thermal energy generation. The geometry in which the variation of temperature in the direction of heat transfer will be linear is (a) plane wall (b) cylindrical shell (c) spherical shell (d) all of them (e) none of them

Short Answer

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(a) Plane wall (b) Cylindrical shell (c) Spherical shell Answer: (a) Plane wall

Step by step solution

01

Recall the governing equations for heat conduction in different geometries

Recall the heat conduction equation for each of the three geometries: 1. Plane wall: \(\frac{\partial^2T}{\partial x^2} = 0\) 2. Cylindrical shell: \(\frac{1}{r}\frac{\partial}{\partial r}\left(r \frac{\partial T}{\partial r}\right) = 0\) 3. Spherical shell: \(\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial T}{\partial r}\right) = 0\) These are the simplified heat conduction equations for each geometry under steady state and one-dimensional conditions.
02

Solve the heat conduction equation for each geometry

Now, we'll solve the heat conduction equations for each geometry to find the temperature distribution in the direction of heat transfer. 1. Plane wall: Since the second derivative of temperature with respect to x is zero, the temperature will vary linearly with x. Solution: \(T(x) = Ax + B\) 2. Cylindrical shell: Solving the cylindrical equation, we have, \(\frac{1}{r}\frac{\partial}{\partial r}\left(r \frac{\partial T}{\partial r}\right) = 0 \Rightarrow r\frac{\partial T}{\partial r} = C\) Integrate once, we obtain, \(\int \frac{\partial T}{\partial r} dr = \int \frac{C}{r} dr \Rightarrow T(r) = C \ln(r) + D\) 3. Spherical shell: Solving the spherical equation, we have, \(\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial T}{\partial r}\right) = 0 \Rightarrow r^2\frac{\partial T}{\partial r} = E\) Integrate once, we obtain, \(\int \frac{\partial T}{\partial r} dr = \int \frac{E}{r^2} dr \Rightarrow T(r) = -\frac{E}{r} + F\)
03

Identify the geometry with linear temperature distribution

Analyzing the temperature distribution solutions for each geometry: 1. Plane wall: \(T(x) = Ax + B\), which is linear with respect to x. 2. Cylindrical shell: \(T(r) = C \ln(r) + D\), which is not linear with respect to r. 3. Spherical shell: \(T(r) = -\frac{E}{r} + F\), which is also not linear with respect to r. From the above solutions, only the plane wall geometry exhibits a linear variation of temperature in the direction of heat transfer. Therefore, the correct answer is: (a) plane wall

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Most popular questions from this chapter

Consider a 20-cm-thick large concrete plane wall $(k=0.77 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})$ subjected to convection on both sides with \(T_{\infty 1}=22^{\circ} \mathrm{C}\) and $h_{1}=8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\( on the inside and \)T_{\infty 22}=8^{\circ} \mathrm{C}$ and \(h_{2}=12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) on the outside. Assuming constant thermal conductivity with no heat generation and negligible radiation, \((a)\) express the differential equations 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 temperatures at the inner and outer surfaces of the wall.

Consider a solid cylindrical rod whose ends are maintained at constant but different temperatures while the side surface is perfectly insulated. There is no heat generation. It is claimed that the temperature along the axis of the rod varies linearly during steady heat conduction. Do you agree with this claim? Why?

How do you recognize a linear homogeneous differential equation? Give an example and explain why it is linear and homogeneous.

What is heat generation? Give some examples.

A cylindrical nuclear fuel rod \(1 \mathrm{~cm}\) in diameter is encased in a concentric tube \(2 \mathrm{~cm}\) in diameter, where cooling water flows through the annular region between the fuel rod $(k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})$ and the concentric tube. Heat is generated uniformly in the rod at a rate of \(50 \mathrm{MW} / \mathrm{m}^{3}\). The convection heat transfer coefficient for the concentric tube surface is $2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. If the surface temperature of the concentric tube is \(40^{\circ} \mathrm{C}\), determine the average temperature of the cooling water. Can one use the given information to determine the surface temperature of the fuel rod? Explain.

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