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Chapter 2: Problem 153

Consider a cylindrical shell of length \(L\), inner radius \(r_{1}\), and outer radius \(r_{2}\) whose thermal conductivity varies in a specified temperature range as \(k(T)=k_{0}\left(1+\beta T^{2}\right)\) 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 the heat transfer rate through the shell.

Short Answer

Expert verified
Answer: The final relation for the heat transfer rate is given by: $$\frac{dQ}{dt}=\frac{2 \pi k_{0}L}{\ln\left(\frac{r_{2}}{r_{1}}\right)}\left[\frac{\arctan\left(\sqrt{\beta}T_{2}\right)-\arctan\left(\sqrt{\beta}T_{1}\right)}{\sqrt{\beta}}\right]$$

Step by step solution

01

Write down Fourier's Law for heat conduction

Fourier's Law of heat conduction for a cylindrical shell can be written as: $$\frac{dQ}{dt} = -k(T)A\frac{dT}{dr}$$ where \(\frac{dQ}{dt}\) is the heat transfer rate, \(k(T)\) is the thermal conductivity, \(A\) is the surface area of the cylindrical shell, and \(\frac{dT}{dr}\) is the temperature gradient with the radius \(r\).
02

Calculate surface area A

The surface area of a cylindrical shell of length \(L\) and radius \(r\) can be calculated as: $$A = 2\pi rL$$
03

Substitute the given conductivity variation

Now we need to substitute the given conductivity variation \(k(T) = k_{0}(1+\beta T^{2})\) in the Fourier's Law: $$\frac{dQ}{dt} = -k_{0}(1+\beta T^{2})\cdot 2\pi rL \frac{dT}{dr}$$
04

Solve the differential equation

The equation we derived in step 3 is an ordinary differential equation, to solve it, we can separate the variables as follows: $$\int_{T_{1}}^{T_{2}}\frac{1}{1+\beta T^{2}}\,dT = -\frac{k_{0}L}{\frac{dQ}{dt}}\int_{r_{1}}^{r_{2}}\frac{1}{r}\,dr$$
05

Evaluate the integrals

Evaluate the left-hand side (LHS) and the right-hand side (RHS) of the equation obtained in step 4: $$\int_{T_{1}}^{T_{2}}\frac{1}{1+\beta T^{2}}\,dT =\frac{1}{\sqrt{\beta}} \arctan{\left(\sqrt{\beta} T\right)}\Bigg|_{T_{1}}^{T_{2}}$$ and $$-\frac{k_{0}L}{\frac{dQ}{dt}}\int_{r_{1}}^{r_{2}}\frac{1}{r}\,dr = -\frac{k_{0}L}{\frac{dQ}{dt}} \ln(r)\Bigg|_{r_{1}}^{r_{2}}$$
06

Obtain the final relation for heat transfer rate

Equate the LHS and RHS evaluation and simplify the equation obtained in step 5 to get the final relation for the heat transfer rate \(\frac{dQ}{dt}\): $$\frac{dQ}{dt}=\frac{2 \pi k_{0}L}{\ln\left(\frac{r_{2}}{r_{1}}\right)}\left[\frac{\arctan\left(\sqrt{\beta}T_{2}\right)-\arctan\left(\sqrt{\beta}T_{1}\right)}{\sqrt{\beta}}\right]$$

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Most popular questions from this chapter

Consider a short cylinder of radius \(r_{o}\) and height \(H\) in which heat is generated at a constant rate of \(\dot{e}_{\text {gen. }}\). Heat is lost from the cylindrical surface at \(r=r_{o}\) by convection to the surrounding medium at temperature \(T_{\infty}\) with a heat transfer coefficient of \(h\). The bottom surface of the cylinder at \(z=0\) is insulated, while the top surface at \(z=H\) is subjected to uniform heat flux \(\dot{q}_{H}\). Assuming constant thermal conductivity and steady two-dimensional heat transfer, express the mathematical formulation (the differential equation and the boundary conditions) of this heat conduction problem. Do not solve.

In a manufacturing plant, a quench hardening process is used to treat steel ball bearings \((c=500 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=\) \(60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=7900 \mathrm{~kg} / \mathrm{m}^{3}\) ) of \(25 \mathrm{~mm}\) in diameter. After being heated to a prescribed temperature, the steel ball bearings are quenched. Determine the rate of heat loss if the rate of temperature decrease in the ball bearing at a given instant during the quenching process is \(50 \mathrm{~K} / \mathrm{s}\).

A spherical container of inner radius \(r_{1}=2 \mathrm{~m}\), outer radius \(r_{2}=2.1 \mathrm{~m}\), and thermal conductivity \(k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is filled with iced water at \(0^{\circ} \mathrm{C}\). The container is gaining heat by convection from the surrounding air at \(T_{\infty}=25^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(h=18 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming the inner surface temperature of the container to be \(0^{\circ} \mathrm{C},(a)\) express the differential equation and the boundary conditions for steady one- dimensional heat conduction through the container, \((b)\) obtain a relation for the variation of temperature in the container by solving the differential equation, and \((c)\) evaluate the rate of heat gain to the iced water.

Consider a spherical container of inner radius \(r_{1}\), outer radius \(r_{2}\), and thermal conductivity \(k\). Express the boundary condition on the inner surface of the container for steady onedimensional conduction for the following cases: \((a)\) specified temperature of \(50^{\circ} \mathrm{C},(b)\) specified heat flux of \(45 \mathrm{~W} / \mathrm{m}^{2}\) toward the center, (c) convection to a medium at \(T_{\infty}\) with a heat transfer coefficient of \(h\).

Consider a long rectangular bar of length \(a\) in the \(x-\) direction and width \(b\) in the \(y\)-direction that is initially at a uniform temperature of \(T_{i}\). The surfaces of the bar at \(x=0\) and \(y=0\) are insulated, while heat is lost from the other two surfaces by convection to the surrounding medium at temperature \(T_{\infty}\) with a heat transfer coefficient of \(h\). Assuming constant thermal conductivity and transient two-dimensional heat transfer with no heat generation, express the mathematical formulation (the differential equation and the boundary and initial conditions) of this heat conduction problem. Do not solve.
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