Chapter 2: Problem 135
What kind of differential equations can be solved by direct integration?
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chapter 2: Problem 135
What kind of differential equations can be solved by direct integration?
These are the key concepts you need to understand to accurately answer the question.
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Get started for freeA 2-kW resistance heater wire whose thermal conductivity is \(k=10.4 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot \mathrm{R}\) has a radius of \(r_{o}=0.06\) in and a length of \(L=15\) in, and is used for space heating. Assuming constant thermal conductivity and one-dimensional heat transfer, express the mathematical formulation (the differential equation and the boundary conditions) of this heat conduction problem during steady operation. Do not solve.
A long homogeneous resistance wire of radius \(r_{o}=\) \(0.6 \mathrm{~cm}\) and thermal conductivity \(k=15.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is being used to boil water at atmospheric pressure by the passage of electric current. Heat is generated in the wire uniformly as a result of resistance heating at a rate of \(16.4 \mathrm{~W} / \mathrm{cm}^{3}\). The heat generated is transferred to water at \(100^{\circ} \mathrm{C}\) by convection with an average heat transfer coefficient of \(h=3200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming steady one-dimensional heat transfer, \((a)\) express the differential equation and the boundary conditions for heat conduction through the wire, \((b)\) obtain a relation for the variation of temperature in the wire by solving the differential equation, and \((c)\) determine the temperature at the centerline of the wire.
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 steam pipe of length \(L=35 \mathrm{ft}\), inner radius \(r_{1}=2\) in, outer radius \(r_{2}=2.4\) in, and thermal conductivity \(k=8 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\). Steam is flowing through the pipe at an average temperature of \(250^{\circ} \mathrm{F}\), and the average convection heat transfer coefficient on the inner surface is given to be \(h=\) \(15 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\). If the average temperature on the outer surfaces of the pipe is \(T_{2}=160^{\circ} \mathrm{F},(a)\) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the pipe, \((b)\) obtain a relation for the variation of temperature in the pipe by solving the differential equation, and \((c)\) evaluate the rate of heat loss from the steam through the pipe.
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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