Chapter 2: Problem 121
Why do we often utilize simplifying assumptions when we derive differential equations?
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chapter 2: Problem 121
Why do we often utilize simplifying assumptions when we derive differential equations?
These are the key concepts you need to understand to accurately answer the question.
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Get started for freeA spherical vessel has an inner radius \(r_{1}\) and an outer radius \(r_{2}\). The inner surface \(\left(r=r_{1}\right)\) of the vessel is subjected to a uniform heat flux \(\dot{q}_{1}\). The outer surface \(\left(r=r_{2}\right)\) is exposed to convection and radiation heat transfer in a surrounding temperature of \(T_{\infty}\). The emissivity and the convection heat transfer coefficient on the outer surface are \(\varepsilon\) and \(h\), respectively. Express the boundary conditions and the differential equation of this heat conduction problem during steady operation.
A circular metal pipe has a wall thickness of \(10 \mathrm{~mm}\) and an inner diameter of \(10 \mathrm{~cm}\). The pipe's outer surface is subjected to a uniform heat flux of \(5 \mathrm{~kW} / \mathrm{m}^{2}\) and has a temperature of \(500^{\circ} \mathrm{C}\). The metal pipe has a variable thermal conductivity given as \(k(T)=k_{0}(1+\beta T)\), where \(k_{0}=7.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\beta=0.0012 \mathrm{~K}^{-1}\), and \(T\) is in \(\mathrm{K}\). Determine the inner surface temperature of the pipe.
A spherical metal ball of radius \(r_{o}\) is heated in an oven to a temperature of \(T_{i}\) throughout and is then taken out of the oven and dropped into a large body of water at \(T_{\infty}\) where it is cooled by convection with an average convection heat transfer coefficient of \(h\). Assuming constant thermal conductivity and transient one-dimensional heat transfer, express the mathematical formulation (the differential equation and the boundary and initial conditions) of this heat conduction problem. Do not solve.
The variation of temperature in a plane wall is determined to be \(T(x)=52 x+25\) where \(x\) is in \(\mathrm{m}\) and \(T\) is in \({ }^{\circ} \mathrm{C}\). If the temperature at one surface is \(38^{\circ} \mathrm{C}\), the thickness of the wall is (a) \(0.10 \mathrm{~m}\) (b) \(0.20 \mathrm{~m}\) (c) \(0.25 \mathrm{~m}\) (d) \(0.40 \mathrm{~m}\) (e) \(0.50 \mathrm{~m}\)
Heat is generated in a long \(0.3-\mathrm{cm}\)-diameter cylindrical electric heater at a rate of \(180 \mathrm{~W} / \mathrm{cm}^{3}\). The heat flux at the surface of the heater in steady operation is (a) \(12.7 \mathrm{~W} / \mathrm{cm}^{2}\) (b) \(13.5 \mathrm{~W} / \mathrm{cm}^{2}\) (c) \(64.7 \mathrm{~W} / \mathrm{cm}^{2}\) (d) \(180 \mathrm{~W} / \mathrm{cm}^{2}\) (e) \(191 \mathrm{~W} / \mathrm{cm}^{2}\)
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