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Chapter 2: Problem 3

Does a heat flux vector at a point \(P\) on an isothermal surface of a medium have to be perpendicular to the surface at that point? Explain.

Short Answer

Expert verified
Answer: Yes, a heat flux vector at a point P on an isothermal surface has to be perpendicular to the surface at that point, because heat flow occurs only in the direction of a temperature gradient, which is normal to the isothermal surface.

Step by step solution

01

Concept of heat flux vector

A heat flux vector represents the amount of heat transferred per unit area per unit time. It mainly depends on the flow direction of the heat and its thermal gradient.
02

Understanding isothermal surfaces

An isothermal surface is a surface with a constant temperature throughout. No heat flows between points on the isothermal surface since they all have the same temperature.
03

Heat flux direction with respect to isothermal surfaces

Heat flow on a medium occurs from high temperature regions to low temperature regions. Since the temperature is constant along the isothermal surface, the heat cannot flow within the surface itself. Therefore, heat flow only occurs perpendicular to the surface, moving in the direction of the temperature gradient which is normal to the isothermal surface.
04

Conclusion

Yes, a heat flux vector at a point P on an isothermal surface of a medium has to be perpendicular to the surface at that point. The reason being that heat flow occurs only in the direction of a temperature gradient, and the temperature gradient is always normal to the isothermal surface.

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Most popular questions from this chapter

Does heat generation in a solid violate the first law of thermodynamics, which states that energy cannot be created or destroyed? Explain.

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 toward the sky and disregarding radiation exchange with the surrounding 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) \(-k \frac{d T}{d x}=h\left(T-T_{o c}\right)-\alpha_{s} \dot{q}_{s}\) (d) \(h\left(T-T_{\infty}\right)=\alpha_{s} \dot{q}_{s}\) (e) None of them

Consider a steam pipe of length \(L\), inner radius \(r_{1}\), outer radius \(r_{2}\), and constant thermal conductivity \(k\). Steam flows inside the pipe at an average temperature of \(T_{i}\) with a convection heat transfer coefficient of \(h_{i}\). The outer surface of the pipe is exposed to convection to the surrounding air at a temperature of \(T_{0}\) with a heat transfer coefficient of \(h_{\sigma}\). Assuming steady one-dimensional heat conduction through the pipe, \((a)\) express the differential equation and the boundary conditions for heat conduction through the pipe material, (b) obtain a relation for the variation of temperature in the pipe material by solving the differential equation, and (c) obtain a relation for the temperature of the outer surface of the pipe.

What is the difference between an ordinary differential equation and a partial differential equation?

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 \(h=18 \mathrm{~W} / \mathrm{m}^{2}, \mathrm{~K}\). 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.
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