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Consider one-dimensional heat conduction through a large plane wall with no heat generation that is perfectly insulated on one side and is subjected to convection and radiation on the other side. It is claimed that under steady conditions, the temperature in a plane wall must be uniform (the same everywhere). Do you agree with this claim? Why?

Short Answer

Expert verified
Answer: Yes, the temperature in the wall will be uniform under steady conditions. This is because the heat flow rate through the wall must be constant, and since it is zero at the insulated side, it must be zero everywhere within the wall. This requires the temperature gradient to be zero everywhere, resulting in a uniform temperature throughout the wall.

Step by step solution

01

Understand the basic principles of heat conduction

Heat conduction is the transfer of heat through a material due to a temperature difference within the material. The rate of heat conduction through a material is given by Fourier's law of heat conduction: q = -kA(dT/dx), where q is the rate of heat transfer, k is the thermal conductivity of the material, A is the cross-sectional area through which heat is transferred, and dT/dx is the temperature gradient in the direction of heat flow.
02

Recall the conditions for steady-state conduction

A steady-state condition refers to the situation where the temperature and heat flow inside the material do not change with time. For one-dimensional heat conduction in a plane wall with no heat generation, the conditions for steady-state are: 1. The temperature gradient (dT/dx) remains constant. 2. The heat flow through the wall (q) remains constant.
03

Analyze the boundary conditions of the wall

The wall has two boundary conditions. One side of the wall is perfectly insulated, which means that no heat can pass through it. In other words, the rate of heat transfer (q) at this insulated side is zero. On the other side of the wall, convection and radiation are taking place: 1. Convection: The rate of heat transfer by convection is given by: q_conv = hA(T_surface - T_ambient), where h is the convection heat transfer coefficient, A is the area, T_surface is the temperature of the wall surface, and T_ambient is the temperature of the surrounding environment. 2. Radiation: The rate of heat transfer per unit area by radiation is given by: q_rad = σε(T_surface^4 - T_ambient^4), where σ is the Stefan-Boltzmann constant, ε is the emissivity of the wall surface, and temperatures are in Kelvin.
04

Determine if the temperature is uniform under steady-state conditions

Under steady-state conditions, the temperature gradient (dT/dx) and heat flow (q) must remain constant throughout the wall. Since the heat flow rate at the insulated side is zero, following this condition, the heat flow should be zero throughout the wall. This would indicate that the temperature gradient must be zero everywhere. If the temperature gradient is zero, it means that the temperature is uniform (i.e., the same everywhere) within the wall.
05

Conclusion

Based on the analysis, we can conclude that under steady-state conditions, the temperature in the plane wall with one perfectly insulated side and the other subjected to convection and radiation will indeed be uniform. The claim that the temperature in the wall must be uniform under steady conditions is correct. This is because the heat flow rate through the wall must be constant, and since it is zero at the insulated side, it must be zero everywhere within the wall. This requires the temperature gradient to be zero everywhere, resulting in a uniform temperature throughout the wall.

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Most popular questions from this chapter

The temperatures at the inner and outer surfaces of a \(15-\mathrm{cm}\)-thick plane wall are measured to be \(40^{\circ} \mathrm{C}\) and $28^{\circ} \mathrm{C}$, respectively. The expression for steady, one-dimensional variation of temperature in the wall is (a) \(T(x)=28 x+40\) (b) \(T(x)=-40 x+28\) (c) \(T(x)=40 x+28\) (d) \(T(x)=-80 x+40\) (e) \(T(x)=40 x-80\)

Consider a plane wall of thickness \(L\) 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.

A spherical shell with thermal conductivity \(k\) has inner and outer radii of \(r_{1}\) and \(r_{2}\), respectively. The inner surface of the shell is subjected to a uniform heat flux of \(\dot{q}_{1}\), while the outer surface of the shell is exposed to convection heat transfer with a coefficient \(h\) and an ambient temperature \(T_{\infty}\). Determine the variation of temperature in the shell wall, and show that the outer surface temperature of the shell can be expressed as $T\left(r_{2}\right)=\left(\dot{q}_{1} / h\right)\left(r_{1} / r_{2}\right)^{2}+T_{\infty}$

The conduction equation boundary condition for an adiabatic surface with direction \(n\) being normal to the surface is (a) \(T=0\) (b) \(d T / d n=0\) (c) \(d^{2} T / d n^{2}=0\) (d) \(d^{3} T / d n^{3}=0\) (e) \(-k d T / d n=1\)

In a food processing facility, a spherical container of inner radius $r_{1}=40 \mathrm{~cm}\(, outer radius \)r_{2}=41 \mathrm{~cm}$, and thermal conductivity \(k=1.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is used to store hot water and to keep it at \(100^{\circ} \mathrm{C}\) at all times. To accomplish this, the outer surface of the container is wrapped with a \(500-\mathrm{W}\) electric strip heater and then insulated. The temperature of the inner surface of the container is observed to be nearly \(100^{\circ} \mathrm{C}\) at all times. Assuming 10 percent of the heat generated in the heater is lost through the insulation, \((a)\) 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 material by solving the differential equation, and (c) evaluate the outer surface temperature of the container. Also determine how much water at \(100^{\circ} \mathrm{C}\) this tank can supply steadily if the cold water enters at \(20^{\circ} \mathrm{C}\).

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