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The temperature of a plane wall during steady onedimensional heat conduction varies linearly when the thermal conductivity is constant. Is this still the case when the thermal conductivity varies linearly with temperature?

Short Answer

Expert verified
Answer: No, the temperature of a plane wall during steady one-dimensional heat conduction does not vary linearly when the thermal conductivity varies linearly with temperature. Instead, the relationship between temperature and position is exponential.

Step by step solution

01

Heat Conduction Equation

The heat conduction equation is given by Fourier's law, which states that the heat flux (q) through a solid is proportional to the temperature gradient: q = -k * (dT/dx) where q is the heat flux, k is the thermal conductivity, and dT/dx is the temperature gradient.
02

Steady State Condition

Since the problem states that the wall is at steady state, this means that the heat flux (q) is constant throughout the wall and does not change over time.
03

Linear Variability of Thermal Conductivity

We are given that the thermal conductivity (k) varies linearly with temperature (T): k = k鈧 + 尾T, where k鈧 is the initial thermal conductivity and 尾 is the constant of proportionality.
04

Substitute Thermal Conductivity Equation

Substitute the equation for the thermal conductivity (Step 3) into Fourier's law (Step 1): q = -(k鈧 + 尾T) * (dT/dx)
05

Solve the Differential Equation

Now we need to solve the above differential equation to determine the relationship between temperature and position (x): (q / (-k鈧 - 尾T)) = dT/dx Now, perform integration to solve for T: 鈭(dT / (k鈧 + 尾T)) = -鈭玠x Let's substitute k鈧 + 尾T = u and d(u) = 尾(dT), then the equation becomes: -鈭(1/尾) * du/u = 鈭玠x Now integrate: (-1/尾) * Ln(u) = -x + C鈧 Ln(u) = 尾x + C鈧 (C鈧 = 尾C鈧) u = e^(尾x) * e^C鈧 k鈧 + 尾T = A * e^(尾x) (A = e^C鈧)
06

Express Temperature as a Function of Position

We already have T in the equation. We just need to express it as a function of x: T(x) = (A * e^(尾x) - k鈧) / 尾
07

Conclusion

The relationship between temperature and position (T(x)) is not linear when thermal conductivity varies linearly with temperature, as evident from the exponential nature of the solution. Thus, the temperature of a plane wall during steady one-dimensional heat conduction does not vary linearly when the thermal conductivity varies linearly with temperature.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Steady State Heat Transfer
In steady state heat transfer, the key idea is that the system's thermal conditions do not change over time. Though heat might be flowing through the material, the temperature at any given point remains constant. This means that the heat flux, or the rate at which heat passes through a surface, is unchanging with respect to time.

  • Steady state implies thermal equilibrium is reached.
  • Heat entering a point equals the heat leaving, maintaining constant temperature distribution.
This type of analysis is crucial for understanding how materials behave under prolonged exposure to heat, and is relevant in designing systems like electronic devices or insulating materials to ensure safety and functionality.
Thermal Conductivity Variation
Thermal conductivity is a measure of a material's ability to conduct heat. In many cases, it is assumed to be constant. However, in reality, it can change with temperature. Linear variation of thermal conductivity means that conductivity changes in direct proportion to temperature.

In the equation used in the solution, thermal conductivity (\(k\)) is related to temperature (\(T\)) by:\[ k = k_0 + \beta T \]where \(k_0\) is the initial thermal conductivity and \(\beta\) is a constant that describes how much \(k\) changes with \(T\). This relationship alters the basic assumptions of heat transfer, making the calculations and predictions more complex. Engineers need to consider these variations to accurately predict how materials will respond under different thermal conditions.
Temperature Gradient
The temperature gradient is a measure of how temperature changes across a distance in a material. It quantifies how quickly temperature rises or falls in a given direction, and is key to understanding heat transfer. According to Fourier's Law, the heat flux is directly proportional to the temperature gradient.

  • A larger gradient means a more rapid change in temperature over distance.
  • In steady state, despite changes in properties like thermal conductivity, the gradient remains constant.
Recognizing the temperature gradient helps in determining the direction and rate of heat flow inside materials, which is crucial for effective thermal management and in preventing potential overheating or freezing in a variety of applications, such as in buildings and electronics.
Fourier's Law
Fourier's Law is a fundamental principle of heat transfer. It states that the rate of heat transfer through a material is directly proportional to the negative of the temperature gradient and the area through which heat is flowing. The law is expressed by the equation:\[ q = -k \frac{dT}{dx} \]where \(q\) is the heat flux, \(k\) is the thermal conductivity, and \(\frac{dT}{dx}\) is the temperature gradient.

  • The negative sign indicates heat flows from high to low temperatures.
  • This linear relationship simplifies many calculations in heat transfer problems.
By applying Fourier's Law, engineers and scientists can design systems that efficiently manage heat flow, critical for improving energy efficiency and safety in a multitude of settings.

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Most popular questions from this chapter

The temperatures at the inner and outer surfaces of a 15 -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 large 3 -cm-thick stainless steel plate \((k=\) \(15.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) in which heat is generated uniformly at a rate of \(5 \times 10^{5} \mathrm{~W} / \mathrm{m}^{3}\). Both sides of the plate are exposed to an environment at \(30^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(60 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Explain where in the plate the highest and the lowest temperatures will occur, and determine their values.

Heat is generated in a \(3-\mathrm{cm}\)-diameter spherical radioactive material uniformly at a rate of \(15 \mathrm{~W} / \mathrm{cm}^{3}\). Heat is dissipated to the surrounding medium at \(25^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(120 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The surface temperature of the material in steady operation is (a) \(56^{\circ} \mathrm{C}\) (b) \(84^{\circ} \mathrm{C}\) (c) \(494^{\circ} \mathrm{C}\) (d) \(650^{\circ} \mathrm{C}\) (e) \(108^{\circ} \mathrm{C}\)

A flat-plate solar collector is used to heat water by having water flow through tubes attached at the back of the thin solar absorber plate. The absorber plate has an emissivity and an absorptivity of \(0.9\). The top surface \((x=0)\) temperature of the absorber is \(T_{0}=35^{\circ} \mathrm{C}\), and solar radiation is incident on the absorber at \(500 \mathrm{~W} / \mathrm{m}^{2}\) with a surrounding temperature of \(0^{\circ} \mathrm{C}\). Convection heat transfer coefficient at the absorber surface is \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while the ambient temperature is \(25^{\circ} \mathrm{C}\). Show that the variation of temperature in the absorber plate can be expressed as \(T(x)=-\left(\dot{q}_{0} / k\right) x+T_{0}\), and determine net heat flux \(\dot{q}_{0}\) absorbed by the solar collector.

Heat is generated uniformly at a rate of \(4.2 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\) in a spherical ball \((k=45 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) of diameter \(24 \mathrm{~cm}\). The ball is exposed to iced-water at \(0^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(1200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the temperatures at the center and the surface of the ball.

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