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Chapter 2: Problem 112

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. The wall surface at \(x=0\) is maintained at a constant temperature of \(T_{1}\), while the surface at \(x=L\) is maintained at \(T_{2}\). Assuming steady one-dimensional heat transfer, obtain a relation for the heat transfer rate through the wall.

Short Answer

Expert verified
Question: Determine the heat transfer rate through a plane wall of thickness L with a temperature-dependent thermal conductivity k(T) = k_0(1+尾T虏). The temperatures at the two surfaces of the wall are given as T鈧� and T鈧�, assuming steady one-dimensional heat transfer. Answer: The heat transfer rate (q) through the wall can be determined using the following expression: $$q = -\frac{k_{0} \Big[(T_{2} - T_{1}) + \beta \frac{1}{3}(T_{2}^3 - T_{1}^3)\Big]}{L}$$

Step by step solution

01

Fourier's law of heat conduction

Fourier's law of heat conduction states that the heat transfer rate (\(q\)) is proportional to the temperature gradient and the material's thermal conductivity: $$q = -k(T) \frac{dT}{dx}$$ Since the thermal conductivity is temperature-dependent in this problem, we cannot directly integrate this equation to obtain the heat transfer rate.
02

Modify the Fourier's law equation

Let's rearrange the equation in step 1 to separate the \(x\) terms from the \(T\) terms: $$\frac{dx}{dT} = -\frac{k(T)}{q}$$ Now, substitute the expression for \(k(T)\): $$\frac{dx}{dT} = -\frac{k_{0}(1+\beta T^{2})}{q}$$ This equation represents a separable ordinary differential equation with variables \(x\) and \(T\).
03

Integrate the modified Fourier's law equation

Integrate both sides of the equation from step 2 to obtain the relationship between \(x\) and \(T\): $$\int_{x=0}^{x=L} dx = -\frac{k_{0}}{q} \int_{T=T_{1}}^{T=T_{2}} (1+\beta T^{2}) dT$$ Since the integral of \(dx\) from \(0\) to \(L\) is simply \(L\), the equation becomes: $$L = -\frac{k_{0}}{q} \Big[\int_{T=T_{1}}^{T=T_{2}}dT + \beta\int_{T=T_{1}}^{T=T_{2}} T^{2} dT\Big]$$
04

Evaluate the integrals and solve for the heat transfer rate

Evaluate the two integrals on the right-hand side of the equation: $$L = -\frac{k_{0}}{q} \Big[(T_{2} - T_{1}) + \beta \frac{1}{3}(T_{2}^3 - T_{1}^3)\Big]$$ Now, solve for the heat transfer rate \(q\) by rearranging the equation: $$q = -\frac{k_{0} \Big[(T_{2} - T_{1}) + \beta \frac{1}{3}(T_{2}^3 - T_{1}^3)\Big]}{L}$$ This is the final expression for the heat transfer rate through the wall with temperature-dependent thermal conductivity.

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Key Concepts

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

Fourier's Law
One of the foundational principles of heat transfer is Fourier's Law. This law provides a way to calculate the rate at which heat is conducted through a material. At its core, Fourier's Law asserts that the rate of heat transfer ( \(q\) ) is proportional to the temperature gradient and the thermal conductivity of the material. A simplified form of the equation is shown as: \(q = -k(T) \frac{dT}{dx}\). Here,
  • \(q\) stands for the heat transfer rate,
  • \(k(T)\) is the thermal conductivity as a function of temperature,
  • \(\frac{dT}{dx}\) signifies the temperature gradient.
The negative sign indicates that heat flows from the higher temperature region to the lower temperature region. In our exercise scenario, \(k(T)\) is temperature-dependent and requires additional steps to find the solution, making the application of Fourier's Law slightly more complex.
Thermal Conductivity
Thermal conductivity (\(k\)) is a critical parameter in the field of heat transfer. It quantifies a material's ability to conduct heat. High thermal conductivity materials, such as metals, efficiently transfer heat, making them suitable for applications like heat sinks. Conversely, materials with low thermal conductivity, such as wood or foam, act as insulators.
In the exercise, thermal conductivity does not remain constant but varies with temperature, specified by the equation: \(k(T) = k_{0} (1 + \beta T^{2})\), where,
  • \(k_{0}\) represents the base thermal conductivity,
  • \(\beta\) is a constant defining how much the thermal conductivity changes with temperature squared.
As the temperature increases or decreases, the thermal conductivity adjusts accordingly. This property complicates calculations since the value of \(k\) changes along with the temperature, requiring careful integration and analysis to determine heat transfer rates throughout materials with temperature-dependent thermal conductivity.
Separable Differential Equations
In mathematics, separable differential equations are a helpful tool for solving problems where variables can be separated and integrated individually. The basic idea is to rearrange the equations such that one side of the equation only contains one variable, and the other side contains the other.
In the given problem, the modified Fourier's law equation becomes: \(\frac{dx}{dT} = -\frac{k_{0}(1+\beta T^{2})}{q}\). This arrangement allows us to separate variables, leading to an equation where all \(x\)-terms are on one side and all \(T\)-terms on the other,
  • \(\int_{x=0}^{x=L} dx\) for the x-variable,
  • \(\int_{T=T_{1}}^{T=T_{2}} (1+\beta T^{2}) dT\) for the temperature.
By integrating both sides, we can then solve for the relationship between \(x\) and \(T\),allowing us to find the heat transfer rate through the wall. This mathematical process simplifies and provides a clearer understanding of the temperature-dependent behavior of materials.

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Most popular questions from this chapter

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.

Hot water flows through a PVC \((k=0.092 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) pipe whose inner diameter is \(2 \mathrm{~cm}\) and outer diameter is \(2.5 \mathrm{~cm}\). The temperature of the interior surface of this pipe is \(50^{\circ} \mathrm{C}\) and the temperature of the exterior surface is \(20^{\circ} \mathrm{C}\). The rate of heat transfer per unit of pipe length is (a) \(77.7 \mathrm{~W} / \mathrm{m}\) (b) \(89.5 \mathrm{~W} / \mathrm{m}\) (c) \(98.0 \mathrm{~W} / \mathrm{m}\) (d) \(112 \mathrm{~W} / \mathrm{m}\) (e) \(168 \mathrm{~W} / \mathrm{m}\)

Why do we often utilize simplifying assumptions when we derive differential equations?

Heat is generated in a 10 -cm-diameter spherical radioactive material whose thermal conductivity is \(25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) uniformly at a rate of \(15 \mathrm{~W} / \mathrm{cm}^{3}\). If the surface temperature of the material is measured to be \(120^{\circ} \mathrm{C}\), the center temperature of the material during steady operation is (a) \(160^{\circ} \mathrm{C}\) (b) \(205^{\circ} \mathrm{C}\) (c) \(280^{\circ} \mathrm{C}\) (d) \(370^{\circ} \mathrm{C}\) (e) \(495^{\circ} \mathrm{C}\)

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) \(0.50 \mathrm{~m}\)
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