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

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Question: Determine the heat transfer rate through a plane wall of thickness L with a temperature-dependent thermal conductivity k(T) = k_0(1+βT²). The temperatures at the two surfaces of the wall are given as T₁ and T₂, 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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