Consider a large plane wall of thickness \(L=0.3 \mathrm{~m}\), thermal conductivity \(k=2.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and surface area \(A=\) \(12 \mathrm{~m}^{2}\). The left side of the wall at \(x=0\) is subjected to a net heat flux of \(\dot{q}_{0}=700 \mathrm{~W} / \mathrm{m}^{2}\) while the temperature at that surface is measured to be \(T_{1}=80^{\circ} \mathrm{C}\). Assuming constant thermal conductivity and no heat generation in the wall, \((a)\) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the wall, \((b)\) obtain a relation for the variation of temperature in the wall by solving the differential equation, and \((c)\) evaluate the temperature of the right surface of the wall at \(x=L\).

Step by step solution

01

Write the governing equation for heat conduction

The governing equation for steady one-dimensional heat conduction with no heat generation is given by the following expression: \[\dfrac{d^2T}{dx^2} = 0\]

02

Write Fourier's Law of heat conduction

Fourier's law states the relationship between heat flux and the temperature gradient as follows: \[\dot{q} = -k \dfrac{dT}{dx}\]

03

Integrate the governing equation once

Integration of the governing equation (\(\dfrac{d^2T}{dx^2} = 0\)) with respect to x gives: \[\dfrac{dT}{dx} = C_{1}\] where \(C_{1}\) is the integration constant.

04

Apply the boundary condition at the left surface (x = 0)

At the left surface (\(x=0\)), the net heat flux \(\dot{q}_{0}\) and surface temperature \(T_{1}\) are given. Using Fourier's law, we can find the integration constant \(C_{1}\). \[\dot{q}_{0} = -k \dfrac{dT}{dx} |_{x=0} = -k C_{1} \Rightarrow C_{1} = -\dfrac{\dot{q}_{0}}{k}\]

05

Substitute the value of C1 into the equation for the temperature gradient

Substituting the calculated value of \(C_{1}\) into the equation for the temperature gradient, we have: \[\dfrac{dT}{dx} = -\dfrac{\dot{q}_{0}}{k}\]

06

Integrate the temperature gradient to obtain the temperature variation

Integration of the temperature gradient equation with respect to x gives the temperature variation: \[T(x) - T_{1} = -\dfrac{\dot{q}_{0}}{k}x + C_{2}\] Since the temperature at the left surface is \(T_{1}\), we have the boundary condition \(T(0) = T_{1}\). Applying this condition, we can find the value of \(C_{2}\) to be: \[C_{2} = 0\] So, the temperature variation in the wall can be expressed as: \[T(x) = T_{1} - \dfrac{\dot{q}_{0}}{k}x\]

07

Calculate the temperature at the right surface (x = L)

Now, we can evaluate the temperature at the right surface of the wall (\(x=L\)) using the temperature variation equation: \[T(L) = T_{1} - \dfrac{\dot{q}_{0}}{k}L\] Substitute the given values for \(T_{1}\), \(\dot{q}_{0}\), \(k\), and \(L\): \[T(L) = 80^{\circ} \mathrm{C} - \dfrac{700 \mathrm{~W} / \mathrm{m}^{2}}{2.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}} \cdot 0.3 \mathrm{~m}\] Solve for \(T(L)\): \[T(L) = 44^{\circ} \mathrm{C}\] The temperature of the right surface of the wall at \(x=L\) is \(44^{\circ} \mathrm{C}\).

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