Question

Consider a 2-m-long and \(0.7-\mathrm{m}\)-wide stainless steel plate whose thickness is \(0.1 \mathrm{~m}\). The left surface of the plate is exposed to a uniform heat flux of \(2000 \mathrm{~W} / \mathrm{m}^{2}\), while the right surface of the plate is exposed to a convective environment at $0^{\circ} \mathrm{C}\( with \)h=400 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. The thermal conductivity of the stainless steel plate can be assumed to vary linearly with temperature range as \(k(T)=k_{o}(1+\beta T)\) where $k_{o}=48 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\( and \)\beta=9.21 \times 10^{-4 \circ} \mathrm{C}^{-1}$. The stainless steel plate experiences a uniform volumetric heat generation at a rate of $8 \times 10^{5} \mathrm{~W} / \mathrm{m}^{3}$. Assuming steady-state one-dimensional heat transfer, determine the temperature distribution along the plate thickness.

Short answer

Question: Determine the temperature distribution along the thickness of a stainless steel plate given the provided information and conditions. Answer: The temperature distribution along the thickness of the stainless steel plate in terms of \(x\) is given by the equation: $$T_x = \frac{1}{9.21 \times 10^{-4 \circ} \mathrm{C}^{-1}}(\ln \frac{9.21 \times 10^{-4 (1+9.21 \times 10^{-4} \cdot 5)} e^{-9.21 \times 10^{-4} \cdot 48\mathrm{~W}/\mathrm{m} \cdot \mathrm{K} \cdot x /8 \times 10^{5} } }{9.21 \times 10^{-4}}+ 5)$$

Step-by-step solution

Establish the energy balance equation

We will express the energy balance equation for a differential control volume of the stainless steel plate in the one-dimensional steady-state heat transfer process.$$q = -k(T) \frac{dT}{dx}$$where \(q\) is the uniform volumetric heat generation rate, \(k(T)\) is the temperature-dependent thermal conductivity, and \(dT/dx\) is the temperature gradient with respect to the thickness of the plate.

Integrate the equation for temperature gradient

Now we need to integrate the equation with respect to \(T\) and \(x\) to find the temperature distribution along the plate.$$q = -k_{o}(1+\beta T) \frac{dT}{dx}$$Divide both sides by \(k_{o}(1+\beta T)\) and rearrange terms:$$\frac{dT}{1+\beta T} = -\frac{q}{k_{o}} dx$$Integrate both sides with the appropriate limits: For \(T\): From \(T_{L}\) (Left surface temperature) to \(T_x\) (Temperature at any position \(x\)) For \(x\): From \(0\) (left surface) to \(x\) (position in the thickness of plate)$$\int_{T_{L}}^{T_x} \frac{dT}{1+\beta T} = -\frac{q}{k_{o}} \int_{0}^{x} dx$$

Solve the integrals and find the equation for Tx

Solve both integrals and rearrange terms to find the equation for \(T_x\).$$\frac{1}{\beta} \ln{\frac{1+\beta T_x}{1+\beta T_{L}}} = -\frac{q}{k_{o}} x$$Now, we have an equation for the temperature, \(T_x\), at any point \(x\) in the thickness of the plate. Note that the left surface temperature, \(T_{L}\), is still unknown.

The left surface temperature

We're given that the left surface experiences a heat flux of \(2000 \mathrm{~W}/\mathrm{m}^2\). To find \(T_{L}\), we need to apply Fourier's Law of conduction using the given heat flux and the thermal conductivity relationship.$$q'' = -k(T_{L}) \frac{dT}{dx}\Big|_{x=0}$$Due to the symmetry of the problem and the uniform heat generation, \(dT/dx\) at \(x=0\) (left surface) is equal to zero. Therefore, there is no heat conduction at \(x=0\). So, the temperature at the left surface, \(T_{L}\), must be equal to the heat flux divided by the convective heat transfer coefficient.$$q'' = h (T_L - T_{\infty})$$Solving for \(T_L\):$$T_L = T_{\infty} + \frac{q''}{h}$$Substitute the given values, \(T_{\infty} = 0^{\circ} \mathrm{C}\), \(q'' = 2000 \mathrm{~W}/\mathrm{m}^2\), and \(h = 400 \mathrm{~W}/\mathrm{m}^2 \cdot \mathrm{K}\), in the equation.$$T_L = 0^{\circ} \mathrm{C} + \frac{2000 \mathrm{~W}/\mathrm{m}^2}{400 \mathrm{~W}/\mathrm{m}^2 \cdot \mathrm{K}}$$So, we have $$T_L = 5^{\circ} \mathrm{C}$$

Final temperature distribution equation

Now that we have the value of \(T_{L}\), we can substitute it back into the equation for the temperature distribution we found in step 3.$$T_x = \frac{1}{\beta}(\ln \frac{\beta (1+\beta T_{L}) e^{-\beta k_{o} x / q}}{\beta} + T_{L})$$Substitute the given values, \(T_{L} = 5^{\circ} \mathrm{C}\), \(\beta = 9.21 \times 10^{-4 \circ} \mathrm{C}^{-1}\), \(k_{o} = 48 \mathrm{~W}/\mathrm{m} \cdot \mathrm{K}\), and \(q = 8 \times 10^{5} \mathrm{~W}/\mathrm{m}^{3}\),$$T_x = \frac{1}{9.21 \times 10^{-4 \circ} \mathrm{C}^{-1}}(\ln \frac{9.21 \times 10^{-4 (1+9.21 \times 10^{-4} \cdot 5)} e^{-9.21 \times 10^{-4} \cdot 48\mathrm{~W}/\mathrm{m} \cdot \mathrm{K} \cdot x /8 \times 10^{5} } }{9.21 \times 10^{-4}}+ 5)$$This is the final equation for the temperature distribution along the plate thickness in terms of \(x\).