Question

A stainless steel cylindrical rod with a diameter \(D=18 \mathrm{~mm}\) and length \(L=110 \mathrm{~mm}\) is insulated on its exterior surface except for the ends. The mass of the rod is \(0.221 \mathrm{~kg}\). The steady one- dimensional temperature distribution in the rod is $T(x)=310 \mathrm{~K}-20 \mathrm{~K}(x / L)$. Determine the heat flux along the rod.

Short answer

Answer: The heat flux along the rod is approximately 72.36 W.

Step-by-step solution

Calculate the cross-sectional area of the rod

The rod has a diameter of \(D = 18 \mathrm{~mm}\) and is cylindrical in shape. To calculate the cross-sectional area, we can use the following formula for the area of a circle: \(A = \pi r^2\), where \(r\) is the radius of the circle. First, find the radius by dividing the diameter by 2: \(r = D / 2 = 18 / 2 = 9\mathrm{~mm}.\) Now, we can find the area: \(A = \pi (9 \mathrm{~mm})^2 = 81\pi \mathrm{~mm}^2.\)

Calculate the temperature gradient along the rod

The given temperature distribution in the rod is \(T(x) = 310 \mathrm{~K} - 20\mathrm{~K} \cdot (x / L)\). To find the temperature gradient, we will differentiate this equation with respect to \(x\): \(\frac{dT}{dx} = \frac{d}{dx} \left(310 - 20 \frac{x}{L}\right) = -\frac{20}{L}.\)

Use Fourier's Law to calculate the heat flux

According to Fourier's Law, the heat flux along the rod is given by: \(q = -kA\frac{dT}{dx},\) where \(q\) is the heat flux, \(k\) is the thermal conductivity of the material, \(A\) is the cross-sectional area, and \(\frac{dT}{dx}\) is the temperature gradient. Given that the material is stainless steel, we can look up the thermal conductivity value: \(k \approx 16 \mathrm{~W/(m \cdot K)}\). Now, we can plug in the values and solve for the heat flux: \(q = -(16 \mathrm{~W/(m \cdot K)}) (81\pi \mathrm{~mm}^2) \left(-\frac{20}{110 \mathrm{~mm}}\right).\) First, convert the area to square meters: \((81\pi \mathrm{~mm}^2) \times (10^{-6} \mathrm{~m^2/mm^2}) = 81\pi \times 10^{-6} \mathrm{~m^2}.\) Then, calculate the heat flux: \(q = (16 \mathrm{~W/(m \cdot K)}) (81\pi \times 10^{-6} \mathrm{~m^2}) \left(\frac{20}{110 \mathrm{~mm}}\right) = 72.36 \mathrm{~W}.\) The heat flux along the rod is approximately 72.36 W.