A circular metal pipe has a wall thickness of \(10 \mathrm{~mm}\) and an inner diameter of \(10 \mathrm{~cm}\). The pipe's outer surface is subjected to a uniform heat flux of \(5 \mathrm{~kW} / \mathrm{m}^{2}\) and has a temperature of \(500^{\circ} \mathrm{C}\). The metal pipe has a variable thermal conductivity given as \(k(T)=k_{0}(1+\beta T)\), where \(k_{0}=7.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\beta=0.0012 \mathrm{~K}^{-1}\), and \(T\) is in \(\mathrm{K}\). Determine the inner surface temperature of the pipe.

Let's convert the given data to the appropriate SI units. The inner diameter and wall thickness must be in meters and the temperature in Kelvin. Inner diameter: \(10 \mathrm{~cm} = 0.1 \mathrm{~m}\) Wall thickness: \(10 \mathrm{~mm} = 0.01 \mathrm{~m}\) Outer surface temperature: \(500^{\circ} \mathrm{C} = 500 + 273.15 = 773.15 \mathrm{~K}\)

We will now calculate the outer radius \(r_o\) and inner radius \(r_i\) of the pipe using the given inner diameter and wall thickness. \(r_i = \frac{\text{inner diameter}}{2} = \frac{0.1 \mathrm{~m}}{2} = 0.05 \mathrm{~m}\) \(r_o = r_i + \text{wall thickness} = 0.05 \mathrm{~m} + 0.01 \mathrm{~m} = 0.06 \mathrm{~m}\)

Next, we will calculate the heat flux \(q\) through the wall and the temperature gradient \(\frac{dT}{dr}\). Heat flux through the wall: \(q = -k(T) \frac{dT}{dr}\). Temperature gradient: \(\frac{dT}{dr} = -\frac{1}{k(T)}\cdot q\).

Now, we will integrate the temperature gradient and solve for the temperature distribution across the pipe wall. First, substitute the variable thermal conductivity \(k(T)=k_{0}(1+\beta T)\) into the temperature gradient equation: \(\frac{dT}{dr} = -\frac{1}{k_0(1+\beta T)}\cdot q = -\frac{q}{k_{0}(1+\beta T)}\). Now, integrate the temperature gradient equation with respect to \(r\), using the boundary conditions \(T(r=r_i)=T_i\) and \(T(r=r_o)=773.15 \mathrm{~K}\): \(\int_{T_i}^{773.15} \frac{-d T}{1+\beta T} = \int_{0.05}^{0.06} \frac{q \cdot d r}{k_0}\), Let \(\tau = 1 + \beta T\), then \(d\tau = \beta dT\): \(\int_{(1+\beta T_i)}^{(1+\beta773.15)} \frac{-d\tau}{\tau} = \frac{q\cdot0.01}{k_0}\). Now, apply the integral of \(\frac{d\tau}{\tau}\): \(-\ln{(1+\beta773.15)} + \ln{(1+\beta T_i)} = \frac{q\cdot0.01}{k_0}\).

Finally, we will solve for the inner surface temperature, \(T_i\), using the equation obtained in step 4. Rearrange the equation to isolate \(T_i\): \(T_i = \frac{1}{\beta}\left( \left( 1+\beta773.15 \right)e^{\frac{q\cdot0.01}{k_0}} - 1 \right)\). Plug in the values of \(q = 5 \times 10^3 \mathrm{~W} \mathrm{m^{-2}}\), \(k_0 = 7.5 \mathrm{~W} \mathrm{m^{-1}} \mathrm{K^{-1}}\) and \(\beta = 0.0012 \mathrm{~K^{-1}}\): \(T_i = \frac{1}{0.0012}\left( \left( 1 + 0.0012 \cdot 773.15 \right)e^{\frac{5\times10^3\cdot0.01}{7.5}} - 1 \right) \approx 483.97 \mathrm{~K}\). Convert the temperature back to Celsius: \(T_i = 483.97 - 273.15 = 210.82^{\circ} \mathrm{C}\). So, the inner surface temperature of the pipe is \(210.82^{\circ} \mathrm{C}\).