A spherical vessel has an inner radius \(r_{1}\) and an outer radius \(r_{2}\). The inner surface \(\left(r=r_{1}\right)\) of the vessel is subjected to a uniform heat flux \(\dot{q}_{1}\). The outer surface \(\left(r=r_{2}\right)\) is exposed to convection and radiation heat transfer in a surrounding temperature of \(T_{\infty}\). The emissivity and the convection heat transfer coefficient on the outer surface are \(\varepsilon\) and \(h\), respectively. Express the boundary conditions and the differential equation of this heat conduction problem during steady operation.

The heat conduction equation in spherical coordinates is given by: \(\frac{1}{r^2}\frac{\partial}{\partial r}\left( k r^2 \frac{\partial T}{\partial r}\right) = 0\) Since we are asked for steady-state operation, we can ignore the time-dependent part of the equation. Now, we need to find the boundary conditions at the inner (\(r = r_1\)) and outer (\(r = r_2\)) surfaces of the vessel.

At the inner surface of the vessel, we are given a uniform heat flux \(\dot{q}_1\). Heat flux is related to the temperature gradient by Fourier's law: \(\dot{q}_1 = -k \frac{\partial T}{\partial r}\Big|_{r=r_1}\) Thus, the boundary condition at \(r = r_1\) is: \(-k \frac{\partial T}{\partial r}\Big|_{r=r_1} = \dot{q}_1\)

At the outer surface, the vessel loses heat through convection heat transfer to the surrounding fluid and radiation heat transfer. The total heat transfer can be expressed as: \(\dot{q}_2 = -k \frac{\partial T}{\partial r}\Big|_{r=r_2}\) Heat transfer due to convection can be calculated using Newton's law of cooling: \(\dot{q}_{conv} = h \left(T\left(r=r_2\right) - T_\infty\right)\) Heat transfer due to radiation is given by the Stefan-Boltzmann law: \(\dot{q}_{rad} = \varepsilon \sigma \left(T^4\left(r=r_2\right) - T_\infty^4\right)\) Since the total heat transfer should be equal to the sum of convection and radiation heat transfer, we can write the boundary condition at \(r = r_2\) as: \(-k \frac{\partial T}{\partial r}\Big|_{r=r_2} = h \left(T\left(r=r_2\right) - T_\infty\right) + \varepsilon \sigma \left(T^4\left(r=r_2\right) - T_\infty^4\right)\) Now we have the steady-state heat conduction equation and the boundary conditions at both the inner and outer surfaces:

1. Differential equation: \(\frac{1}{r^2}\frac{\partial}{\partial r}\left( k r^2 \frac{\partial T}{\partial r}\right) = 0\) 2. Boundary conditions: (a) At \(r = r_1\): \(-k \frac{\partial T}{\partial r}\Big|_{r=r_1} = \dot{q}_1\) (b) At \(r = r_2\): \(-k \frac{\partial T}{\partial r}\Big|_{r=r_2} = h \left(T\left(r=r_2\right) - T_\infty\right) + \varepsilon \sigma \left(T^4\left(r=r_2\right) - T_\infty^4\right)\)