Steam at \(320^{\circ} \mathrm{C}\) flows in a stainless steel pipe \((k=\) \(15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) ) whose inner and outer diameters are \(5 \mathrm{~cm}\) and \(5.5 \mathrm{~cm}\), respectively. The pipe is covered with \(3-\mathrm{cm}\)-thick glass wool insulation \((k=0.038 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). Heat is lost to the surroundings at \(5^{\circ} \mathrm{C}\) by natural convection and radiation, with a combined natural convection and radiation heat transfer coefficient of \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Taking the heat transfer coefficient inside the pipe to be \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat loss from the steam per unit length of the pipe. Also determine the temperature drops across the pipe shell and the insulation.

We can model the heat transfer process as a series of thermal resistances. We are given the inner and outer diameters of the stainless steel pipe, the thickness of the glass wool insulation, the thermal conductivities of the materials, as well as the heat transfer coefficients for the inside of the pipe and the natural convection. Using this information, we can calculate the thermal resistance for each layer (inside pipe, pipe wall, insulation, and convection at the outer surface). Let \(R_i\) be the inside heat transfer resistance, \(R_s\) be the steel pipe resistance, \(R_i\) be the insulation resistance, and \(R_o\) be the outside heat transfer resistance. For the inside heat transfer resistance: $$R_i = \frac{1}{h_i A_i}$$ For the steel pipe resistance: $$R_s = \frac{\ln(\frac{D_o}{D_i})}{2 \pi k_s L}$$ For the insulation resistance: $$R_i = \frac{\ln(\frac{D_o + 2 t_i}{D_o})}{2 \pi k_i L}$$ For the outside heat transfer resistance: $$R_o = \frac{1}{h_o A_o}$$ Where \(h_i\) is the inside heat transfer coefficient, \(h_o\) is the outside heat transfer coefficient, \(A_i\) and \(A_o\) are the inside and outside surface areas of the pipe, \(D_i\) and \(D_o\) are the inner and outer diameters of the pipe, \(t_i\) is the insulation thickness, \(k_s\) is the thermal conductivity of the stainless steel pipe, \(k_i\) is the thermal conductivity of the insulation, and \(L\) is the length of the pipe.

The overall resistance of the system can be obtained by adding the thermal resistances of each layer: $$R_{total} = R_i + R_s + R_i + R_o$$ Using the given values for heat transfer coefficients and thermal conductivities, as well as the inside and outside surface areas of the pipe, we can calculate the overall resistance. Now, the heat loss rate per unit length of the pipe can be calculated using the temperature difference between the steam and the ambient and the overall resistance: $$q = \frac{T_{steam} - T_{ambient}}{R_{total}}$$ Where \(q\) is the heat loss rate per unit length, \(T_{steam}\) is the steam temperature, and \(T_{ambient}\) is the ambient temperature.

We can now determine the temperature drops across the pipe shell and the insulation using the heat loss rate and the individual thermal resistances. Temperature drop across the pipe shell (\(\Delta T_s\)): $$\Delta T_s = q \cdot R_s$$ Temperature drop across the insulation (\(\Delta T_i\)): $$\Delta T_i = q \cdot R_i$$