Exposure to high concentration of gaseous ammonia can cause lung damage. To prevent gaseous ammonia from leaking out, ammonia is transported in its liquid state through a pipe \(\left(k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, D_{i}=2.5 \mathrm{~cm}\right.\), \(D_{o}=4 \mathrm{~cm}\), and \(L=10 \mathrm{~m}\) ). Since liquid ammonia has a normal boiling point of \(-33.3^{\circ} \mathrm{C}\), the pipe needs to be properly insulated to prevent the surrounding heat from causing the ammonia to boil. The pipe is situated in a laboratory, where the average ambient air temperature is \(20^{\circ} \mathrm{C}\). The convection heat transfer coefficients of the liquid hydrogen and the ambient air are \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Determine the insulation thickness for the pipe using a material with \(k=\) \(0.75 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) to keep the liquid ammonia flowing at an average temperature of \(-35^{\circ} \mathrm{C}\), while maintaining the insulated pipe outer surface temperature at \(10^{\circ} \mathrm{C}\).

Given the parameters for the pipe, we can calculate the overall heat transfer coefficient (U) for the pipe following the expression: $$ \frac{1}{U A}= \frac{\ln{(D_{o}/D_{i})}}{2 \pi k L}+\frac{1}{h_{i} A_{i}}+\frac{1}{h_{o} A_{o}} $$ Where: \(D_{o}\) = outer diameter (4 cm) \(D_{i}\) = inner diameter (2.5 cm) \(k\) = pipe's thermal conductivity (25 W/(m·K)) \(L\) = pipe's length (10 m) \(h_{i}\) = heat transfer coefficient for ammonia (100 W/(m²·K)) \(h_{o}\) = heat transfer coefficient for ambient air (20 W/(m²·K)) In order to calculate \(U\), we also need the inner area \(A_{i}\) and outer area \(A_{o}\), which are given by: $$ A_{i}= \pi D_{i} L \\ A_{o}= \pi D_{o} L $$

Considering the mechanical energy balance equation: $$ \underbrace{q}_{\text { heat flow }}=U A \Delta T $$ We want to keep the liquid ammonia at an average temperature of -35°C, while maintaining the insulated pipe outer surface temperature at 10°C. Therefore, $$ \Delta T = 10^{\circ} \mathrm{C} - (-35^{\circ} \mathrm{C}) = 45 K $$

Now, we can use the heat flow equation to determine the required insulation thickness (thickness of the insulation layer). We know the insulation material's thermal conductivity (\(k_{ins}\)) is 0.75 W/(m·K). To find the thickness, we can use the following relationship between heat flow through the insulation and the temperature difference across it: $$ q = k_{ins} A_\text{ins} \frac{\Delta T}{L_\text{ins}} $$ Where: \(A_\text{ins}\) = effective heat transfer area of the insulation \(L_\text{ins}\) = insulation thickness Note that the heat flow \(q\) is the same through the pipe and the insulation layers. We can relate the insulation thickness to the pipe dimensions, using the area of the insulation (assuming cylindrical): $$ A_\text{ins} = 2 \pi (r_\text{ins} - r_{o}) L $$ Where: \(r_\text{ins}\) = insulation layer's radius \(r_{o}\) = pipe outer radius Since we already have the heat flow \(q\), we can use this along with the given parameters to find the insulation thickness (\(L_\text{ins}\)): $$ L_\text{ins} = \frac{k_{ins} A_\text{ins} \Delta T}{q} $$ Calculating all the needed parameters, we can find the required insulation thickness to maintain the desired temperature conditions.