Question

Exposure to high concentrations 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 ammonia and the ambient air are \(100 \mathrm{~W} / \mathrm{m}^{2}, \mathrm{~K}\) and \(20 \mathrm{~W} / \mathrm{m}^{2}, \mathrm{~K}\), 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 \(10^{\circ} \mathrm{C}\).

Short answer

Based on the given temperature conditions and using concepts of conduction and convection heat transfer, the required insulation thickness for the pipe carrying liquid ammonia is 18.4 cm.

Step-by-step solution

Calculate the heat transfer rate through conduction in the pipe wall

First, we need to calculate the heat transfer rate through conduction in the pipe wall by using the formula for heat conduction in a cylinder: \(q_{c}=\frac{2 \pi k L(T_i-T_o)}{\ln \frac{D_o}{D_i}}\) Where \(q_c\) is the heat transfer rate, \(L\) is the length of the pipe, \(k\) is the thermal conductivity of the pipe, \(D_i\) and \(D_o\) are the inner and outer diameters of the pipe, and \(T_i\) and \(T_o\) are the temperatures of the inner and outer surfaces of the pipe. Substitute the given values into the above equation: \(q_{c}=\frac{2 \pi \times 25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \times 10 \mathrm{~m}(-35^{\circ} \mathrm{C}-10^{\circ} \mathrm{C })}{\ln \frac{4 \mathrm{~cm}}{2.5 \mathrm{~cm}}}\) Solving for \(q_c\), we get: \(q_{c}=2561.65 \mathrm{~W}\)

Calculate the heat transfer rate through convection at the outer surface of the pipe insulation

We now need to calculate the heat transfer rate through convection at the outer surface of the pipe insulation using the formula: \(q_{h} = h_o A (T_o - T_{\infty})\) Where \(q_h\) is the heat transfer rate, \(h_o\) is the convection heat transfer coefficient of the ambient air, \(A\) is the area of the outer surface of the pipe insulation, and \(T_{\infty}\) is the ambient air temperature. Since the heat transfer rate through the pipe wall should equal the heat transfer rate through convection, we set \(q_c = q_h\) and rearrange the formula to solve for the area \(A\): \(A = \frac{q_c}{h_o(T_o - T_{\infty})}\) Substitute the given values and the previously calculated \(q_c\) into the formula: \(A = \frac{2561.65 \mathrm{~W}}{20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}(10^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C})}\) Solving for \(A\), we get: \(A = 12.808 \mathrm{~m^2}\)

Calculate the insulation thickness

Now we can determine the insulation thickness by using the area \(A\) we calculated above and the outer diameter of the pipe insulation \(D_{ins}\): \(A = \pi D_{ins} L\) Simplify and solve for \(D_{ins}\): \(D_{ins} = \frac{A}{\pi L} = \frac{12.808 \mathrm{~m^2}}{\pi \times 10\mathrm{~m}} = 0.408\mathrm{~m}\) Lastly, we need to calculate the insulation thickness \(t\) by subtracting the outer diameter of the pipe from the outer diameter of the insulation: \(t = \frac{D_{ins}-D_o}{2} = \frac{0.408 \mathrm{~m} - 0.04\mathrm{~m}}{2} = 0.184\mathrm{~m}\) So, the insulation thickness required for the pipe to maintain the given temperature conditions is 0.184 meters or 18.4 cm.