Question

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 \((k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), $D_{i \text {, pipe }}=2.5 \mathrm{~cm}, D_{a \text {, pipe }}=4 \mathrm{~cm}$, and \(\left.L=10 \mathrm{~m}\right)\). 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 air at \(20^{\circ} \mathrm{C}\) is blowing across it with a velocity of \(7 \mathrm{~m} / \mathrm{s}\). The convection heat transfer coefficient of the liquid ammonia is $100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Calculate the minimum 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

Answer: The minimum insulation thickness required is approximately 8.2 mm.

Step-by-step solution

Calculate the pipe metal resistance

Let's calculate the thermal resistance of the pipe's metal due to conduction: Pipe metal thermal resistance: \(R_{pipe} = \frac{\ln{\frac{D_a}{D_i}}}{2 \pi k_{pipe} L}\), where \(D_i\) - inner diameter of the pipe (0.025 m), \(D_a\) - outer diameter of the pipe (0.04 m), \(k_{pipe}\) - pipe's thermal conductivity (25 W/m·K), \(L\) - pipe's length (10 m) Plug in the values: \(R_{pipe} = \frac{\ln{\frac{0.04}{0.025}}}{2 \pi \cdot 25 \cdot 10} ≈ 1.682 \times 10^{-4} \mathrm{~m}^2 \cdot \mathrm{K} \cdot \mathrm{W}^{-1}\).

Calculate the convection resistance

Now, let's calculate the convection resistance of the ammonia's surface: Convection resistance: \(R_{conv} = \frac{1}{h_{conv} A}\), where \(A = \pi D_i L\), \(h_{conv}\) - convection heat transfer coefficient (100 W/m²·K). Plug in the values: \(R_{conv} = \frac{1}{100 \cdot \pi \cdot 0.025 \cdot 10} ≈ 1.273 \times 10^{-4} \mathrm{~m}^2 \cdot \mathrm{K} \cdot \mathrm{W}^{-1}\).

Set up heat transfer formula and integrate insulation resistance

Next, set up the heat transfer equation in terms of the temperature difference: \(Q = \frac{T_{air} - T_{liq}}{R_{pipe} + R_{conv} + R_{ins}}\), where \(T_{air}\) - temperature of laboratory air (20°C = 293.15 K), \(T_{liq}\) - temperature of liquid ammonia (-35°C = 238.15 K), \(R_{ins}\) - insulation resistance. Now, we need to include the insulation resistance: \(R_{ins} = \frac{\ln{\frac{D_{ins}}{D_a}}}{2 \pi k_{ins} L}\), where \(D_{ins}\) - outer diameter of insulation, \(k_{ins}\) - insulation thermal conductivity (0.75 W/m·K). We are given that the temperature of the insulated pipe's outer surface should be 10°C (283.15 K). So, \(Q = \frac{T_{air} - T_{outer}}{R_{conv} + R_{ins}}\). Plugging the values known values into the equation, we get: \(\frac{293.15 - 238.15}{R_{pipe} + R_{conv} + R_{ins}} = \frac{293.15 - 283.15}{R_{conv} + R_{ins}}\).

Solve for insulation thickness

We will solve for \(D_{ins}\): First, simplify the equation: \(R_{ins} = \frac{T_{outer} - T_{liq}}{T_{air} - T_{outer}} (R_{pipe} + R_{conv}) - R_{conv}\). Now, plug in the values: \(R_{ins} = \frac{283.15 - 238.15}{293.15 - 283.15} (1.682 \times 10^{-4} + 1.273 \times 10^{-4}) - 1.273 \times 10^{-4} ≈ 6.545 \times 10^{-4} \mathrm{~m}^2 \cdot \mathrm{K} \cdot \mathrm{W}^{-1}\). Rearrange the insulation resistance formula to solve for \(D_{ins}\): \(D_{ins} = D_a \cdot \exp{\left(\frac{2 \pi k_{ins} L \cdot R_{ins}}{\ln}\right)}\). Now, plug in the values: \(D_{ins} = 0.04 \cdot \exp{\left(\frac{2 \pi \cdot 0.75 \cdot 10 \cdot 6.545 \times 10^{-4}}{\ln}\right)} ≈ 0.04816 \mathrm{~m}\). Finally, find the minimum insulation thickness: \(Thickness = D_{ins} - D_a = 0.04816 - 0.04 = 0.00816 \mathrm{~m} ≈ 8.2 \mathrm{~mm}\). Therefore, the minimum insulation thickness required to keep the liquid ammonia flowing at an average temperature of -35°C, while maintaining an insulated pipe outer surface temperature of 10°C, is approximately 8.2 mm.