The boiling temperature of nitrogen at atmospheric pressure at sea level ( 1 atm pressure) is \(-196^{\circ} \mathrm{C}\). Therefore, nitrogen is commonly used in low-temperature scientific studies since the temperature of liquid nitrogen in a tank open to the atmosphere will remain constant at \(-196^{\circ} \mathrm{C}\) until it is depleted. Any heat transfer to the tank will result in the evaporation of some liquid nitrogen, which has a heat of vaporization of \(198 \mathrm{~kJ} / \mathrm{kg}\) and a density of \(810 \mathrm{~kg} / \mathrm{m}^{3}\) at 1 atm. Consider a 3-m-diameter spherical tank that is initially filled with liquid nitrogen at 1 atm and \(-196^{\circ} \mathrm{C}\). The tank is exposed to ambient air at \(15^{\circ} \mathrm{C}\), with a combined convection and radiation heat transfer coefficient of \(35 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The temperature of the thin-shelled spherical tank is observed to be almost the same as the temperature of the nitrogen inside. Determine the rate of evaporation of the liquid nitrogen in the tank as a result of the heat transfer from the ambient air if the tank is \((a)\) not insulated, \((b)\) insulated with 5 -cm-thick fiberglass insulation \((k=0.035 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), and (c) insulated with 2 -cm-thick superinsulation which has an effective thermal conductivity of \(0.00005 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

Step by step solution

01

(Step 1: Calculate the heat transfer rate for each case)

First, let's calculate the heat transfer rate for each case. For that, we need to know the temperature difference between the liquid nitrogen and the ambient air, and the surface area of the tank. The heat transfer rate formula for each case will be different depending on the insulation. For the non-insulated case (a), we will apply the combined convection and radiation heat transfer coefficient. For the insulated cases (b) and (c), we'll need to apply the formula for heat conduction through the insulation.

02

(Step 2: Find the temperature difference and tank's surface area)

Determine the temperature difference between the liquid nitrogen (\(T_N= -196°C\)) and the ambient air (\(T_A = 15°C\)), which is the driving force for heat transfer: \(\Delta T = T_A - T_N = 15 - (-196) = 211 K\) Now calculate the surface area of the 3-m diameter spherical tank: \(A = 4 \pi r^2\) Where \(r\) is the radius of the tank, \(r = \frac{D}{2} = \frac{3}{2} = 1.5 \thinspace m\). Calculate the surface area: \(A = 4 \pi (1.5)^2 = 28.27 m^2\)

03

(Step 3: Calculate the heat transfer rate for Case A)

For the non-insulated case (a), the heat transfer rate (\(\dot{Q}\)) will be given by: \(\dot{Q}_A = h \cdot A \cdot \Delta T\) Where \(h = 35 \frac{W}{m^{2} \cdot K}\). Calculate the heat transfer rate for case (a): \(\dot{Q}_A = 35 \cdot 28.27 \cdot 211 = 208012.95 \thinspace W\)

04

(Step 4: Calculate the heat transfer rate for Case B)

For the fiberglass insulated case (b), the heat transfer rate will be given by: \(\dot{Q}_B = \frac{k \cdot A \cdot \Delta T}{L}\) Where \(k = 0.035 \frac{W}{m \cdot K}\) and \(L = 0.05 m\). Calculate the heat transfer rate for case (b): \(\dot{Q}_B = \frac{0.035 \cdot 28.27 \cdot 211}{0.05} = 4205.17 \thinspace W\)

05

(Step 5: Calculate the heat transfer rate for Case C)

For the superinsulated case (c), the heat transfer rate will be given by: \(\dot{Q}_C = \frac{k \cdot A \cdot \Delta T}{L}\) Where \(k = 0.00005 \frac{W}{m \cdot K}\) and \(L = 0.02 m\). Calculate the heat transfer rate for case (c): \(\dot{Q}_C = \frac{0.00005 \cdot 28.27 \cdot 211}{0.02} = 73.85 \thinspace W\)

06

(Step 6: Determine the rate of evaporation for each case)

Finally, determine the rate of evaporation for each case by dividing the heat transfer rate by the heat of vaporization of liquid nitrogen: \(\dot{m} = \frac{\dot{Q}}{h_v}\) Where \(h_v = 198 \frac{kJ}{kg} = 198000 \frac{W \cdot s}{kg}\). Calculate the mass evaporation rate for case (a), (b), and (c): \(\dot{m}_A = \frac{208012.95}{198000} = 1.051 \thinspace kg/s\) \(\dot{m}_B = \frac{4205.17}{198000} = 0.021 \thinspace kg/s\) \(\dot{m}_C = \frac{73.85}{198000} = 0.00037 \thinspace kg/s\) The rates of evaporation of liquid nitrogen in the spherical tank for each case are \(\dot{m}_A = 1.051 \thinspace kg/s\) (non-insulated), \(\dot{m}_B = 0.021 \thinspace kg/s\) (fiberglass insulation), and \(\dot{m}_C = 0.00037 \thinspace kg/s\) (superinsulation).

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