Question

The boiling temperature of nitrogen at atmospheric pressure at sea level $(1 \mathrm{~atm})\( is \)-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 remains constant at $-196^{\circ} \mathrm{C}$ until the liquid nitrogen in the tank is depleted. Any heat transfer to the tank results 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 \(1 \mathrm{~atm}\). Consider a 4-m-diameter spherical tank initially filled with liquid nitrogen at \(1 \mathrm{~atm}\) and \(-196^{\circ} \mathrm{C}\). The tank is exposed to \(20^{\circ} \mathrm{C}\) ambient air with a heat transfer coefficient of $25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. The temperature of the thin- shelled spherical tank is observed to be almost the same as the temperature of the nitrogen inside. Disregarding any radiation heat exchange, determine the rate of evaporation of the liquid nitrogen in the tank as a result of the heat transfer from the ambient air.

Short answer

Answer: The rate of evaporation of liquid nitrogen in the tank is approximately 1.375 kg/s.

Step-by-step solution

Calculate the tank's surface area

The question states that we should disregard any radiation heat exchange. Therefore, we can consider only the heat transfer by convection. To do this, let's calculate the surface area of the 4-m-diameter spherical tank, using the formula \(A = 4 \pi r^2\). Given the diameter (d) of the tank is 4 m, the radius (r) would be: \(r=\dfrac{d}{2} = \dfrac{4}{2} = 2\,\mathrm{m}\) Now we can calculate the surface area: \(A = 4 \pi r^2 = 4 \pi (2)^2 \approx 50.27\,\mathrm{m^2}\)

Calculate the heat transfer rate

To calculate the heat transfer rate due to convection, we can use the formula \(Q = hA\Delta T\), where \(Q\) is the heat transfer rate, \(h\) is the heat transfer coefficient, \(A\) is the surface area, and \(\Delta T\) is the temperature difference between the ambient air and the liquid nitrogen. Given in the exercise: \(h = 25 \,\mathrm{W/m^2\cdot K}\) \(\Delta T = (20 + 196) = 216^\circ \mathrm{C} = 216\,\mathrm{K}\) Now we can calculate the heat transfer rate (Q): \(Q = hA\Delta T = 25 \cdot 50.27 \cdot 216 \approx 272\,363\,\mathrm{W}\)

Calculate the rate of evaporation

With the heat transfer rate determined, we can now calculate the rate of evaporation \(\left( \dfrac{dm}{dt} \right)\) using the heat of vaporization equation: \(\dfrac{dm}{dt} = \dfrac{Q}{L}\), where \(L\) is the heat of vaporization. Given in the exercise: \(L = 198\,\mathrm{kJ/kg} = 198,000\,\mathrm{W\cdot s/kg}\) Now we can calculate the rate of evaporation: \(\dfrac{dm}{dt} = \dfrac{272\,363}{198\,000} \approx 1.375\,\mathrm{kg/s}\) Therefore, the rate of evaporation of liquid nitrogen in the tank as a result of the heat transfer from the ambient air is approximately 1.375 kg/s.