The boiling temperature of nitrogen at atmospheric pressure at sea level (1 atm pressure) is -196°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°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 kJ/kg and a density of 810 kg / m ^{3} at 1 atm. Consider a 3-m-diameter spherical tank that is initially filled with liquid nitrogen at 1 atm and -196°C. The tank is exposed to ambient air at 15°C, with a combined convection and radiation heat transfer coefficient of 35 W / m ^{2} \cdot{ }^{\circ} C. 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 W / m \cdot{ }^{\circ} C), and (c) insulated with 2-cm-thick superinsulation which has an effective thermal conductivity of 0.00005 W / m \cdot{ }^{\circ} C.
Question 3.81: The boiling temperature of nitrogen at atmospheric pressure ...
A 3-m diameter spherical tank filled with liquid nitrogen at 1 atm and -196°C is exposed to convection and radiation with the surrounding air and surfaces. The rate of evaporation of liquid nitrogen in the tank as a result of the heat gain from the surroundings for the cases of no insulation, 5-cm thick fiberglass insulation, and 2-cm thick superinsulation are to be determined.
Assumptions 1 Heat transfer is steady since the specified thermal conditions at the boundaries do not change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the midpoint. 3 The combined heat transfer coefficient is constant and uniform over the entire surface. 4 The temperature of the thin-shelled spherical tank is said to be nearly equal to the temperature of the nitrogen inside, and thus thermal resistance of the tank and the internal convection resistance are negligible.
Properties The heat of vaporization and density of liquid nitrogen at 1 atm are given to be 198 kJ/kg and 810 kg / m ^{3}, respectively. The thermal conductivities are given to be k = 0.035 W/m⋅°C for fiberglass insulation and k = 0.00005 W/m⋅°C for super insulation.
Analysis (a) The heat transfer rate and the rate of evaporation of the liquid without insulation are
A=\pi D^{2}=\pi(3 m )^{2}=28.27 m ^{2}R_{o}=\frac{1}{h_{o} A}=\frac{1}{\left(35 W / m ^{2} \cdot{ }^{\circ} C \right)\left(28.27 m ^{2}\right)}=0.00101{ }^{\circ} C / W
\dot{Q}=\frac{T_{s 1}-T_{\infty 2}}{R_{o}}=\frac{[15-(-196)]^{\circ} C }{0.00101^{\circ} C / W }=208,910 W
\dot{Q}=\dot{m} h_{f g} \longrightarrow \dot{m}=\frac{\dot{Q}}{h_{f g}}=\frac{208.910 kJ / s }{198 kJ / kg }= 1 . 0 5 5 kg / s
(b) The heat transfer rate and the rate of evaporation of the liquid with a 5-cm thick layer of fiberglass insulation are
A=\pi D^{2}=\pi(3.1 m )^{2}=30.19 m ^{2}R_{o}=\frac{1}{h_{o} A}=\frac{1}{\left(35 W / m ^{2} \cdot{ }^{\circ} C \right)\left(30.19 m ^{2}\right)}=0.000946^{\circ} C / W
R_{\text {insulation }}=\frac{r_{2}-r_{1}}{4 \pi k r_{1} r_{2}}=\frac{(1.55-1.5) m }{4 \pi\left(0.035 W / m \cdot{ }^{\circ} C \right)(1.55 m )(1.5 m )}=0.0489^{\circ} C / W
R_{\text {total }}=R_{o}+R_{\text {insulation }}=0.000946+0.0489=0.0498^{\circ} C / W
\dot{Q}=\frac{T_{s 1}-T_{\infty 2}}{R_{\text {total }}}=\frac{[15 – (-196)]^{\circ} C }{0.0498^{\circ} C / W }=4233 W
\dot{Q}=\dot{m} h_{f g} \longrightarrow \dot{m}=\frac{\dot{Q}}{h_{f g}}=\frac{4.233 kJ / s }{198 kJ / kg }= 0 . 0 2 1 4 kg / s
(c) The heat transfer rate and the rate of evaporation of the liquid with 2-cm thick layer of superinsulation is
A=\pi D^{2}=\pi(3.04 m )^{2} = 29.03 m ^{2}R_{o}=\frac{1}{h_{o} A}=\frac{1}{\left(35 W / m ^{2} \cdot{ }^{\circ} C \right)\left(29.03 m ^{2}\right)}=0.000984{ }^{\circ} C / W
R_{\text {insulation }}=\frac{r_{2}-r_{1}}{4 \pi k r_{1} r_{2}}=\frac{(1.52-1.5) m }{4 \pi\left(0.00005 W / m \cdot{ }^{\circ} C \right)(1.52 m )(1.5 m )} = 13.96^{\circ} C / W
R_{\text {total }}=R_{o}+R_{\text {insulation }}=0.000984+13.96 = 13.96^{\circ} C / W
\dot{Q}=\frac{T_{s 1}-T_{\infty 2}}{R_{\text {total }}}=\frac{[15-(-196)]^{\circ} C }{13.96^{\circ} C / W } = 15.11 W
\dot{Q}=\dot{m} h_{f g} \longrightarrow \dot{m}=\frac{\dot{Q}}{h_{f g}}=\frac{0.01511 kJ / s }{198 kJ / kg }=0.000076 kg / s
