How do I go about calculating final temperature of tank draining at saturation?

Question

I am working on calculating pressure in a tank where the fluid is sitting at its vapor pressure. For example, imagine a 12 in diameter 100 L tank of nitrous oxide at room temperature (745 psi vapor pressure). This tank would be drained at 1 L/s from the liquid side, and some of the nitrous oxide would boil off in order to maintain vapor pressure, but would come at the cost of reducing the temperature and vapor pressure in the fluid.

How do I find what the final fluid temperature at drain would be (assuming no heat being input into the system)?

I was thinking there would be some combination of Antoine's, Clausius-Clapeyron, ideal gas law, and some derivation of an energy balance equation, but I am not too sure.

The experimental situation would literally be a tank (assume perfect insulation) that is filled with nitrous oxide, say a 100 L tank because it is arbitrary. This tank would then be drained through a choked orifice into the atmosphere. The flowrate at the start would be 1 L/s, but as pressure drops, would decrease in flowrate. I want to find out the ending pressure and temperature when the liquid would run out. Experimental data points to the tank draining from liquid before running out, similar to how a propane barbeque runs out of liquid and becomes cold with lower pressures.

Answer 1

Let m be the mass of N2O in the tank at any time, x be the mass fraction liquid N2O,V be the tank volume, v be saturated volume per unit mass, u be saturated internal energy per unit mass, and h be saturated enthalpy per unit mass. At any time, the internal energy of the tank contents will be: $$U=m[xu_L(T)+(1-x)u_V(T)]$$If we apply the open system version of the 1st law of thermodynamics to the contents of this adiabatically insulated tank, we obtain: $$dU=h_L(T)dm$$or$$[xu_L(T)+(1-x)u_V(T)]dm-m(u_V-u_L)dx+m\left[x\frac{du_L}{dT}+(1-x)\frac{du_V}{dT}\right]dT=(u_L+Pv_L)dm$$or $$-(u_V-u_L)dx+\left[x\frac{du_L}{dT}+(1-x)\frac{du_V}{dT}\right]dT=-(1-x)(u_V-u_L)d\ln{m}+Pv_Ld\ln{m}$$This equation is subject to the constraint that the total volume is constant: $$m[xv_L+(1-x)v_V]=V$$or$$x=\frac{v_V-\frac{V}{m}}{v_V-v_L}$$These equations provide the basis for determining x(m) and T(m).