Final pressure and temperature knowing only air mass outflow

[Anachronist]

This seems simple, but I have some confusion. I'm trying to refine a numerical simulation to get away from my isothermal assumption, to account instead for isentropic temperature and pressure changes in pressure tank with an orifice.

Given an insulating air tank of volume $V$, containing air at absolute pressure $P_0$ (Pascals) and temperature $T_0$ (degrees K), a mass of air $\Delta m$ (kg) is allowed to escape rapidly into the atmosphere at ambient pressure $P_a$ over a short interval $\Delta t$ (seconds). What is the tank's internal pressure and temperature $P_1$ and $T_1$ at the end of the interval?

Whether the air flow is choked or unchoked, or the orifice geometry, shouldn't matter; this is accounted for by the fact that we already know the mass flow rate.

I have this so far:

The initial density of air would be $\rho_0 = \frac{P_0}{R T_0}$

The initial mass of air would be $m_0 = \rho_0 V$

So after the change in mass, the new air density would be $\rho_1 = \frac{m_0 - \Delta m}{V}$

...and that's kinda where I get stuck, getting a final temperature and pressure from there. The adiabatic relationship
$$\frac{T_1}{T_0} = \left( \frac{P_1}{P_0} \right)^{1-\frac{1}{\gamma}}$$
would need to be used, but I need to know if there's a relationship that connects density ratios to pressure ratios? I suspect it's something like this (I can't find a source yet):
$$\gamma \frac{\Delta \rho}{\rho} = \frac{\Delta P}{P}$$

Is there enough information to solve this problem?

 

[Anachronist]

I might have found what I need. From http://www.engineeringtoolbox.com/compression-expansion-gases-d_605.html equation 2:
$$\frac{P}{\rho^\gamma} = \text{constant}$$
So
$$\frac{P_1}{\rho_1^\gamma} = \frac{P_0}{\rho_0^\gamma}$$
that is:
$$P_1 = P_0 \left(\frac{\rho_1}{\rho_0}\right)^\gamma$$
Once I have $P_1$ I can calculate $T_1$ from the adiabatic expansion formula.
Does that solution make sense? Am I missing anything?

 

[Chestermiller]

It looks to me like your approach is correct. You should be able to get the answer from your development in the first post. Let P1 be an unknown. Then you know T1 in terms of P1. Then you can use the ideal gas low to solve for P1.

 

[Anachronist]

Thanks for confirming I was on the right track.

Using the density relationship in my second post to calculate the new pressure at each time step, along with adjusting the speed of sound based on the temperature formula from my first post, resulted in a more conservative (and likely more realistic) estimate of my water rocket's performance than my original constant-temperature calculations.

You and I had a conversation in the past in which we concluded that $\gamma=1.34$ is reasonable to fudge the ideal gas formulas for humid air expanding rapidly to push water out of a bottle. I started this thread as the next stage: what to do with the pressurized air left in the bottle after the water is all gone. The pressure and temperature were nicely modeled during the water thrust phase as a result of that previous thread, but I was never happy with the fact that my air-thrust calculations ignored any further temperature changes. Taking temperature into account, as described above, made a small but significant difference in altitude achieved by the rocket (about 6% less with the new calculations).