Looks like you're assuming an allowable pressure of about 0.5 bar, since you assumed choked flow.
Neither of the cases you did represents the true situation.
In fact you have a mixture of steam and air so I attempted this problem conservatively and got flows less than yours.
First, I assumed no airflow until the pressure inside was o.5 bar (correct me if that is wrong)
Next , using the steam tables, the initial partial pressure of steam was 8.5 psi and the air 6.2, T=185F
It is clear to me that the energy equation for water injection results in condensing steam, with only second order heat exchange from the air.
So, I followed the steam tables down to 2 psi partial pressure of steam and 5+ for the air at 129F (forgive me for using these units since the steam tables I have are in BTU/F units.
At this point I allow air to enter, but conservatively DIO NOT let it mix. to sustain the pressure
The energy equation is with good acuracy
Ww(129-25)=1070Ws
So
Ww=0.097Ws
Looking for a rate of temperature change so I can get the change in volume as the mixture cools.
From the tables, I got
for 65M^3 (1950FT^3) of mixture in cooling 3.7F I condensed 1950(1/173-1/191)=1.16lb steam
Frrom the ratio of Ww/Ws=.097, that means the amount of water needed
1.16/.097=11.95 lb cooling water
Since you are pumpinmg about
10lb/sec
we need 1.195 seconds to do it
and the rate of temperature drop becomes
3.77/1.195=3.15F/sec
The volumetric change is then
3.15/(460+126)*1950=10.4 FT^3/sec
Which is the inlow air conservatively needed to sustain the inside pressure.
Subsequent cooling is less demanding on airflow.
