Okay, here it comes.
If you make the following assumptions ...
Airflow into your engine of 600 CFM (at boost)
Intake air temperature of around 70 C (post Turbo)
Water temperature of 17 C
Propane self pressurized at 17 C
Boost of 14.7 psi
Flowing at equal mass rates, propane blows water away... I'm not exactly sure what a sane mass rate is, but .01 kg/s is .6 kg a minute, or about a pound a minute. One of those little propane bottles is a pound.
That would drop you about 10 degrees C on your intake air temperature.
If you were to use water at the same rate, 1 lb is about 1/8th of a gallon or 1 pint, or 2 cups.
That would give you a drop in temperature of about 5 degrees C.
Now, if you added enough propane to change your mixture ratio from 14:1 to say 10:1 at the above conditions, that would be about:
Mass flow of the air at that point is about 1 lb of air per second. At 14:1 that means that for every 1 lb of air you have ~1/14 lb of fuel. To add enough fuel to that to increase it to 10:1 then you get the delta of 1/14 and 1/10 which is about .03 lbs.
So .03 lbs/sec of propane would be needed. That puts you at about .014 kg/s of propane.
According to my model... that would lower your intake temperature about 13 degrees C. That's not bad...
I probably made some errors in here as I was doing it in my head. Please point them out
I will take a closer look at my thermodynamic model tomorrow and see if it makes sense... but I think it does.