Assuming you're starting the LP50s completely empty and 3450psi in the hp120..
P1V1=P2V2
3450*120=P2*220
P2=1880 psi
Hi guys,
sorry but the calculations upthread are incorrect. Yes, P1V1 = P2V2 using your notation. This is writing conservation of mass, however the V's here refer to the physical volumes of the tanks, while in your calculation you have used the rated capacity of the tanks in cf, which is not an actual volume, but the number of moles of gas contained in each cylinder. If the rated pressure of the tanks was the same, you could do that and find the correct answer. The pressure you derived would be that obtained after transfilling an empty HP100 from a full HP120.
Here the rated pressure of the double LP50's is 2640, and their physical volume is larger than that of an HP100 cylinder. Therefore, more gas will transfer to those tanks than to an HP100, but the overall end pressure will be lower.
If V1 is the volume of your double 50's, and V2 that of the HP120, the equation used should be:
P(V1+V2) = P1V1 + P2V2 where P is the final pressure of all tanks after transfilling, P1 is the pressure in your double 50's after dive 1, and P2 is the pressure in your HP120 before the transfill.
Hence the final pressure is P = (P1V1+P2V2)/(V1+V2) = P1*V1/(V1+V2) + P2*V2(V1+V2)
So you have to determine the percentage of the total volume of the combined tanks made up by each tank.
The actual volume in liters can be found from the manufacturer's specs, but you don't need it here.
V1 is proportional to 100/2640 and V2 is proportional in the same manner to 120/3440.
So you'll find that V1/(V1+V2) ~ 0.52 and V2/(V1+V2) ~ 0.48
The message here is that 52% of the total gas amount combined between the tiny doubles and HP120 will partition to the double 50's, while 48% will remain in the HP120 after equilibrium is reached.
The final pressure P is P = P1*0.52 + P2*0.48.
Assuming you start with empty doubles (P1 = 0) and a full 120 (P2 = 3440), your doubles will be refilled to P = 1651 psi, and not 1880 psi as calculated above.
An easier way to think about this is to think in gas amount (cf) instead of pressures. When you transfill, you will have 52% of the combined gas available in your tiny doubles after the operation. So for a practical case scenario, if you come back with 20 cf of gas left in your doubles after dive 1 (a little over 500 psi), and have approx 110 cf of gas available in your HP120 (about 3100 psi, which is more realistic than 3440 given than fills aren't always perfect and even if they are after cooling down to room temp, once at the dive site in cold winter conditions, pressure will drop), you would have for dive #2 an amount of usable gas equivalent to 52% of 130 cf ~ 67.6 cf.
Now whether or not that is sufficient for your second dive is for you to decide.
If you brought a second HP120 (or even an HP100), you could do a second transfill which in this case would yield (assuming 110 cf usable again in the second HP120) a total of 52% of 177.6 cf ~ 92.35 cf, which should suffice, but then you have to lug two additional large tanks with you and you are using another 2 fills for your second dive.
If as hinted above, you close the isolation valve on your manifold and transfill the LP50's one by one from a single HP120, you will have ~ 35% of the total gas in each LP50 after transfilling, which allows you to transfer a little more gas overall and is a bit more efficient. In this way, you would be able to transfer ~58% of a full HP120 to completely empty doubles vs 52% otherwise.
In my other example above, with 20 cf left after dive 1, and 110 cf available in your HP120, you would be able to refill to ~72.8 cf your tiny doubles doing it sequentially before reopening the isolator valve, vs ~67.6 cf doing it in one go, thus gaining about 5 cf of available gas for your second dive.
