# Pressure questions filling smaller tank from larger

1. Feb 25, 2012

### Jynx18

Ok I am wondering how to solve this or go about figuring it out. This is what I am trying to do.

I have a 253.859 cu inch tank (tank A) filled to 4500psi. I want to fill 68 cubic inch tanks (tanks B) with it. I know that after each fill tank A will decrease in pressure by a little amount and obviously tank B can only be filled to the same pressure as tank A since they would equal out. How can I go about this?

I know it is not as simple as tank A is ~3.7 times larger so it should fill 3.7 tanks
253.859/68=3.733

Could I do that after each fill it should lose 26.79% (1/3.733) of its pressure (since essentially 26.79% of volume is gone)?

Fill 1 3294.45 psi
Fill 2 2411.87 psi
Fill 3 1765.73 psi
Fill 4 1292.69 psi
Fill 5 946.34 psi
etc....

Last edited: Feb 25, 2012
2. Feb 26, 2012

### Hassan2

I don't think you can go around this without using a compressor for filling the tanks with the required pressure.

3. Feb 26, 2012

### Jynx18

i understand that but i want to try and figure out what each fill would be with no compressor. i know it wont make sense to do this in real life but now it is bugging me because it seems so simple so figure out the pressures yet i am not 100% sure if the math i am doing is right.

4. Feb 26, 2012

### Hassan2

If the content of the high pressure tank is a condensed or partially condensed gas, you are lucky as the pressure doesn't drop as long as the some liquid has remained in the tank.

However if the the content is all gaseous, then we can easily calculate the pressure at after each filling. I think your calculation is wrong. I calculate it for the first filling ( I hope I am not wrong in my calculation!):

First suppose the empty tank is really empty ( vacuum) . After connecting the tanks, the content of changes volume from 253.859 to 253.859+68=321.589 . Hence the volume has increased by factor321.589/253.859= 1.268 . The pressure drops by the same factor. The initial pressure is 4500+15 pis ( I added the atmospheric pressure to get the absolute pressure). The absolute pressure after the first filling then becomes 4515/1.268 =3556.782 psi absolute ≈ 3541.782 psi gauge. Here I assumed that the temperature remains constant. ( of course the temperature tends to drop by expansion but in case of a metallic tank, the it gets energy from the ambient and the temperature remains the same as the ambient temperature.

If the empty tank is not vacuum but atmospheric pressure, the pressure would be slightly higher. If you need it, i can calculate it.

5. Feb 26, 2012

### Hassan2

This part is wrong. First, the volume is gone is not the right statement. Perhaps you mean 26.79 % of the mass ( moles) has gone. Again this is not correct. The mass expands into both tanks first, then 68/(253.859+68)=.211 ( 21.%) of it is gone.

6. Feb 27, 2012

### Jynx18

I should have mentioned it is compressed air so all gas. Wow I know boyles law but didn't think to look at it as the volume is increasing as it is connected during filling but was thinking it is "removed" since it is leaving one tank. That clears it up a bunch.

Yea I knew the wording was wrong but that was the best I thought to describe it using intuition which usually causes problems as you can see. Like before I was looking in terms of air leaving the tank thats why I said removed incorrectly. Thanks for the help.

7. Feb 27, 2012

### Jynx18

ok using p1v1=p2v2 i did this

p1 = 4500psi + 14.7 psi - 4514.7 psi
v1 = 253.859 ci
p2 = ?
v2 = 253.859 ci + 68 ci = 321.859 ci

so p2 = 3560.9 psi and that is 3546.2 psi guage.

Last edited: Feb 27, 2012
8. Feb 28, 2012

### cjl

Yes, although that's assuming the temperature doesn't change. In reality, the temperature will drop as the gas is expanded, so the pressure will drop even farther than that (though the proportion of the gas transferred should still be the same).