volume of free air in a pressure vessel

[DM2]

I've got a pressure vessel which is 53 cubic feet and has a pressure of 500 psig. I'm trying to determine the volume of free air if the pressure vessel release the air to atmosphere.

Is this formula correct?

Volume x Pressure = Free Volume
53 cubic feet x 500 psig = 26500 cubic feet

Regards,
Dan Marr

"Real world Knowledge isn't dropped from a parachute in the sky but rather acquired in tiny increments from a variety of sources including panic and curiosity."

 

[PVDean]

No. It would basically be setting two pv=nrt equations to each other, if you have the same temperature (but you'd different temps the same way). So p1v1=p2v2. p1=500psi, v1=53ft^3, p2=14.7psi. v2=(p1/p2)v1

 

[PVDean]

You were right, I missed the psig thing. You needed to divide by 1 psig to get the units to cancel, and essentially you did the same thing I was saying. I was thrown off by the 'Is this formula correct?'. Your units don't work out, but for psig your answer did.

So my 'No' was partially correct answering the formula question, but your answer did 'handwavingly' work out.

 

[DM2]

PVDean,
Thanks for you input. I've got to sketch this up and and do a writeup so I'll use your methods in your first post for the writeup.

This is for a fire protection system. When a pressure vessel is installed within a protected space, the amount of agent needed to extinguish a fire needs to compensate for the possibility that the safety relief opens and vents the compressed air into the protected space. Normally we qualify the quote to assume that the vent will discharge the compressed are outside of the protected space, but the customer didn't do that in this project.

The code is actually written incorrectly as it speaks in terms of volume. The protected space volume remains the same, so in reality the added agent should be based on pressure. The net result is the same, but when people hear about compensating for added volume in a room that's constructed of steel, they start scratching their head and asking how does the volume increase just because the vessel relief valve opened.

Regards,
Dan Marr

"Real world Knowledge isn't dropped from a parachute in the sky but rather acquired in tiny increments from a variety of sources including panic and curiosity."

 

[jamesl]

You should use "Absolute Pressure" in the formula for ideal gas law. Air expands from 514.7 psia to 14.7 psia.

 

[DM2]

James,
Can you show me in terms of a formula i'd use in Excel?

Regards,
Dan Marr

"Real world Knowledge isn't dropped from a parachute in the sky but rather acquired in tiny increments from a variety of sources including panic and curiosity."

 

[jamesl]

See the first post by PVDean.

 

[zdas04]

The first post is fine if you ignore the numbers. He divided 500 psig by 14.7 psia the answer is meaningless (but at that big of a difference it is numerically pretty close).

PVDean,
It is a good practice to never use "psi" or "kPa" or even "bar" without qualifying it. Saving that digit just leads to confusion. Your later post that you have to divide by 1 psig is just wrong if I understand what you are saying. That would give you (psia)^-1 which doesn't mean much. The correct response was provided by James1.

You can't divide psig by psia, but your statement of the problem implies that you can. In America you should use psia or psig. In the rest of the world you still need to clarify, but it is messier. kPa(g) and kPa(a) and bar(g) and bar(a) are sometimes used, but that representation is far from universal. Their are purists who say that "bar" can only be in absolute terms, and there are gages that show zero bar at local atmospheric pressure.

David

 

[DM2]

Ok, if i follow PVDean, and...

P1 = 500 psi
P2 = 14.7
V1 = 53

(P1/P2)V1
(500/14.7)V1
(34)V1
1802.72

Excel Formula =(500/14.7)*43 = 1802.721 ????? Is this right??

The result is different than my math.



Regards,
Dan Marr

"Real world Knowledge isn't dropped from a parachute in the sky but rather acquired in tiny increments from a variety of sources including panic and curiosity."

 

[chicopee]

DM2, as stated above you must add 14.7 psi to the 500 psig so your answer s/b V2=((500+14.7)*144)*53)/(14.7*144)=1855.7 cu.ft.. The 144 in the numerator and denomitor will cancel out, so don't bother putting that value(144) in your spreadsheet.
What I don't understand is that this is basic chemistry taken by all engineering students.

 

[zdas04]

DM2,
This is way past absurd. I will not respond to another one of your posts that uses "psi" without qualification. STOP IT. And where the heck did the "43" come from? V1 is 53 ft^3. You have an answer why do you still need your hand held?

David