Tank Volume: Find Volume at 3,000 psi, 80 F

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SUMMARY

The discussion focuses on calculating the interior volume of a scuba diving tank designed to hold 50 standard cubic feet of air at a pressure of 3,000 psi and a temperature of 80°F. The ideal gas law (PV=nRT) is applied, with corrections for absolute pressure and temperature. The final calculated volume is approximately 438.4 in³, which aligns with the expected volume for an "aluminium-80" cylinder. Additionally, the importance of using the compressibility factor for high-pressure scenarios is highlighted.

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  • Understanding of the Ideal Gas Law (PV=nRT)
  • Knowledge of unit conversions between cubic feet and cubic inches
  • Familiarity with absolute pressure and temperature concepts
  • Basic principles of gas compressibility at high pressures
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Engineers, physicists, and scuba diving professionals involved in gas calculations and tank design will benefit from this discussion.

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Homework Statement


A tank for scuba diving is designed to contain 50 standard cubic feet of air when filled to a pressure of 3,000 pounds per square inch (gage) at an ambient temperature of 80 F. Calculate the interior volume of the tank. A standard cubic foot occupies one cubic foot at T=15 C and 101.3 kPa.

Homework Equations


PV=nRT

The Attempt at a Solution


I used PV/RT = PV/RT . I made sure all the units were correct but the volume I get is way too large.
 
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Hi Matt766. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

Please show your arithmatic so readers can look for mistakes.
 
Last edited by a moderator:
Sure. The underline is for division, and the temperatures go on the left and right of the equal sign respectively. Is there an equation editor or something?

(V * 3,000 lb/in^2) = (14.7 lb/in^2)*(86,414.7 in^3
(539.7 deg R) (518.7 deg R)​

V= 441.4 in^3. This seems weird to me because its such a small volume for a tank.
 
Is Rankine going to work there?

Yes, looks like it can.
 
Last edited:
NascentOxygen said:
Is Rankine going to work there?

Yes, looks like it can.
One point here: the pressure in the tank is supposed to be 3000 psi gage. The ideal gas equation uses absolute pressure and absolute temperature.

Second point: where did 86,414.7 in.3 come from? After all, 1 foot = 12 inches, so 1 ft.3 = ? in.3
 
Didn't convert the right pressure to gage pressure, otherwise I can't see anything amiss. Yes, SK, should go with absolute temps though I see using R doesn't change the result here.
 
I made the tank pressure into an absolute pressure and corrected the volume for the cubic feet. For some reason I added 14.7 to the volume? Thanks haha


(V)*(3,014.7 lb/in^2)
= (86,400 in^3)*(14.7 lb/in^2)
(539.7 deg R) (518.7 deg R)

V=438.4 in^3.
 
From Wikipedia:
An especially common cylinder available at tropical dive resorts is an "aluminium-80" which is an aluminium cylinder with an internal volume of 0.39 cubic feet (11 L) rated to hold about 80 cubic feet (2,300 L) of atmospheric pressure gas at its rated pressure of 3,000 psi (210 bar).
 
NascentOxygen said:
Is Rankine going to work there?

Yes, looks like it can.
Sure.
 
  • #10
Matt766 said:
I made the tank pressure into an absolute pressure and corrected the volume for the cubic feet. For some reason I added 14.7 to the volume? Thanks haha


(V)*(3,014.7 lb/in^2)
= (86,400 in^3)*(14.7 lb/in^2)
(539.7 deg R) (518.7 deg R)

V=438.4 in^3.
In this problem, considering how high the pressure is, we should also have used the compressibility z factor. The pressure is 200 atm., which corresponds to a reduced pressure on the order of about 6. In any event, giving the volume to 4 significant figures is not justified.

Chet
 

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