adiabatic index of air in receiver tank sizing

[chrizofer]

I have a question for katmar based on a response to a question on air receiver sizing. The below response was in response to a sizing question and I have an acquaintance that is referencing an adiabatic index of air which I did not see in your response. Can you or anyone tell me why or why not this index is not referenced in these calculations?

Thank you in advance

Reference: thread378-250669

katmar (Chemical) 29 Jul 09 04:46

quark, the 40 psi is not an actual pressure - it is a differential pressure.

When the air pressure in the accumulator is decreased from 120 psi to 80 psi it must have released 125.88 gallons of free air (but at 80 psi).

Let the accumulator volume be Vacc. Before the valves are actuated the air in the accumulator in terms of free air (i.e. at atm press) is Vacc x 134.7/14.7

After the accumulator pressure is let down to 80 psi the remaining air, in terms of free air, is Vacc x 94.7/14.7

The change in air contained between these 2 conditions is 125.88 gallons i.e.

(Vacc x 134.7/14.7) - (Vacc x 94.7/14.7) = 125.88

Vacc / 14.7 x (134.7 - 94.7) = 125.88

Vacc / 14.7 x (40) = 125.88 (This is where the 40 psi comes in)

Vacc = 46.3

 

[Latexman]

chrizofer,

Please provide the link to the thread next time, like thread378-250669.

Good luck,
Latexman

To a ChE, the glass is always full - 1/2 air and 1/2 water.

 

[bimr]

In the particular example, the air is used as storage in case the air supply is cut-off:

"I forgot to say that the tank will supply air to handle the valves in a air supply cut off."

The air in the receiver is presumably at room temperature or atmospheric temperature depending on where the receiver is located.

As the differential pressure decreases from 120 to 80 psi, the air flow is low and the temperature drop is small. Most people would assume the process to be adiabatic (without the gain or loss of any heat) to simplify the air volume calculation.