Pressure Decay Equation: P1-P0 e-(A/V)t + P0 Explained

[Jackstraw]

I am trying to understand an equation that I found in an old document concerning pressure decay of a sealed assembly. The assembly is pressurized and over time decays to 1 atmosphere ambient pressure.
The equation P = (P1-P0)e(-(A/V)t) + P0 is used but not all the terms are defined

P = pressure at time t in psia
P1 = starting pressure in psia
P0 = ambient pressure in psia
t = time in hours

I have assumed V = assembly volume in cubic inches

The A/V term in the exponent is referred to as a time constant. I can use the equation for my data analysis but would like a better understanding of "A". Its units appear to be volume over time and is related to leak rate.

Hoping someone can shed some light. Thanks
Jackstraw

 

[Simon Bridge]

Welcome to PF;

You'd probably guess that the rate the inside pressure drops at time t would be proportional to the inside-outside pressure difference at the same time t. The way you say this in math is:

$\small{\dot P=-k(P-P_0)}$ ... k is a constant of proportionality.

You can see that k has to have dimensions of 1/T for the equation to balance.

The equation you've found is the solution with k=A/V and P(0)=P₁

What was it you needed to understand?

 

[Jackstraw]

Thank you Simon, that helps. The original work (this is a set of hardcopy, old presentation charts from 1991) makes reference to a relationship between the leak rate expressed in atm cc/s and the A/V term. I have not been able to work out the math.
In a table, A/V term 0.93931 is associated with a leak rate of 4.5 X 10^-5 atm cc/s
(1272 psi cubic inches/yr) and this is stated to be the specification.
A second A/V term, 0.13238 is associated with 4.7 X 10^-6 atm cc/s (133 psi cubic inches/yr).
The volume of the assembly is 177 cubic inches in the first case and 220 in the second.
P₁ is 19.3 psia and P₀ is 14.696 psia.
Temperature is constant at 25^oC.
The assembly contains dry N₂ with a He tracer for leak testing.
The 0.13238 was found by fitting the curve to the data. The author (haven't been able to track him/her down) equates 0.13238 to the 4.7 X 10^-6 atm cc/s but doesn't show the math. That mathematical relationship is what I'm trying to work out.
Given the units of leak rate, it appears the gas constant is part of the equation which would mean the volume of gas in moles may be part of it as well.
Thanks,

Jackstraw

 

[Simon Bridge]

The 0.13238 was found by fitting the curve to the data. The author (haven't been able to track him/her down) equates 0.13238 to the 4.7 X 10-6 atm cc/s but doesn't show the math. That mathematical relationship is what I'm trying to work out.

Well it was found by regression analysis from data right?
He'd have plotted log-pressure against time to get a line with slope A/V then used least-squares.

Off the units - gas volume in length-units is all that is needed.

In a table, A/V term 0.93931 is associated with a leak rate of 4.5 X 10-5 atm cc/s

So if r is this specific leak rate, notice [r]=[volume][pressure][time]^-1
Then A=r(P-P₀) is (modeled) volume of gas escaping the equipment per unit time and A/V is the proportion of the overall volume that escapes per unit time.

If all those figures come from one bit of equipment then it may be safe to say the volume is the same each time. Then you can find out the individual A and P values by simultaneous equations.

The A value will depend on a great many more fundamental variables like the molecular structure of the gas, it's temperature, the type of seal... so it's something you measure rather than calculate. i.e. the gas constant, the molar mass etc. is already a part of the value of A.