Pressure vs. Time for air tank with a slow leak
Pressure vs. Time for air tank with a slow leak
(OP)
Is there a "standard" equation for pressure versus time for a vessel pressurized with air which has a slow leak?
I'm sure it's exponential in nature, or, at least, "exponentialish". I've tried some basic stuff using ideal gas law, conservation of mass, and Bernoulli, and I get an "exponentialish" equation that seems to make some physical sense, but mine is not an exponential result. (Not unless I assume constant density and a linear flow-to-pressure relationship.) However, a web search yielded a couple of examples for tank blowdown scenarios, where the leak was choked flow. In the blowdown examples I found, pressure was shown to be an exponential function of time. I'm not certain how well the blowdown equations apply to the case of a slow leak.
Note: By "slow", I mean a volume of about 50 gallons of air depressurizing from 25 psig to 0 psig over the course of a few days.
I'm sure it's exponential in nature, or, at least, "exponentialish". I've tried some basic stuff using ideal gas law, conservation of mass, and Bernoulli, and I get an "exponentialish" equation that seems to make some physical sense, but mine is not an exponential result. (Not unless I assume constant density and a linear flow-to-pressure relationship.) However, a web search yielded a couple of examples for tank blowdown scenarios, where the leak was choked flow. In the blowdown examples I found, pressure was shown to be an exponential function of time. I'm not certain how well the blowdown equations apply to the case of a slow leak.
Note: By "slow", I mean a volume of about 50 gallons of air depressurizing from 25 psig to 0 psig over the course of a few days.





RE: Pressure vs. Time for air tank with a slow leak
You can easily perform a stepwise, piecewise linear material balance using a spreadsheet calculations.
Given enough steps it will converge to the analytical solution
RE: Pressure vs. Time for air tank with a slow leak
Anyway you must first of all make an assumption on the flow rate vs. pressure differential relationship to attack this problem. Flow through a porous medium is a possibility, besides the classical assumption of a circular orifice: I'm not sure, but I would expect a linear relationship to be the correct one in that case.
Another assumption you need to make is on the type of transformation: an isothermal one is probably good for very slow rates.
Also do not forget, if relevant to your problem, that, for such a long time, thermal effects related to ambient conditions may become of importance.
prex
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RE: Pressure vs. Time for air tank with a slow leak
The leak will be small, but it's through a small area, so...???
Crane TP410, Page A-22, has a table that allows you to determine choked flow through a pipe, but only if you know the hydraulic resistance factor (K) across the flowpath. Would a small leak have a high or low K value? My first instinct would be to say "high", but if your reference frame is across the area of flow itself - say, a valve seat and not the nominal pipe size - maybe it's not that high?
RE: Pressure vs. Time for air tank with a slow leak
Found this in an article from Assembly Magazine, which seems to suggest that even modest pressure differences (5 psid) will cause choked flow (or nearly choked flow) in very small diameter openings:
RE: Pressure vs. Time for air tank with a slow leak
RE: Pressure vs. Time for air tank with a slow leak
The internal volume was 12 ft3 and the start pressure was about 40" water gauge and the presssure was measured against time in minutes.
The resulting graphs gave
pressure = 48.65 * exp(-0.175 * time)
pressure = 46 * exp(-0.186 * time)
pressure = 40.9 * exp(-0.178 * time)
I pass it on for what it's worth!
RE: Pressure vs. Time for air tank with a slow leak
Marty
RE: Pressure vs. Time for air tank with a slow leak
RE: Pressure vs. Time for air tank with a slow leak
(1) dP/dt = d(rho)/dt*R*T
where,
dP/dt = pressure rate of change
d(rho)/dt = gas denstiy rate of change
R = ideal gas constant divided by gas molecular weight
T = gas temperature (absolute scale: K or R).
You mentioned the isothermal assumption, therefore T = constant.
The density as a function of time, rho(t), can be restated as a function of the leak rate:
rho = mass/vol; d(rho)/dt = (dm/dt)/Vol
the mass rate of change dm/dt is simply the mass flow rate through the leak:
(2) dm/dt = w(t) = rho(t)*A*u
where,
w(t) = mass flow rate at time t
rho(t) = gas density at time t
A = leak area
u = gas velocity
So,
(3) d(rho)/dt = -rho(t)*A*u/Vol
The negative sign is included because gas is leaving the system and the density is decreasing as a fucntion of time.
If the gas is choked, then for an ideal gas under isothermal conditions, the choked velocity is
(4) u = sqrt(R*T)
where sqrt is the mathematical symbol for square root.
In equaqtion 3, the density can be represented by the ideal gas law as,
(5) rho(t) = P(t)/R/T
By substituting equation 5 and 4 into equation 3, we get
(6) d(rho)/dt = -P(t)*A*sqrt(R*T)/R/T/Vol
Equation 6 can now be subtituted into equation 1:
(7) dP/dt = -P(t)*A*sqrt(R*T)/Vol
Upon integration and solving for P, we arrive at the final solution for pressure as a function of time,
(8) P(t)= P0*exp[-A/Vol*sqrt(R*T)*t]
where,
exp = exponential function
P0 = initial gas pressure
A = leak area
Vol = tank volume
R = ideal gas constant divided by gas molecular weight
T = constant gas temperature (absolute scale: K or R)
t = time
The assumptions of choked flow and constant temperature will eventually break down as the pressure approaches 0.