Air leak rate conversion
Air leak rate conversion
(OP)
how do I convert 0.0027 cubic inch p/second to psi p/second?
help please.
appreciate it thanks
b
help please.
appreciate it thanks
b
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RE: Air leak rate conversion
another day in paradise, or is paradise one day closer ?
RE: Air leak rate conversion
RE: Air leak rate conversion
I don't know the atmospheric / temp conditions.
isn't there a more direct way to convert?
RE: Air leak rate conversion
would give me psi?
RE: Air leak rate conversion
Fixed load being held by a variable volume cylinder - like a hydraulic cylinder?
Fixed temperature but variable volume - as in leaking tank with a gas & liquid inside?
Fixed temperature but losing volume of gas from a constant volume vessel - like an underground pipe with a liquid left inside
Fixed volume but variable temperature and mass of gas inside - like in leaking pipe up in the sunshine?
COME ON! Be a professional and describe YOUR problem!
RE: Air leak rate conversion
it is a fixed temperature, measuring a volume air leak - as in leaking tank of air.
T=298.15 k
n= 0.01154
R= 8.314
RE: Air leak rate conversion
In short
Is there any air input into your vessel during this leaking period? If so how much?
What is the start pressure and temperature of your tank?
Most importantly - What is the internal volume of your tank / vessel??
without that basic data you have no chance of calculating the impact of a leak in terms of internal pressure drop over time.
In any event, the effect of the volume leak will change over time as the amount of air decreases.
Remember to do all your calcs once you get this info in absolute pressure and temperature
Is your leak volume rate at atmospheric conditions?
My motto: Learn something new every day
Also: There's usually a good reason why everyone does it that way
RE: Air leak rate conversion
1) if you have a static pressure vessel, PV=constant will help you. if you know your initial pressure then you're GTG, since someone has told you n, R, T [is it just me or does this sound like a student problem ? in practice, would you know nRT or PV ?]
guessing that your volume leakage is measured at atmospheric conditions, then you know how much "stuff", the "n", is leaving the pressure vessel, then you can calculate the pressure drop.
2) if it's a pipe problem, then you're looking for the pressure drop across the leak. and you need more info (pipe static pressure, flow rate) ... so it sounds like it's 1) ...
another day in paradise, or is paradise one day closer ?
RE: Air leak rate conversion
What the responders are trying to help you with is also known as the "perfect gas law".
Wikipedia has a page that may provide more insight into your problem, and how these folks are trying to help.
http://en.wikipedia.org/wiki/Ideal_gas_law
RE: Air leak rate conversion
My motto: Learn something new every day
Also: There's usually a good reason why everyone does it that way
RE: Air leak rate conversion
RE: Air leak rate conversion
RE: Air leak rate conversion
have a pressure loss of .008 psi on a leak tester was told that it equated to 0.0027 cubic inch p/second but I need to prove this.
The people who originally set this up are no longer here and the Leak test company no longer in biz.
leak tester (provides external pressure on a unit) testing a unit for leaks, the machine will fail a unit with a loss of .008 psi.
what about PL at .008 psi?
Calculated knowns...
L.R. = PL V / t
n = 0.011548824 (at 68 deg F)
PL = .008 psi
V = 178.2
atm 14.7+8.6 = 23.3
t = 8 secs
Pa Test = 157953.7673
R = 8.134
T = 293.15 K
RE: Air leak rate conversion
Questions -
why is atm 14.7 PLUS 8.6?? where are you doing this? what does the +8.6 represent? I've seen lower than sea level atm pressure at higher altitudes but not higher unless you're in a dive chamber?
Volume is 172.2 cubic inches?
Pa test - is this in absolute units of Pascals or what?
Also I would love to know how what you use to measure 0.008 psi pressure drop over 8 seconds with any repeatability or accuracy. This sounds like serious lab type equipment.
My motto: Learn something new every day
Also: There's usually a good reason why everyone does it that way
RE: Air leak rate conversion
a chamber (alum box, with a unit in it) is filled with 8.6 psi for testing a unit (so 1 atm + 8.6), V in cc.
Pa absolute Pascal.
machine is fairly repeatable, again this was handed down to me.
RE: Air leak rate conversion
you know n,R,T and P (pressure = 8.6psi) so you can calc V and compare with known to validate other parameters.
btw ... use consistent units !! we're mixing all sorts.
how do you know the volume leakage rate ? i mean you're not collecting the leakage in a balloon and measuring it ??
but you can calculate the amount of "stuff" (air) from pv = (n1)*Rt ... p = 1 atm, v = leakage, t room temp? >>> calc n1
then back to the testing tank to calc p (due to leakage) = (n1)*RT/V
... no?
another day in paradise, or is paradise one day closer ?
RE: Air leak rate conversion
which I need to know in psi.
the PL of .008 psi is what the machine is capable of catching known fails at.
i'll review... thx rb1957
RE: Air leak rate conversion
Now I could have got that wrong, but it's a benchmark to work from. I worked tat on the basis that your leakage rate was measured at atmospheric conditions not the pressurised one and we're talking such small numbers any error on a big number means an error on the much smaller one.
I would love to know what this is for though...
My motto: Learn something new every day
Also: There's usually a good reason why everyone does it that way
RE: Air leak rate conversion
157953.7673 (sic) Pa = 1.559atm = 22.9psi = 8.2psig
(i'm using 1atm = 101325Pa)
with data provided PV = 28.15 (with the correct R = 8.314; i'm guessing the units match)
(how the heck do you know "n" ?)
with P = 56629, V = 0.000497m3 = 497cc ... not your 178cc ... must be doing something wrong, haven't done this type of chem in 30+ years
with your PV = 56629*0.000178 = 10.1 ... no?
n = 10.1/(293*8.314) = 0.004146 ?
another day in paradise, or is paradise one day closer ?
RE: Air leak rate conversion
n = 293*.0001782(Vm3) / 8.314*293.13
The part is communication amplifier device (is sealed but hermitically).
RE: Air leak rate conversion
29356629*.0001782(Vm3) / 8.314*293.13another day in paradise, or is paradise one day closer ?
RE: Air leak rate conversion
"part (is NOT sealed hermitically). Not that it matters.... been a long day..
rb1957... I follow you on the atm, psig etc... but maybe im tired
I don't see how you derived at P = 56629 ?
Thanks for help by the way.
RE: Air leak rate conversion
thinking about it further i think i should've used the absolute pressure ... then n = 157964*0.000178/(8.314*293) = 0.01154 ... harumph ... your number ... oops !
another day in paradise, or is paradise one day closer ?
RE: Air leak rate conversion
question, why wouldn't Ptest not be 160603.6Pa ?
RE: Air leak rate conversion
and using it gave the "n" you quoted
another day in paradise, or is paradise one day closer ?
RE: Air leak rate conversion
Thanks
RE: Air leak rate conversion
Converting: .00027 cubic inch into Psi - using Ideal Gas Law. All constant variables reduced.
P1V1 = P2V2
P2=P1V1/V2 = ((23.3) (1in3))/((.9973 in3)) - knowns (14.7+8.6 absolute pressure)
P2 = 23.36 psi - calculate variable
ΔP = 23.36 – 23.3 - subtract initial pressure to get ΔP after 1 second.
ΔP = 0.063 psi / sec
RE: Air leak rate conversion
RE: Air leak rate conversion
The V in this case is your fixed volume of the chamber.
you need to use the PV=nRT route.
I've done it again using 10 seconds worth to eliminate a few of the errors, but still come out with 0.0037 psi/second for your 0.0027 in^3/second
The volumes and pressure here seem minute and so I'm not surprised you're getting some strange answers
My motto: Learn something new every day
Also: There's usually a good reason why everyone does it that way
RE: Air leak rate conversion
determining how much air is in the leakage is easy nL = P(VL)/RT, P = 1 atm, V = leakage, T = 293
then return to your original PV = nRT with n = n(original)-nL.
i do wonder how this question is being asked. a leakage meter is going to measure the pressure decay, isn't it ? it won't readily let you the laekage volume rate, would it ? the most sense i can up to the question is the meter tells you one (pressure drop or volume rate) and you've got a client spec that expresses itself with the other ?
i guess the volume leakage is directly proportional to the psig insde the tank, so that as the tank pressure falls the leakage becomes less, asymptoting to 0 as the psig approaches 0 ?
another day in paradise, or is paradise one day closer ?
RE: Air leak rate conversion
So to add to the confussion I have attached my thinking
RE: Air leak rate conversion
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RE: Air leak rate conversion