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Air leak rate conversion

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

RE: Air leak rate conversion

by equating the lost volume as lost pressure (Charle's law, Boyle's law ... PV = constant) ?

another day in paradise, or is paradise one day closer ?

RE: Air leak rate conversion

PV=nRT

RE: Air leak rate conversion

(OP)
Hmmm I only know the 0.0027 cubic inch p/second to psi p/second leakrate...
I don't know the atmospheric / temp conditions.

isn't there a more direct way to convert?

RE: Air leak rate conversion

(OP)
actually I don't see how PV=nRT
would give me psi?

RE: Air leak rate conversion

What is your situation?

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

(OP)
hi I was given more more info...
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

You still need much more information before you can start to calculate this.

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

do you have a static pressure vessel or a dynamic pipe flow problem ?

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

bonll,
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

We're also trying to point out that he hasn't got enough data yet to da a calculation (no initial volume or pressure!)

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

How were you able to determine the leak rate at 0.0027 cu. in./sec, another word tell us the scope of the project in which you determine that leak rate.

RE: Air leak rate conversion

Smells like homework to me.

RE: Air leak rate conversion

(OP)
not homework really,
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

Now we're getting somewhere - why did you not give us all this info at the start??
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

(OP)
littlrich...

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

ok, so PV = nRT.

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

(OP)
Hmmm, the vol leak rate (0.0027) is actually the max LR spec.
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

I did this just for a laugh and for a leakage rate of your .0027 in^3 equates to .0035 psi.

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

we're using such hyper-precise numbers as Pa Test = 157953.7673, but 8.6psig = 1.585atm = 160603.6Pa
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

(OP)
errr I keep getting pulled off this to do other things sorry.

n = 293*.0001782(Vm3) / 8.314*293.13

The part is communication amplifier device (is sealed but hermitically).

RE: Air leak rate conversion

n = 29356629*.0001782(Vm3) / 8.314*293.13

another day in paradise, or is paradise one day closer ?

RE: Air leak rate conversion

(OP)
Meant to say...
"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

Ptest-Patm = 157953.7674-101325 ... i think it's right (like i said it's been a while)

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

(OP)
hi rb1957
question, why wouldn't Ptest not be 160603.6Pa ?

RE: Air leak rate conversion

'cause you said it was 157953.7674 (i calc'd it'd be 160603 Pa for 8.2psig)
and using it gave the "n" you quoted

another day in paradise, or is paradise one day closer ?

RE: Air leak rate conversion

(OP)
OK got it.
Thanks

RE: Air leak rate conversion

(OP)
rb1957 looking at it another way... (as in a bubble of 0.0027in3 p/sec)

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

(OP)
rb1957 so am I crazy??? or does this look right (ΔP = 0.063 psi / sec) to you?

RE: Air leak rate conversion

P1V1=P2V2 in this case doesn't work because you're not changing the volume, only the pressure and the mass of air in your fixed volume.

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

agreed ... you can't equate the laekage with the original condition.

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

I have tried to follow this thread as I some times need to evaluate the leakage rate for a generator gas space,which I see as the same test bonll is trying to perform, but on a much smaller scale. between the mixing of units, etc. I have been lost completely. Unfortuantely I am currently in such a hectic state I am also having trouble deriving the bases for the equations I have used for years so I just started with them and applied to this situation.

So to add to the confussion I have attached my thinking

RE: Air leak rate conversion

That's nice, but I'm a bit disappointed that you didn't use Mathcad's built-in units. That would avoid some of the unit conversion shenanigan you employed to convert the problem. Can you post the actual worksheet?

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RE: Air leak rate conversion

I never got where I liked mathcad. the version I have is 4.0 and the generator air test sheet was something I used long time ago. the equations are actually published in the t/g manual. I did not save the changes I made for this situation to my original file. attached is that first equation with a couple more for determining H2 equivalant and correcting if test done at reduce pressure. the varables used are where I wanted to show if the test duration was just 1 hour, what thier expected error could be

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