Air leak rate conversion

[bonll]

how do I convert 0.0027 cubic inch p/second to psi p/second?

help please.

appreciate it thanks
b

 

[bonll]

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).

 

[rb1957]

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

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

 

[bonll]

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.

 

[rb1957]

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 ?

 

[bonll]

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

 

[rb1957]

'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 ?

 

[bonll]

OK got it.
Thanks

 

[bonll]

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

 

[bonll]

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

 

[LittleInch]

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

 

[rb1957]

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 ?

 

[byrdj]

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

 

[IRstuff]

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?

TTFN
faq731-376

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Of course I can. I can do anything. I can do absolutely anything. I'm an expert!

 

[byrdj]

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