Recent content by bonll

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

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...

OK got it. Thanks

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

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.

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).

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

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.

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...

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

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

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?

how do I convert 0.0027 cubic inch p/second to psi p/second? help please. appreciate it thanks b