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On Dec 7, 2013, at 8:21 PM, rjensen@ualberta.ca wrote:
Interesting that they are filling cylinders by pressure. I read (no
first hand experience with this) that cylinders are normally filled by
mass to address the thermal problem. The internal volume of the
cylinders likely has a tolerance of less than 5 %, so the final
pressure is within 5 %. Filling by mass also ensures the customer gets
what they paid for. The added initial pressure is not a problem:
cylinders have a safety factor of between 3 and 5.
To answer your question, you don't need to work in theory here, you
have experimental data! By knowing the pressure drop, you can
calculate the internal temperature using the combined gas equation.
After cancellation for constant volumes and moles, you have P1/T1 =
P2/T2. Given your data, the gas temperature from filling is 83 °C.
Applying this formula again tells you that you must pressurize the
cylinder to 5470 psi so that the cooled pressure is 4500 psi. (To whom
do I send the consulting bill?)
Please, no-one mention Boyle, Charles, etc. The combined gas equation
is much more convenient.
Note: at 300 bar, nitrogen has a compressibility of 1.1, so it doesn't
behave exactly ideally, but the above calculation will work
approximately.
Dr. Roy Jensen
(==========)-----------------------------------------¤
Lecturer, Chemistry
E5-33F, University of Alberta
780.248.1808
On Sat, 7 Dec 2013 15:34:29 -0500, you wrote:_______________________________________________
A friend of mine asked me about the following issue they are having. They
fill cylinders of compressed air. The next day they have to top the
cylinders back up. His questions were can we estimate how large the drop is
and ways to reduce it?
I reason as follows. We can easily estimate an upper and lower bound on the
pressure drop overnight.
The lower bound is if we fill the cylinder so slowly that it is isothermal.
In that case, there is no temperature rise of the gas and hence there is no
subsequent cooling of the compressed gas and hence the pressure drop is
zero.
The upper bound is we fill the cylinder quickly enough that we can treat it
as adiabatic (but slowly enough to be treated reversibly). By this, I mean
there is no heat loss to the surroundings during the filling. Treat air as
an ideal diatomic gas so the ratio of heat capacities is g = 5/3. Then we
have T proportional to p^0.4.
Now he gave me some numbers. They fill the cylinder to about 4500 psi.
That's about a factor of 300 increase from atmospheric. Hence we would
estimate a temperature rise of the air by about a factor of 10. Then the
pressure should drop to about one-tenth overight, to about 450 psi. This is
the worst case.
But the actual answer is the pressure drops from about 4500 psi to 3700
psi. So yes, it's between zero and one-tenth. But much closer to zero than
to 1/10. I'm surprised. I thought the adiabatic case would be closer to the
answer than the isothermal case. Seeking a reason, I asked him about how
long it takes to fill - about 20 minutes. And about how long it takes for
the pressure to drop - about 3 hours. So part of the answer is the filling
is neither fast nor slow.
Can I better estimate the pressure drop? Suggestions welcome.
As far as reducing the pressure drop more, the two options I suggested are:
(1) fill it even more slowly - since they have to fill 15 cylinders, they
could fill them all part way and then go back and fill them the rest of the
way for instance; (2) cool the cylinders - for example immerse them in
water which he said is doable. Other suggestions again welcome. -Carl
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