DISCHARGING OF VESSEL AIR TEMPERATURE
DISCHARGING OF VESSEL AIR TEMPERATURE
(OP)
I am trying to determine the temperature in a high pressure air vessel after it has blown down to 30 seconds. Starting pressure is 5000 psig 80 deg F. I have used the first law: m1u2-m2u2=dmhe and determined a temperature after being -56 deg F. The first law states it is isentropic. However, another person believes will be more like irreversible process where 2nd law should apply. Does anyone have any good test data to support how this problem should be modeled. thanks much





RE: DISCHARGING OF VESSEL AIR TEMPERATURE
RE: DISCHARGING OF VESSEL AIR TEMPERATURE
RE: DISCHARGING OF VESSEL AIR TEMPERATURE
You could also look at it this way. Consider a control mass (a control volume around a given mass) that remains inside the vessel during the venting process. Consider also no heat transfer for this control mass. As the mass expands, it is doing the same thing it would do if it were expanding against a piston. Hence, it is an isentropic expansion.
Either way - using the first law or assuming an isentropic expansion - gives the same result.
Regarding heat transfer, just do a simple calculation to convince yourself that heat transfer from ambient isn't going to be a significant factor. Look at how much air is left in the vessel when it blows down to the lower pressure isentropically. Find the temperature the air is at. Then calculate what temperature the vessel would come to assuming the vessel were perfectly insulated. In other words, you have air at some low temperature and it comes to equilibrium with the vessel. I think you'll find the temperature of the vessel drops very slightly compared to the temperature rise of the air. So the vessel isn't going to get very cold because it has so much thermal mass.
So if the vessel doesn't get cold, then a first order estimate of the heat transfer can be done assuming the temperature of the vessel doesn't change (or only changes very slightly) and only convective heat transfer acts between the vessel at constant temperature and the air.
If the time is only 30 seconds, there won't be a whole lot of heat transfer, I agree. But if you want to get a bit closer to reality, that would help. Otherwise, the transient heat transfer analysis becomes quite complex.
RE: DISCHARGING OF VESSEL AIR TEMPERATURE
What does this meaan??
Also, the process must be irreversible and can be approximately adiabatic.
Your equation should be after pushing dm out of the chamber with mass M
dm=-dM
d(Mu)=-h*dm=h*dM
mdu+udM=RTdM+udM
Mdu=RTdM
MCvdT=RTdM
dT/T=R/Cv*dM/M
ln(T2/T1)=R/Cvln(M2/M1)
T2=T1*M2/M1)^(R/Cv)=T1*(M2/M1)^(k-1)
RE: DISCHARGING OF VESSEL AIR TEMPERATURE
The equations you wrote out are correct of course, but they're for an ideal gas. A better, more accurate way to do this is to use actual properties from a database like NIST REFPROP. Using a database allows you to use actual values for internal energy, enthalpy and entropy. For example, you can determine the entropy of air at 5000 psi and 70 F where the compressibility factor is ~1.15 and then assuming an isentropic expansion, determine the state at a lower pressure simply from the value of pressure. It's actually much easier, more accurate and lends itself well to computerized analysis. I use a proprietary data base but REFPROP is very similar and can be purchased for a few hundred $ here:
http://www.nist.gov/srd/nist23.htm
RE: DISCHARGING OF VESSEL AIR TEMPERATURE
If I needed it more accurately I would do it numerically.
Then, the basic equation
dT/T=R/Cv*dM/M
is better rewritten as
dM/M=dRho/Rho=-dR/R+dP/P-dT/T]
since Rho=P/RT
and M=Rho* VOLUME
assuming R a function of T (although can include others)
dM/M=-dR/dT/R*dT+dP/P-dT/T
dT/T=R/Cv*[-dR/R+dP/P-dT/T]
[(1+R/Cv)/T+dR/dT/R]*dT=R/Cv*dP/P
which is easily solved with the computer using the data you mentioned.
RE: DISCHARGING OF VESSEL AIR TEMPERATURE