DISCHARGING OF VESSEL AIR TEMPERATURE

[loganevh]

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

 

[iainuts]

You have the right equation. The temperature of the gas that's inside the vessel follows a line of constant entropy (isentropic) with the exception that there is considerable heat transfer between the vessel walls and the air. The vessel has much more thermal mass than the air so generally, there is no need to consider heat transfer from ambient. The isentropic expansion is easy enough to model, but the heat transfer is much tougher and will result in the temperature of the air being warmer than it would otherwise.

 

[loganevh]

thank you very much iainuts for replying. I tried to model heat transfer into the equation originally by m1u1-m2u2=dmhe+Q. I found that the heat added was small, but maybe because is was modeling it as conduction through the vessel wall with outside and inside convection, I cant remember exactly what equations were, but something like 1/Ahoutside+ln(radii)/kAsteel+1/Ahinside. YOu get the point, but maybe that isnt what I should have used. Do you think should be purely convective heat transfer Q=Ah(deltaT). I figured that it is happening so fast (only 30 seconds) that heat added from the 1" wall vessel would not have time to heat the air. Another thing I am confused about is how do we know the process is reversible. I suppose I dont have a firm grasp on the concept. Thanks again all.....

 

[iainuts]

The equation dU = Hout gives the same answer as dS=0.

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.

 

[zekeman]

"I am trying to determine the temperature in a high pressure air vessel after it has blown down to 30 seconds. "

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)

 

[iainuts]

Hi zekeman,
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

 

[zekeman]

You are essentially correct, since the integration I performed assumed constant R, Cv and Cp.

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.

 

[loganevh]

thanks Zekeman for the response, however like iainuts says the air is no longer ideal at that pressure. I have a calculator that calculates the properties of air at high pressures. But what really happens in this vessel is after 6 seconds, 10 lbm is removed. This process happens 5 times total, which amounts to the 30 seconds total. I used the lumped capacitance method to estimate the heat that the air would receive from the vessel, but came up with like under 10 BTU. A college used the second law and figures after 6 seconds, the temperature of the air should be roughly 60 degrees F. I used the first law as stated above with Q=0 and calculated 50 F. However, to acheive a temp of 60 F, I believe heat input to the air after the 6 seconds should be roughly 230 BTU. I am not sure the air can receive that amount of heat in such a small time. Convection coeff. I believe is around 10 BTU/hr/sq ft R. Anyway, thank you guys for responding....