Isentropic Expansion for a Blast Calc 4

[wallace24]

I am designing a blast chamber to contain potential failure of pressure vessels tested with air. In order to do this, "the available energy released by isentropic expansion of the fluid from rupture conditions to one atmosphere" must be calculated. I'm more than a little rusty on thermo!
The largest vessel that will be put into the chamber would be 400 gallons, at ambient temp(60degF) and at a test pressure of 404.7 psia.
Can anyone help me?
Any other info on this would be helpful too.
Thanks.

 

[israelkk]

The energy release = Pressure difference X Volume meaning=400 gallons X (404.7-14.7) psig = 4072000 Joules

 

[sailoday28]

The approximation of the blowdown from the pressurized vessel can be approximated as isentropic in the blowdown vessel as long as heat transfer from the surrounding metal is neglected.

For a perfect gas and const specific heats, the relation of the blowdown temp to mass in the blowdown vessel is
T/Ti=(M/Mi)^(gamma-1) =(M/Mi0^0.4 for air.
where Mi is inital mass in blowdown vessel and Ti is initial temp (absolute)
The energy blow down from the blowdown vessel=
integral h dm where h is the stag enthalpy and m, the mass of this pressurized vessel.
After integration,
Cp*Mi*Ti[(M/Mi)^1.4-1] EQUATION 1
This blowdown energy is received by the vessel designed for 1 atm. This energy then increases the internal energy of the receiver------ assuming no heat transfer----
The change of internal energy for the receiver is
Cv(m2T2-m2oTi) EQUATION 2
or
Cv/R(P2-Pi)Vreceiver EQUATION 3
Where Cv is spec heat and m2 and m2o the final and inital mass of the receiver. T2 is final temp of receiver.
Equation 1 must be equated to equation 2 or 3 .
Equation 1 is not equal to 0
Therefore P2, the final press of the receiver cannot equal its initial pressure.

 

[Latexman]

Work done on the environment = integral of PdV = 14.7 psia x [400 gallons x 404.7/14.7 - 400 gallons] = 14.7 lbf/in.^2 x (144 in.^2/ft.^2) x 400 gallons x (ft.^3/7.4805 gallon) x (404.7/14.7 - 1) = 3003008 ft-lbf = 4071532 J = 3859 Btu. Same as israelkk.

Good luck,
Latexman

 

[25362]

Have you considered, or is it at all possible, to do the pressurizing tests with water, in order to markedly drop the stored energy levels ?

 

[sailoday28]

Haste makes waste.
With reference to my previous post equation 1 should be
-Cp/gamma*Mi*Ti[(M/Mi)^1.4-1] EQUATION 1
or Cv*Mi*Ti[1-(M/Mi^)1.4]
For the isentropic source pv^gamma =const
(M/Mi)^gamma= P/Pi Equation 1A
where pressures refer to source
and subscript i to initial condition of source

And EQUATION 1 BECOMES
Cv*Mi*Ti[1-P/Pi] eq 1B
for perfect gas this results in
Cv*[Pi-P]*Vsource/R eq 1c

Cv/R(P2-Pi)Vreceiver EQUATION 3
equating equations 1c and EQUATION 3

press increase in receiver/pressure decrease in source=
Vreceiver/Vsource
Determine final pressure of source. With know volume of source and receiver, the change in receiver pressure is obtained.
Note volume of receiver = vol of room receiver is in minus volume of source.
Sorry for being long winded.

 

[sailoday28]

Another approach to determine the final pressure in the receiver.
Assume perfect gas,- constant specific heats-no heat transfer from receiver-neglect heat capacity of metal.

Let the source disintegrate
Then internal energy of mixed contents will equal inital internal energy of source + receiver.
Since initial temp of source and receiver are equal, then final temperature of mixed contents is equal to initial temp.

Final mixed room pressure=(sum of initial masses)*R*T/V

Where V=volume of receiver+volume of source.

Ratio of final receiver pressure to inital receiver pressure=
Vreceiver/V*[1+Msi/Mri]
where Msi=mass of source initial
Mri=mass or receiver initail