Thermodynamic problem, with time factor

[nastran2002]

Hello All,

I have a thermodynamic question. If you have steam at 365 F, 150 psi gage, total volume 2.3434 ft3. There are two vales on the tank 2" diameter pipe. I want to know how long it will take for the pressure to drop to 10 psi gage if the vales are opened. I really need a procedure to calculate this. I have looked at USUF processes etc in thermo books, but there are no equations trhat have time in it.

Please help

Thanks

...SPK

 

[pmover]

since u know the tank volume and steam data, you can determine the initial mass. then determine the amount of mass when steam pressure is 10 psig.

now, just what is the downstream pressure - atmospheric or 10 psig? also, what is the final temperature? those two factors need to be considered!

you calc is straightforward, u just need to clarify the rest of the problem.

as far as a formula, any thermodynamics textbook has the energy formula.

if this is your homework, suggest you learn the definitions and learn how to apply them. perhaps i already provided too much info.
-pmover

 

[nastran2002]

Thanks Pmover. This is not a homework problem. I work in industry. If I know the final temperature then ofcourse I know the final mass. So I know how much mass has left the tank. How do I know how much time it took ? Let us say the outside pressure is atmospheric. I would appreciate if you could elaborate.

 

[Latexman]

nastran2002,

It's less of a thermodynamic question, and more of the classical tank "blowdown" problem. Close the thermo book, and open the fluid flow book, and find one on chemical processes (I use Felder and Rousseau) or dynamic modelling. You need to write an unsteady-state mass balance around the tank. Characterize the mass in the tank and the flow out of the tank as a function of the tank pressure and the outlet valve and piping system. Combine these, solve the differential equation, and bada bing bada boom, you got the answer. If the 2" valves go straight to atmosphere, I'd guess the blowdown takes less than 10 seconds on that small tank.


Good luck,
Latexman

 

[MortenA]

more down to earth:

Use a spreadsheet and cacl. stepvise. Decrease timesteps until you have a reliable result.

Best regards

Morten

 

[TD2K]

Ditto on Morten's suggestion, it's how I've done this in the past. I would say for your volume with 2, 2" valves 'snapping' to the full open position, it will only take a few seconds to depressure it.

 

[nastran2002]

Thanks All,

I am not a process engineer. You mention about spread sheet calculations. Are there any available on the web that would do something like this.

Thanks

 

[Latexman]

nastran,

Sounds like you are in over your head on this one. I recommend you farm it out.



Good luck,
Latexman

 

[chicopee]

nastram2002, I hope that I am not to late due to other commitments but I may have a solution to your problem. In summary I determined that .7887 lbm of vapor was released within a time span of .128 sec. Some assumptions were made
such as isentropic condition to estimate steam release within one control volume(ie) the vessel; the second control volume (ie) two gate valves -half opened- and 2" piping, 10'long discharging to atmosphere was considered isenthalpic in nature to apply the first law of thermo. and in which I used initial and final states of the vapor to determine the time element.

 

[imok2]

Nastran2002:
Could this be solved with the application of Bernoulli equation for solving unsteady state problems? If so go to

http://www.svce.ac.in/~msubbu/FM-WebBook/Unit-II/EffluxTime.htm

 

[davefitz]

for zero heat transfer , constant volume, one has:

0=d(mu)/dt + Wo*(u+Pv), where Wo= flow out = -dm/dt

for a perfect gas, Pv= ZRT, and u=Cp(T-T')

0= [2Cp(T-T')+ZRT]*dm/dt + mdT/dt

dT/dt / [2Cp(T-T')+ZRT] = -(1/m) dm/dt

the flow rate out can be calcualted via choked flow if the vale is the primary restriction, Wo=63.4*0.67*Cv*SQRT(Xt*Pi/sv,i), or if the flow is frictionally choked thru a long pipe then use the Fanno relationships.

 

[chicopee]

SORRY DAVEFITZ, THIS IS NOT A CONSTANT VOLUME PROBLEM AND ALTHOUGH WE ARE BORDERING ALONG THE SATURATED VAPOR LINE WE SHOULD NOT USE THE EQUATION OF STATE; AND YES TO NASTRAM2002, IT IS SOLVABLE W/ BERNOULLI' EQUATION WITHIN MY SECOND CONTROL VOLUME AND BEARING IN MIND THAT DENSITY(OR SPECIFIC VOLUME),PRESSURE AND VELOCITY CHANGE BETWEEN INTAKE AND OUTLET OF MY 2ND CONTROL VOLUME. THE KEY POINTS ARE THE ISENTROPIC CONDITION WITHIN THE FIRST C.V. AND ISENTHALPIC CONDITION WITHIN THE SECOND C.V.

 

[davefitz]

chicopee:
The problem statement is a tank volume of 2.343 ft3; this is a constant volume problem.

If the steam is saturated, then the perfect gas equations are not used and the calculation procedure would need to use the steam tables to determine the amount of steam condensed during the depressurization.

 

[chicopee]

You are right davefitz,it is a constant volume problem when you are to analyze the residual steam mass after.Dont know what came over me.

 

[chicopee]

Davefitz,after reevaluating my answer about the constant volume, i realize that is not not a constant volume problem because there are two valves which are briefly opened lowering the internal pressure, therefore my original evaluation stands.

 

[poetix99]

Transient problems have been discussed on this site before. Search on keyword "Fliegner"; choose "Search Posts (keyword)" from the pull-down. You should get at least four other threads.