Could someone verify that I am doing this correctly.  The number calculated seems excessive.  I am new to the industry and have been looking over equations and have nothing to compare my answer.  In advance, I appreciate the help.

Problem Statement:  An 8" nominal Rockwell dynamic balance plug valve was left open approximately 19 minutes.  The inlet pressure was 825 psig and the valve was open to atmosphere.

Find: How much gas was lost to the atmosphere.

Knowns:

- Valve properties -
Cv=2680
Critical pressure drop ratio=0.42

- Supply Line -
8" NPS schedule 80

- Fluid Properties -
SG=0.595

- Choked flow conditions -
Expansion coefficient (Y)=0.667
Compressibility Factor (Z)=0.8728

Comments: I have read the post at Milton Beychok's site at
http://www.air-dispersion.com/feature2.html, which is very good by the way.  I understand the equation, but is there a way to relate the flow coefficient (Cv) to the discharge coefficient (Cd)?  All most all of the information I deal with has everything based on the flow coefficient.

My Solution: I researched the ANSI/ISA-75.01.01 standard, "Flow Equations for Sizing Control Valves".  Using Eq. 14B

Cv=(Q/Y*1360*Pi)*sqrt[(SG*T1*Z)/(Fy*xT)]  (English Units)

Where,

Cv=2680 (from manufacturers data)
Q=? cf/h
Y=0.667
Pi=825+14.73=839.73 psia
SG=0.595
T1=60F=520R
Z=0.8728
Fy=k/1.4=1.27/1.4=0.9071
xT=0.42 (from manufacturers data)

Calculated flow rate, Q=1,279,847 scf/min. 

In terms of 1000 of cf:

Q/1000 = 1,280 Mscf/min

Total lost over a 19 min period:

Total lost=Q*t --> 1,280*19min=24,317 Mscf

Questions:

Is this the correct equation to use for this situation?
If not, which equations are appropriate?  (in terms of the flow coefficient (Cv))
How accurate is this equation?
If this is wrong could someone do an example of how to properly calculate this situation?

you have choked flow, so the flow rate is constant once the outlet pressure is below 420 psi.  You calculation is correct for that rate.  Next, a few feet of 8" pipe has about 50 psig drop to atmosphere, so all the drop is through the valve.

You have a typical PSV scenerio with 4 - 8T10 relief valves all opening up at once.

dcasto, it depends on whether the 825 psig is at the valve inlet or at the header.

If it is at the header, then the majority of the pressure drop is NOT though the valve. The bulk of the pressure drop is through the 50ft inlet pipe and the choke at the pipe outlet to atm.

It would only be the same as 4-8T10 (or did you mean 2) relief valves if he had 4x8" pipes to his valve and 4x10" pipes from his valve and the valve xT matched that of the relief valve.

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