Relief Rate Through a Full Open Globe Bypass Valve.

[Pavan Kumar]

Hi All,

I want to calculate the flow through a full open 1 1/2" globe valve for pressure relief valve sizing . This globe valve is the bypass valve to a control valve and I want to calculate the relief rate for in-advertant opening of this bypass valve. I have evaluated this scenario and found that in-advertant opening of this bypass valve can lead to over-pressure of the vessel. The process conditions upstream and downstream of the globe valve are as below:

Inputs:
Valve Size = 1 1/2"
Line size upstream and downstream of the valve = 1 1/2" Sch 40
Pipe ID = 1.610 inches = 0.13416 feet.
PSV Set Pressure = 72.5 psig
Relief Stream = Water
Relief Temperature = 248 Deg F
Upstream pressure, P1 = 116 psig = 116+14.7 = 130.7 psia
Downstream Pressure,P2 = 1.1* PSV Set pressure + 14.7 = 1.1*72.5 +14.7 = 79.75+14.7 = 94.45 psia

Calculations:

As an intial guess assuming the flow to be turbulent, I got the Darcy friction factor, fD from Pipe friction data table Crane Paper from Appendix A-26.

for 1 1/2" pipe, fT = 0.021

The equivalent Length, L/D for full open globe valve = 340 ( from A-27)

so , K factor is K = fT(L/D)= 0.021*340= 7.14

From the formula relating K to Cv, I calculated CV for the full open Globe valve as below.

CV = 29.9 *(d^2)/ SQRT(K) - Equation 3-16 from Page 3-4 of Crane TP410M

CV= 29.9*(1.610^2)/SQRT(7.14) = 29.9*2.5921/2.6720 = 29.005 US gpm/psi^0.5

Density of water ( relief stream ) at conditions of globe valve

is Rho = 58.8982 lbm/ft3
Specific gravity , G = 58.8982/62.4 = 0.9438

I calculated the flow through the full open globe valve using the formula below:

Q = CV * SQRT(DP/G)

DP = P1-P2 = 130.7-94.45 = 36.35 psi
CV = 29 ( calculated above)
G = 0.9438

Q = 29*SQRT(36.25/0.9438) = 179.7 US gpm

Relief Rate = 179.7*0.133681*60*58.8982= 84905.3 lbm/hr

Now that we know the flow, I calculated the friction factor through 1 1/2" Sch 40 line

Pipe ID,D = 1.610" = 0.13416 feet
Flow Area,A = (PI()/4)*(0.13416^2) = 0.01414 feet^2
Velocity, V = Q/ A = (179.7*0.133681/60)/0.01414 = 28.3 feet/sec
Density, Rho = 58.8982 lbm/ft3
Viscosity, Mu = 0.232 cP = 1.558X10^-4 lbm/ft-sec

Pipe roughness = 0.0018 inches

NRe = 0.13416*28.3*58.8982/1.558X10^-4 = 1434432.45

Using Chruchill Eqn

1/SQRT(f) = -4*Log(0.27e/D + (7/Nre)^0.9), f calculated using this eqn is fanning friction factor.

ffanning = 0.00511
fdarcy = 4* 0.00511 = 0.02044

This is close to the 0.021 I assumed initially, so the K factor and CV calculated is going to stay.

My question is, if the flow capacity of full open globe valve calculated above is correct. If not what corrections need to be made?. Any reference material that you can provide will be great help to me.

Thanks and Regards,
Pavan Kumar

 

[shvet]

@Pavan Kumar
Note that in your case ρv2 ~100 kPa. Many companies prohibits ρv2 > 25-50 kPa for DN < 2", actual limit depends on company practice and diameter/thickness ratio.
The same situation with erosion velocity, for example in your case Norsok std. P-001 prohibits velocity > 3 m/s. Note that many companies apply API 14E to erosion prediction of all process piping and as per API 14E your velocity is well above safe limit.
In addition to some risk (not obvious to personnel) an inspection/audit is recommended after every such relief or proactive replacement of the critical piping segment.

It may be not a best idea to place a regular piping into extreme conditions even those are sporadic. Piping elements are designed more on economic basis than on absolute safety. This situation is not able to be called an inherently safer design. A dedicated device would fit better.

 

[ReonArd]

Your question: is the flow capacity of full open globe valve calculated above correct?
As a reference I used this CV sizing spreadsheet:

https://sitepad1.wordpress.com/

-> select "Control Valve" (see the screenshot attached to this post)
Conclusion: yes, your calculation is correct, I arrived at a pretty similar result.