Pavan Kumar
Chemical
- Aug 27, 2019
- 333
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
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