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Practice Problem - Valve Sizing

1. A valve for an 8" liquid hydrocarbon line was sized with a Cv of 568 to control 1000 gpm at a pressure drop of 10 psi, checking for cavitation. A 688 Cv Fisher valve was selected. 2. A valve for a 3" line controlling 200 gpm at 20 psi was sized with a Cv of 112 to prevent cavitation, but the available pressure drop was only 11 psi. A 136 Cv Fisher valve with a trim of % was selected. 3. The document provided details on sizing two control valves for liquid hydrocarbon lines, including checking for cavitation based on process parameters and selecting appropriately sized control valves from a Fisher catalog.

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321 views5 pages

Practice Problem - Valve Sizing

1. A valve for an 8" liquid hydrocarbon line was sized with a Cv of 568 to control 1000 gpm at a pressure drop of 10 psi, checking for cavitation. A 688 Cv Fisher valve was selected. 2. A valve for a 3" line controlling 200 gpm at 20 psi was sized with a Cv of 112 to prevent cavitation, but the available pressure drop was only 11 psi. A 136 Cv Fisher valve with a trim of % was selected. 3. The document provided details on sizing two control valves for liquid hydrocarbon lines, including checking for cavitation based on process parameters and selecting appropriately sized control valves from a Fisher catalog.

Uploaded by

RONO
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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Practice Problem – Valve Sizing

1. Size a valve to control the flow of liquid hydrocarbon.


The process data is:
Liquid hydrocarbon
Flowing temperature 65C
8” line
Gf = 0.804
P1 = 50 psig
Pvap = 11 psia
Pc = 950 psia
Cf = 0.8
nominal flow = 1000 gpm
size for 100% overcapacity
valve pressure drop at the nominal flow = 10 psi
line pressure drop at the nominal flow = 5 psi
valve trim is linear, fail closed
Calculate the Cv and select the correct sized valve. Be sure to check for cavitation. Select
and ED Fisher valve, vendor data is attached.
2. Size a valve to control the flow of liquid hydrocarbon.
The process data is:
Liquid hydrocarbon
Flowing temperature 65 C
3” line
Gf = 0.85
P1 = 30 psig
Pvap = 25 psia
Pc = 950 psia
Cf = 0.7
nominal flow = 200 gpm
size for 100% overcapacity
valve pressure drop at the nominal flow = 20 psi
line pressure drop at the nominal flow = 25 psi

1
Practice Problem – Valve Sizing

valve trim is =%, fail closed


valve theoretical rangeability = 33
Calculate the Cv and select the correct sized valve. Be sure to check for cavitation. Select
and ED Fisher valve, vendor data is attached.

Fisher Design ED Classes 125 – 600, Cage Guided

2
Practice Problem – Valve Sizing

SOLUTIONS

Question 1
Data:
8” line
Gf = 0.804
P1 = 50 psig
Pvap = 11 psia
Pc = 950 psia
Cf = 0.8
nominal flow = 1000 gpm
size for 100% overcapacity
valve pressure drop at the nominal flow = 10 psi
line pressure drop at the nominal flow = 5 psi
valve trim is linear, fail closed
Check for cavitation,

Pallow  C f 2 Ps

To calculate Ps,

If Pv < 0.5P1 then use Ps = P1 – Pv

 PV 
If Pv >= 0.5P1 then use Ps  P1   0.96  0.28  PV
 PC 

Calculate PS,
P1 = 50 psig + 14.7 = 64.7 psia
It is the case Pvap < 0.5P1 or 11 < 0.5(64.7).
So use the following equation to calculate Ps,

Ps = P1 – Pv = 64.7 – 11 psi = 53.7 psi

Sub intoPallow,

Pallow = 0.8253.7 = 34 psi

3
Practice Problem – Valve Sizing

In this system,

PV,available < PV,allow

So there will be no cavitation if we size the valve for Pv = 10 psi.


The valve sizing equation is,

Pv
F  Cv
Gf

The nominal Cv is,

Gf 0.804 gpm
C V  Cv,nominal  F  1000  284
Pv 10 psi

The valve coefficient sized for 100% overcapacity is,


gpm gpm
Cv,max  2  Cv,nominal  2  284  568
psi psi

Select a Fisher ED valve, linear trim,


Cv,selected = 688
8” linear valve, full port, 2” travel

Question 2
Data:
Liquid hydrocarbon
Flowing temperature 65C
3” line
Gf = 0.85
P1 = 30 psig
Pvap = 25 psia
Pc = 950 psia
Cf = 0.7
nominal flow = 200 gpm
size for 100% overcapacity
valve pressure drop at the nominal flow = 20 psi
line pressure drop at the nominal flow = 25 psi
valve trim is =%, fail closed
valve theoretical rangeability = 33

4
Practice Problem – Valve Sizing

Check for cavitation,


P1 = 30 psig + 14.7 = 44.7 psia
Pv > 0.5P1 so use,
 P 
Ps  P1   0.96  0.28 v  Pv
 Pc 

 25 psia 
Ps  44.7   0.96  0.28  25  44.7psi  22.9 psi  21.8psi  22 psi
 950 psia 

Pallow = Cf2Ps
Pallow = 0.7221.8 = 11 psi
Pv, available > Pallow
So there will be cavitation if we size the valve for Pv = 22 psi, therefore size the valve using
11 psi.
The valve sizing equation is,
Pv
F  Cv
Gf

The nominal Cv is,


Gf
C v,nominal  F
Pv
0.85 gpm
 200  56
11 psi
The valve coefficient sized for 100% overcapacity is,
gpm gpm
Cv,max  2  Cv,nominal  2  56  112
psi psi
Select a Fisher ED valve, =% trim.
3”, 3 7/16” port, 1 1/2” travel, Cv,selected = 136

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