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Dear All,
I was just going through some of basic equations for control valve. I have come across something for which I dont find any answer to that.
We all know the control valve sizing equation as
Q = Cv * (dP/sp.gr)^0.5 ----- (1)
where Q is in US GPM, and dP in PSI.
And we also know if we write the equation for SI unit. It is
Q = 0.856* Cv * (dP/sp.gr)^0.5 ---- (2)
where Q is in m3/hr and dP in bar.
Now if we derive the eq. (2) from eq (1) by converting GPM into m3/hr and dP into bar from PSI, it is coming
Q = 1.156*Cv * (dP/sp.gr)^0.5 ---- (3)
where Q is in m3/hr and dP in bar
So, it is observed that there is some discrepancy between eq. (2) and eq (3). So, any body can cite it out whats the problem in this ????
And one more thing 1.156 is exactly inverse of 0.856.
Thanking you in anticipation.
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Control Valve Sizing Equation
Started by gunjan, Jan 27 2007 04:07 AM
5 replies to this topic
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#2
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Posted 28 January 2007 - 01:18 PM
gunjan,
You did your unit conversions "upside down," an easy mistake to make.
By definition (and by inspection of your fist equation), Cv has units of gpm / psi^0.5. If you want to use bar and get a result of cubic meter per hour, do the following unit conversion:
1 m^3/hr (14.5 psi)^0.5 m^3-psi^0.5
----------- * ----------------- = 0.865 -------------------
4.4 gpm bar^0.5 gpm-bar^0.5-hr
Hope that helps.
This is one reason why I don't like equations that have combined numerical conversion factors, particularly since these conversion factors almost never carry proper units, and they often hide aspects of the underlying fundamental equation. Unless it is an empirical equation, just use the equation as is and keep careful track of all of your units. You will make far fewer mistakes in the long run.
Mike
You did your unit conversions "upside down," an easy mistake to make.
By definition (and by inspection of your fist equation), Cv has units of gpm / psi^0.5. If you want to use bar and get a result of cubic meter per hour, do the following unit conversion:
1 m^3/hr (14.5 psi)^0.5 m^3-psi^0.5
----------- * ----------------- = 0.865 -------------------
4.4 gpm bar^0.5 gpm-bar^0.5-hr
Hope that helps.
This is one reason why I don't like equations that have combined numerical conversion factors, particularly since these conversion factors almost never carry proper units, and they often hide aspects of the underlying fundamental equation. Unless it is an empirical equation, just use the equation as is and keep careful track of all of your units. You will make far fewer mistakes in the long run.
Mike
#3
proinwv
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Posted 28 January 2007 - 07:43 PM
Be careful using this equation. What you have shown does not account for the liquid recovery factor FL which is used in accounting for pressure drops in excess of critical.
See ISA S75.01 and similar publications. Several of the major valve manufacturers also use this theory, or their slight modifications to it.
See ISA S75.01 and similar publications. Several of the major valve manufacturers also use this theory, or their slight modifications to it.
#4
gunjan
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Posted 28 January 2007 - 11:48 PM
Thanx Mike..
You are right about what you said. Cv has a unit gpm / psi^0.5 and after converting into SI it will become 0.865 m3/bar^0.5. But if you again form the equation using the later unit it will be 0.865*Q = Cv*(dp/Sp.Gr)^0.5 which is different from correct equation in SI unit i.e. Q = 0.865*Cv*dp/Sp.Gr)^0.5. I know there is some problem of understanding.
Regards,
Gunjan
You are right about what you said. Cv has a unit gpm / psi^0.5 and after converting into SI it will become 0.865 m3/bar^0.5. But if you again form the equation using the later unit it will be 0.865*Q = Cv*(dp/Sp.Gr)^0.5 which is different from correct equation in SI unit i.e. Q = 0.865*Cv*dp/Sp.Gr)^0.5. I know there is some problem of understanding.
Regards,
Gunjan
#5
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Posted 07 February 2007 - 01:42 PM
Gunjan,
Why did you put the 0.865 on the left-hand side of the equation? We developed a conversion factor for Cv, so that's where you should put it, not next to Q. Keep the units with it ( 0.865 m^3-psi^0.5 / gpm-bar^0.5-hr ) and work through a random example (again, rigorously keeping track of units, including gpm / psi^0.5 for Cv), and it will be obvious.
Always rigorously carry your units and you will make many, many, many fewer mistakes.
Mike
QUOTE (gunjan @ Jan 28 2007, 11:48 PM) <{POST_SNAPBACK}>
But if you again form the equation using the later unit it will be 0.865*Q = Cv*(dp/Sp.Gr)^0.5
Why did you put the 0.865 on the left-hand side of the equation? We developed a conversion factor for Cv, so that's where you should put it, not next to Q. Keep the units with it ( 0.865 m^3-psi^0.5 / gpm-bar^0.5-hr ) and work through a random example (again, rigorously keeping track of units, including gpm / psi^0.5 for Cv), and it will be obvious.
Always rigorously carry your units and you will make many, many, many fewer mistakes.
Mike
#6
shivaprasad
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Posted 07 February 2012 - 04:53 AM
Gunjan,
please tell the refernces of equation 1 & 2.
Please send the article/paper.
give me reply a.s.ap
shiva
please tell the refernces of equation 1 & 2.
Please send the article/paper.
give me reply a.s.ap
shiva
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