## Cv Cd & effective valve area

## Cv Cd & effective valve area

(OP)

hi everybody , i have a curve for a butterfly valve , it's (perc of Cv) Vs. (perc opening) is there an equation for converting Cv to Cd ?

is there an equation for the effective area of the valve , for example for 90 degree rotation Area = Area max ,, i need an equation to give me different areas for different rotational degree.

I found an equation that says : flow through valve (Q) = Cd * Av * sqrt (2*g*H) ,,, if that H is the headloss across the valve and my valve is fully open the equation will give me flow = 0 ,cause H across valve = 0 ( fully open ,no headloss ) is that correct ??

Plz, if anyone knows interent sites that talks about this plz mention it

thanks alot

is there an equation for the effective area of the valve , for example for 90 degree rotation Area = Area max ,, i need an equation to give me different areas for different rotational degree.

I found an equation that says : flow through valve (Q) = Cd * Av * sqrt (2*g*H) ,,, if that H is the headloss across the valve and my valve is fully open the equation will give me flow = 0 ,cause H across valve = 0 ( fully open ,no headloss ) is that correct ??

Plz, if anyone knows interent sites that talks about this plz mention it

thanks alot

## RE: Cv Cd & effective valve area

Butterfly valve opening area will differ depending on the type of disc construction and type of butterfly valve. Doubtful you will find an equation; check with manufacturers that make brand-name Centerline or Keystone (Tyco). Those flow curves for volume vs %opening are based on emperical data, or perhaps modeling software.

Also, you are not correct in assuming headloss of zero. If the valve is fully open your headloss is you upstream gauge pressure. Re-think it.

## RE: Cv Cd & effective valve area

## RE: Cv Cd & effective valve area

Kevin and you are saying the same thing. Kevin is saying, if I understand him right, that for 0 psig downstream pressure, the upstream or inlet pressure to the valve is simply the valve's head loss. Which is what you said.

Back to your original post, why do you want/care about the Cd for the valve? Cd is not a common term to use for a valve's capacity, your formula looks more like what I've seen for flow through an orifice where you have a discharge coefficient.

If you have the Cv curve for your valve, then you can just directly calculate the flow using Q = Cv (dP/SG)^0.5 where Q is the flow in gpm, Cv is in this case off your curve, dP is the pressure drop across the valve, psi and SG is the specific gravity which = 1 for water.

## RE: Cv Cd & effective valve area

i thought it is the losses that is caused by the valve when fluid flows through it ,so if the valve is fully open there would not be any restriction on the flow from the valve ,as if the valve is not there ,,if we started closing the valve the head loss will start to rise and it will reach it's maximum when the valve fully closed.

is that completly wrong ??

what is the difference between the head loss and the pressure drop across the valve ,,i think i'm mixing the two.

i'm using EPAnet software (water modelling ) if i modelled an open valve it will give me a 0 headloss reading , it will give me the maximum headloss value if the valve is closed . plz exaplain the contradiction ? frankly i'm still convinced that the headloss is 0 for an open valve cause the valve controls the flow by imposing headloss ,if it's not controlling the flow then it can't be imposing any headloss ????

the only reason i can think of is that you are talking about somthing differnet than what i and EPAnet is calculating ? maybee ?!!

## RE: Cv Cd & effective valve area

Head loss normally does increase as you close a valve but it's really a function of the piping system.

If you had a piping system from a constant pressure source going to atmosphere through a 'short' length of piping (so piping pressure drop is negligble), the head loss by the valve would be constant as you opened/closed it. But, since you are changing the Cv as you open/close it, the flow will still correspondingly change.

I think your program is basically defaulting to a zero value. The head loss/pressure drop through a fully opened butterfly valve or gate valve or other on/off valve will be essentially zero (assusming it's not a 4" butterfly in a 10" line of course ) so it's not a bad default. You can do a quick check by seeing what the dP will be for your wide open flow using the equation Q = Cv (dP/SG)^2 I posted earlier.

## RE: Cv Cd & effective valve area

I fully agree with you. The flow through the system, if the valve is fully open, is only determined by the pressure drop across the other components in the system (most notably the piping). What waseem needs is a curve of Cv as a function of valve position, so kevinf's suggestion should help him.

There is a small typo in your last post, which could add to confusion: the equation is Q = Cv (dP/SG)^0.5 such as posted earlier.

Regards,

Joerd

## RE: Cv Cd & effective valve area

## RE: Cv Cd & effective valve area

I often covert between Cv and Cd. Here's how to do it. I have included a conversion factor in case you are working in gpm and psi.

First, the governing equations for volumetric flow, Q, in gpm:

Q = Cd*A*sqrt(2*dP/rho)*F

where,

Cd is the discharge coefficient

A is the flow area

sqrt is the mathematical symbol for square root

dP is the pressure drop

rho is the fluid density

F is the conversion factor to get gpm assuming the pressure is in psi, area in ft^2 and density in lb/ft^3; note that F is outside of the sqrt. F = sqrt(32.2*144)*7.48*60 = 30561.

Given a valve flow coefficient, Cv, the volumetric flow rate is,

Q = Cv*sqrt(dP)

Q is in gpm

dP is in psi

Cv is in gpm/sqrt(psi)

Setting the two equations equal to each other and solving for Cd yields,

Cd = Cv*sqrt(rho/2)/F/A

If using gpm and psi, then F = 30561.

If SI units are used, for which P is in Pa, flow rate is m^3/s and Cv is (m^3/s)/sqrt(Pa), the conversion factor is F=1.