×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Log In

Come Join Us!

Are you an
Engineering professional?
Join Eng-Tips Forums!
  • Talk With Other Members
  • Be Notified Of Responses
    To Your Posts
  • Keyword Search
  • One-Click Access To Your
    Favorite Forums
  • Automated Signatures
    On Your Posts
  • Best Of All, It's Free!
  • Students Click Here

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

Students Click Here

Jobs

Flow and Loss Coefficients
3

Flow and Loss Coefficients

Flow and Loss Coefficients

(OP)
Can someone explain the difference between Cv and Cd?  Also, is there an equation to convert between Cd and Cv?  I have a reference to this equation for finding the Cv of a valve:

Cv ~ 38*A*Cd

A=cross-sectional area

Cd was assumed to be 0.6.  I do not know where this equation comes from or if Cd=0.6 is reasonable.  Is anyone else familiar with this?

Any help is much appreciated.

RE: Flow and Loss Coefficients

I haven't seen this link before between the two, you would have to get some information on various valves and compare them.  What does the A apply to?  I would assume it's the port area.

Cd is the discharge coefficient for an orifice and is essentially a factor for how close your orifice/nozzle comes to a perfectly smooth nozzle.

Cv for a valve is an indication of its capacity.  For water, a Cv of 1 means it will flow 1 gal/min for a dp of 1 psi (Q = Cv (dP/SG)^.5)

RE: Flow and Loss Coefficients

TreeEng,

This topic has come up for discussion previoulsy.  See Thread408-46925


In that thread, I responded as follows:

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.


TreeEng,

To answer your basic question, Cv and Cd both express the hydraulic resistance of a pipeline component such as a valve, orifice, or nozzle.  The Cv is typically used for valves and is avaiable from manufacturers.  As pointed out by TD2K, Cd is typically used to indicate variations from the ideal nozzle or orifice.

As for your equation, Cv ~ 38*A*Cd, I do not think that is valid.  I cannot get there from the equations I laid out above.

RE: Flow and Loss Coefficients

All depends on the units - cv is unit dependent and is usually in FPS units, whereas cd is non-dimensional.

Cv = k x A x cd

Where k is a constant dependent on the units used.

my quick calc for water in SI unts gives me

Cv = 25909 x A x cd

where Cv = 1.156 x Q x (G/dp)^0.5

Q = m3/hour
G = specific gravity water
dp = presure drop in bar.

But back to the question. cd = 0.6 is reasonable for an orifice where the orifice diameter is less than about 55% of the pipe diameter. It is not reasonable for a valve. For a valve Cd (fully open) will depend on the type of valve.  
 



RE: Flow and Loss Coefficients

TreeEng

Regarding the equations that you posted:

Cv ~ 38*A*Cd

A=cross-sectional area

please see Crane Technical Publication 410.

In Crane, you will find that Cv = 29.9 d^2 /sqrt(K), where
(pi)d^2/4 = A, i.e. the effective cross-sectional area.

Therefore Cv/A = 38/sqrt(K).

Hence, 1/sqrt(K) = Cd, where
K is the loss coefficient of the physical cross-sectional area.

Substituting and solving, you will get

Cv ~ 38*A*Cd

A=cross-sectional area

I hope this helps.

4carats

Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Reply To This Thread

Posting in the Eng-Tips forums is a member-only feature.

Click Here to join Eng-Tips and talk with other members!


Resources