×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Are you an
Engineering professional?
Join Eng-Tips Forums!
• Talk With Other Members
• Be Notified Of Responses
• Keyword Search
Favorite Forums
• Automated Signatures
• Best Of All, It's Free!

*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.

# waseem19: I almost forgot. The

## waseem19: I almost forgot. The

(OP)
waseem19:

I almost forgot.  The effective flow area is Cd*A, where A is the actual, full open flow area.
Replies continue below

### RE: waseem19: I almost forgot. The

tremlo ,thanks for your help ,, but r u sure about this equation ? , the reason i asked my original q was to get a value of (tau),(tue) i'm not sure how is it written ,,but the equation i have for it is
tau= ((Cdi)(Ai))/((Cdfo)*(Afo)) , varies from 1( fully open) to 0 fully closed)
Cdi : Cd at a certain valve area ,
Ai  : Area of valve at certain time,
Cdfo : Cd for fully open valve
Afo : area for fully open valve ,

so Ai is the effective area at a cetain time ,and it is independent of Cdi . there should be some tables that gives effective area Vs rotation angle for disk, or disk travel, or whatever valve motion is. Cd as i understand it is to tell how much of the head is converted into velocity ,so if all your dp is converted into velocity Cd=1 ,but this is an ideal situation,and cd values are always less than 1.

plz correct me if i'm wrong , i can't be 100% sure of what i'v said untill i understand everything else about valve theory 100% and i'm a bit far at the moment.

### RE: waseem19: I almost forgot. The

(OP)
waseem,

What you said about Cd seems correct.  As the valve is closing, the flow area is reduced and the head loss increases.

Often times, you know neither the Ai or the Cdi, but you do know the product Cdi*Ai.  So, we usually hold the area constant at its full open value then simply change the value of Cd (according to the correlation from my previous post) to represent both the decrease in area and the increased head loss.  Then, the flow can be calcualted as,

Q = Cd*Afo*sqrt(2P/rho)

In this equation, you can say Cd*Afo = Cdi*Afi

I usually refer to the product Cd*Afo as the effective flow area.

In any case, I typically use the product Cd*Afo and am not interesed in the Cd or Afo by themselves.  Is it important in your application to know each individual quantity or is it sufficient to know the product only?

Going back to your equation for Tau:

Tau = (Cdi*Ai/(Cdfo*Afo)

substitute Cd*Afo for Cdi*Ai and you get,

Tau = Cd*Afo/(Cdfo*Afo) = Cd/Cdf0

As defined in my previous post, Cd varies linearly as a fucntion of Cv.  This implies that Tau can also be written in terms of the Cv:

Tau = Cvi/Cvfo

I hope this is helping and not making things more confusing.  Please write back and let me know.

Tremolo.

#### 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.

Close Box

# Join Eng-Tips® Today!

Join your peers on the Internet's largest technical engineering professional community.
It's easy to join and it's free.

Here's Why Members Love Eng-Tips Forums:

• Talk To Other Members
• Notification Of Responses To Questions
• Favorite Forums One Click Access
• Keyword Search Of All Posts, And More...

Register now while it's still free!