Discharge Coefficient of Valve equation

AI Thread Summary
The discussion centers on the Discharge Coefficient of Valve equation from the book "Advanced Water Distribution Modeling and Management." A participant argues that the term V^2 in the discharge equation is unnecessary and suggests a revised equation for Cv. Others point out that the two equations involve different contexts and velocities, indicating that the equations may not be in direct conflict. The distinction between the velocities in the equations is emphasized, suggesting that one represents flow velocity while the other pertains to the descent rate in a reservoir. The conversation highlights the importance of understanding the physical setup and the meaning of each variable in the equations.
Mikealvarado100
Messages
52
Reaction score
0
I have a problem with Discharge Coefficient of Valve equation which is described in book 'advanced water distribution modeling and management'
let me say:

I think the expression V^2 in Discharge Equation (P. 618 of the book mentioned) is extra and the equation must be changed to:
I) Cv=V/[(2gh)^.5]
because other equation for Cv is:
II) Q=Cv (2gh)^0.5 3.14D^2/4
and these two equations are equal.
Additionally the equation P. 402 (equation 9.3) confirms my thought.
What is your idea?
thanx
 
Engineering news on Phys.org
So as not to restrict potential respondents to those who posssess a copy of the book, I suggest you quote the whole of the equation in question and explain the context in which each equation applies.
One possibility that occurs to me is that the v is a change in velocity of the flow before it descends height h to the valve. This might apply to the context of one equation but not the other.
 
haruspex
One equation says that Cv=V/[(2gh+V^2)^.5].
According to another equation: Q=Cv (2gh)^0.5*3.14*D^2/4
These two equations must be equal. Therefore insert Cv from second equation in fist equation. Then you will see the sentence V^2 in first equation is extra. Also you can compare it with another equation: Q=Cf Cv D^2 (P)^0.5. which Cf=constant.
In both of them you can see that V^2 is extra in first equation. Is not it?
Thanx
 
Mikealvarado100 said:
haruspex
One equation says that Cv=V/[(2gh+V^2)^.5].
According to another equation: Q=Cv (2gh)^0.5*3.14*D^2/4
These two equations must be equal. Therefore insert Cv from second equation in fist equation. Then you will see the sentence V^2 in first equation is extra. Also you can compare it with another equation: Q=Cf Cv D^2 (P)^0.5. which Cf=constant.
In both of them you can see that V^2 is extra in first equation. Is not it?
Thanx
As I posted, you need to explain the context: the physical set-up and what each variable means within that set-up.
It is not immediately clear that the two equations are in conflict. One of them involves Q (a volumetric flow rate) and a cross sectional area, the other does not.
Are you quite sure the two Vs in the first equation above are the same? It seems to me they should represent two different velocities.
Can you post an image of the text and any diagrams?
 
Hi
Cv is Discharge Coefficient of a valve which defined the relation between Q and H.
Have a look at image attached. have a look at below page too:
http://www.valvias.com/discharge-coefficient.php
No. Both V are velocity of flow.
Insert Cv from second equation into fist equation and make it simple. It is very easy to do. Then you can see V^2 is extra.
Another equation for Cv is Q=Cf Cv D^2 (P)^0.5. which Cf is constant. You can use this equation too be sure of what I believe.

http://Hi Cv is Discharge Coefficient of a valve which defined the relation between Q and H. Have a look at image attached. have a look at below page too: http://www.valvias.com/discharge-coefficient.php[/PLAIN] http://s6.uplod.ir/i/00777/ofsxrg0ftfbe.jpg
 
Last edited by a moderator:
Mikealvarado100 said:
Hi
Cv is Discharge Coefficient of a valve which defined the relation between Q and H.
Have a look at image attached. have a look at below page too:
http://www.valvias.com/discharge-coefficient.php
No. Both V are velocity of flow.
Insert Cv from second equation into fist equation and make it simple. It is very easy to do. Then you can see V^2 is extra.
Another equation for Cv is Q=Cf Cv D^2 (P)^0.5. which Cf is constant. You can use this equation too be sure of what I believe.

http://Hi Cv is Discharge Coefficient of a valve which defined the relation between Q and H. Have a look at image attached. have a look at below page too: http://www.valvias.com/discharge-coefficient.php[/PLAIN] http://s6.uplod.ir/i/00777/ofsxrg0ftfbe.jpg
The equation without the V2 is an approximation that is valid where the cross sectional area of the reservoir is much greater than that of the valve. In this case, we can ignore the velocity with which the level descends in the reservoir.
The equation with the V2 is correct if that V is taken as referring to the rate of descent of the level in the reservoir. It looks to me as though the distinction between the two velocities has been lost somewhere in copying from an original source.
 
Last edited by a moderator:
I need some assistance with calculating hp requirements for moving a load. - The 4000lb load is resting on ball bearing rails so friction is effectively zero and will be covered by my added power contingencies. Load: 4000lbs Distance to travel: 10 meters. Time to Travel: 7.5 seconds Need to accelerate the load from a stop to a nominal speed then decelerate coming to a stop. My power delivery method will be a gearmotor driving a gear rack. - I suspect the pinion gear to be about 3-4in in...
Thread 'Calculate minimum RPM to self-balance a CMG on two legs'
Here is a photo of a rough drawing of my apparatus that I have built many times and works. I would like to have a formula to give me the RPM necessary for the gyroscope to balance itself on the two legs (screws). I asked Claude to give me a formula and it gave me the following: Let me calculate the required RPM foreffective stabilization. I'll use the principles of gyroscopicprecession and the moment of inertia. First, let's calculate the keyparameters: 1. Moment of inertia of...
Thread 'Turbocharging carbureted petrol 2 stroke engines'
Hi everyone, online I ve seen some images about 2 stroke carbureted turbo (motorcycle derivation engine). Now.. In the past in this forum some members spoke about turbocharging 2 stroke but not in sufficient detail. The intake and the exhaust are open at the same time and there are no valves like a 4 stroke. But if you search online you can find carbureted 2stroke turbo sled or the Am6 turbo. The question is: Is really possible turbocharge a 2 stroke carburated(NOT EFI)petrol engine and...

Similar threads

Back
Top