3 replies to this topic

[villivord]

Hello everyone,

 

I'm a bit confused about liquid flowing through a control valve. Suppose for example a centrifugal pump sending a certain flow through a pipeline in which a control valve is present. I perfectly understand how this valve reduces flow as it creates an increased pressure drop (by going from for example from 90% opening to 30% opening) which increases the upstream pressure (suppose downstream pressure constant) on which the pump, according to it's pump curve, provides a lower flow rate. 

 

As I am now studying choked flow in liquid CV applications, I'm a bit confused why the flow rate increases as the pressure drop increases according to this figure:

(http://kb.eng-softwa...99054000&api=v2)

 

Does this figure just shows how the flow varies through the valve at a certain fixed position? Do I need to interpret this as the flow rate increasing as the pressure drop increases because the valve opening becomes smaller and smaller (for higher pressure drop) and so the velocity becomes bigger and bigger at the vena contracta (Bernoulli) until a maximum is reached (choked flow)? I don't see the link between my two different insights. What am I missing in understanding it properly? 

 

Thanks in advance. 

[rychurek]

Borpe
If you would like to manipulate Volumetric flow you have to change control valve opening (CV - flow coefficient = flow area) or/and driving force = pressure drop. Remeber that in flow rate equation volumetric flow is proportional to valve opening (assume linear characteristic) and square root of pressure drop). So to decrease flow on pump discharge side you decrease valve opening but pressure drop doesn't increase so much to keep flow constant. You are right that velocity increase, and depend of valve opening and pressure drop you can reach flashing point, where further pressure drop increasing don't affect flow rate.

[Francisco Angel]

Dear Borpe, there are a few concepts intertwined here:
First, and as also mentioned by rychurek, the valve equation is:
Q=Cv*x* (DP/s)^0.5

 

Cv: Valve coefficient.

x: Valve aperture.

DP: Pressure drop through valve.

s: Specific gravity.

 

So the graph you provided is correct, as pressure drop at the valve increases, keeping valve aperture constant, the flow through the valve increases.

 

 

 

Does this figure just shows how the flow varies through the valve at a certain fixed position? Do I need to interpret this as the flow rate increasing as the pressure drop increases because the valve opening becomes smaller and smaller (for higher pressure drop) and so the velocity becomes bigger and bigger at the vena contracta (Bernoulli) until a maximum is reached (choked flow)? I don't see the link between my two different insights. What am I missing in understanding it properly? 

 

Thanks in advance. 

 

Yes the figure correspond to fixed valve position. Another fact is that at greater pressure drop the pump is capable of delivering a reduced flow. So what happens when the valve aperture is reduced?, pressure drop at the valve increases, flow decreases (we know it from the pump), flow at the valve decreases overall because the effect of reduced aperture surpass the effect of increased pressure drop.

How can you get something similar to the graph at operation?, if you want to increase both flow and pressure drop at the valve at the same time, you need to get another pump, because simultaneously increasing pressure drop and flow can't be achieved with the model pump we use regularly

Best regards.

Edited by Francisco Angel, 24 January 2017 - 11:04 AM.

[Francisco Angel]

Yes, the graph corresponds to a valve "in isolation".
In your case, there is no way to make the pump delivers more flow at increased pressure drop (we assume a pump curve in which the flow delivered decreases with increased pressure drop).
You increase pressure drop at the valve by decreasing valve aperture, the flow, calculated using the valve equation, also decreases.