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How Does A Valve Work?

valve pressure pressure drop

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#1 oyakobi

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Posted 09 September 2013 - 07:55 PM

Hi,

 

I am a little confused about how a valve works exactly. 

 

I am considering a simple scenario of water flowing through a pipe. From the principle of continuity we know that the in any steady state process, the rate at which mass enters a system is equal to the rate at which mass leaves the system. Meaning, the flow rate is equal everywhere throughout the pipe.

 

For example, If we have a certain flow through a pipe that converges into a smaller pipe the flow rate will

 

remain constant since the velocity will compensate for the reduction in pipe diameter.

 

My question is, why is a valve any different since essentially all it is doing is changing the area in which the

 

flow is passing through. I am sure the answer is simple but I just need to wrap my mind around this.

 

Thanks in advance,

 

Yakobi



#2 fallah

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Posted 10 September 2013 - 12:44 AM

Hi,

 

It was better the question submitted in another forum e.g. in student Forum...

 

However a valve will reduce the flow by pressure loss due to local reduction in pipe cross section area, but the continuity will still remain for each passing flowrate...



#3 oyakobi

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Posted 10 September 2013 - 01:22 PM

Hi Mr. Fallah,

 

You are right I should have posted it in the student forum.

 

That kind of answers my question. However, if you have pump pumping a flow of 2m3/s passing through a 2'' pipe and the pipe converges into a 1'' pipe, the flow rate will remain the same and there will be a pressure loss due to reduction in pipe cross section area. If we have that same pump pumping 2m3/s and we make the flow pass through a valve, the valve can and will cause the flow rate to decrease. This is where I am confused, what does a valve do that's different?

 

Thanks again,

 

O.Yakobi



#4 demank

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Posted 11 September 2013 - 04:36 AM

Dear Oyakobi,

 

To understand what you curious is by reading a pump curve.

For One Pump with 2 m3/s and valve fully open.

Then, when you throttle the valve, the pump flowrate will decrease.



#5 JosvantWestende

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Posted 12 September 2013 - 02:52 AM   Best Answer

With a valve the flow velocity increases because the valve body decreases the flow through area. Downstream of the valve, the flow velocity decelerates again creating extra turbulence. This extra turbulence is the cause of the pressure drop of the valve, and can be regulated by changing the flow velocity under the valve body (ie by changing the blockage by the valve). In this way, a valve can be used for regulating flow rates (ofcourse some valves are better for this than others).

 

Note: in a diffuser the flow velocity is also increased at the throat, a deceleration of the flow. However, due to the design of diffusers, the deceleration is much more gentle, resulting in a smaller creation of extra turbulence. Therefore, the pressure loss is much smaller.

 

In other words: a valve creates extra turbulence which results in an effective pressure loss.



#6 oyakobi

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Posted 12 September 2013 - 12:42 PM

Thank you all for the answers. Picture is clear now. (special thanks to JosvantWestende)



#7 adewumi_bj

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Posted 13 September 2013 - 04:19 PM

Now I'm confused about this: if the valve element causes pressure drop in flow based on valve coeficient and the square of flow then where does the pressure dropped go? I have nearly broken my closet tap because what I'm observing is actually disobeying bernoulli and mass continuity equations. Somebody help.

#8 ChemEng01

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Posted 14 September 2013 - 12:50 AM

adewumi

 

The energy loss from pressure drop through a valve or restriction is mainly in the form of temperature and noise. 

 

Generally your fluid will drop in temperature across a valve (Joule thomson effect). A few fluids increase in temperature such as helium. 

 

Never the less. Bernoullis eqn assumes conservation of energy. 

 

In reality there will be energy loss through temperature and noise. 

 

hope this helps



#9 JosvantWestende

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Posted 14 September 2013 - 06:14 AM

Indeed, the loss in pressure is actually a 'loss' in energy. Ofcourse energy cannot truly be lost, but will be converted to another form of energy (turbulence -> heat (via the turbulence cascade), vibration/noise).

How the medium reacts on this in terms of temperature, depends on the type of medium and its phase (gas, liquid, 2phase). For this you need a phase diagram. Gasses in general have strong temperature effect (expansion, joule thomson), whereas liquids do not. For a2phase fluid, the quality of the medium can also change.

Edited by JosvantWestende, 15 September 2013 - 04:05 AM.


#10 adewumi_bj

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Posted 20 September 2013 - 10:20 PM

Thanks to you josevand and jruud. Very clear abt it now




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