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Flow Coefficient for fluid other than water 2

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jfaucher

jfaucher

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May 18, 2004
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I'm currently doing pump sizing calculation, so I have to determine the total pressure loss of my piping system. Most of commercial components like ball valves, check valves, strainers, etc. have experimental data from suppliers indicating the flow coefficient (Cv) for water at 60degF. If I want to calculate the resulting pressure loss, I get it from following formula:

deltaP (psi) = GPM^2 * Specific Gravity / Cv^2

My problem is that the fluid in my system is oil. How to evaluate the Cv factor for other fluid than water? I assume it should depend at least on fluid viscosity?

If it is not possible to determine the Cv factor for other fluids without practical experimention, how to determine the pressure loss through the valves and other components?

Thanks for your help!

Jean-Pierre Faucher, ing.
 
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Thats why the SG term is there, indicating that this type of head loss is dependent more on mass * velocity change rather than fluid friction * length (viscosity).

**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)
 
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Viscosity does have an effect on flow. However, for fully turbulent flow, the effect is very minor. Check out the the Moody friction factor chart. For high Reynolds numbers, the friction factor is essentially a flat line. In these cases, only the fluid density is needed to calculate the pressure drop given the resistance (k), or the equivalent feet.

If the flow is not turbulent, all bets are off. Even your equation above may not work well for water.

--Mike--
 
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Thanks for your response! Know I have a better understanding of my problem... My system is laminar flow, since I'm pumping about 6 GPM of engine oil through 3/4 pipes. Am I correct if I guess a loss coefficient factor "K" for each component, in order to evaluate the pressure loss following this equation:

deltaP = K * rho * V^2 * 1/2

Jean-Pierre Faucher, ing.
 
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I wouldn't exactly call turbulent viscous friction effects minor. Overcoming its effects represents a rather large portion of total energy consumed in the world.

The Cv equation is valid for laminar or turbulent flow because it depends only on change in velocity, not on friction factor. Viscosity is shearing force per unit length, hence you have to integrate over distance to get force and pressure drop. A valve is far too short a length for friction to make any appreciable difference.



**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)
 
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That means if you have high viscosity oil with a specific gravity of 1.0, you will have the exact same pressure lost than if you have the same flow of clear water going through a gate valve, for instance?

Jean-Pierre Faucher, ing.
 
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That's what that equation assumes.

But even if you decide to use a very detailed procedure for sizing control valves, you will still find there is no correction for viscosity, although it is mentioned in other parts of the text that viscous or corrosive fluid applications might fall outside the normal application.


See. Nothing.

I would suggest that, if you have a very viscous fluid and you are simply not going to be happy with a Cv pressure drop procedure that ignores viscosity, then use an equivalent head loss method for fittings and valves, where you calculate the friction loss for 1 foot of pipe and multiply by 10 or 20 or some other "equivalent length factor". You can usually find those equivalent lengths for valves that are open, or "paritally closed", etc. If you need to apply viscous head loss to a particular valve position corresponding to a known Cv and include viscos effects, then do something like ratio the calculated value from the Cv equation to the Cv of the valve at the "full open" position and multiply that by whatever the "equivalent length" is for the valve at full open.

**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)
 
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if you are concerned about laminar flow IEC 534-2 (ISA S75 1) gives a correction term fr which takes in account the viscosity of fluids (laminar flow), with that term the resulting formula would be

deltaP (psi) = GPM^2 * Specific Gravity / (fr * Cv) ^2

I guess the software from Masoneilan, Fisher or other manufacturers include that correction, I can provide details, if required,
Paolo

there is also a free software here, contact Prode to get the full version,
 
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I think that's worth including in Fisher's handbook.
In fact I'm quite disillusioned that its not.

**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)
 
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One more question:

In most fluid mechanics handbook, they explain how to calculate minor loss from the loss coefficient factor "K", going like this:

deltaP = K * rho * V^2 * 1/2

However, as for the Cv factor that we have seen previously, there is no correction here for viscosity. How to evaluate the minor loss (valve, elbow, reducer, etc.) for a viscous laminar flow? Should we apply the same correction than for the Cv factor (as a divider instead of multiplicator)?

Jean-Pierre Faucher, ing.
 
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jfaucher,

This is a good question. As valves represent a minor loss we should extend the considerations above to other items which represent minor losses too.

The minor loss can be converted to a length equivalent to the length of a pipe that would produce the same pressure loss.
Head loss can be expressed as:
dp = ? (leq / de) (v2/ 2 g)
where:

dp = pressure loss
? = friction coefficient
leq = equivalent pipe length
de = equivalent diameter
v = fluid velocity
g = acceleration of gravity

As the friction coefficient ? depends both on the roughness of the pipe and on whether the flow is laminar, transient or turbulent, it turns that the friction coefficient is related to the Reynolds number and so to the fluid viscosity.

Unfortunately I am not aware about graphs, similar to those I’ve mentioned earlier, which could be applied to elbows, reducers or other minor loss.
 
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jfaucher and ione, there are factors for other fittings. The best known are the 2K and 3K methods by Hooper and Darby respectively. A search of Eng-Tips will reveal many discussions about them.

In texts that discuss Cv's for valves it is never mentioned that the Cv varies with Reynolds number (which in turn is influenced by viscosity etc). As BigInch has pointed out, even the very highly regarded Fisher Handbook does not cover it.

If we use the formula 3-16 from Crane TP410 to convert between Cv and K, and we use Darby's 3K method to estimate K as a function of Reynolds numbers we get a graph very similar to that referred to above by ione. In K value terms, a typical 3" globe valve would have a value of about 6 in fully turbulent flow, but if the Reynolds number decreased to 100 (laminar flow) the K value jumps to 20. In Cv terms, it would decrease from about 110 down to 60.

Katmar Software
Engineering & Risk Analysis Software
 
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there are several empirical methods for estimating pressure drop in fittings with laminar flow but unfortunately I am not aware of correlations validated by extensive tests as for control (and relief) valves, I would add that the equivaled lenght methodology doesn't seem the best with laminar (and critical) flows.
As said in my previous post IEC 534-2 / ISA S75-1 provides a way for calculating a "pseudo" Reynalds number based on CV (and estimated port diameter), very few software tools show the calculated Reynolds number so I mentioned a free tool which shows Re and Fr (which could be of interest).
 
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It would be nice to see how those programs are handling this.

**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)
 
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Katmar,

I’ve followed your suggestion and taken a glance to other threads discussing this matter: you definitely are a stars’ collector in this topic and your posts are very authoritative.

jfaucher,

I suggest you to get a copy of “Chemical engineering fluid mechanics” (Rob Darby). There you will find an explanation of the “3-K” method, which shows a correlation between Reynolds number and a coefficient Kf which enters the formula for minor loss (as already stated by Katmar).

dp = Kf *(v^2/2*g)

Kf is expressed as a function of Reynolds number and three other K values (the “3-K” method)

Kf = f (Re, Kl, Ki, Kd)


In the book abovementioned you will also find a set of values for Kl, Ki and Kd for elbows, tees etc.

Regards
 
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Instrument Engineers' Handbook, Fourth Edition, Volume Two- Process Control and Optimization (by Bela G. Liptak) in chapter 6.15 explains an excellent procedure for sizing control valve in laminar regime, with a graph for Reynolds number/correction factor applicable for several types of valves, and in my opinion more precise than those reported in the link BigInch has previously posted.
 
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Fortunately I very rarely see anything operating in the laminar flow region ... well, not for very long anyway.

**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)
 
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