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nightcrawler
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Comprehending Resistance Coefficient K If 1 & 0

04/02/2020 12:40 PM

This seems a fairly trivial question but I am dealing with an application help manual where 'K' for a shut check valve is input as 0 & K for a frictionless control valve (or a fitting) is taken as 1

Because K values are largely standardized e.g http://www.metropumps.com/ResourcesFrictionLossData.pdf

what do you comprehend if this coefficient for any given fitting is 1 or 0?

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SolarEagle
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#1

Re: Comprehending Resistance Coefficient K If 1 & 0

04/02/2020 2:11 PM

Well to me it seems like you are deferring the calculation until you have the relevant field data...

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nightcrawler
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#2
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Re: Comprehending Resistance Coefficient K If 1 & 0

04/02/2020 2:46 PM

No, not like that. It is just that I am conversant with such calculations but its particular use as 1 or 0 may be only relates to this particular application.

What do you think if I have K=0 then fitting is frictionless?

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SolarEagle
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#3
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Re: Comprehending Resistance Coefficient K If 1 & 0

04/02/2020 3:06 PM

No, it means to me anyway, that the calculation is deferred or irrelevant...

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Tornado
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#4

Re: Comprehending Resistance Coefficient K If 1 & 0

04/02/2020 3:29 PM

Scanning through the table(s), it appears that low K values are low friction, and high K values are greater friction, with values even greater than 1.

From that, it appears that K = 0 would be no friction, K of infinity would be tightly closed. K = 1 seems like some default fitting chosen as the basis for comparison.

That is the most comprehensive such table I've seen. Thanks for sharing it.

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nightcrawler
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#5
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Re: Comprehending Resistance Coefficient K If 1 & 0

04/02/2020 7:39 PM

Thanks for helping out

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PWSlack
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#6

Re: Comprehending Resistance Coefficient K If 1 & 0

04/03/2020 6:10 AM

Of course the K-value for a shut valve is <...0...>: as v is also 0, the head loss due to flow is also 0, because there is no flow through a closed valve.

Next!

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#7

Re: Comprehending Resistance Coefficient K If 1 & 0

04/03/2020 7:30 AM

Difficult to say without knowing how the manual is written. Is it possible K is entered as 1 to indicate there is that type of element, but still need to enter the resistance coefficient? For a closed check valve doing its job I'd expect loss factor = ∞.

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PWSlack
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#8
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Re: Comprehending Resistance Coefficient K If 1 & 0

04/03/2020 7:52 AM

<... loss factor = ∞...> multiplied by [V = 0, squared], divided by 2g is, er....tap, tap, tap, press equals button....zero! So it doesn't really matter what K equals for a closed valve, as the head loss due to flow is zero, as there is no flow.

The original poster was only after what the interpretation is.

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#9
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Re: Comprehending Resistance Coefficient K If 1 & 0

04/03/2020 8:35 AM

No, that's wrong, 0*∞ is not (necessarily) zero, however much you tap, tap. It's indeterminate and can be anything from 0 to ∞. This was discussed on CR4 a few years ago.

I was trying to put things in context, obviously (to some people) you wouldn't calculate the headloss across a closed NRV.

If you want to help the original poster, what do you think the interpretation is?

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Jpfalt
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#10

Re: Comprehending Resistance Coefficient K If 1 & 0

04/03/2020 10:05 AM

Is this perhaps a fluidics issue where the fluid circuit is being used as a binary computer?

At one time Parker was pushing fluidics for control circuits and NASA even considered using fluidics for spacecraft logical controls. They had a large catalog of fluidics plug in modules with AND, OR, NAND and NOR modules as well as other combinations, but lost out to electronics logic control due to size considerations though the technology was promising.
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Rixter
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#11

Re: Comprehending Resistance Coefficient K If 1 & 0

04/03/2020 5:11 PM

The velocity head h is a measure of the kinetic energy of flowing water:

The dimension is length and represents the distance the water could climb if directed upward against the force of gravity.

https://en.wikipedia.org/wiki/Hydraulic_head

K is a (dimensionless) measure of the loss of head of water flowing through a valve, pipe elbow, etc..

Fitting Head Loss = Kv² / 2g
where:
K= K factor associated with a specific type and size of fitting
v = fluid velocity (m/s or ft/s)
g = 9.806 m/s² or 32.174 ft/s²

hf = (1/2)Kv2/g

where g = 32.174 ft/s2 or 9.806 m/s2 and hf is "head" in feet or meters. (Pressure drop is related to "head loss" by factor g.)

https://www.pipeflow.com/pipe-flow-wizard-software/pipe-flow-wizard-fittings-database

what do you comprehend if this coefficient for any given fitting is 1 or 0?

K in a hydraulic circuit is analogous to resistance in an electrical circuit. Fluid velocity squared is analogous to current and pressure drop to voltage.

For a shut valve (0) is analogous to infinite resistance or an open circuit. K is infinite.

For a frictionless open valve, the value of K is zero.

********

Your K sounds more like fluidic logic.
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Codemaster (2); Jpfalt (1); nightcrawler (2); PWSlack (2); Rixter (1); SolarEagle (2); Tornado (1)

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