Valve flow-pressure relationship, a theoretical question

[DavidButler]

I created a spreadsheet to calculate hydronic loop pressure drops. To make the spreadsheet more functional, I regressed several AquaPex friction tables for the velocity range of interest. AquaPex tables are unusual in that they account for water temperature, providing a higher degree of accuracy in pressure-sensitive designs. What I discovered is that the curves regress nicely to a first order exponential equation of the form:

H[sub]L[/sub] = R * Q[sup]1.75[/sup]

where Q is flow rate (gpm) and R is hydraulic resistance.

For the sizes I checked, the exponent was always very close to 1.75 (+/-.01). This should not be surprising since this is exactly the relationship Siegenthaler describes in various treatments on head loss in hydronic circuits (e.g., Idtronics issue #16, Section 4). Siegenthaler's resistance formula is a special case based on D'Arcy-Weisbach (turbulent flow in relatively smooth pipes). But D-W has a velocity exponent of 2, not 1.75. So I asked John about the derivation of the 1.75 exponent. He referred me to the formula for what he refers to as the Fluid Properties Factor α:

α = (D/u)[sup]-0.25[/sup]

where D is density (lbm/ft[sup]3[/sup]) and u is dynamic viscosity (lbm/ft-sec). The Fluid Properties Factor α is essentially a stand-in for the more complex Friction Factor (f) in D'Arcy-Weisbach. In effect, the -0.25 exponent pulls the pipe pressure-flow exponent from 2 down to 1.75.

If we know a pipe's roughness, we can calculate the D'Arcy Friction Factor using one of the Colebrook equations, and then use D-W to calculate the pressure drop or head loss for a given length of pipe. As it turns out, this still yields a curve with an exponent of 1.75 (at least for relatively smooth pipes such as pex).

When I graph the pressure drop for a hydronic loop (pipes, fittings, valves), the curve regresses to an exponent greater than 1.75, typically in the 1.80's. The reason for this is obvious when you consider that the formula for calculating a valve's pressure drop has an exponent of 2 (ΔP = Q[sup]2[/sup]/C[sub]v[/sub]). The pressure-flow exponent for the overall loop thus depends on the valve's relative contribution to the loop's pressure drop.

This exercise led me to question why the pressure-flow exponent for valves should be 2. We know that the C[sub]v[/sub] is empirically determined by the manufacturer, but we then use C[sub]v[/sub] as an anchor point to extrapolate the ΔP at different flow rates using an exponent of 2. Essentially what that says is that valves are not subject to the same fluid properties as a pipe, which doesn't seem correct. Yet every reference I've seen on a valve's pressure-flow relationship uses an exponent of 2 (or 1/2, depending on which variable you're solving for).

Would anyone care to explain or defend why a valve's pressure-flow relationship should have an exponent of 2?

 

[SPDL310]

I would argue that the special case rules for the 1.75 exponentiation that you describe above are not met by a valve. The interiors of most valves are not straight or smooth. There are geometry changes across seals both entering and leaving the valve as well as varying degrees of flow directional change (I.E. angle valve and globe valves). Valve surfaces can also be rougher than extruded pipe due to machining, forging or casting, therefore fluids flowing through a valve must be described by classical D-W behavior.

 

[DavidButler]

Thanks for your reply. I didn't mention in my op but I also regressed the fan coil pressure drop curves based the manufacturer's modeling tool. I checked several models & sizes and the exponent comes out to 1.88 +/- 0.01. For similar reasons that you posit for valves, we can expect fan coils to have a higher exponent than straight pipe (due to lots of U-turns). In fact, if we were to empirically test an elbow or Tee-branch, we would likely find that the "equivalent length" varies with flow/velocity, which would be evidence that the exponent for disruptive fittings is different (higher) than the 1.75 exponent for a reference pipe.

But I would also wager than empirical testing of a valve would reveal an exponent lower than 2, although I agree it would be higher than 1.75. Fun stuff!

 

[LittleInch]

"that valves are not subject to the same fluid properties as a pipe," Not very surprising because a valve, especially a control valve is nothing like a pipe.

Control valves have been studied for hundreds of ears and the equations built up from there. There are many inputs into a pie flow equation which interact, but valves are much more brutal in terms of pressure drop which occurs over a very short section of the valve.

You can get for a specific and quite low range of flows certain simplifications to work, but you seem to have proven that valves are different to pipe.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[MTES93]

Is it possible that a valve, like a fitting is a dynamic (kinetic loss) meaning the V squared thing is still valid, whereas a pipe/tube/heat exchanger/coil also has the 'pipe friction component (ie... a length) which varies as described by f value and others???

 

[DavidButler]

@11241, my original post questions this assumption. But your point, which echoes what SPDL310's wrote, properly recognizes that friction (due to roughness) doesn't contribute much to overall valve resistance. I'm satisfied that the velocity exponent for a valve is probably close to 2.