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# Converting from Cv to GPM

Member Posts: 7
I am looking for a somewhat easy formula to calculate a Cv value to GPM's. I have to admit, I am not the smartest guy in the world, but the formulas I keep coming up with are way over my head. Any help would be appreciated.

• Member Posts: 31
Cv

Cv is the amount of flow required to create 1psi of loss through a device(typically a valve). Cv is mathematically the sqrt of delta P. I am not sure what you are asking to do when you say that you want to convert Cv to gpm?

Ted
• Member Posts: 669

In other words a valve with an 8CV rating will flow 8 GPM at 1 psi pressure drop. The same valve will flow 4 GPM at 1/2 psi pressure drop.
• Member Posts: 2,542
Close...

but not quite. A 50 % decrease in flow will result in 1/3 the pressure drop. A doubling of flow will increase the pressure drop by 3.3 times. It's not a lineal function.

ME
• Member Posts: 111
A valve with a Cv of 8

will have a head loss of 0.25 psi at a flow of 4 gpm.

In order to derive flow from a Cv, you also need to know the pressure drop. The formula is easy - GPM = Cv x the square root of the pressure drop in psi where the specific gravity of the fluid is 1.0. The hard part is getting two of the three pieces of information so that you can calculate the third.

Scale No. 5 on a Bell & Gossett System Syzer allows you to do it without the mathematical gymnastics.

• Member Posts: 7

• Member, Moderator, Administrator Posts: 15,977
That's

correct. Thanks, Jerry.
Retired and loving it.
• Member Posts: 6,106
Try this demo

A hydronic circuit simulator as well as a fluid properties module is in the Hydronics Design Studio. It will allow you to enter various valves and CVs and determine fluid numbers.

The Fluid modules shows properties of various fluids at various temperatures. (the missing numbers

www.hydronicpros.com will lead you to a demo of this software.

Better yet get the Modern Hydronic Heating textbook with the software for an in depth explanation how all this fits together.

hot rod

• Member Posts: 171
Kv to Cv

Anyone have the multiplier to convert Kv to Cv. Those Europeans are pesky.

Regards,

PR
• Member Posts: 71
Kv

Kv X 1.16 = CV
• Member Posts: 171
Thanks

Thanks

PR
• Member Posts: 549
Most Manufacturers...

...have tables for their valves showing flows for various pressure drops and fluid services. Depending upon what you're doing, it can be GPM for liquid, lbs/hr for steam, CFM for compressed air, etc. Most of the time, you can select a valve from these tables, without actually doing the calcs for Cv. I do it for steam control valves all the time. Once I determine how many lbs/hr the valve needs to pass, I look at the inlet pressure available for the valve, and what outlet pressure I figure will be suitable. The outlet pressure is the bit where art kind of meets science, so you need to do a few to get a feel for it. The big thing to keep in mind, is that the outlet pressure from the valve you pick is the inlet pressure for everything downstream. The bigger the pressure drop you can take across the valve, the smaller it can be for the same flow. However, it is possble to get "too much of a good thing". You need to blend this with the fact that the biggest single problem (by far) I see with control valves (and check valves), in all kinds of services and applications, is OVERSIZING.

I can look at the chart in the valve manufacturers catalogue, for my inlet pressure, and whatever I've picked for my outlet pressure, and just go across until I find the flow number that's the same or bigger than what I need. Then it's up the chart to see what valve size corresponds. Usually, the Cv is listed in the same box as the valve size. It's not much more complicated than reading the "distance between cities" chart on a map. When I first started doing this, I found it gave me a much better feel for calculating Cv's. Once I had the valve size, I also had the Cv. Now, I could run the Cv calcs to see if I got the same (or lower) number as the catalogue chart. After you do a few, and get the numbers to match both ways, you'll feel a lot more comfortable.

• Member Posts: 71
CV

If you have a basic knowledge of Excel, set up a small spread sheet exactly like the attachment.

Enter your CV, actual or desired flow and it will calculate your pressure drop in PSI and Ft. of water.
• Member Posts: 40
ME's Non-Square Flow Function

Mark:

You posted:

"Subject: Close...
but not quite. A 50 % decrease in flow will result in 1/3 the pressure drop. A doubling of flow will increase the pressure drop by 3.3 times. It's not a lineal function. "

I've always understood friction to be an exponent of the flow rate. Friction being the square of new flow. I don't understand why doubling the flow would increase friction by 3.3 times, instead of 4.0 times as a square function. Would you elaborate please...

Tom Anderson
• Member Posts: 1
Siegenthaler says

it's 3.4. Mark likes 3.3 so he can tell his students to remember something about a dog and a tree.

According to Siegenthaler, it's exponential, but the exponent is 1.75 rather than 2.0. 2^1.75 = 3.34.
• Member Posts: 111
We're getting into some deep

math here! The general equation for calculating the pressure drop in any element in a circuit includes a system coefficient (exponent). For steel pipe, the value of this exponent is 1.85. For turbulent flows, the value of this exponent is assumed to be 2, so that for general use the head loss will vary as the square of the flow.

I have a computer program that analyzes piping systems, so I just did a quick analysis of a piece of 1" steel pipe 100' long, with water at 175 F. At 10 gpm, the head loss is 2.572 psi, and at 5 gpm the head loss is 0.687 psi. Using the general assumption that the head loss varies as the square of the flow, I would have come up with 2.572 x 0.25 = 0.643 psi, so it's pretty close.

If you want to get into how to calculate the value of the exponent, get out some hip waders and maybe a snorkel!
• Member Posts: 2,542
Tree-n- a turd...

Tom, As Nick said, I am using Siegenthalers Modern Hydronic Heating book as the core book for the hydronics course.

When teaching students from the field, I find it important to make it as simple as possible. These guys are mechanics, not engineers. They have a tendency to get smothered by the logistics of "how to".

To keep it as simple as possible, I tell them when determining the pressure drop for a system resistance curve, to plot the pressure drop of the known flow rate, and plot that on the pump manufacturers curve. Then, double the flow rate, which increases the pressure drop by 3.3 times, and plot that point. Then take the original flow rate and cut it in half, and plot the 1/3 pressure drop for that point. The point of all of this is the graphic I draw on the board to make them memorize what the pressure drop does if you double the flow. I draw a picture of an old oak tree, with a smelly dog turd underneath it. Tree-n- a turd. Get it!

Not one student has ever missed that question on final, and I've had numerous students come up to me LONG after graduating and tell me they remember that trick all the time. Just a graphic reminder.

ME
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