# Calculating Flow Rate Help Please

1. Jul 30, 2014

### npc214

Hey everyone, sorry I posted this in two different forums. I think this forum was more applicable so I wanted to move it.

I have a fluid system of water coming from a pressure source. Then there is a converging nozzle, which I have calculated the resistance coefficient using Crane's Manual and it has an outlet to the atmosphere.
I have tried to simplify the system. The full system is a pressure source, converging nozzle which has outlet into original diameter at converging nozzle inlet followed by hose to outlet to atmosphere. (this is like a fire hose system - slightly different for my actual application)
The pressure of the pressure source is my independent variable, I am doing a sweep in excel. So if it helps for simplicity, you can assume the pressure is 100 psi.

How do I calculate the flow rate (in GPM)? Do I need to take away the outlet of the system?

Currently I am thinking using:

hL = [k(v)^2]/(2g)
ΔP = (ρ*hL)/144
then what for q??

Or I am thinking q = K*A*(2*144*g*ΔP/ρ)
but then what is ΔP?? is it the pressure drop across the converging nozzle or is it the pressure drop from the inlet to outlet at atmospheric?

2. Jul 30, 2014

### Travis_King

The water will flow with exactly the amount of volume/capacity required to create a pressure drop which, in the length of pipe between the "main" (which is usually treated as just a pressure source and not a pipe) and the atmosphere, including the hose, will drop the pressure from 100 psig to atmospheric. Or, more directly, what volumetric flow rate will, in this specific system, result in a pressure head loss of 100 psig, or roughly 231 feet of head.

The first equation looks to me like an open channel fluid equation (though, where did the 144 come from?). The second looks closer, though I haven't used resistance coefficients in a while, so I forget how the formula is supposed to look. If there is a pressure source, I assume this is a piped flow, in which case you'd want to use the Darcy-Weisbach equation or the simpler Hazen-Williams simplification.

http://www.engineeringtoolbox.com/hazen-williams-water-d_797.html
http://www.engineeringtoolbox.com/darcy-weisbach-equation-d_646.html

These equations require iteration when solving for flow, but that's pretty much par for the course when doing fluid dynamics.

If you have K, though, you're pretty much good to go. Just check your equations, because they seem a little off to me.

This page has some good information on mathematically calculating flow through nozzles.
http://www.pipeflowcalculations.com/pipe-valve-fitting-flow/flow-in-valves-fittings.php

Last edited: Jul 30, 2014
3. Jul 30, 2014

### npc214

Here is a picture of the system. Thank you, I think your reply helps, but I am not concerned about frictional losses right now. Just the loss from the nozzle and what my flow rate will be. I feel like I'm missing one piece of the puzzle. Some input or assumption I don't have right quite yet.

4. Jul 30, 2014

### Travis_King

Ok, yea, so if you know the pressure immediately before the nozzle and immediately after, you can use that as your delta P. If the nozzle is open to the atmosphere, then that pressure is 0 psig, or ~14.7 psia (at STP). Understand that if there is a hose after the nozzle, that will contribute to the losses, and hence will increase the pressure at the point just after the nozzle to above atmospheric. Basically, as I said in the first post, your flow comes from the fact that the high pressure system seeks to get rid of all it's excess energy (in the form of pressure). So the fluid flows from the pipe with enough capacity to lose all it's excess energy (pressure) by the time it exits to atmosphere.

I just want to reiterate that if your line prior to point 0 is not a close-by water/liquid main, then any piping between the main and the nozzle will contribute to losses from that 100 psi and will consequently lower the delta-P across the nozzle, thus lowering flow rate. With that said...take a look at that third link, that should give you a good idea of what you're looking at.

How did you find K? Did you use the Cv?

Last edited: Jul 30, 2014
5. Jul 30, 2014

### npc214

Thanks Travis_King, your second post helped me understand exactly what you're saying. I found K (from smaller diameter = 0.2072, from larger diameter = 13.8082), our source is a fire hydrant, so I suppose there are long pipes from the main to the hydrant, but we get consistently 75-80 psi at the hydrant. The system pictured above is an eductor or water jet pump attached to the hydrant typically there is a hose with a gun at the end, but sometimes we take the hose off and shoot the eductor right off the hydrant. There is some short pipe length (diffuser) after the converging nozzle in the eductor, but I think that can ignore that short length for now.

Could I simply use Bernoulli's... ? Q = A2*SQRT(2*144*g*dP/(rho(K-A2/A1+1))

and here dP would be source P - atm P?

6. Jul 30, 2014

### Travis_King

Oh, ok. Yea, using the crane tables and plugging into those formulas with Beta; gotcha.

And yea, if it's a hydrant you're probably ok to treat the upstream pressure as that same 75-80 psi. There shouldn't be much losses prior to the nozzle. You'll likely lose a couple psi in the fire hose, so you can account for that later.

I think you'd be better off using the equations in the third link. I do a lot of hydraulic calculations and almost never use Bernoulli. It may work, give it a try with both.

To answer your question though, yes, dP is Psource-Patm. Be sure that if your source is 80 psig (gauge pressure) then you subtract 0 psig. If it is 80 psia (absolute) you subtract the 14.7 (or whatever it is at your altitude). Also note that in most equations, the dP is in feet H20, not psi.