flowrates of 1 input 5 output system as a function of pipe diameter. 4

[Jfoland]

I am trying to model a spray nozzle.

There is 1 pressureized stream in of which I know the volumetric flowrate, and 5 streams out to atmosphere of varying Cross sectinal areas.

I need to know the velocities and flowrates of each outlet stream. The fluid is essentially water.

___________________
Qi --->| |-->A1,v1
| |-->A2,v2
| |-->A3,v3
| |-->A4,v4
| |-->A5,v5
|__________________|

I believe the interaction is similar to 5 parallel resistors in an electrical circuit, but I'm not sure how to calculate the equivalent resistance. I'm sure it's inversely proportional to the cross sectional area, but I'm not exactly sure how to do it.

All outlets are at the same height.

It's been 8 years since my last brush with Fluid Dynamics in college. I've had to purge that memory for stuff more related to my field.
Can somebody help?

-Jeremy Foland

 

[rbcoulter]

The pressure in the pipe is not constant.
The supply pressure (P1) to the pipe is constant.
The pressure at the end of the pipe (P2) (just before the thumb) is not constant. The flowrate Q, does change,
when you apply the thumb. It gets smaller. The pressure behind the thumb gets bigger as you cover up more of outlet area of the pipe/hose.

Look at it this way. The pipe has a definite K value that doesn't change. The applying of the thumb increases the total K value of the pipe/thumb system. The inlet pressure to the pipe is the same. The outlet pressure (after the thumb) does not change. So, the flowrate has to go down. Because of the lower flowrate in the pipe, the delta P in the pipe has to be smaller. This means the pressure just before the thumb has to be higher.

 

[poetix99]

I get the impression that we are now (mostly) talking about "the garden hose" in particular.

I think that some simple fluid mechanical concepts are being misapplied in some of these posts. Boundary layers and separated flow occur in real flows, but do not need to be considered here to accurately understand what is going on. For the circumstances described in this thread, continuity ALWAYS "applies". Continuity cannot ever be IMPOSED. If I use my thumb to cover progressively more area until I have blocked it entirely, does anyone believe that the flowrate remains constant until it abruptly goes to zero?

You must be certain that mass flow is constant, and from that circumstance, if it is present, one may then make conclusions based upon A1*V1 = A2*V2

"rbcoulter" is making some good points. You must consider the resistance of the entire fluid circuit.

Back to the spray nozzle arrangement: if the resistance of each of the five "legs" is essentially only the nozzle itself and that the resistance of any supply piping is nil (i.e. there is the same value of pressure feeding each nozzle), then any change in nozzle area will change the flowrate, not the velocity.

 

[Jfoland]

Ever the pragmatist, just to be sure, I drilled two holes in the bottom of a 1 gallon bottle. 1 was 3/8 the other around 1/8 or so.

When I filled the bottle up, the trajectories of the water streams were the same.

I'm satisfied now.

Thanks for the help, folks.

 

[mucour]

rbcouletr, I agree with your analysis, the pressure in the pipe is not constant. But jfoland is creating another confusion with his experiment. It contradicts known norms of velocity downstream a constriction, when he said the two holes he drilled have the same trajectories. My understanding is that with a reduction in the hole area, the velocity will increase not the same irrespective of the hole area as suggested. May be I am getting something wrong.

 

[rbcoulter]

Don't see how the velocity can increase in the bottle experiment. The pressure drop across each orifice, regardless of size, is the same. The velocities are the same; however, the flowrates are significantly different.

 

[BobPE]

mucour:

It is tough to comprehend, the tank problem is a classic. With the tank, velocity is dependent on the head in the tank. Its related directly tby the fundumental formula used to derive the orifice equation, h=v^2/2g, solve for v and you get the theoretical velocity for any given depth in the tank for any size hole (theorietically!!) Flow is entering the system and need not be conserved and as such becomes a function of the hole size.

That said, now add energy gradients and hydraulic gradients to address energy losses by entrances, exits and friction. You can get an increase from observed velocity by modifying conditions at the discharge that alter the HGL's and make the observed velocity approach the theoritical velocity, determined by pressure, like was discussed before in the hose problem, which is also a classic. The hose must be looked at as a tube when connected to a static source of pressure which has pressure loss unlike an orifice that has entrance and exit.

It is fun to think about....LOL

BobPE

 

[rbcoulter]

BobPE - How did you insert the LOL face in the last response? Where is the list for the other faces?

 

[BobPE]

rbcoulter:

you knew how to insert the face the whole time without ever knowing you knew!!!!!!! LOL

Isn't is scary being an engineer at times!!!!!

I don't know of a list, I just use the letters or face punctuation to make them....

Take care

 

[rbcoulter]

That was just luck

 

[mucour]

Thanks all for the insight.

 

[inControl]

Dear all,

A lot of words for a simple pressure drop problem, just go to:

http://www.pressure-drop-calculator.com/

(freeware)

or, even better:

http://www.pipeflow.co.uk/

(30-day trial)

there you will find software + equations which do the trick ...

Good luck & Happy New Year!