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]

On page 3-2 of the Crane handbook there is
a formula that shows that square of the flow rate varies with the fifth power of the diameter. Or the flow is proportional
to the 2.5 power of th diameter. So if each branch has the same friction factor and length and outlet pressure (e.g. atmospheric discharge), this would be the case.

 

[Jfoland]

By that method there is no interplay between the flows. I think that this system has to be treated like a series of parallel resistors. If I constrict one hole, the flows through the other holes will increase.

I found a formula for Equivalent pipe resistance, but the variables aren't explained. It was something like (8 pi mu)/A, but that doesn't jive.

 

[ve7brz]

Your original analogy is, in my opinion, correct. Qi is your power source feeding a series/parallel resistor network to ground (atmosphere). Every length of pipe feeding the nozzles acts as a series resistor, and ever orifice a parallel resistor to ground.

The following site might be of some assistance in your calculations:

http://www.efunda.com/formulae/fluids/calc_orifice_flowmeter.cfm

 

[rbcoulter]

Is the pressure supply to the system not constant? If the pressure supply to the system was constant then restricting a branch would not affect the flow in the other branches. Is the flow consant? Unless a positive displacement pump is feeding the system, then this is not the case. What type of pump is feeding the system? If a centrifugal pump then these is a pressure vs. flowrate relationship that figures into the calculation which is makes it more complicated but still easily solvable with a spreadsheet using the data from the pump curve and the equation on page 3-2 of the Crane Handbook.

 

[Jfoland]

rbcoulter,

Sorry, all. Even though by diploma I am a Chemical Engineer, I'm currently employed in the Electrical field, so I think like an Electrical Engineer (..Gasp!)

You know, I think you are right. The Constant Pressure source is like a constant voltage source, where a constant flowrate system would be like a constant current source.

If I hook ten resistors up to a battery (Assuming I don't load down the system) the current through each resistor obeys Ohms Law. However if there is a constant current, then it gets a little more complicated, and interplay results.

I'm going to try to find a Crane's handbook around here. Otherwise, I'll use that website that ve7brz suggested. Thanks all for your help.

Regards,

Jeremy Foland

 

[Jfoland]

rbcoulter,

Could you please give me the complete title and publisher of that Crane's Handbook. I searched Google and just got a handbook about Cranes (Construction machinery )

 

[rbcoulter]

Here is a link to their website:

http://www.cranevalve.com/tech.htm

 

[poetix99]

http://www.cranevalve.com/tech.htm

BTW, I don't know how REAL "constant" voltage and current sources work, but "rbcoulter" just scratched the surface by pointing out that you might have to take the flow/pressure source into account. A centrifugal pump has a characteristic curve of discharge pressure vs. flow and is therefore not a constant anything. Pos. displ. pumps are fairly close to constant flow. This could be an iterative solution if you must consider significant "turndown" to your system due to changes in total discharge area.

 

[rbcoulter]

That's true. A centrifugal pump is not a constant anything; however, even for a centrifugal pump there is a definite pressure vs. flow relationship. It is most likely an iterative solution unless the pump curve is approximated by a line or other simple curve. Most spreadsheets have a "goal seek" function that would greatly assist in the solution. The iteration would go something like this:

1. Assume a total flow rate, Q
2. Calculate discharge pressure of pump (from pump curve)
3. Calculate resistances of each pipe branch (from
Crane formula above).
4. Determine flow in each branch.
5. Add all flows in each branch and compare to
Q above.
6. Adjust initial guess of Q and repeat steps above until there is an agreement.

 

[Jfoland]

Fortunately in my case, I can be certain that I have a constant pressure source.

There is no pump, and the fluid is driven by a nitrogen line on the backside of a buffer tank, and can be considered very stable.

Thanks everyone for all of the help. I've reccomended this site to several of my engineering buddies. I hope I can give as much as I've received!

If I still have problems, I'll post here later.

Thanks again.

BTW, I got ahold of the Crane Book, but can't fine the formula specified. My version was reprinted on 7/96

 

[Jfoland]

Ok, I think I've got it now.

Thanks ve7brz for the websote you suggested.

Using Bernoulli's law
r = Density
g = Gravitational Constant
z = head
P = Pressure
v = velocity


rgz + Pin + 1/2rVline^2 = rgz+Patm+1/2rV1^2
^Negl ^I'll say buffer tank is stagnant and drops

Pin = rgz + 1/2rV1^2 + patm
^ negligible

V1 = (2(Pin-Patm)/r)^0.5

I'm happy with that solution. EXCEPT, everyone keeps asing me the same question.

"When I stick my thumb over the end of a garden hose, the velocity coming out of the hose increases (Shoots farther). Why does your equation suggest that velocity is independent of cross sectional area?!"

And I say....Hmm, I'm not sure.

Does it have something to do with friction loss being less for the turbulent flow which is produced by reducing the reynolds number?

 

[Roach]

To answer your question concerning the water hose,

"A diverging passage.... should not be modeled using the Bernoulli equation. Adverse pressure gradients cause rapid growth of boundary layers, severely distorted velocity profiles, and POSSIBLE FLOW SEPARATION." ........

"Flow separation that can occur at inlets with sharp corners and in abrupt bends, causes the flow to depart from that predicted by a one-dimensional model and the Bern. eq."


Introduction to Fluid Mechanics
5th Edition
Fox and Mcdonald
page 256

Streamlines must be indentified for use of Bernoulli's, and having a finger over the hose distorts these lines. I think. Correct me if I am wrong.


Roach

 

[rbcoulter]

The velocity increases at the end of the water hose because the water pressure increases (behind the thumb and in the hose) when you put your thumb over the end. The water pressure is not constant at the end of the hose (or any significant run of pipe) even if the water pressure supply at the inlet is constant if you vary the outlet resistance (e.g. with your thumb).
By putting your thumb over the end of the pipe you slow that water flow down enough in the hose/pipe so that the pressure drop (lost to friction) in the hose is significantly reduced. Now the pressure at the end of the hose/pipe is very close to the inlet pressure of the hose. The result is higher pressure and thus higher velocity from the end of the hose but lower flowrate.

 

[rbcoulter]

Add should add that the velocity head term can usually be ignored when dealing with water flows in most common situations. Most of the energy (from the pressure drop) is lost to friction.

Air flows are usually a different matter where the velocity term is significant in many cases.

 

[bhwtam]

HI Folks,

I think you missed out the continuity equation, which is:

A1V1 = A2V2

where A1 & A2 are the area of pipe before and after reduction, V1 & V2 is the corresponding velocity of fluid.

As a result, when you place your thumb at the hose end, you reduced the area of the pipe (similar to adding a jet nozzle) thus making the velocity increase.

 

[Jfoland]

Now that I think about it...

You're right. The continuity equation applies for the hose because it's a restriction of an existing pipe flow.

Why does it not apply to a Draining tank?

 

[bhwtam]

Hi Jfoland,

I think the continuity equation also applies to draining tank.

However, with a very large surface area of the tank compare with the pipe cross-section area, the velocity in the tank will be very small and will be taken as negligible for ease of calculation. But this does not mean it does not exist at all.

Perhaps you can consider a large hydraulic piston with a very small oil infeed pipe. When oil is feed into the piston, the piston will move very slow, but it still move.

 

[rbcoulter]

The continuity equation only applies to a fixed point in time. For it to apply before and after applying the thumb to the hose then the flowrate needs to be constant in the hose (which we know it isn't) because we know this is a constant pressure situation at the hose supply. (Pressure is not constant at the hose outlet, before and after the thumb).

For example, let's take the hose out of the problem, and imagine a half inch hose punctured at the water supply (where the pressure is constant). As long as there is enough "capacitance" , such as a large water tank, then the velocity coming out of the half inch hole is the same before and after the thumb is applied to the hole.
This is because the pressure does not change upstream of the hole unlike the situation at the end of a long hose where the pressure does change when you apply your thumb.

 

[mucour]

If the pressure in the flowing pipe is constant, and we know that water in a pipe is incompressible, then by placing your thumb on the hose nozzle, in order to maintain the flowrate, Q, because the area has reduced, the velocity has to increase. If you remove your thumb, the area will increase and the velocity will reduce. This is to maintain the same flow rate.