Flow Rate vs Flow Area 1

[metootoo]

I'm pretty far removed from flow dynamics, so please excuse the elementary question.

A technician is testing a oil pump at my work place and taking flowrate data at a given output pressure. The output pressure is being set by a needle valve.

These are some of the points:
0psi ---- 39GPM
200psi -- 39GPM
400psi -- 26GPM
600psi -- 6 GPM
from 800 to 2500 psi flow drops incrementally from 6 to 5 GPM

In reality field use, the flow path will be short and no smaller than 3/8" in diameter. I believe the problem is the needle valve, he thinks it is the pump.

I've been looking at flow equations and found that fundamentally flow rate is area times velocity:

Q=AV , where V ^2= P/(density[1+loss/2]) and A = pi x r^2

This says to me that if you have a given pressure and density, flowrate will decrease as flow area decreases. Am I correct?

 

[metootoo]

There's also another thing I'm trying to figure out. I that equation....

Q=AV , where V ^2= P/(density[1+loss/2]) and A = pi x r^2

..... Q is volume/time. Where did the time come from? Velocity, yes, but the equation for velocity has Density and Pressure, neither of which have a unit of time. So where does the time come from?

 

[hydtools]

What kind of oil pump?

Density is mass density. weight/unit volume/g
g is gravitational constant, distance/time^2

Ted

 

[ione]

The lower the pipe diameter the higher the velocity for a given flow rate.
Your pump has a performance curve (flow rate vs head) and your system has its own curve.
The working point of the system is given by the intersection of the above curves.

Velocity is velocity, that is distance/time. Check Bernouilli theorem.

 

[ciise]

I am not an expert here but if Q=AV, reducing the flow area (A) increases the Velocity(V) meaning that flow (Q) will not change at all!

Q in=Q out

 

[BigInch]

Why? Nothing in the equation itself says that the value of V is conditional on the value of A.

According to the same equation Q = AV, Reducing A has no effect on V, thus Q decreases in the same amount as did A.

If you defined steady state flow where Q is NOT ALLOWED to change then, V does become conditional on A. Then, if A decreased, V would have to increase, but only to continue to hold your defined steady state condition as true.

 

[zdas04]

MeTooToo,
I don't understand your question at all. The power that a pump needs is a function of both flow rate and differential pressure across the pump (head). If your input power is held constant, then raising the discharge pressure has to lower the flow rate, just like your data is showing.

In order to get a constant flow rate at changing head, you have to add power for increasing pressure and reduce power for decreasing pressure.

Also, the "equation" you sort of included in your second post is just nonsense. It has its roots in the Bernoulli equation, but the Bernoulli equation does not allow losses (no work done on or by the system, no irreversible changes) so I don't know what the heck you're talking about. Yes "velocity" must have both a time component, a distance component, and a direction component.

David

 

[tr1ntx]

"I believe the problem is the needle valve, he thinks it is the pump."

What is the pump's rated capacity? What type of pump is it?

 

[metootoo]

Q=AV , where V ^2= P/(density[1+loss/2]) and A = pi x r^2 was taken from my "Fundamentals of Fluid Mechanics" book published by John Wiley & Sons Inc. (see attached). Please feel free to call and correct them, zdas04. There may be a flaw. I certianly don't see a time unit source in the equation.

I will correct one thing though, I by "loss", I meant "loss coeficient." Pressure is really differential pressure. In my case P2 is zero.
______________________________________________


As for my second post, see the attachment. Notice where they solved the equation with units, they stuck in N*s^2 seeminly from nowhere. Where did that come from? There is not one variable in equation 6 that has time as any part of its unit. Loss coeffient is unitless.

 

[hydtools]

metootoo, go to your referenced text and look up the definition of density rho. Verify that gravitational constant g is used in the definition.

Do a dimensional check on your equation.

Ted

 

[chicopee]

Where are your pressure measurements taken from. If the pressure measurements are velocity pressures, then your velocity formula is correct; if the pressure values are from line pressures, then these values are static pressures and the velocity formula does not apply.

 

[btrueblood]

" I believe the problem is the needle valve, he thinks it is the pump."

What problem? I see a pump curve in your OP, and would guess it's for a 2-stage hydraulic pump, given the somewhat sudden shift from high flow to lower flow rate at a "medium" pressure of 600 to 800 psi.

 

[hydtools]

metootoo, look at the whole term. [1N*s^2/kg*m], not just N*s^2.

1 N = 1 kg*m/sec^2

The term converts kg*m/s^2 to N, force unit.

Ted

 

[zdas04]

MeTooToo,
Snarky responses get you snarky responses. Then they get you no responses. You've been here since September and the only threads you've "contributed" to have been threads you've started. To me that spells "leech".

I don't need to contact the publishers of your text book because they simply made the assumption that anyone taking "Fundementals of Fluid Mechanics" could do simple unit analysis. Maybe they set the bar too high.

David

 

[hydtools]

btrueblood, that makes sense. The pressure v flow does look like that of a two-stage, hi-lo pump.

Ted

 

[btrueblood]

hydtools,

Dunno, it was a guess. A second possibility is the unreported 3-way valve that the tech was opening, or the pressure relief valve that was installed upstream of the needle valve. But, sometimes, I'm good at reading the OP posts where they've written the critical information in white text.