Pump power and efficiency

1. Sep 20, 2006

hanson

Hi all!
How do we define the power of a pump?
I read a formula from a book which defines the power of the pump as follows:

Power of the pump = Maximum back pressure * maximum flow rate / 2

Can anyone explain why the right hand side equals the left hand side?

I don't understand why the LHS is divided by a factor of 2.

And, is the power of a pump always the same value, no matter at any flow rate/back pressure?

It seems to be that at maximum back pressure, the flow rate is zero and the pump power is zero then?

2. Sep 20, 2006

Q_Goest

Hi hanson,
If you draw a control volume around a pump you'll note the following.
1. There is power going in, in the form of electricity, shaft power or other.
2. There is enthalpy going in. That's your fluid entering the control volume.
3. There is enthalpy leaving. That's your fluid leaving the control volume at a higher pressure.

There could be other forms of energy entering or leaving, but those are the basic ones. For a pump, we generally don't consider aftercoolers part of the pump itself because a pump is for liquids which are relatively incompressible.

Now if you apply the first law to the control volume, you find that the power is equal to the difference in enthalpy. The enthalpy out minus the enthalpy in is equal to pump power.

You could also derive pump power making the assumption that liquid is completely incompressible. In that case, power is work done per unit time so power equals pressure times area times distance divided by time. Here, pressure is the difference in pressure between inlet and outlet, area is inside pipe diameter (ID) and distance divided by time is fluid velocity.

H = W/t = F s / t = dP A s / t = dP A V = dP vdot

Where H = power
W = Work
t = time
F = force
s = distance fluid moves per unit time
dP = pressure increase (discharge minus suction pressure)
A = Area at which distance is measured such as pipe ID
V = s / t = velocity
vdot = A V = volumetric flow rate

I suspect your equation with the 1/2 also provides units and also makes some assumptions. If it were a reciprocating pump for example, we might have a 1/2 put in there by defining flow rate as how much is discharged during the discharge stroke. The 1/2 could be a conversion factor if units were provided. It should explain what assumptions are being made when they write the equation. If not, throw it out, it's not worth it.

3. Sep 20, 2006

LURCH

Yeah, it sure sounds like a conversion factor. In the Fluid Power course I'm curently taking, we use the same formula of pressure times flow rate, but we have to multiply by .000583, because pressure is in "pounds per square inch", and flow rate is in "gallons per minute", and multiplying by .000583 converts the whole mess into Horsepower. (Appearently, it takes .000583 of a horse to move one gallon a minute against one pound per square inch of pressure; though I'd surely like to see him try it!)

4. Sep 20, 2006

hanson

But the 1/2 seems not to be a conversion factor since the maximum back pressure and maximum flow rate are used below. I guess it is some reasons behind this because this definiton is to be used for evaluating pump performance in general.

Power of the pump = Maximum back pressure * maximum flow rate / 2

There is an assumption, it assumed that the flow rate is a linear function of the load pressure. But I don't know what to do with this assumption..

5. Sep 20, 2006

Staff: Mentor

Where are you getting this? In general, that isn't true (pressure is a square function of velocity), but it doesn't have a lot to do with the problem here.

Can you elaborate on what the book says? Units involved?

6. Sep 20, 2006

hanson

Actually that is a paper,
it simply says:
In this paper, we recount efficiency
for micropumps for which measured values are specifically
reported. For micropump papers which do not report η but
do report Qmax, delta pmax and P, we use these values to calculate
estimated thermodynamic efficiency, ηest, by assuming that
pump flow rate is an approximately linear function of load
pressure. Estimated thermodynamic efficiency ηest is then
0.25Qmax*delta pmax/P.

7. Sep 21, 2006

Q_Goest

Hi Hanson.
It looks to me like they're providing some test data, not a method to calculate pump power in general. If you have a paper, feel free to post it. I don't think there's enough information yet to understand what it's all about.