No

Why?

As an approximation, the current is the vector sum of an inductive magnetizing component (Inoload= no-load current) and a resistive load component (Iload).
Itot=Iload + j*Inoload

For powers near to nameplate power for high-speed motors, the ratio of actual current to nameplate current is reasonable approximation of the fraction of full load, since Inoload will be somewhat less than Iload and it’s contribution is reduced even further by the vector addition.

At lower powers, or for lower-speed motors, the magnetizing component becomes more significant.   If we want a more exact solution, we can estimate:
Iload=sqrt(Itot^2-Inoload^2)
P = Iload/Iload100 = sqrt (Itot^2-Inoload^2)/Iload100
  where P is the fraction of full power
  Iload100~Inameplate but more exactly  Iload100=sqrt (Inameplate^2-Inoload^2)

we have neglected changes in power factor, and efficiency.

Also we neglected leakage reactances in the equivalent circuit.

Are you trying to figure the actual HP less the drag of the seal and others that are adding to the total?

Steve - I'm not sure if your question is directed at me or the original poster. I'll assume it's directed at me:

I was determining the power associated with the load branch of the equivalent circuit. That would include shaft output plus friction plus rotor losses, and definitely not core losses. (Stator losses and stray losses are a little muddier to associated with a branch in this model).  For simplicity I'd like to think of them as output power.

The reason that we cannot always predict output load based on the ratio of no-load current to FLA becomes obvious by looking at an unloaded motor which can draw 20%-60% of full load current under no-load conditions (toward the low end for high-speed motros and towrad the high end for low-speed motors).  Let's say I measure 30% of FLA when uncoupled... clearly I am not operating at 30% of output power.  That points out the need to correct the calculation by vector subtraction to remove the effect of the inductive no-load current.

Forgive me if I have misunderstood your question or told you what you already knew.

My mistake Pete, It was for Stan. I asked that question since he asked "why" and figured he was trying to exclude any other variables that add to the motors load and gives him a less than accurate picture of what the pump is actually doing. Especially if he is plotting the HP vs GPM on a curve and nothing adds up the way it should.

if all you are trying to do is a load test at the site, i have seen it done using a watt-hour-meter.

motorman's comment reminds me that the most straightforward way to determine real power input is an instrument which measures phase angle between voltage and current. That is a much more direct and precise method than the approximate method I identified above.

I guess that if a watt-hour meter is available, then it can serve the same function (divide watt-hours by measurment time to determine watts).  Good point.

Suggestion to stankap (Mechanical) Aug 5, 2002 marked ///\\\
I was wondering if it is necessary to subtract no load amps from measured amps of a motor/ pump combination
///No.\\\
 to determine pump hp draw in the following formula:

HP = (1.73 x amps x voltage x P.F. x eff.)/746
///Please, notice that the efficiency and power factor take care of that since they are both smaller than one. However, there are relationships or equations that lead to the HP determination over the mechanical power delivered to rotor over the motor air gap. They are essentially used in textbooks to explain the motor functioning or principles, only. E.g.
Pmech=nu x Re[p x ms x Ir' x Ems* / (2 x ws)], in Watts
nu .. rotor speed in rad/sec
ws=2 x pi x f .. stator angular synchronous frequency in rad/sec
f .. frequency in Hz, e.g. 60Hz
p .. number of poles
ms .. number of stator phases
Ir' .. rotor current transferred to the stator side in Amps
Ems* .. complex conjugate (phasor) of internal stator voltage
Then, 1HP = 746 Watts converts the Pmech to HP.\\\

I guess I never read the question carefully. IF power factor is known (it appears that is the case in the original question), THEN there is no need to subtract no-load current.

Quite often only current is known. That is the situation I addressed in my 8/5/ message.

Sorry for the confusion.

Thanks Guys
Stan K.