## motor amps / pump power

## motor amps / pump power

(OP)

Hi,

I was wondering if it is necessary to subtract no load amps from measured amps of a motor/ pump combination to determine pump hp draw in the following formula:

HP = (1.73 x amps x voltage x P.F. x eff.)/746

Thanks,

Stan K.

I was wondering if it is necessary to subtract no load amps from measured amps of a motor/ pump combination to determine pump hp draw in the following formula:

HP = (1.73 x amps x voltage x P.F. x eff.)/746

Thanks,

Stan K.

## RE: motor amps / pump power

## RE: motor amps / pump power

## RE: motor amps / pump power

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.

## RE: motor amps / pump power

## RE: motor amps / pump power

## RE: motor amps / pump power

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.

## RE: motor amps / pump power

## RE: motor amps / pump power

## RE: motor amps / pump power

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.

## RE: motor amps / pump power

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.\\\

## RE: motor amps / pump power

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

Sorry for the confusion.

## RE: motor amps / pump power

Stan K.