pump power
pump power
(OP)
we have measured a three phase pump power with a tongue power meter and result is as follows.
R Y B
current 191.4 197.7 194.5 AMPS
power fac. 0.7 0.96 0.27
POWER 33.8 48.2 14.6 KW
VOLTAGE IS 440V
WHER I WENT WRONG IN MEASURING, OR IS IT POSSIBLE LIKE THAT.
R Y B
current 191.4 197.7 194.5 AMPS
power fac. 0.7 0.96 0.27
POWER 33.8 48.2 14.6 KW
VOLTAGE IS 440V
WHER I WENT WRONG IN MEASURING, OR IS IT POSSIBLE LIKE THAT.





RE: pump power
You didn't mention the power rating of the pump but i assume it should be around 120KW.
Pls brief how did u derived the power in the KW and p.f as per in your mail.
Something is not right so do provide further detail.
RE: pump power
THE POWER METER USED WAS TONGUE POWER METER IN WHICH ALL THREE PHASES WERE CONNECTED FROM BUS FOR VOLTAGE AND TONGUE USED FOR EACH INDIVIDUAL PHASE.
IT STRAIGHT AWAY READS CURRENT, POWER AND POWER FACTOR OF THAT PHASE ON SELECTION.
THE SUM OF THREE PHASES POWER IS TOTAL POWER MEASUREMENT.
RE: pump power
The power factors are unreasonable, and that throws the power calculation off.
Is it possible that you have created a phase shift through the connection of the voltage taps? ie maybe the instrument expects phase-to-ground voltage and you are inputting phase-to-phase? There must be a voltage input terminal for ground or else there is no way the instrument could pretend to calculate power on each phase. If there are no problems here, are you reading on the secondary of potential transformers which might be creating a phase shift (delta-wye or wye-delta?).
RE: pump power
The numbers certainly don't add up correctly. If you do the vector summation, you end up with a residual current of around 184.7A at an angle of 46.6 deg (relative to the R phase voltage), which is very close to the measured R phase current. I am sure that there is not that amount of unbalance current flowing in your circuit.
As I understand your tester, you connect the voltage leads to the 3-phase source and then clamp around each phase lead in turn, selecting the appropriate voltage for the phase under test. If this is the case, then what may have happened is an error in clamping on to the phase wires for the current - can you confirm that you clamped on to the correct phase cable for each measurement, with the same polarity on the tester? Also, is the phase sequence of the motor leads the same as the voltage measurement (are the phases correctly lined up)?
RE: pump power
As I mentioned above, it's important to check for proper ground reference to your instrument (there may be one ground input or three ground inputs) and verify no delta-wye connection on the plant potential transformers, if used. If you happen to have an ungrounded system, then you would at least need access to a neutral.
As peterb mentioned, there is a polarity of a clamp-on probe. There are two ways to attach it, they give current 180 degrees apart. This in itself doesn't seem to explain your results... a 180-degree phase shift would give the correct magnitude of power factor but the wrong sign (i.e. negative power factor... which the instrument probably automatically recognizes and corrects, since negative power factor in a mathematical sense would represent real power flowing out of the motor). But it can't hurt to check this.
In my first response I assumed that you don't have current transformers on your 440 volt system. But if you have them and are using them to sense current, then recognize that the connections of current transformers (delta or wye) can also cause a phase shift.
RE: pump power
RE: pump power
my night shift tech (one of those real old timers) had a read of this and reckons "yeh real hard to stuff up the amps with the tong testers but the volts probes aint plugged in right or somethin's broke".
How ever we both agree if the amps are close to the plate values then check with a standard meter the volts phase to phase. if they are all balanced you haven't got a motor problem.
The concensus is that if the motor was only developing that sort of power then it probably wouldn't turn.
Let us know how it goes
Don
RE: pump power
1. To meassure the power by a proper wattmeter; or
2. Establish a neutral to obtain the Line (or phase) to neutral voltages since:
2a. Line current will be measured
2b. cos(fia)=PFa will be between the Valn and Ia
cos(fib)=PFb will be between the Vbln and Ib
cos (fic)=PFc will be between the Vcln and Ic
3. Finally, input power of the pump motor is:
Pin=Vlna x Ia x PFa + Vbln x Ib x PFb + Vcln x Ic x PFc, in Watts
If the efficiency is known than the output or shaft power is:
Pout=Pin x EFF, in watts
Else, one must measure the motor watt losses.