Use this formula; KW= (volts(avg) x amps(avg) x power factor x 1.732) divided by 1,000. If you don't have a power factor use 0.9. Also make sure you use the larger voltage of the system voltages, ex) if it's 480/277 use 480 in the formula...208/120 use 208...240/120 use 240. If possible use a volt meter to read the actual voltage between each phase. Use the average of the three voltage readings (V1+V2+V3 / 3). Take the amperage reading from each phase and average them (A1+A2+A3 / 3). To find KWh, just multiply KW by the amount of hours your system will be serving the load.
An operating three phase motor has voltages measured with a voltmeter on each phase of 453, 458, and 461 volts, amperage measured on each phase with an ammeter are 14.1, 13.9, and 13.8 amps, power factor was measured as 0.82.
(453 + 458 + 461) / 3 = 457V
(14.1 + 13.9 + 13.8) / 3 = 13.9A
(457V x 13.9A x .82pf x 1.732) / 1000 = 9.02 Kwatts
To calculate the kWh for one days use:
9.02 Kw x 24hrs = 216.48 kWh
If you are serving unbalanced loads and the amps vary between phases, you can calculate each phase separately. In this case, you are calculating three separate single-phase values, so the 1.73 multiplier is not used, and you use the phase to neutral voltage.
Phases measured with a voltmeter at 272, 264.4, and 266.2 volts (A-N, B-N, C-N), amperage measured on each phase with an ammeter are 10, 25, and 16 amps, power factor is assumed as 0.9:
(272V x 10A x .9pf) / 1000 = 2.45kw
(264.4V x 25A x .9pf) / 1000 = 5.95kw
(266.2V x 16A x .9pf) / 1000 = 3.83kw
2.45 + 5.95 + 3.83 = 12.23kw total
To calculate the kWh for one days use:
12.23 Kw x 24hrs = 293.52kWh (IF the load remains constant!)
This method also works for balanced loads, it just takes longer and is more work, but you get the same answer (actually slightly more accurate, since you are not averaging) as the first method!
Modification of the example above for an accurate total kw:
If you are serving unbalanced loads and the amps vary between phases, you can calculate each phase separately, however you must still average the phases, and multiply by the square root of 3.
Phases measured with a voltmeter at 272, 264.4, and 266.2 volts (A-N, B-N, C-N). We will assume 480v phase to phase. Phase to neutral voltages will not be used in the calculation. Amperage measured on each phase with an ammeter are 10, 25, and 16 amps, power factor is assumed as 0.9:
(10 + 25 + 16) / 3 x 480 x 1.732 x 0.9 / 1000 = 12.71 kw total
To calculate the kWh for one days use:
12.71 Kw x 24hrs = 305.26 kWh (IF the load remains constant!)AnswerANSWER">ANSWERThe above answers are confusing because they use incorrect terminology. The three energised conductors supplying a three-phase load are called LINE CONDUCTORS, which is why the voltages across them are called LINE VOLTAGES and the currents through them are called LINE CURRENTS. The PHASES are represented by the three loads, and the voltages across individual loads, and the currents through those individual loads are, therefore, called PHASE VOLTAGES and PHASE CURRENTS. If there is a neutral conductor, then the voltage measured between any LINE CONDUCTOR and the NEUTRAL CONDUCTOR are called PHASE VOLTAGES.
So if, in your question, you are referring to the currents flowing in the supply conductors, then your question should read, 'LINE CURRENTS', not 'phase currents'! Incidentally, from the practical point of view, it's quite difficult to access the phases of a machine in order to measure phase current, whereas it's very easy to measure a line current.
Having got the terminology out of the way, let's answer the question. For a BALANCED LOAD (i.e. each load being identical in all respects), the total power is given by the following equation:
P = 1.732 UL IL xpower factor
...where the suscript, L, represents LINE quantities.
If the loads are unbalanced, you cannot use this equation. Instead, you have to measure the phase currents (IP) -i.e. the currents in the individual loads or PHASES, and the voltages across them (UP) -i.e. the phase voltages. To find the total power, you must apply the following equation to each phase, and add them together:
P = UP IP x power factor.
Finally, to find the energy consumed, in kilowatt hours, you must multiply the total power (expressed in kilowatts) by the time for which the load is operating (expressed in hours).
Measuring the current in each phase (or do you mean 'line'?) will not give you sufficient information to work out what you are asking for.
You simply add the power of each phase together to determine the total power.
In three-phase systems, we have to measure the current in each of the three line conductors individually. There is no 'overall' current for a three-phase system. So if you know the current in each phase, then that's it! There's nothing else to calculate.
It means 225 amps on each phase.
In a standard 3 phase system in North America, 7kVa would be equivalent to 19.5 amps on each phase. The equation is: 7kva*1000/208v/1.73=19.45 amps (3 phase)
each phase is within 2% of amps per phase
Assuming it is a 208-volt line voltage (as normal in 3-phase) the phase voltage is that divided by sqrt(3), or 120 volts. Each phase has to supply 10 kW so the current on each phase is 83.3 amps.
To answer this question the voltage of the generator must be given. That is because for each phase the kVA figure is equal to the line-neutral kV times the amps.
For a single phase circuit, the equation you are looking for is I = W/E. Amps = Watts/Volts.
thong each wire using amp meter
There are two ways: If the amps are the same on all three phases (voltspp is the phase to phase voltage): amps * voltspp * 1.73 = watts If the amps are different on each phase (voltspn is the phase to neutral voltage): (ampsA + ampsB + ampsC) * voltspn = watts In a 480V system, 480 is phase to phase and 277 is phase to neutral, likewise 208/120, etc. Example - 12A on all phases, 480/277V system: 12 * 480 * 1.73 = 9965W Example - phase A = 4A, phase B = 6.3A, phase C = 2.2A, 208/120V system: (4 + 6.3 + 2.2) * 120 = 1500W
Yes. In a 3-phase motor, all 3 phases have the same current.
Yes - a 3 phase load (in a balanced network) that consumes 270A would have a flow of 90A down each phase
If the mains is a 200 amp two pole breaker each leg can carry up to 200 amps in relationship to the common neutral of the service.
Electrical load is measured in amps. You measure each single conductor with an ammeter.
If you have three adjacent houses each with a single-phase supply taken from different phases in a three-phase cable in the street, the total power is equal to the sum of the powers in each of the three phases.
For a balanced load, you don't have to worry about phase values when you want to determine the power (or, in this case, the energy), whether delta or wye. Rather, you always use line values:P = 1.732 VL IL cos (phase angle)For an unbalanced load, however, you need to measure the phase voltage and phase current and power factor for each of the three phases, and add them together:P = [VpIp cos (phase angle)]phase A +[VpIpcos (phase angle)]phase B+[VpIp cos (phase angle)]phase CTo then calculate the energy expended in kilowatt hours, you need to multiply the total power (as calculated above), expressed in kilowatts, by the time for which the load is operating, expressed in hours.
Each hot leg to the neutral wire of the service has the ampacity of 200 amps, that is why 3/0 wire is required. A 3/0 copper wire with an insulation factor of 90 degrees C is rated at 210 amps.
135 A at 120 v single-phase is 16.2 kVA. With a 208 v three-phase supply you get three single-phase 120 v supplies, so the same kVA is produced with a balanced load of 45 amps on each phase.
1 HP=746 watts 15 HP=11,190 watts Ohm's Law says Current (in Amps) = Power (in Watts) divided by Voltage (in Volts) 11,190 watts divided by 415 volts = 26.96 amps. <<>> For a three-phase motor each phase supplies one third of the power, so that is 5 HP on each phase. A 415 v supply has a line-to-neutral voltage of 240 v, and 3730 watts would therefore required 3730/240 amps, or 15.54 amps. However due allowance must be made for (i) the power factor and (ii) the conversion efficiency. This would increase the current by an estimated 20% so the current is therefore estimated to be 19 amps.
One horse power for a electric motor is defined as 746 watts. Figure out how many watts each winding is pulling, assume they are balanced (they should be - if not, something is wrong), multiply by three, and then divide by 746. Don't forget that power factor will enter into the calculation, so if you are measuring volt-amps instead of watts, then you will need to divide by the power factor (which is the cosine of the phase angle between voltage and current) in order to determine watts.
96 kW means 32 kW from each phase. If the load is star connected each resistor has 277 volts across it and carries 32,000/277 amps, 115.5 amps. The load resistors are 277/115.5 ohms or 2.40 ohms If the load is delta connected each resistor has 480 volts across it and carries 66.7 amps. The load resistors are 480/66.7 ohms or 7.20 ohms. In both cases the line current is 115.5 amps.
You will need to determine the power per phase, and add them up to give the total power of the three-phase load. To do this, you will need to multiply the phase-voltage by the phase current by the power factor -for each phase.
Total KVA of the transformer divided by (square root of 3 times the voltage). This will give the individual phase currents. These individual phase currents will be 120 degrees out of phase with each other.
50 HP is 37.3 kW so allowing for a 90 % coversion efficiency the motor draws 41.4 kW. It might have a power factor of 0.8 so the VA drawn would be 51.8 kVA which is 17.3 kVA on each phase. On a 440 v system the line to neutral voltage is 254 v therefore the line current on each phase is 68 amps. That is the estimated answer, 68 amps.