The (ideal) voltage is 480V line to line. The line to neutral voltage (if it is a grounded system - 4 wires) is 480 / 1.732 = 277 volts.
The actual measured voltage may be slightly higher or lower than this.
50 kVA is 16.667 kVA per phase and you divide that by the phase voltage. Current = 16667 / 277 = 60 Amps
746 Watts per horsepower / 480 volts x power factor x efficiency x 1.73 = amps assuming that the motor is three phase. 746 x 60 =44760 watts divided by 480 x 1.73 = 53.9 amps ( If the pf and eff. information is not known this will put you in the ball park)
It doesn't 'equate' to current (amperes) because they are two quite different quantities. However, you can find out how much line current is flowing (for a balanced load only) if you divide the number of volt amperes by (1.732 x VL), where VL is the line voltage.
To calculate the current per phase, use the formula: Current (I) = Power (P) / (Square Root(3) x Voltage (V)). For an 18kW motor at 415V, the current per phase will be: 18,000W / (1.732 x 415V) ≈ 24.5A per phase.
If you have 208 between legs, you have 120 from each leg to neutral (208 divided by 1.73). Assuming a balanced load, this would be 1 KW per leg - 1000 divided by 120 = 8.33 amps per leg. Your question contradicts itself. If you have 208v, the answer above is correct, 8.33 amps. If you have 480v, you have 3.6 amps.
90 kW on 480 volts single phase would be 187.5 amps. On 480 v 3-phase it would be 108 amps.
50 kVA is 16.667 kVA per phase and you divide that by the phase voltage. Current = 16667 / 277 = 60 Amps
746 Watts per horsepower / 480 volts x power factor x efficiency x 1.73 = amps assuming that the motor is three phase. 746 x 60 =44760 watts divided by 480 x 1.73 = 53.9 amps ( If the pf and eff. information is not known this will put you in the ball park)
1.73*480*22
I(per Phase)=4000/(230*3) = 5.8A
By 'volts per phase', I assume you mean 'phase voltage' as opposed to 'line voltage'?It depends on the country in which you live. In North America, for example, the secondary output of a three-phase distribution transformer typically delta connected, 240 V corresponds to both the phase and line voltages.In other countries, where a three-phase distribution transformer's secondary is wye connected, 240 V is typically a phase voltage and 415 V is a line voltage.
Normally, it is zero.Except in very special cases, the neutral and ground (earth) conductors in a building are tied together at one point in the system, so ideally the voltage difference would be zero. The reason that it might not be zero is there is current flowing in the neutral and, thus, voltage drop in the neutral conductor. Since the ground conductor normally never has current flow (unless there is a fault), there will be a difference in voltage equal to the voltage drop across the neutral conductor, which varies with load (current).It should be 0V , but as per our earth pit maintaining that voltage will be varied even also not exceed 5V.
It doesn't 'equate' to current (amperes) because they are two quite different quantities. However, you can find out how much line current is flowing (for a balanced load only) if you divide the number of volt amperes by (1.732 x VL), where VL is the line voltage.
The maximum single phase HP motor listed in the CEC is 10 HP. At 115 volts 100 amps and 230 volts 50 amps.
Homes in Canada use 240/120 volts at 60 cycles per second (60Hz) single phase. It is a split-phase system that delivers 240 volts to large domestic appliances (e.g. washing machines, dryers, air-conditioners, etc.) and 120 volts to lights and general socket outlets used for small appliances (coffee machines, electric shavers, televisions, etc.).
Current is about 18 amps per phase. Allowing for power factor and startup, I'd size the wire for 40 amps per phase, which is #5 wire. This wire has a resistance of 0.39 ohm per 1000 ft, which is 0.39 ohms per 305 meter, or 0.09 ohms per 70 meter. At 40 amps, there is a voltage drop on the wire of 40*.09 or 3.6 volts, less than 1% loss, so that is OK. Use 4 conductor #5 copper wire insulated for 480 volts. <<>> A #12 copper conductor will limit the voltage drop to 3% or less when supplying 13 amps for 230 feet on a 415 volt system.
To calculate the current per phase, use the formula: Current (I) = Power (P) / (Square Root(3) x Voltage (V)). For an 18kW motor at 415V, the current per phase will be: 18,000W / (1.732 x 415V) ≈ 24.5A per phase.