The 208 volt configuration is one phase of a three phase source at 240 volts, where the 208 volt circuit is connected between the center tap of one 240 volt phase (usually a grounded neutral, in the style of a standard 120/240 split phase system) and the high delta connection on either of the other two phases.
208 circuit would consist of two phases of a 208 volt wye system or could be all three phases. The voltage between conductors would be 208 volts. The voltage to ground from any phase would be 120 volts. A 240 volt delta system would give you a high leg to ground, somewhere around 190 volts and the other two would be 120 volts to ground.
A wire size of 250 MCM will limit the voltage drop to 3% over a distance of 200 feet.
If you're asking about load sizing, I think you already have the answer (at least 150A, although you could go up to 208 for full load, and maybe higher if you can overload the transformer). If you're talking about short circuit rating, you have entirely the wrong information. Short circuit rating will tell you if the fuse (if you're using fusing) mounting is rated to interrupt the available fault current.
It depends on your definition of efficiency. 480 can certainly push more power than 208 through the same size conductors, but it would not be efficient to wire a motor for 480 when 208 was all that was needed.
An open circuit or a short-circuit (if that circuit is complete).
Single phase power from a L-L voltage of 208 volts is calc'd: P = V(phase to phase) * I (individual phase current) / sqrt(3) Total power from all three legs is the above P * 3. P is in watts; continuous sampling will result in watt hour measurements; 1000 x this is kWh's.
Because its designed to operate at 230-460, not 120-208.
You don't. Every circuit needs a return wire back to the source.
1,040 watts or 1.04 KW.
Yes, with the proper control equipment ahead of the motors.
208 v is a three-phase supply voltage used in North America, carried on a 4-wire system. The voltage between one of the three live wires and the neutral wire is 120 v and a single-phase supply can be taken by using one live and the neutral. If in doubt about the connection consult an electrician.
To calculate the approximate power consumed by a 10 kW heat strip rated at 230 V when connected to a 208 V circuit, we can use the formula for power: ( P = V^2 / R ). Since the heat strip is designed for 230 V, its resistance ( R ) can be calculated as ( R = V^2 / P = (230^2) / 10000 = 5.29 , \Omega ). When connected to 208 V, the power consumed will be ( P = 208^2 / 5.29 \approx 8.04 , kW ). Thus, the heat strip will consume approximately 8.04 kW when connected to a 208 V circuit.
For a 225 amp 208 volt 3 phase circuit, you would typically need to use a 3/0 AWG copper wire or a 250 kcmil aluminum wire. Be sure to consult with local electrical codes and regulations to ensure compliance with specific requirements for your installation.
A 208/230 volt single phase appliance requires a dedicated electrical circuit with a voltage of 208 to 230 volts and a single phase power supply. The appliance should be connected to a properly rated outlet or wiring to ensure safe and efficient operation.
20-amp breakers can be found for many different voltage levels. The higher-voltage ones tend to be more expensive. A breaker can be used in a circuit of equal or lower voltage than the rated voltage of the breaker.
208/10 = 20.8 208/100 = 2.08 208/1,000 = 0.208
60% of on 208 = 40% left of 208 = 208*40/100 = 83.2
It is: 208 times 208 = 43,264 square feet