5000 volt
No, a 4160 volt motor with a high inertia load will use a resistor bank in the starting of the motor.
In a 4160VAC system, the voltage of one leg (phase) is measured as 4160 volts between that leg and the neutral point in a wye configuration, or between that leg and the ground in a delta configuration. However, the phase-to-phase voltage is 4160V, meaning each leg carries the same voltage when measured against the other phases. If you divide the line-to-line voltage by the square root of 3 (approximately 1.732), the phase voltage in a wye system would be about 2400 volts.
Wiring a 4160-volt motor involves connecting the motor's terminals to a compatible power supply and ensuring proper grounding. Start by ensuring that all power sources are turned off, then connect the three-phase power lines to the motor's terminals, typically labeled U, V, and W. Make sure to follow the motor's wiring diagram for correct connections and phase orientation. Finally, secure all connections, double-check for any loose wires, and ensure proper grounding before restoring power to the system. Always consult local electrical codes and regulations.
Wow your in some serious stuff. Are you worried about rotation. 4160 is not something to be messed with if your not a qualified electrician or lineman. If you mess up you can have a big fire ball on your hands.
To answer this question voltage, power factor, efficiency, and run configuration must be known, none of which was given in the question. I will make some assumptions, to shown how the calculation is done. Given: 50 HP, three phase. Assumed: 480 volts delta, 0.92 PF, 0.95 EF, and WYE-DELTA start-run configuration. 1 HP (electric motor) is 746 watts, so 50 HP is 37.3 KW. Divide by PF 0.92 and by 0.95 EF to get 42.7 KVA. Divide by 480 V and by 3 to get 29.6 A per winding. Multiply by 1.732 (square root of 3) to get 51.3 A per phase. Now, 51.3A is a lot of current, requiring large windings and, therefore, a large motor. I would consider a higher voltage motor, such as 4160. Rerun the calculations and you get 5.9A per phase. Does this justify a 4160 service? Its marginal, but you begin to see the issues involved in choosing a motor.
To calculate the amperage of a 1250 HP motor at 4160 volts in a three-phase system, you can use the formula: Amperage (I) = (Horsepower × 746) / (Voltage × √3 × Power Factor). Assuming a power factor of around 0.9, the calculation would yield approximately 174.5 amps. Therefore, a 1250 HP motor at 4160 volts would draw around 174.5 amps in a three-phase system.
You can use an insulation rating 5000 volt to run 4160 volt supply.
No, a 4160 volt motor with a high inertia load will use a resistor bank in the starting of the motor.
4160 - 800 = 3360
317.025280 KILOWATTS = 317,025.28 WATTS
Answer: 4160 mi. = 6694.871 km
Answer: 4160 mi. = 6,694.871 km
In a 4160VAC system, the voltage of one leg (phase) is measured as 4160 volts between that leg and the neutral point in a wye configuration, or between that leg and the ground in a delta configuration. However, the phase-to-phase voltage is 4160V, meaning each leg carries the same voltage when measured against the other phases. If you divide the line-to-line voltage by the square root of 3 (approximately 1.732), the phase voltage in a wye system would be about 2400 volts.
64 x 65 = 4160
4160 ÷ 21 = 198 with remainder 2
There should be no voltage to ground on a delta system. That is the reason that these types of systems have to have a set of grounding lights to warn when the system inadvertently grounds. A delta system is a three wire connection, no neutral. A voltage to ground is only available on a three phase four wire (star or wye) connection system. The fourth wire being a neutral which is grounded thereby giving a voltage from each leg to neutral.
The Wye (also know as Star - especially in the motor rewind industry) is a 4-wire system which provides two different supply voltages. The center-point of the Wye is the system neutral and is usually solidly grounded. Where it is desirable to limit the phase-to-ground fault magnitude the center-point of the Wye may be connected to ground through and neutral grounding resistor or a current limiting reactor. Because the system is tied to ground it is easy to provide system ground fault protection. Three-phase loads can be connected phase-to-phase and singlephase loads can be connected from any phase to the system neutral. On a wye system, the phase unbalance currentis carried by the system neutral. On a Wye system the line current is equal to the phase current i.e. ( ILine 1 = IPhase A) and the line-to-line voltage is equal to the vector sum of two individual phase voltages i.e. (E Line1-2 = E PhaseA + E PhaseB' ). In a Wye system the phase-to-phase voltage is 1.732 x the phase-to-ground voltage. Some typical Wye system voltages are: 120/208Y, 277/480Y, 2400/4160Y, 4160/7200Y, 7200/12470Y, 7620/13200Y,and 19920/34500Y