The equation for amperage is Amps = Watts/Volts = 10000/240 = 41.6 amps. It is not stated if this is a motor load or not. If it is a motor load then the feeder has to be rated at 125% x 41.6 = 52 amp wire. Also if it is a motor load then the breaker will have to be 250% x 41.6 = 104 amps. A 100 amp breaker will work and still be within the electrical code ruling. Otherwise for 41.6 amps a 50 amp breaker will be sufficient.
I = W/E. Amps = Watts/Volts = 6500/240 = 27 amps. A #10 copper conductor with an insulation factor of 90 degrees C is rated at 30 amps. If the load is continuous the conductor can only be loaded to 80% of its capacity. 30 x 80% = 24 amps which is too low for your load. Move to the next size wire which is a #8 which is rated at 40 amps x 80% = 32 amps. As long as no long distance runs are involved this wire size will work well.
Generators are sized by the electrical load that is to be applied to it. If you want to pick up the total house load then a 25 to 30 kW generator will do the job. At 240 volts this will give you just over 100 amps to work with. Amps = Watts/Volts.
Amps * Volts = Watts So, Watts / Volts = Amps 2000 / 240 = 8.333 Amps You should run the circuit on a two pole 15 Amp breaker, using 14 AWG, 2 conductor (plus ground) wire, just so you have a little safety factor in the circuit size.
The sizing of a breaker is dependant upon the size of the wire it protects. The size of the wire is dependant upon what the load current in amps is. Without the amperage and wattage and the voltage of the appliances an answer can not be given. W = A x V, Amps = Watts/ Volts.
5 HP is equal to 3.7 KW. 415 Volts 3 phase 50 hz system takes 6.29 amps. Same 3.7 KW motor with 230 volts, single phase 50 hz takes 18.9 amps. Same 3.7 kw motor with 230 volts three phase 50 Hz takes 11.1 amps. These are full load current with 0.85 assumed power factor. To arrive at over load, you may 5 to10% over load and set the over load setting.
To answer this question a voltage needs to be stated. Wire is sized by the amount of amperage the load takes. W = Amps x Volts. Amps = 650/ volts.
The equation for amperage is Amps = Watts/Volts = 10000/240 = 41.6 amps. It is not stated if this is a motor load or not. If it is a motor load then the feeder has to be rated at 125% x 41.6 = 52 amp wire. Also if it is a motor load then the breaker will have to be 250% x 41.6 = 104 amps. A 100 amp breaker will work and still be within the electrical code ruling. Otherwise for 41.6 amps a 50 amp breaker will be sufficient.
Read the specification plate on the motor to determine the amount of current that the motor uses, as well as the voltage and phase. Most likely, it will be a 3 phase motor, so you can't simply connect it to your house wiring, but without sufficient information, it is impossible to tell you all of the specifics. <<>> The code book rates the amperage of a three phase 40 HP motor by different voltages. at 200 volts - 120 amps, 230 volts - 104 amps, 460 volts - 52 amps and 575 volts - 52 amps. A breaker for a motor has to be sized to 250% of the motors full load amps. Also the wire size for a motor has to be 125% of the motor full load amps.
I = W/E. Amps = Watts/Volts = 6500/240 = 27 amps. A #10 copper conductor with an insulation factor of 90 degrees C is rated at 30 amps. If the load is continuous the conductor can only be loaded to 80% of its capacity. 30 x 80% = 24 amps which is too low for your load. Move to the next size wire which is a #8 which is rated at 40 amps x 80% = 32 amps. As long as no long distance runs are involved this wire size will work well.
This depends solely on the load you might need a 15 amp service or a 1200 amp service. To find out look at the installation instructions and it will tell you what size the service supply needs to be. Watts = Amps times Volts Amps = Watts divided by Volts Volts = Amps divided by Watts Example 1500 Watt unit will pull 12.5 Amps at 120 Volts or 1500 Watt unit will pull 3.26 Amps at 460 volts or A 460 volt unit rated for 34 amps will be using 15.64 kWh or 15,640 Watts Also the higher the voltage, the lower the Amperage, however, the wattage will stay the same <<>> The question is asking about two different values. Voltage is an insulation value whereas amperage is a conductor size. The basic insulation values are 300, 600 and 1000 volts. In the question on this service, a insulation factor of 600 volts will be used for the conductors. The amperage of a service is based on the connected load that needs to be supplied. The load can be an individual load or as in a distribution panel a variety of loads. At 460 volts, the service can be either a three phase service or a single phase service. Once the load amperage has been established, the wire size for the service can be calculated. Once the wire has been sized, the breaker used to protect the wire from being overloaded will be calculated.
The formula you are looking for is Watts = Amps x Volts. Amps = Watts/Volts. This comes to 4 amps load. Minimum size fuse would be 5 amps.
I = W/E. Amps = Watts/Volts = 6500/240 = 27 amps. A #10 copper conductor with an insulation factor of 90 degrees C is rated at 30 amps. If the load is continuous the conductor can only be loaded to 80% of its capacity. 30 x 80% = 24 amps which is too low for your load. Move to the next size wire which is a #8 which is rated at 40 amps x 80% = 32 amps. As long as no long distance runs are involved this wire size will work well.
The size of conductor needed will be a #14. It should be copper and have an insulation factor of 90 degrees C. The fact that it is a three phase load does not enter into the calculation of the wire size.
The size of the generator is based on the size of the load you want to supply. Size up the load in watts or amps along with what phase (single or three) and voltage that the load requires. These are needed to give a complete answer.
Generators are sized by the electrical load that is to be applied to it. If you want to pick up the total house load then a 25 to 30 kW generator will do the job. At 240 volts this will give you just over 100 amps to work with. Amps = Watts/Volts.
A # 14 copper conductor will be fine to carry 8 amps at 120 volts. This size conductor is rated at 15 amps.