The formula you are looking for is I = W/E.
For a 7000watts or 7kw a 240V then it is 30 Amp
If the load is resistive (heater) then the breaker size would be a two pole 40 amp breaker. If the load is inductive (motor) then the breaker size would be a two pole 70 amp breaker
9000/191=63A
The demand load would be 12 kW. When you talk about line and neutral you are referring to the voltage that the equipment operates on. In North America, an electric kitchen range operates on two voltages 120/240 volts. The 240 volts is obtained across two breakers in the distribution panel and the 120 volts is obtained between either 240 volt leg and the neutral. On a range of this size the 120 volts is usually used for the clock and controls for the range. The range at maximum would draw 50 amps, fed by a two pole 50 amp breaker using a #6 copper conductor with an insulation factor of 90 degrees C. The range will have an internal breaker or fuse to protect the 120 volt circuit.
2.3 kw per hour on a 110-120 volt circuit.
1KW is 1000w. one watt is equals to product of volt and current. current=watt/volt =1000/240 =4.1667 amps
If your power source is 120 V then 8000/120 = 66.7 Amps. If operated at 240 V then it is 33.3 Amps. In the first case you would need 3 AWG and in the second 8 AWG.
P = E^2 / R P = (240 x 240)/ 8 = 7200 watts If this is coninuously energized for 24 hours: 7.2 kw/hr x 24 = 172.8 kw/hrs
AWG #3 copper.
The demand load would be 12 kW. When you talk about line and neutral you are referring to the voltage that the equipment operates on. In North America, an electric kitchen range operates on two voltages 120/240 volts. The 240 volts is obtained across two breakers in the distribution panel and the 120 volts is obtained between either 240 volt leg and the neutral. On a range of this size the 120 volts is usually used for the clock and controls for the range. The range at maximum would draw 50 amps, fed by a two pole 50 amp breaker using a #6 copper conductor with an insulation factor of 90 degrees C. The range will have an internal breaker or fuse to protect the 120 volt circuit.
120 15 amp service ? 210 7 amp service ?
2.3 kw per hour on a 110-120 volt circuit.
1KW is 1000w. one watt is equals to product of volt and current. current=watt/volt =1000/240 =4.1667 amps
10 mm2 cross section should be sufficient (#6 wire?)
A breaker protects the wire size of the feeder that is connected to it. The amperage of the load must be found. Without a voltage stated the amperage from the wattage given can not be calculated. The equation for amperage when the kw is given is A = kW x 1000/1.73 x volts x pf. The pf constant to use is .9.
If your power source is 120 V then 8000/120 = 66.7 Amps. If operated at 240 V then it is 33.3 Amps. In the first case you would need 3 AWG and in the second 8 AWG.
P = E^2 / R P = (240 x 240)/ 8 = 7200 watts If this is coninuously energized for 24 hours: 7.2 kw/hr x 24 = 172.8 kw/hrs
I=270000/380/1.732 I=410A USE: 500A CIRCUIT BREAKER
the given kw Divide by the your voltage
3000 / 240 = Amps. You de-rate a breaker by 20 % for continuous load like an oven. You could get by with a 20 Amp breaker and 12 AWG wire. However, I would recommend 30 Amps and 10 AWG for an oven for the long run.