The formula you are looking for is I = W/E.
For a 13.8 kW electric heater at 208 volts, you would need a 3-phase circuit with a minimum of a 50-amp breaker to handle the load safely. This calculation is done by dividing the power (13,800 watts) by the voltage (208 volts) and then dividing by the square root of 3 (since it's a 3-phase system).
The formula you are looking for is I = W/E, Amps = Watts/Volts. Amps = 5000/230 =21.7 amps. The wire size to run this heater would be a #10 copper conductor. The supply breaker would be a two pole 30 amp breaker.
A 3 kW immersion heater circuit should be provided with a dedicated 15 Amp circuit that is protected by a 15 Amp circuit breaker and wired with appropriately sized cables (typically 14 AWG for residential wiring in the US). Additionally, a Ground Fault Circuit Interrupter (GFCI) protection may be required for safety.
The heater element is a coiled wire resistor that draws enough current to supply the intended amount of power, which might be 1.5 - 3 kW. Quickly the temperature of the wire rises until it reaches an equlibrium where the heat power conducted away by convection is the same as that draw from the power supply.
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.
The formula you are looking for is I = W/E.
AWG #3 copper.
The use of a breaker in a circuit is to protect the wire size used in the circuit from becoming overloaded. Using the wattage of the load does not help for breaker sizing because the breaker operates on amperage. Amperage can be found from wattage by using the following equation. I = W /E. Amps = Watts / Volts. As you can see the amperage can not be calculated because there is no voltage stated.
For a 13.8 kW electric heater at 208 volts, you would need a 3-phase circuit with a minimum of a 50-amp breaker to handle the load safely. This calculation is done by dividing the power (13,800 watts) by the voltage (208 volts) and then dividing by the square root of 3 (since it's a 3-phase system).
The formula you are looking for is I = W/E, Amps = Watts/Volts. Amps = 5000/230 =21.7 amps. The wire size to run this heater would be a #10 copper conductor. The supply breaker would be a two pole 30 amp breaker.
Yes, the total amperage load of a 2000 watt heater at 240 volts is 8.3 amps. Keep in mind that the wire feeding the heater must be a #10 because the breaker is rated at 25 amps. A wire's ampacity rating can be larger that the breaker amperage rating but never smaller. Example, a #14 rated at 15 amps or a #12 rated at 20 amps can not be connected to a 25 amp breaker. The 25 amp breaker does not trip until it reaches 25 amps well over the allowable amperage of the #14 amd #12 wire. This is why a #10 wire must be used as its rating is 30 amps.
10 kW at 220v will use 45.45 Amps. I'd recommend a 60A circuit, with a 60A breaker, but the heater or electric furnace should have "maximum fuse amps" rated on its nameplate. Above all, neverconnect wiring to a breaker that is rated higher than the maximum current capacity of the wire. In the case of 60A, use #6 AWG wire.The HVAC Veteran
To answer this question the voltage of the system is needed.
To answer this question the output voltage of the generator must be stated. Breakers are sized to protect the wire that is connected to it. Wire is sized by the amperage that it can carry. Once the voltage is stated use the following equation Amps = kW x 1000/1.73 x Volts x pf. for three phase and Amps = kW x 1000/Volts x pf. In both cases use .9 for a power factor value.
I=270000/380/1.732 I=410A USE: 500A CIRCUIT BREAKER
the given kw Divide by the your voltage