watts = amps x volts
400amps x 480V = 192,000 watts
192,000/1000 = 192KW
You would need 192KW generator at 277/480 <<>> The formula for kilowatts is, kW = I x E x 1.73 x pf/1000. kW = 400 x 480 x 1.73 .90 = 298944/1000 = 298.9 or 300 kW generator.
For a 3-phase service that can supply 400A, a 333kVA transformer would be needed. If you aren't planning on using that much power, a smaller transformer can be used. Also, if load is not constant, you can overload Transformers to a degree.
A #600 MCM wire with an insulation factor of 75 and 90 degrees C is rated at 420 and 455 amps respectively.
Most 400 amp services disconnects will use parallel wires feeds on each leg to feed the circuit. Parallel conductors will be #3/0
A #3/0 wire with an insulation factor of 75 and 90 degrees C is rated at 200 and 210 amps respectively.
The size/mass of the transformer is set by the power that is being put through the transformer. One for 1 kW would be a lot smaller than one for 100 kW.
Type your answer here... a very big one
50KVA
10 AWG in copper.
14 AWG is fine for this application.
What size breakers are needed for a 30kva transformer 208 volt feed 600 volt out put
Fuses are rated in Amps. Although the physical size of a fuse is to do with volts; the further the terminals are apart the less likelihood there is of 'sparkover' between them.
10 AWG.
The size of grounding wire is based on the amperage output of the transformer. The voltage of the transformer needs to be stated. Without this voltage a calculation can not be made. Amps = Watts/Volts = 30000/?.
#8 copper
On a 1kva you have 1000 watts capacity. To fine the current the formula is I = W/E. The secondary side of the transformer has the capacity of 1000/120 = 8.3 amps. In your question you do not put the amps across the secondary you draw amps from it. Using the transformer to its maximum, without overloading it, the primary will be 4.16 amps at 240 volts and the secondary will be 8.33 at 120 volts. <<>> voltage times amps equals wattage
10 AWG in copper.
Depending on size of Fridge. But AVERAGE is 12 volts for fridge, circuit necessity 15 amps 15 amps X 120 Volts=1800 watts minimum...I'm LEARNING myself
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.
14 AWG is fine for this application.
A # 14 copper conductor will be fine to carry 8 amps at 120 volts. This size conductor is rated at 15 amps.
What size breakers are needed for a 30kva transformer 208 volt feed 600 volt out put
There is not enough information provided to answer. KVA is short for "Kilo Volt Amperes". That is, thousands of Volt Amps. In order to determine how many Amperes are flowing, you must know at what voltage it is operating. Amperes = 45,000 ÷ volts Bill Slugg
The V stands for volts and A is amps. If for example you have a 12kVA device and are running off a voltage of 120 volts then Amps = 12000/120 = 100. You then use the calculated amps in a wire size table to get the correct size.
Each baseboard heater will draw a little over 4 amps at 120 volts or 2 Amps at 240 volts. The total number of baseboards on a circuit will draw the sum of these amps. Keep the load under 80% of the amperage rating of the breaker.