For that current you would need five 0000 (4/0) wires in parallel for each side of the circuit, and they would heat to 90 degrees C if used for any length of time according to the NEC tables. Those wires have a diameter of 0.46 inch exactly.
There are better ways of transmitting 153 kW if there is a higher voltage available.
To calculate the wire size, a system voltage is needed.
10 AWG in copper.
A #10 copper conductor with an insulation factor of 90 degrees C is rated at 30 amps.
Canada and US - Standard size breaker in home panel is 40 amps for stove. If not standard range open Discuss Question page.
14 AWG is fine for this application.
To calculate the wire size, a system voltage is needed.
#8 copper
10 AWG in copper.
A #10 copper conductor with an insulation factor of 90 degrees C is rated at 30 amps.
Canada and US - Standard size breaker in home panel is 40 amps for stove. If not standard range open Discuss Question page.
A parallel run of 750 MCM AWG conductors will handle 1000 amps. if we want 1000amps to flow, 250sqmm cable is enough.
To answer this question the pump voltage is needed.
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
14 AWG is fine for this application.
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.
What size breakers are needed for a 30kva transformer 208 volt feed 600 volt out put
Conductor can be loaded to only 80% capacity. 312/ 80% = 390 amp wire is needed for 312 amps. A 500MCM conductor with an insulation factor of 90 degrees C is rated at 395 amps.