i=v/R and R=rho.L/A . by analysing these fourmulas we find the size . here i is electric current in the conductor to flow or load current to flow . v is the voltage across the conductor , R is the resistance of the conductor . L is the length of the conductor ,rho is a constant of a material called specific resistance here the the material is the material of the conductor , A is the area of cross section of the conductor .
refer to ANSI lor..
350mcm
120mm
To answer this question the size of the cable or the amperage of the load is needed to calculate the correct wire size and connectors to fit the cable.
Yes I could. How?
The Full load current of that amp is approx. 17A, but you base you fuse size between the Full load current of your load and the current carrying capacity of the supply cable, if the size or the CCC of the cable is unknown, I would recommend sticking as close to the FLC as possible
If you know the size of the load to be served, multiply it by 125% ( times 1.25) and choose a conductor that is rated for that ampacity or higher.
Assuming being fed by c32 breaker for start load minimum cable size SWA is 70mm which will go to 456 meters
100 kw is the power drawn by the load. to calculate the cable size you need to know the voltage. From that you can calculate the current. this decides the cable size. for example if the voltage is 400Volts then the current flowing in the circuit when the load is 100 KW will be 250 amps. (100,000/400). for 250 amps to flow with out causing excessive heating of the cable the cross sectional area of the copper cable should be 150 sq mm.for a round cross section the diameter will be roughly 15mm.
35 KVA generator using for load 100 meter length which cable need to used
4sqmm
To answer this question a voltage is needed.