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1kw=1.25kvaby this relation 75kva=60kwp=1.732*380*I*1I=60000/1.732*380*1I=91AmpsCommentPower factor is irrelevant to this question. You don't need to convert the kV.A rating to kilowatts, simply divide 75 000 V.A by (1.732 x 380). The answer is 114 A.This, of course, assumes that the question refers to a three-phase generator and that 380 V is the line voltage.
A 15 KVA - Generator operating at 380 Volts can cater to only 22.79 Amps. It cannot cater to a 60 Amps breaker
depends on the amperage. 14 AWG for 15 amps, 12 AWG for 20 amps, 8 AWG for 50 amps.
The formula you are looking for is Amps = kW x 1000/1.73 x E x pf. Use .84 for a power factor value.
volts = watts divided by amps amps = watts divided by volts watts = amps times volts so 266,000 watts divided by 380 volts = 700 amps and I might also point out that whatever it is you are talking about is very dangerous and can kill you in less than a heartbeat. I'd be sure to talk with an electrician if I were you if you plan on going anywhere near that.
Watts = Volts times Amps. Therefore, if the voltage was 220 volts, the motor would draw 500 amps. If the voltage was 4,000 volts, the motor would draw 27.5 amps. The voltages for large powerful motors tend to be relatively high, for example in the 380 Volts to 11,500 Volts range.
380
hb 063 360 to 380 cold crank amps...
The equation that you are looking for is; amps when kilowatts are known. kW x 1000/1.73 x Volts x pf.45 x 1000 = 45,000/208 x 1.73 x pf. Power factor to use will be .9. 45000/ 324 = 139 amps.
A 228 sq mm conductor equates to 450 MCM. A conductor of 450 MCM is not a standard AWG wire size. A standard 400 MCM will carry 380 amps. A standard 500 MCM will carry 430 amps. Difference between 400 and 500 MCM amperage's is 50 amps. Transposing between the two amperage's of 50 amps will be 380 + 25 = 405 or 430 - 25 = 405 amps. This is a very rough calculated answer for the question.
There are 3800 tenths in 380.
You would need to know how FAR the load is from the service point. Then use an online calculator or the voltage-drop formula to determine the minimum diameter of the conductors necessary to provide 95 percent of the supply voltage to the device at the other end. <<>> If distance is not a factor, the wire size is determined by the amperage the wire has to carry. Using the formula I = W/E, Amps = Watts/Volts = 3000/380 = 7.98 amps. A #14 copper conductor with an insulation factor of 75 or 90 degrees C is rated at 15 amps. This size wire has nearly double the capacity that you need.