The kV.A (not 'kva') rating is the total apparent power of the machine. So a 75 kV.A machine is 25 kV.A per phase.
It depends upon the Generator system voltage. For 3 Phase, 600 Volt system, it will be 73 Amps For 3 Phase, 480 Volt system, it will be 90 Amps For 3 Phase, 208 Volt system, it will be 208 Amps
A transformer requires a 75-kVA minimum load in order to be set.
200 and 100
100 is to 800 as 75 is to X. Cross multiply, 100 x X and 800 x 75, 100X = 60000. To get rid of the 100 divide it into both sides of the equation. X = 600. If the copper loss is linear, at 75 percent the copper loss will be 600 watts.
The kV.A (not 'kva') rating is the total apparent power of the machine. So a 75 kV.A machine is 25 kV.A per phase.
The formula you are looking for is , A = kva x 1000/Volts.
It depends upon the Generator system voltage. For 3 Phase, 600 Volt system, it will be 73 Amps For 3 Phase, 480 Volt system, it will be 90 Amps For 3 Phase, 208 Volt system, it will be 208 Amps
Multiply by Amps.
12HP is approximately 10.8 KVA. You would want to use a 15KVA transformer to supply this motor. KW = HP * .75 KVA = KW * 1.2 (These formulas are approximate)
Rating for DG set and any of electrical machines is calculated in KVA. KVA is calculated as KW/pf. One can calculate the required KVA for DG set with this formulation: (KW/pf)/load rate. For example KW=110, pf=0.8 and one loads the DG at 75%, so KVA= (110/0.8)/0.75=185 KVA.
At full load a new 400 kva genset will use approx 80 Litres /hr, at 75% load approx 62 L/hr and at 50% approx43 L/hr...assume 0.8 pf
speculation is that it consumes about .75 ltr of diesel in an hour
A transformer requires a 75-kVA minimum load in order to be set.
About 55,970 watts.
200 and 100
75 Amps theoretically Need to know if the generator is 3 phase or single phase.