Answer / mohan

Formula:1.73 X V X I/1000

180X1000/1.73X415 =I

Full load current 250A.

But we should have take only 80%

Answer / santosh jagale.

Generator rating is 180KVA.
Actually all electrical data is made avialable on genrator
name plate by manufacturer.

If you consider the PF is 0.8 Lag.

Then KW=KVA*PF
i.e. KW=180*0.8
=144 KW
P(KW)= 1.732*V*I*PF
Then,

144=1.732*415*I*0.8

I= (144/1.732*415*0.8)*1000
I= 250.42 Amps.

This is the full load current of generator with 0.8 PF &
415V.

In case of DG sets the generator O/P power is restricted by
power of driving engine, beyond that limit you can,t load
the particular generator. You can put upto 125% of to
genrator but not for too long period.

Answer / ranjith18

BETTER AS THUMB RULE
I = kVA X 1.4
ie
180 X 1.4 = 252A (AS FULL LOAD)

Answer / gins

Now a days DG manufactures assuring that the DGs can be
loaded up to 120% of its rated capacity. But it is always
good to design DG for a maximum load of 80% of the rated
capacity to get the maximum efficient running. In your case
You can load a maximum load of 125.8 kW at pf 0.85. So you
have to give an incomer of MCCB with rating suitable for
125.8kW.

W=root3 x VI x pf
so I = W/root3 x V x pf
I think 250A, 16kA, 4Pole MCCB with 80% plug setting may be
suitable for you.

Accordingly select your incomer with suitable kA rating
with respect to your total load.

Answer / mubashar ali

I = KVA/ 1.732*V
or
I = KW/ 1.732*pf*V

u can also use the fector as mantion Ranjith

I = KVA * 1.4

Answer / ahmed umar

Since the Genset Rating is given in KVA as 180KVA.

I think you should calculate the full load Amperage as
follows:

Formula
I=KVAx1000/Root3x400V

ie, 180x1000/1.732(400)
=180,000/692.82
=259A as full load Amperage.

Answer / deepak kapil

VA=1.732xVxI
180x1000=1.732x415xI
I=250A

Answer / ashvini kumar meena

People are unnecessarily trying to guess pf. Its not required.

To calculate KVA quantity you just need voltage and current.

Power factor comes into picture when you are talking about
real power(in KW.)

KVA = √3 * V * I

Here V is line Viltage and I is Line current. K will be for
Kilo. Here V line is 415=√3 * 240(phase Voltage) and I to be
fount.

Above Mr. Deepak Kapil in answer have solved it correctly.

Answer / hitesh

(Watt*1000)/(Root3*V*Pf)

Answer / adnan khan

180kva x1.44=259.2 ampere/ phase

prove that power consumed in 3 phase delta system is 3 times as compared to 3 phase star system

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