Answer / prakash veer singh

IT is very sample . Normaly DG's power factor is about to
0.8 .
We multipli the pf from kva , we get KWH of DG.

KWH= KVA*pf
KWH= 725 * 0.8
= 580 KWH

Answer / satish s.nanal

The Term Ratio of KWH itself is incorrect.

KWH means Energy Consumption.It is multiplication of Watts
( Or Kilowatts) consumed over a period of time.Now
considering Power Factor of 0.8 , KW shall be = 725*0.8
= 580 KW.
Now if 580 KW are consumed for say 2 Hours then Energy
Consumption will be 580*2= 1160 KWH

Answer / ramali

In any DG sets, power factor should be mentioned.Either it
is 0.7 to 0.8.
so, first we calculate the KW,

KVA*pf=kw
725*0.7=507.5kw
725*0.8=580Kw
now KWH is according to power consumption in your DG.
Hence, your 725KVA DG should gives 580Kw per hour max.
this calculation contains temp,cable laying method,cable
quality,lugs, crimping,fixing bolt nuts etc.,

Answer / bala

power=1.73*V*I*p.f kwh

V- Average voltage per hour
I- Average current drawing per hour
P.F- Average power factor per hour

Answer / girish chandra adhikari

The term of KWh is itself incorrect. Rather we should say
how to calculate the DG. Ratio ?

For calculating DG.ratio we should calculate the total unit
consumption ( Kwh ) and also calculate the total HSD(Fuel)
consumption. Then the DG. ratio would be Total KWh/Total
HSD Consumption in one hrs.

Suppose your 725 KVA DG Unit consumption in one hrs is 400
and HSD.consumption is 110 Lts. DG.ratio = 400/110 =3.63

Answer / jagjeet

power=1.73*V*I*p.f kwh*efficiency

V- Average voltage per hour
I- Average current drawing per hour
P.F- Average power factor per hour

Answer / amit sonmale

Its verry Simle 1stlly I want to know The Current
CApacity Of d.g Set It means I=Kva/P.f
725/0.80
=906.25 Amps.
Then Know Load Capacity Kw =V*1.732*.80*I/1000
=415*1.732*0.80*906.25/1000
=521.1155Kw

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