Hi to all!
I do not know if this is the right forum to ask this question. I'm new here and this is my first post.
I work for a small power generating plant which most of the time runs at 85% power factor. Like any other power plant, fuel consumption is computed in g/KWh or liter/KWh. I wish to know if it is correct to equate fuel consumption to kVA output? If so, does this mean correcting the line pf would translate to reduction in fuel consumption?
Thanks!

Pfc And Fuel Consumption
Started by krone, Oct 30 2007 02:52 PM
2 replies to this topic
#1
Posted 30 October 2007 - 02:52 PM
#2
Posted 30 October 2007 - 06:01 PM
Hello krone
Welcome to the forum.
The fuel consumption of the engine is related to the KW produced by the engine. It is not a linear function, but increasing the KW consumed will increase the fuel used.
A poor power factor will increase the KVA load on the alternator, but will not increase the KW consumed by the load. The effect of the increased KVA however, will be that there will be an increase in the losses in the alternator and distribution system. The additional copper losses will increase the KW loading on the engine and will therefore increase the fuel consumption.
If your copper losses are say 10% (alternator and distribution), and you increase the current by 40% for the same KW loading, then the losses will increase to twice what they were and become 20%, so you can see the advantage on reducing the current by pf correction.
Best regards,
Welcome to the forum.
The fuel consumption of the engine is related to the KW produced by the engine. It is not a linear function, but increasing the KW consumed will increase the fuel used.
A poor power factor will increase the KVA load on the alternator, but will not increase the KW consumed by the load. The effect of the increased KVA however, will be that there will be an increase in the losses in the alternator and distribution system. The additional copper losses will increase the KW loading on the engine and will therefore increase the fuel consumption.
If your copper losses are say 10% (alternator and distribution), and you increase the current by 40% for the same KW loading, then the losses will increase to twice what they were and become 20%, so you can see the advantage on reducing the current by pf correction.
Best regards,
Mark Empson | administrator
Skype Contact = markempson | phone +64 274 363 067
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#3
Posted 30 October 2007 - 10:58 PM
Thanks for welcoming me and shedding light on the matter.
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