If we have 15000kw connected load in a factory than how much
power demand sanction should be taken from electricity board.
Pls provide complete calculation.
Answers were Sorted based on User's Feedback
Answer / ashish negi
the above calculation is correct but if we have to give our
demanded load to electricity dept. than the load should be
greater than the load as on starting motors , motors take 2-
3 times the full load cureent and at that time your MDI
(maximum demand index)will go beyond the calculated load.
so in that case your have to pay the penalty. so your
demanded load should be between 2500-3000 KVA.
Is This Answer Correct ? | 5 Yes | 0 No |
Answer / rakesh maharjan
Remember the formula
KW=1.732 x V xI x PF
or
KW= KVA x PF, where PF = 0.8
connected load KW= Full load kw x 80%
Here in your case, connected load KW= 15000kw
so full load KW= 15000/80%=18750 KW
KVA= KW/.8
KVA=18750/.8=23437 KVA, this is the demand sanction.
I am junior Engineer.
Please verify by senior designer also.
Is This Answer Correct ? | 5 Yes | 3 No |
Only the connected is mentioned and not the peak load.
Maximum demand is registered only if that apparent load
lingers for more than 15 to 30 minutes (as set).You have to
consider load factor(?) also .Apply for 12000 KVA MD
initially and you can apply for higher ,later.Once you get
the MD sanctioned, you can not get it reduced during the
agreement period.
Is This Answer Correct ? | 1 Yes | 0 No |
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