Answer / m p narayanan

The question is about the KW saving if pf is improved from
0.92 to 0.98. Answer #1 is silent on the KW saving, but
logically says that 80 KVAR is needed for the pf
improvement (say, by installing a local capacitor bank).
Here , there is a calculation mistake. The value should be
77.99 KVAR instead of 80 KVAR.
Answer #2 also has calculated the same thing and wrongly
declared that it is the KW saving.
The fact is that there is no direct saving in KW by
improving the pf. If 350 KW is needed by the load, it has
to be supplied by the grid. Pf improvement is done for
reducing the KVAR requirement from the grid, not the KW. A
350 KW load at 0.92 pf takes 149.07 KVAR from the grid
whereas the same load of 350 KW at 0.98 pf takes only 71.08
KVAR from the grid, the difference of 77.99 KVAR is
generated locally by the Capacitor bank. The corresponding
KVA at 0.92 pf and 0.98 pf are 380.43 KVA and 357.14 KVA
respectively.
The industries install capacitor banks to improve the pf
with the specific intention of saving electricity bill
because power utilities have a two part tariff – part-1 for
the maximum demand (ie; KVA) and part-2 for the KWH
consumed. This is necessary because the line current is
directly proportional to the KVA and hence the overhead
line conductor size, KVA rating of the transformers
upstream, size of the circuit breakers etc. are dependent
on this. The less the line current, the less will be the
transmission losses. Saving in KW is the result of
reduction in line losses. Hence, as indicated above, there
is no direct saving in KW on account of improvement in pf,
but indirectly the power utility company save KW by
reducing the line losses and also, they save money by
investing less on the size of transformers etc. The quantum
of saving depends on the resistance of the transmission
line, resistance of the transformer windings etc. Hence it
can not be calculated with the available data given in the
question.

Answer / pillay

if the load is 350kw at o.92 and also at 0.98 than there is
no kw saving unless you are talking about the losses you
reduce as what Mr MS Narayanan has mentioned.

Answer / joel q.

You need to connect a power factor corrector and the size
of the capacitor is 80KVAR.

350tan23 - 350tan11 = 80KVAR

Answer / engr mimkhan

IF your system losses is 5 %
Then loss reduction =1- (squr(.92)/squr(.98))
=0.13
=0.13 x 0.05 (5%)
=0.0065
Thus saving in kw =0.0065 x 350
=2.27 kw
This is your saving if system losses increase than kw saving will also increase..
Note squr mean Squre not root.

Answer / adil naseem(pakistan)

cos(pi1)=0.92 //old power factor
cos(pi2)=0.98 //desired power factor
the forlmula for saving of Kw is:
total Kw= 350kw



Saving KW = total kw * (tan(pi1) - tan(pi2))
=350 * (0.4259 - 0.203058)
=77.9947Kw is the answer!(100% correct)

if the HR asked to me., you have got 92% in higher secondary but got only 72% in engg whats the reason??what should i have to say ..please help me..

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