Answer / ashish kumar rai

We can improve power factor by:-
1) Capacitor
2) Synchronous motor
3) Synchronous Generator

When Power facto Cos(pi)= .8
KW = KW*Cos(PI) = 240
KVA = KW/ Cos (pi)= 300
KVAR = KW*Sin(pi) = 180

When Power factor Cos(pi)= 1
KW= 300
kVA = 300
KVAR= 0

So we use the capacitor of rating of 180 kvar to neutralise
the inductive 180 kvar.

Effect on consumer:-
1) Lower utility fees by:
a. Reducing peak KW billing demand
b. Eliminating the power factor penalty
2) Increased system capacity and reduced system losses in
your electrical system
3) Increased voltage level in your electrical system and
cooler, more efficient motors
Effect on Utility:-
1) It has to generate less poaer
2) Reduced Current Hance Reduced losses
3) Less condutor for transmisssion
4) Less current hence insulating cost
4) Less conductor and insulation decrease cost of
supporting structure
5) High Efficiency
6) More stable system

Answer / michael

using the formula for cmputing KVAR

KVARcap = KWrating(tan(theta1) - tan(theta2))

where theta 1 is the angel of .8pf and theta 2 is angel of 1.0pf

KVARcap = 240k (tan36.87 + tan 0)
KVARcapacitor = 180KVAR..
that will be the answer.. or using vector solution

Answer / abduljameel

Normally required capacitors will be in KVAR.
Required KVAR = KW (tan pi1 - tan pi2 )
cos pi1= .8
cos pi2 =1

then calculate accordingly

Answer / mohit parmar

first of all u are not supposed to improvethe power factor
to 1.....since, u have to take clearance for sudden
disconnection of loads.any ways if u want to calculate
capaciitance value refer to any hand book of electrical
engineering,,they have the charts telling u the value of
capacitance if u know the present and ur required
value.regarding power factor...it is one of the main
criteria for efficient elctrical system...since it directly
effects the current carried by the system or cables or
equipment at load side.let us suppose there is a motor of
say...15 kw and pf is( case 1:.85and case2:.95)keeping
voltage constant we can see that current drawn in case1 is
1.2 times that of in case 2 so,this extra current for same
wattage generates more heat and hence losses.

Answer / sudheer(denso)

required kvar= 240 kw x 0.750= 180 kvar
0.750 is the multipling factor.
you have to use 180 kvar capacitor.

Answer / furat

A 500KVA 6600/440 transformer has primary and secondary winding resistance of 0,45 ohms and 0,0015 respectively .The iron loss is 2.9KW. Calculate the efficiency of the transformer on a full load and again in half load if the power factor of the load is 0,8 lagging.

we have one pumping station and 4.6 MW 7 no.s synchronous motors.so there is no required HT capacitor to improve power factor.But we have two no.s of 25 Mva power transformer connect to distribute the power for motors connect with 220 kv line.So i have required the HT capacitor to improve power factor at HT side i mean power transformer bus side.So please suggest how much HT capacitor banks required for same.

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