Answer / deepakindi1

Boss what if u dont have the table?

let Cos(@1) be the present PF

@1=cos'(@1) (cos inverse)
let Cos(@2) be the desired PF
@2=cos'(@2) (cos inverse)


KVAR required= KW[ tan(@1)-tan(@2)]

for 65 Hp load
KW= 65HP*0.746 = 48.49KW
Considering Preent P.F=0.50
desired pf =0.99
ie cos(@1)=0.5
@1=cos'(0.5)=60
@2=cos'(.99)=8.1

kvar=48.49*[tan(60)-tan(8.1)]
=48.49*[1.732-.142]
=77.08

kvar=77.08

Answer / mahesh sonawane.

Capacitor range is actually given in Kvar, so to find that
range in kvar

As we know
Power factor = KW/KVA
for 65 Hp load
KW= 65HP*0.746 = 48.49KW
Considering Preent P.F=0.50
Target pf =0.99
Multiplying Factor by standard table= 1.59
Kvar= 48.49 KW * 1.59 = 77.09 Kvar

hence for the load of 65hp, capacitor range will be 77.09
Kvar.

Answer / bala123

please mention the existing powerfactor....

Answer / girish kumar samal, hindalco,h

as pf not giving ,taking pf 0.8
let pf of load =0.8
kw of load= 65 hp=65*0.746=48.5kw
kva =48.5/0.8=60.625kva
kvar =sqrt(60.625^2-48.5^2)=36.375kvar
if we want to improve pf to 0.95 then
improved kva= 48.5/0.95=51.052
improved kvar= sqrt(51.052^2-48.5^2)=15.9411
requard capacitor range = 36.375-15.94=20.43kvar.

Answer / chandru

In This case no other detail is given except the power
rating .

Therefore pf=( KW/KVA)
In order to get incentive and good power factor it is
required to maintain power factor above 0.95

So we take pf=0.95 and kw=(65*0.746)=48.49
Therefore KVA= 48.49/0.95 = 51.04

We know that KVAr= Squareroot( (KVA)^2-(KW)^2)


Therefore KVAr = sqrt(51.04^2-48.49^2)

=15.90 approx..


Capacitor required = 16KVAr.

Answer / manish deswal

let us conside current p.f = .8 & improved to .9,
Load = 65 hp =65 * 746 = 48.49 KW
KVA1 = 48.49/.8= 60.61

KVAR = Squareroot of [ square(60.61)-square(48.49)]=36.36
KVA2 = 48.49/.9= 53.88

KVAR = Squareroot of [ square(53.88)-square(48.49)]=23.49

Increased Capacitor bank capacity = 36.36-23.49 = 12.87 KVAR

Answer / sonu

Deepankandil why we are subtracting initial-final. littele
bit confusion. please throw some light on this issue.
because it will depand on lagging or leading?

Answer / ram lal

power factor is not mention so how can calculate ...........

Answer / naveen kumar mishra

first we should calculate the load(inductive or capacitive
or resistive).
in case of resistive load-----
in case of resistive load the, actual power factor will be
01(maximum),in this case the voltage will linior with
respecte to load current.
so tatal resistive load is(in example 65 hp),
so total load in kw = 65*0.746
=48.49.
As we know
Power factor = KW/KVA
for 65 Hp load(if pure resistive)

Considering Preent P.F=1
Target pf =0.99
Multiplying Factor by standard table= 1.56
Kvar= 48.49 KW * 1.56= 75.64

hence for the load of 65hp, capacitor range will be 75.64
Kvar.(in pure resistor).

Answer / sudhakara

MR. girish
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