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Question
```HOW TO CALCULATE THE CAPACITOR RANGE FOR LOAD FOR PF
IMPROVEMENT?
EXPLE:FOR 65 HP LOAD HOW TO CALCULATE CAPACITOR RANGE?```
Question Submitted By :: Electrical-Engineering
I also faced this Question!!     Answer Posted By

# 1
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)]

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

 Is This Answer Correct ? 49 Yes 3 No
Deepakindi1

# 2
Capacitor range is actually given in Kvar, so to find that
range in kvar

As we know
Power factor = KW/KVA
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.

 Is This Answer Correct ? 66 Yes 23 No
Mahesh Sonawane.

# 3
 Is This Answer Correct ? 28 Yes 9 No
Bala123

# 4
as pf not giving ,taking pf 0.8
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.

 Is This Answer Correct ? 24 Yes 10 No
Girish Kumar Samal, Hindalco,h

# 5
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

 Is This Answer Correct ? 2 Yes 0 No
Manish Deswal

# 6
first we should calculate the load(inductive or capacitive
or resistive).
in case of resistive load the, actual power factor will be
01(maximum),in this case the voltage will linior with
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).

 Is This Answer Correct ? 2 Yes 2 No
Naveen Kumar Mishra

# 7
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.

 Is This Answer Correct ? 8 Yes 10 No
Chandru

# 8
power factor is not mention so how can calculate ...........
 Is This Answer Correct ? 4 Yes 6 No
Ram Lal

# 9
MR. girish
What is meaning of sprt and that calculation

 Is This Answer Correct ? 5 Yes 8 No
Sudhakara

# 10
mr
Deepakindi1 iwant to know what the maximum range of mccb

 Is This Answer Correct ? 3 Yes 6 No
Sivananda Samantara

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