Answer / nilesh kumar singh

Power Factor is defined as the ratio of Real power and
Apparent power.

for calculating Capacitor bank value the formula is :

kVAr = KW {tan(cos^-1 Phi e) - tan(Cos^-1 Phi d)}

Where - Phi e = existing Power Factor.
Phi d = Desired Power Factor.

Answer / rajendra singh

for calculating Capacitor bank value the formula is :

kVAr = KW {tan(cos^-1 Phi e) - tan(Cos^-1 Phi d)}

Where - Phi e = existing Power Factor.
Phi d = Desired Power Factor.

Answer / shivappa

to calculate desired capacitor value, the formula is

KVAR= KW(TAN PHI 1 - TAN PHI 2 )

WHERE PHI 1 = EXISTING POWER FACTOR
PHI 2 = REQUIRED POWER FACTOR

Answer / kim leang

KVAR= KW(TAN PHI 1 - TAN PHI 2 )

WHERE PHI 1 = EXISTING POWER FACTOR
PHI 2 = REQUIRED POWER FACTOR

Answer / midhun

Calculation and selection of required capacitor rating
Qc = P * {tan [acos (pf1)] - tan [ acos (pf2)]}
Qc = required capacitor output (kVAr)
pf1 = actual power factor
pf2 = target power factor
P = real power (kW)

Answer / fancy102

hi..i also interested in it..i am carrying out pf
measurement system and i am headache in how to maintain the
unity pf too..

Answer / ezzuddin

The correct answer and below is the proof of the formula:

KVAR= KW X [√(1–(COSФ)^2)]/COSФ

The proof:
Pythagoras Theorem states that in a right-angled triangle
the square of the hypotenuse is equal to the sum of the
squares of the other two sides.
so that:
Q is the hypotenuse.
S is the opposite.
P is the adjacent.
TanФ= opposite/adjacent = S/P -----> S=P x(TanФ)
Pf=COSФ= adjacent/hypotenuse = P/Q ---->Q=P/COSФ
Pythagoras said that:
Q^2= P^2 + S^2
P^2/(COSФ)^2= P^2 + (P^2 x (TanФ)^2)
by divide P^2:
1/(COSФ)^2 = 1 + (TanФ)^2
by changing the both sides:
(TanФ)^2 = (1/(COSФ)^2) - 1
(TanФ)^2 = (1/(COSФ)^2) - ((COSФ)^2/(COSФ)^2)
by Taking the (1/(COSФ)^2) Common factor:
(TanФ)^2 = {1/(COSФ)^2)} x ( 1 - (COSФ)^2)
by taking the square root:

TanФ = [√(1–(COSФ)^2)]/COSФ

so that:
KVAR = KW x [√(1–(COSФ)^2)]/COSФ

Answer / santosh

react power kvar=v i sin phi, we know p.f=1,i.e cos phi
calculate sin phi,v i values from total load

Answer / vekuva

KVAR= KW(TAN PHI 1 - TAN PHI 2 )

WHERE PHI 1 = EXISTING POWER FACTOR ANGLE i.e., COS^¯1(p.f1)
PHI 2 = REQUIRED POWER FACTOR ANGLE i.e., COS^¯1(p.f2)
p.f 1 = EXISTING POWER FACTOR
p.f 2 = REQUIRED POWER FACTOR

Answer / ezzuddin

The correct answer and below is the proof of the formula:

KVAR= KW X [Sqrt(1–(CosQ)^2)]/CosQ;

The proof:
Pythagoras Theorem states that in a right-angled triangle
the square of the hypotenuse is equal to the sum of the
squares of the other two sides.
so that:
C is the hypotenuse.
S is the opposite.
P is the adjacent.
TanQ= opposite/adjacent = S/P -----> S=P x(TanQ;)
Pf=CosQ;= adjacent/hypotenuse = P/C ---->C=P/CosQ;
Pythagoras said that:
C^2= P^2 + S^2
P^2/(CosQ)^2= P^2 + (P^2 x (TanQ)^2)
by divide P^2:
1/(CosQ)^2 = 1 + (TanQ)^2
by changing the both sides:
(TanQ)^2 = (1/(CosQ)^2) - 1
(TanQ)^2 = (1/(CosQ)^2) - ((CosQ)^2/(CosQ)^2)
by Taking the (1/(COSQ)^2) Common factor:
(TanQ)^2 = {1/(CosQ)^2)} x ( 1 - (CosQ)^2)
by taking the square root:

TanQ; = [Sqrt(1–(CosQ)^2)]/CosQ;

so that:
KVAR = KW x [Sqrt(1–(CosQ)^2)]/CosQ;

hi frenz,its aswathy.i hve aplied for the hpcl test on feb,pls send the test patern & the placement papers as soon as possible. mail-id :kjaswathy@gmail.com

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