Answer / habeebm

The KVAR required is:

KVAR=KW(tan@1-tan@2) (where @1=cos-1(p.f1)& @2=cos-1(p.f2)

Therefore,

KVAR=KW {tan[cos-1(p.f1)]-tan[cos-1(p.f2)]}

=KW {tan[cos-1(0.85)]-tan[cos-1(0.99)]}

=KW {tan[31.79]-tan[8.11]}

=KW {0.48}

=1500 {0.48}

=720

Therefore 720 KVAR is required to improve powerfactor from
0.85 to 0.99 for 1500KW load.

NOTE:-
Excuse me if the answer is wrong.

Answer / alimohmad khatri

KVAR=KVA*SIN(PHI)
KW=KVA*COS(PHI)

HENCE....KVAR/KW=TAN(PHI)

KVAR=KW*TAN(PHI)

NOW, FOR P.F.(1)=COS(PHI(1))=.85 THEN TAN(PHI(1))=.6197
FOR P.F.(2)=COS(PHI(2))=.99 THEN TAN(PHI(2))=.1425

KW=1500 KW IS COMMON FOR BOTH CASE 1 & 2

REQUIRED....
KVAR=KW*[TAN{PHI(1)}-TAN{PHI(2)}]

HENCE KVAR=1500*(0.6197-0.1425)=715.8

YOU NEED TO ARRANGE CAPACITOR BANK OF 715.8 KVAR.

Answer / waqi

first calculate the kva at .99 and 1500kw
(cosp=kw/kva),than put the value of kva in this formula kva=
(kw)2+(kvar)2)^1/2 to caculate kvar at .99 pwr factor

Answer / habeebm

Dear Waqi

I am not understand your answer for the above mentioned question. Will you please solve and give us the answer for required KVAR to improve power factor from 0.85 to 0.99 for 1500KW load.

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