A reactance X_p when connected in parallel to a load Z = R + jX will raise the p

Question

A reactance X_p when connected in parallel to a load Z = R + jX will raise the power factor of the composite load (Z//jX_p) to a new value pf_new. Show that the reactance X_p is given by x_p = V^2_s/P/tan(cos^-1 pf_new) - tan(cos^-1 pf_old) where P is the average AC power and pf_old and pf_new are, respectively, the values of the power factor before and after the connection of reactance in parallel and V_s is the source voltage (rms). An industrial plant is powered from a 220-V (rms), 60-Hz source, and it consumes 50 kW with pf = 0.75, lagging. Using the equation derived in part (i) above, find a suitable parallel capacitor to ensure pf = 0.95, lagging. What current must this capacitor be able to withstand?

Explanation / Answer

let assume the source emf is referrence.

that is source emf=V<0 volts in magnitude and phase format.

(voltage<angle means in complex form: volate*cos(angle) + j*voltage*sin(angle))

let initial power factor angle be A and final power factor angle be B.

i.e. cos(A)=pf_old, cos(B)=pf_new

let initial current be i1 and final current be i2.

as due to Xp, there is no change in average AC power (power is only dissipated through resistive elements)

P=V*i1*cos(A)=V*i2*cos(B)

then initially when only R+jX are connected:

1/(R+jX)=(i1<-A)/V.....(1)

when Xp is added in parallel, remembering that admittances are added in parallel,

(1/(R+jX))+(1/jXp)=(i2<-B)/V...(2)

subtract equation 2 from equation 1:

-1/(j*Xp)=(i1<-A - i2<-B)/V

taking imaginary parts on both sides:

1/Xp=(-i1*sin(A)+i2*sin(B))/V

==>1/Xp=(-i1*tan(A)*cos(A) + i2*tan(B)*cos(B))/V

multiplying V on numerator and denominator on right had side:

1/Xp=(-V*i1*cos(A)*tan(A)+ V*i2*cos(B)*tan(B))/V^2

==>1/Xp=P*(tan(B)-tan(A))/V^2

==>Xp=(V^2/P)/(tan(B)-tan(A))

where B=arccos(pf_new) and A=arccos(pf_old)

hence proved.,

part b:

give that V=220 volts

P=50*10^3 watts

pf_old=0.75,lagging

pf_new=0.95,lagging

then Xp=(V^2/P)/(tan(acos(pf_new))-tan(acos(pf_old)))

==>Xp=1.7497 (taking the absoulte value)

then 1/(w*C)=1.7497

==>C=1/(2*pi*60*1.7497)=1.516 mF

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.