Question

2. A load is connected to a voltage of 1320 volts at 60 Hz. The load dissipates 100 kW with a 0.867 power factor. Specify the capacitance needed to correct the power factor to being 1. 3. Consider a circuit for a single-phase AC power system, where a 200 volt AC voltage source is delivering power to a loads. Find the true power, reactive power and apparent power delivered by 150 mA current (Impedance phase angle 0 = 30°)

Answer 1

Anwen.- Griven; 2) The objective is to determine the capacitance power factor to being ; Page noso needed to correct the - Xc= ettfc. est load is connected to voltage, V=1920 volts. frequency = 60 Ha. load dissipatos (Phloo kw power factor (PF) = 0.867 We know; WN non sienynek la, Ractive power. PF:0.867 P, Real power Power factor , PF = CODE 0.867 - Coso. O- COS“ (0.867) 0 = 29.890. let us calculate the reactive power of the load. а: Рw tом 6 By substituting values in above equation, Q: 100k * tan (29.890) 1 = 100k * 0.5747 Q =57.479 KVAR It is mentioned in the problem, the power factor is required to be unity ine, . And the reactive power at pf=1 is a' Here O'= angle between apparent power and true power.

Page no:2 :: Pf=1. i coso' =1 o's cost (1) 0-0 Similarly; Q = P tano =100k x tano Q=0. power, O . In both cases there is a difference in reactive this power is equals to correction capacitor (a) Qe=Q-Q. By substituting Q and a values, we get Qe=0-57.47. Q = -54.47 KVAR. Now we determine capacitive reactance by cering Pcapacitor - Vez → X.2V - Pcapacitor, By substituting values in it : X 2(1320) (57.47 k) 2640 57.47k Xc -0.0459.12

Page nors 1. By putting all these values in above equation; we get, te - 2 T F C. 0.0459: _! 27T X60XC 17.303 - с. т 19.303 C -0.05779 power and Oo 9 The objective is to determine the true power, reactive apparent power, for a single phase AC power system, Voltage - 200 volts. Power delivered by = 150.mA current Impedance phase angle , 8= 30" We know that, the angle between current and voltage phase angle Now we consider the power triangle angle O= 30°. is impedance share and know that, the anal Apparent power Reactive power seal and apparent power Real power : 0 Angle between - 30° this triangle; Pactive - IV coso. from

Page no: 4 By substituting values in above equation, we get Pactive = (150x1039)(200(COs 30") l Pactive = 25.98w7 And formule for Reactive power is: Preactive = I u sino - (150X10-3) (200)(Sin 3009 Preactive = 15. VAR Finally, for Apparent power ; This is Resultant of active and reactive PApparent = IV. = (150) (200) Permanent = 30 VA De powers;