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Transcript

Power Factor Correction The process of increasing the power factor without altering the voltage or current to the original load is known as power factor correction Most domestic loads (such as washing machines, air conditioners, and refrigerator) and industrial loads (such as induction motors) are inductive and operate at a low lagging power factor. Although the inductive nature of the load can not be changed, but the power factor can be increased. Most loads are inductive, as shown in figure below (a) which has a power factor of cos 1 , a load’s power factor is improved or corrected by installing a capacitor in parallel with load, as shown in Figure below (b) which has a power factor of cos 2. I IL V IL Inductive load V Inductive load (b) (a) Figure Power factor correction: (a) original inductive load, (b) inductive load with improved power factor IC C The effect of adding the capacitor can be illustrated using power triangle or the phasor diagram of the current involved. IC 1 2 V IC I IL The original load has power factor pf 1 cos 1 and the inductive load with improved power factor has pf 2 cos 2 The power factor correction can be looked from another perspective. Consider the power triangle in Figure below. QC S1 Q1 S2 Q2 1 2 P If the original inductive load has apparent power S1, then P S1 cos 1 Q1 S1 sin 1 P tan 1 If desired to increase the power factor from pf cos to pf cos without altering the real power (i.e P S 2 cos 2 ), then the new reactive power is 1 1 2 2 Q2 P tan 2 The reduction in the reactive power is caused by the shunt capacitor, that is QC Q1 Q2 P(tan 1 tan 2 ) ; QC 2 Vrms 2 CVrms XC The value of the required shunt capacitance of capacitor C is determined by C QC P(tan 1 tan 2 ) 2 2 Vrms Vrms Although the most command situation in practice is that of an inductive load, it is also possible that the load is capacitive, that is, the load is operating at a leading power factor. In this case, an inductor should be connected across the load for power factor correction. The required shunt inductance L can be calculated from 2 2 Vrms Vrms QL XL L 2 Vrms L QL Where QL Q1 Q2 , the difference between the new and old reactive powers. Example 1 When connected to a 120 V (rms), 60 Hz power line, a load absorbs 4 kW at a lagging power factor of 0.8. Find the value of capacitance necessary to rise the pf to 0.95. Solution Before improving power factor : If the pf=0.8, than cos 1 0.8 1 36.87 0 The apparent power from the real power S1 P cos 1 400 500 VA 0.8 After improving power factor : The pf is raised to 0.95 cos 2 0.95 2 18.19 0 The real power has not changed, but the apparent power has changed, its new value : S2 P 4000 4210.5 VA cos 2 0.95 The new reactive power is Q2 S 2 sin 2 4210.5 sin( 18.19 0 ) 1314.4 VAR The reactive power due to installation of capacitor is QC Q1 Q2 3000 1314.4 1685.6 VAR and C QC 1685.6 310.5 F 2 Vrms 2 x60 x120 2 Example 2 The pf correction capacitor in a 240 V, 50 Hz fluorescent light unit has broken down and needs replacing. A test on the unit shows that, without the capacitor, the supply current is 0.86 A at pf of 0.5 lagging. The value quoted on the original capacitor have faded and the only other information is that the working pf of the unit should be 0.95. Determine the value of the capacitor needed (look in Figure below). Choke and lamp 0.86 A Pf=0.5 240 V 50 Hz Three phase system The total average power P : P 3V I cos 3VL I L cos The total reactive power Q : Q 3V I sin 3VL I L sin The total apparent power S : S P jQ 3VL I L For a Y-connected load : VL 3V , but I L I For a - connected load : I L 3I V L , V , I L and I , but V L V are all rms value and is the angle between the phase voltage and the phase current Example 3 A three-phase motor can be regarded as a balanced Y-load. A threephase motor draws 5.6 kW when the line voltage is 220 V and the line current is 18.2 A. Determine the power factor of motor. Solution The power factor is : 3V L I L cos P pf cos P 3V L I L 5600 3 x 220 x18.2 0.8075 Example 4 A balanced three phase load draws 50 kW at a lagging power factor of 0.85 from a 480 V supply. a. Find the apparent power S b. Find the line current I L c. Find the reactive power Q Example 5 A three phase industrial load consists of 136 kW at 0.8 power factor lagging. What size capacitor bank is required to correct the power factor to 0.95 lagging?