Lecture 11Power factor correction

Power factor :

:What is the total power factor ?

7-1

80%67,42095%2.3(1)(2)(3)(4)1% 3 1% 1.5(5):2030

[7-5] 11.4KV100A0.80.95

[7-7] 0.810KW0.95

[7-8] 11.4KV4,000KVA0.81200KVAR0.26+j0.72(1)(2)(3) IR=IL1cosL=2030.8=162 IX=IL1sinL=2030.6=122

(1) (2) IL2=162-j122+j61=162-j61 =32143-23345=8798W(3) (4) 72.6%

[7-9] 0.81000KVA0.95

157.9KVA

[7-10] 10000KWH10000 KVARH10000KWH5000KVARH4.2%

(KVAR)C1C2 C3 C4

300KW 720000.7 11.4KV/220v 400KVA 5.5%150KVAR0.9 PF0.7 PF0.9 72002. 300(0.90.7)0.785KW3. =5kW 1(0.70.9)2=0.395 3600KWH 0.3951442KWH 1442 1.37=19754. v%=150 5.54002.1 220220 2.1%224.5v 25000 2500091752.72example

:Apparent powerActive powerReactive power

A 3/4 HP electric motor has a power factor of .85. The nameplate current is 10 Amps at 115 Volts, or 1150 Volt Amps. exampleApparent power = 1150 Volt Amps

Active power = .85 * 1150 = 977.5 Watts

Reactive Power = sqrt(1150^2 - 977.5^2) = 605 VAR

So, we need 605 var of power factor correction.

Calculating the required impedance from Q = E^2/X

605 = 115^2/X => X = 115^2/605 = 21 ohms

C = 1 /( 2 * pi * f *X) = 1/ (377 * 21) = 126 uF

example500KW 0.80.95kVAR ?60hz220v.