Explanations & Calculations
When a circuit is not purely resistive, the power distribution occurs both as reactive & real power. Real power is what accessible by the user whereas the reactive power stores inside the system during the operation. The relationship between these is given by True power = Apparent power × \small \times × power factor
Power factor = cousin of phase angle
Apparent power = V i = 230 V × 60 A = 13800 = 13.8 k V A \small Vi=230V\times60A=13800=13.8 kVA Vi = 230 V × 60 A = 13800 = 13.8 kV A Units are expressed in volt-ampere instead of Watts
True power = 13.8 × 0.9 = 12.42 k W \small 13.8\times 0.9=12.42 kW 13.8 × 0.9 = 12.42 kW Units are expressed in watts
If the phase angle is θ \small \theta θ , cos θ = 0.9 → θ = 0.451 r a d \qquad\qquad \cos\theta=0.9\to\theta=0.451 rad cos θ = 0.9 → θ = 0.451 r a d
Reactive power can be calculated in 2 ways. First, R.Power = Apparent power × sin ( p h a s e a n g l e ) \small \times \sin(phase angle) × sin ( p ha se an g l e ) = 13.8 × sin ( 0.451 ) = 6.015 k V A R \small 13.8\times \sin(0.451)=6.015 \,kVAR 13.8 × sin ( 0.451 ) = 6.015 kV A R
Second, R.Power = ( A p p a r e n t p o w e r ) 2 − ( T r u e p o w e r ) 2 = 13. 8 2 − 12.4 2 2 = 6.015 k V A R \small \sqrt{(Apparent\,power)^2-(True\,power)^2}=\sqrt{13.8^2-12.42^2}= 6.015\,kVAR ( A pp a re n t p o w er ) 2 − ( T r u e p o w er ) 2 = 13. 8 2 − 12.4 2 2 = 6.015 kV A R
Units are expressed as VA reactive.
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