Question #158280

A single phase load draws 60 amps at a power factor of 0.9 lagging from a 230 v 50 hertz supply.

calculate

-apparent power

-true power

phase angle

the reactive power


1
Expert's answer
2021-01-26T08:43:56-0500

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

  • Then,


  • Apparent power = Vi=230V×60A=13800=13.8kVA\small Vi=230V\times60A=13800=13.8 kVA

Units are expressed in volt-ampere instead of Watts


  • True power = 13.8×0.9=12.42kW\small 13.8\times 0.9=12.42 kW

Units are expressed in watts


  • If the phase angle is θ\small \theta,

cosθ=0.9θ=0.451rad\qquad\qquad \cos\theta=0.9\to\theta=0.451 rad


  • Reactive power can be calculated in 2 ways.
  • First, R.Power = Apparent power ×sin(phaseangle)\small \times \sin(phase angle)

= 13.8×sin(0.451)=6.015kVAR\small 13.8\times \sin(0.451)=6.015 \,kVAR

  • Second, R.Power = (Apparentpower)2(Truepower)2=13.8212.422=6.015kVAR\small \sqrt{(Apparent\,power)^2-(True\,power)^2}=\sqrt{13.8^2-12.42^2}= 6.015\,kVAR


Units are expressed as VA reactive.



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