SSE JE Electrical Engineering 24 March 2021 Shift 1 Paper

Question : 200 of 200

Marks: +1, -0

In a factory, a three-phase, 4-kV, 400-kVA synchronous machine is installed along with other induction motors. The following are the loads on the machine:
(i) Induction motors: 500 kVA at 0.8 power factor lagging
(ii) Synchronous motor: 300 kVA at unity power factor
Determine the overall power factor of the factory loads.

${7}/{√{58}}$ lagging
${7}/{√{58}}$ leading
${9}/{√{58}}$ leading
${9}/{√{58}}$ lagging

Validate

Solution:

The power triangle is shown below.

(a) Induction motors:Power = 500 × 0.8 = 400 kW
Reactive Power = 500 × 0.6 = 300 kVAR
(b)Synchronous motor:
Power = 300 kW
Reactive Power = 0.0
Factory :
Power = 400 + 300 = 700 kW
Reactive Power = 300 + 0 = 300 kVAR
Complex Power

= √(700^2 + 300^2) = 100√58

= 762 kVA
Power factor

= 700/(100√58) = 7/√58

= 0.92 lagging

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