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Power Triangle

This component of the power furnishes the energy required to establish and maintain electric and magnetic fields. Our electrical system must handle this component. This is the only “ useful ” work done. Power Triangle. Total or Apparent Power S kVA. Reactive Power Q kVAR. .

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Power Triangle

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  1. This component of the power furnishes the energy required to establish and maintain electric and magnetic fields. Our electrical system must handle this component. This is the only “useful” work done. Power Triangle Total or Apparent Power S kVA Reactive Power Q kVAR  THEREFORE, VARS REDUCE THE “USEFULNESS” OF THE TOTAL ELECTRIC POWER. Real Power P kW cos   Power Factor

  2. Power Triangle  Power triangle in first quadrant (positive ) indicates INDUCTIVE load.

  3. Power Triangle  Power triangle in fourth quadrant (negative ) indicates CAPACITIVE load.

  4. Power Triangle Z S X Q  R R = resistance X = reactance Z = impedance P Power triangle and impedance triangle are SIMILAR TRIANGLES.

  5. Three-Phase Systems Three-phase systems Most common: 3-wire delta and 4-wire wye Mathematical representation Time domain (oscilloscope view) Frequency domain (phasor diagram)

  6. Three-Phase Systems

  7. Three-Phase Systems

  8. Three-Phase Systems Delta-connected loads Line-to-line voltage only No neutral, so best for balanced loads Ideal for motors Wye-connected loads Line-to-line AND line-to-neutral voltage Better for imbalanced loads, particularly mix of three-phase and single-phase

  9. Unbalanced Delta Load A three-phase, three-wire, 240-volt system with A-B-C phase sequencing supplies a delta-connected load as shown below. Determine the line currents and draw the phasor diagram.

  10. ICA 240 /120 VAB IAB = = = 24 /120 ZAB 10 /0 IAB IBC 240 /0 VBC IBC = = = 24 /330 ZBC 10 /30 240 /240 VCA ICA = = = 16 /270 ZCA 15 /–30 Unbalanced Delta Load A three-phase, three-wire, 240-volt system with A-B-C phase sequencing supplies a delta-connected load as shown below. Determine the line currents and draw the phasor diagram.

  11. ICA IAB IBC Unbalanced Delta Load A three-phase, three-wire, 240-volt system with A-B-C phase sequencing supplies a delta-connected load as shown below. Determine the line currents and draw the phasor diagram. IAB = 24 /120 IBC = 24 /330 ICA = 16 /270 IA = IAB – ICA = 24 /120 – 16 /270 =38.69 /108 IB = IBC – IAB = 24 /330 – 24 /120 =46.37 /315 IC = ICA – IBC = 16 /270 – 24 /330 =21.17 /191

  12. Unbalanced Delta Load A three-phase, three-wire, 240-volt system with A-B-C phase sequencing supplies a delta-connected load as shown below. Determine the line currents and draw the phasor diagram. VAB = 240 /120 IA = 38.69 /108 VBC = 240 /0 IB = 46.37 /315 VCA = 240 /240 IC = 21.17 /191

  13. Unbalanced 4-Wire Wye Load A three-phase, four-wire, 208-volt system with A-C-B phase sequencing supplies a wye-connected load as shown below. Determine the line currents, the neutral current, and draw the phasor diagram.

  14. VAN 120 /270 IA = = =20 /270 6 /0 ZAN VBN 120 /30 IB = = =20 /0 6 /30 ZBN VCN 120 /150 IC = = = 24 /105 5 /45 ZCN Unbalanced 4-Wire Wye Load A three-phase, four-wire, 208-volt system with A-C-B phase sequencing supplies a wye-connected load as shown below. Determine the line currents, the neutral current, and draw the phasor diagram. IN = –(IA + IB + IC) = 14.15 / 193

  15. Unbalanced 4-Wire Wye Load A three-phase, four-wire, 208-volt system with A-C-B phase sequencing supplies a wye-connected load as shown below. Determine the line currents, the neutral current, and draw the phasor diagram. IA = 20 /270 VAN = 120 /270 IB = 20 /0 VBN = 120 /30 IC = 24 /105 VCN = 120 /150 IN = 14.15 / 193

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