Power Triangle Our electrical system must handle this

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

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

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

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

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)

Three-Phase Systems

Three-Phase Systems

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

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.

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. ICA IAB IBC 240 /120 VAB IAB = = = 24 /120 ZAB 10 /0 240 /0 10 /30 VBC IBC = ZBC = VCA ICA = ZCA 240 /240 = = 16 /270 15 /– 30 = 24 /330

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 ICA IAB 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

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

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.

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 = IB = IC = VAN ZAN VBN ZBN VCN ZCN = = = 120 /270 6 /0 120 /30 6 /30 120 /150 5 /45 = 20 /270 = 20 /0 = 24 /105 IN = –(IA + IB + IC) = 14. 15 / 193

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. VAN = 120 /270 IA = 20 /270 VBN = 120 /30 IB = 20 /0 VCN = 120 /150 IC = 24 /105 IN = 14. 15 / 193