1Three Phase Systems• Generation of Three Phase Voltages• Three Phase Power• Three Phase Circuit AnalysisThree Phase Systems• Nearly all electric power generated and distributed is in the form of 3 phase AC.• These systems consists of 3 phase generators, transmission lines and loads.• Advantages over single phase systems– More efficient (more power per kg of metal)– Instantaneous power is a constant, not pulsing or oscillatinga) abc sequenceb) acb sequenceCan you show that VA + VB + VC = 0 ?We can also connect negative terminals of the three generator and loads together to form a three phase circuit. With the neutral wire only four wires are required. Note: IN= IA + IB + IC= 0 for a balanced 3 phase generator and balanced load.a) Y-connected generators b) ∆ connected generatorsELEC 316Dr. Barsanti2Y connected generatorEach generator is also called a phase. The voltage and current in a single generator are called the phase voltage (∅)and phase currents (∅)Y connection•Phase Voltages= ∅< 0° = ∅< −120° = ∅< −240°•Phase Currents (assuming resistive loads)= ∅< 0° = ∅< −120° = ∅< −240°Y connection•Line (to Line) Voltages= − = ∅< 0° − ∅< −120° = 3∅< 30° ''= 3∅ Y connection•Phase Currents = Line currents= ∅= ' Y connection= −  = 3∅< 30°∅ −DELTA CONNECTED GENERATOR∆ connection•Line Voltages = Phase voltages= ∅< 0° = ∅< −120° = ∅< −240°•Line Currents (assuming resistive loads)= − = *< 0° − *< 120° = 3*< −30°ELEC 316Dr. Barsanti3∆ connection•Line (to Line) Voltages= ''∆connection •Phase Currents = Line currents= '= 3 *∆ connection-IcaThree Phase Power./ = 2*cos 0/./ = 2*cos 0/ − 120./ = 2*cos 0/ − 2401/ = 2*cos 0/ − θ1/ = 2*cos 0/ − θ − 1201/ = 2*cos 0/ − θ − 2403/ = 2**cos 0/ cos 0/ − θ = **[cos θ − cos 20/ − θ ]3/ = 2**cos 0/ − 120 cos 0/ − θ − 120 = **[cos θ − cos 20/ − θ − 240 ]3/ = 2**cos 0/ − 240 cos 0/ − θ − 240 = **[cos θ − cos 20/ − θ − 480 ]3787/ = 3/ + 3/ + 3/= 3**cos θ= constantPower Equations forphase and line quantities• The power supplied to a balanced three phase load in terms of the phase quantities is:: =3**cos (θ), Q =3**sin θ , S =3**• Or equivalently in terms of the line quantities : = 3'''cos (θ), Q = 3'''sin θ , S = 3''Power Equations forphase and line quantities• All voltages and currents on previous page are RMS values.• Using the line quantities is generally preferred since easier to measure and equations are same for Y or ∆ config. • Note: θ is angle between phase voltage and phase currents and not between line voltage and line currents.ELEC 316Dr. Barsanti4Analysis of Balanced 3 phase Circuits• Use a single phase equivalentcircuit approach.• Requires the use of a neutral return path.• If ∆ configuration is encountered it must be converted to a Y>?=>∆3Three Phase LoadsExample 1…Solution…'=120 < 00.06 + B0.12 + (12 + B9)= 7.94 <-37.1°∅'= 7.94 < −37.1 12 + B9= 119.1 <-0.2°:'8E= 3∅∅cos (θ) = 3119.1 7.94 cos 36.9= 2270 wG'8E= 3∅∅H1I (θ) = 3 119.1 7.94 H1I 36.9= 1702 varJ'8E= 3∅∅= 3 119.1 7.94 = 2839 varExample 2… ∆ connected loadSolution…'=120 < 00.06 + B0.12 + (4 + B3)= 23.4 <-37.5°∅'= 23.4 < −37.5 4 + B3= 117 <-0.6°:'8E= 3∅∅cos (θ) = 3 117 23.4 cos 36.9= 6571 wG'8E= 3∅∅H1I (θ) = 3 117 23.4 H1I 36.9= 4928 varJ'8E= 3∅∅= 3 117 23.4 = 8213 varELEC 316Dr. Barsanti5One Line Diagram One Line DiagramIf the line transmission lines can be assumedto have negligible impedances, then a one line diagram simplification can be made.Example… Find the total Power.∅L=MNOLO= 48P :L= 3∅L∅Lcos 30 = 3 480 48 cos 30 = 59.9 Q0GL= 3∅L∅LH1I 30 = 3 480 48 sin 30 = 34.6 Q.RS ∅T=MNO U⁄W= 55.4P :T= 3∅T∅Tcos −36.87 = 3MNOU55.4 cos −36.87 = 36.8 Q0GT= 3∅T∅Tsin −36.87 = 3MNOU55.4 sin −36.87 = −27.6QvarExample continued … Find the total Power and PF:787= 59.9 + 36.8 = 96.7 Q0G787= 34.6 + 3 − 27.6 = 7.0 Q.RSθ = tanXLYZ= tanXL[\].[= 4.14°:^ = cos θ = 0.997 _R``1I`ELEC 316Dr.