Dr. Sahar Abd El Moneim Moussa 1

Pharos UniversityEE-385

Electrical Power & Machines“Electrical Engineering Dept”

Prepared By:Dr. Sahar Abd El Moneim Moussa

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Three-Phase System

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Balanced Three-Phase SystemBalanced three-phase voltage consists of three sinusoidal voltage having the same amplitude & frequency but are out of phase with each other exactly by 120o

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3 Phase Voltages in Time DomainVa = Vm Sin ωtVb = Vm Sin (ωt-120)Vc = Vm Sin (ωt-240)

Phase

(a)Phas

eb

Phase (c)

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3-Phase Voltages in Terms of Phasors

Va = Vm ∠0Vb = Vm ∠-120Vc = Vm ∠-240 = Vm ∠120

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Types of Connections in 3-phase system

Wye”Y” Delta”∆”

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Wye Connection “Y”Wye Connection: “Y”For Y circuit:

Iline = Iphase

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Delta Connection “∆”For Delta Circuit:Eline = Ephase

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Relationship between three-phase delta-connected and wye connected impedance

Wye connected load

Delta connected load

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Four Different Configurations for the three-phase source and loads Connections

Load Source

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Power in 3-φ System• P(total) = • Q(total) =

S(total) =

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Example 1:A balanced three-phase Y-connected generator with positive sequence has an impedance of 0.2 +j0.5 / and internal voltage 120V/ feeds a -connected load through a distribution line having an impedance of 0.3 +j0.9 /. The load impedance is 118.5+ j85.8 /. Use the a phase internal voltage of the generator as a reference.A. Construct the single-phase equivalent circuit of the 3-

system.B. Calculate the line currents IaA , IbB and IcC.C. Calculate the phase voltages at the load terminals.D. Calculate the phase currents of the load.E. Calculate the line voltages at the source terminals.F. Calculate the complex power delivered to the -

connected load.

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Solution:A. The load impedance of the Y equivalent is

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B. The a-phase line current is

A.Therefore,

IbB=2.4-156.87 A.IcC= 2.483.13 A.

C. because the load is - connected, the phase voltages are the same as the line voltages. To calculate the line voltages,

VA=(39.5 + j28.6)(2.4-36.87) = 117.04-0.96

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The line voltage VAB is

= 202.72 29.04VTherefore,

VBC=202.72 -90.96 VVCA= 202.72 149.04 V

D. The phase currents of the load will be,

= 1.39 -6.87 A.

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Therefore, IBC=1.39-126.87 AICA=1.39113.13 A

E. The line voltage at the source terminals will be,Va=(39.8 + j29.5) (2.4-36.87)

=118.9 -0.32 V.

The line voltage will be = 205.9429.68 V.

Therefore ,Vbc=205.94 -90.32 V.Vca= 205.94149.68 V.

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F. The total complex power delivered to the load will be,

V=VAB= 202.72 29.04 V.I=iAB=1.39-6.87 A.

Therefore,ST= 3 (202.72 29.04) (1.396.87)

= 682.56 +j 494.21 VA