three-phase circuitsfaculty.citadel.edu/potisuk/elec202/notes/3phase1.pdf · ac generator •...
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THREE-PHASE CIRCUITS PART I
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AC GENERATOR
• Single-phase AC generator - designed to generate a single sinusoidal voltage for each rotation of the shaft (rotor).
• Polyphase AC generator - designed to generate multiple out-of-phase sinusoidal voltages for each rotation of the rotor by
increasing the number of coils on the stator.
• The three-phase generator has three induction coils placed 120° apart on the stator.
• The three coils have an equal number of turns. → the voltage induced across each coil will have the same peak value, shape, and frequency. → the three sinusoidal voltages are out of phase by 120°
• Two types of configuration: Wye (Y) or Delta (Δ)
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SINGLE-PHASE AC GENERATOR
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THREE-PHASE AC GENERATOR
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FARADAY’S LAW
• The physical or experimental law governing the operation of and AC generator.
• “The electromotive force (EMF) induced in a circuit is directly proportional to the time rate of change of magnetic flux through the circuit.”
• The EMF can either be produced by changing B (induced EMF)
or by changing the area, e.g., by moving the wire (motional EMF).
It is the relative movement between the coil and the magnet
that matters.
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THREE-PHASE VOLTAGE SOURCES
Y-connected Source Δ-connected Source (uncommon)
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BALANCED Y-CONNECTED VOLTAGE SOURCE
• Balanced phase voltages are equal in magnitude and are out of phase with one another by 120 degrees.
oVV VV0∠= pano120−∠= pbn
o120∠= pcn VV
• Phase voltages sum up to zero, i.e., 0=++ cnbnan VVV • Two possible combinations:
bc (+) acb (−) a positive phase sequence negative phase sequence
0
120
240
an p
bn p
cn p
V V
V V
V V
= ∠ °
= ∠ + °
= ∠ + °
0
120
240
an p
bn p
cn p
V V
V V
V V
= ∠ °
= ∠ − °
= ∠ − °
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• Balanced line voltages are equal in magnitude and are out of phase with one another by 120 degrees.
o303 ∠= VV pab opbc 903 −∠= VV o1503 ∠= pca VV
• Line voltages sum up to zero, i.e., 0=++ cabcab VVV
• The magnitude of line voltages is √3 times the magnitude of the phase voltages.
• Line voltages lead their corresponding phase voltages by 30°
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Y−Δ SOURCE TRANSFORMATION
YYan VV θ∠= o303 +∠= YYab VV θ
o120−∠= Yθ ⎯⎯ →⎯ Δ−Y Ybn VV o903 −∠= YYbc VV θ
o240−∠= YYcn VV θ o2103 −∠= YYca VV θ
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Δ−Y SOURCE TRANSFORMATION
ΔΔ∠= θVVab o303
−∠= ΔΔ θVVan
o120−∠= ΔΔ θVVbc ⎯⎯ →⎯ −Δ Y o1503
−∠= ΔΔ θVVbn
o240−∠= ΔΔ θVVca o2703
−∠= ΔΔ θVVcn
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BALANCED THREE-PHASE LOAD CONFIGURATIONS
• A balanced load is one in which the phase impedances are equal in magnitude and in phase.
• Two possible configurations: Wye or Delta
• Conversion from Y to Δ or Δ to Y
Δ==== ZZZZZY 33211 Z Z Ycba Z Z Z= = = = 3Δ
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Ex. Practice problem 12.1
Given that o30110∠=bnV V, find anV and cnV , assuming a positive sequence.
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THREE-PHASE CONNECTIONS
• Both the three phase source and the three phase load can be connected either Wye or DELTA.
• We have 4 possible connection types.
1) Y-Y connection
2) Y-Δ connection
3) Δ-Δ connection
4) Δ-Y connection
• Balanced Δ connected load is more common.
• Y connected sources are more common.
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BALANCED Y-Y CONNECTION
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Ex. Calculate line currents in the three-wire Y-Y system shown.
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Ex. Practice problem 12.2
A Y-connected balanced three-phase generator with an impedance of 0.4+j0.3 Ω per phase is connected to a Y-connected balanced load with an impedance of 24+j19 Ω per phase. The line joining the generator and the load has an impedance of 0.6+j0.7 Ω per phase. Assuming a positive sequence for the source voltages and that o30120∠=anV V, find the line voltages and line currents.
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BALANCED Y-Δ CONNECTION
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Ex. A balanced abc-sequence Y-connected source with phase o voltage 10100∠=anV V is connected to a Δ-connected
balanced load of 8+j4 Ω per phase. Calculate the phase and the line currents.
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Ex. Practice problem 12.3
One line voltage of a balanced Y-connected source is o20240 −∠=ABV V. If the source is connected to a Δ-connected
load of Ω, find the phase and line currents assuming the abc sequence.
o4020∠
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BALANCED Δ-Δ CONNECTION
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Ex. A balanced Δ-connected load having an impedance of 20−j15 Ω is connected to a Δ-connected, positive sequence generator having o0330∠=abV V. Calculate the phase currents of the load and the line currents.
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Ex. Practice problem 12.4
A positive-sequence, balanced Δ-connected source supplies a balanced Δ-connected load. If the impedance per phase of the load is 18+j12 Ω and o35202.19 ∠=aI A, find ABI and .ABV
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BALANCED Δ-Y CONNECTION
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Ex. A balanced Y-connected load with a phase impedance of 40+j 25 Ω is connected to a balanced, positive sequence Δ- connected source with a line voltage of 210 V. Calculate the phase currents. Use Vab as a reference.
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Ex. Practice problem 12.5
In a balanced Δ-Y circuit, o15240∠=abV V and .1512 Ω+= jZY Calculate the line currents.