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THREE-PHASE CIRCUITS PART I 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 outof-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) SINGLE-PHASE AC GENERATOR THREE-PHASE AC GENERATOR 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). Æ It is the relative movement between the coil and the magnet that matters. THREE-PHASE VOLTAGE SOURCES Y-connected Source Δ-connected Source (uncommon) BALANCED Y-CONNECTED VOLTAGE SOURCE • Balanced phase voltages are equal in magnitude and are out of phase with one another by 120 degrees. Van = V p ∠0o Vbn = V p ∠ − 120o Vcn = V p ∠120o • Phase voltages sum up to zero, i.e., Van + Vbn + Vcn = 0 • Two possible combinations: Van = V p ∠0° Van = V p ∠0° Vbn = V p ∠ − 120° Vbn = V p ∠ + 120° Vcn = V p ∠ − 240° Vcn = V p ∠ + 240° abc (+) positive phase sequence acb (−) negative phase sequence • Balanced line voltages are equal in magnitude and are out of phase with one another by 120 degrees. Vab = 3 V p∠30o Vbc = 3 V p ∠ − 90o Vca = 3 V p ∠150o • Line voltages sum up to zero, i.e., Vab + Vbc + Vca = 0 • The magnitude of line voltages is √3 times the magnitude of the phase voltages. • Line voltages lead their corresponding phase voltages by 30° Y−Δ SOURCE TRANSFORMATION Vab = 3 VY ∠θY + 30o Van = VY ∠θY Vbn = VY ∠θY − 120o Vcn = VY ∠θY − 240o Y −Δ ⎯⎯⎯→ Vbc = 3 VY ∠θY − 90o Vca = 3 VY ∠θY − 210o Δ−Y SOURCE TRANSFORMATION Vab = VΔ ∠θ Δ Vbc = VΔ ∠θ Δ − 120o Vca = VΔ ∠θ Δ − 240o Δ −Y ⎯⎯⎯→ Van = VΔ ∠θ Δ − 30o 3 Vbn = VΔ ∠θ Δ − 150o 3 Vcn = VΔ ∠θ Δ − 270o 3 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 1 ZY = Z1 = Z 2 = Z 3 = Z Δ 3 Z Δ = Z a = Z b = Z c = 3Z Y Ex. Practice problem 12.1 Given that Vbn = 110∠30o V, find Van and Vcn , assuming a positive sequence. 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. BALANCED Y-Y CONNECTION Ex. Calculate line currents in the three-wire Y-Y system shown. 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 Van = 120∠30o V, find the line voltages and line currents. BALANCED Y-Δ CONNECTION Ex. A balanced abc-sequence Y-connected source with phase voltage Van = 100∠10o V is connected to a Δ-connected balanced load of 8+j4 Ω per phase. Calculate the phase and the line currents. Ex. Practice problem 12.3 One line voltage of a balanced Y-connected source is V AB = 240∠ − 20o V. If the source is connected to a Δ-connected load of 20∠ 40o Ω, find the phase and line currents assuming the abc sequence. BALANCED Δ-Δ CONNECTION Ex. A balanced Δ-connected load having an impedance of 20−j15 Ω is connected to a Δ-connected, positive sequence generator having Vab = 330∠0o V. Calculate the phase currents of the load and the line currents. 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 I a = 19.202∠35o A, find I AB and V AB . BALANCED Δ-Y CONNECTION Ex. A balanced Y-connected load with a phase impedance of 40+j 25 Ω is connected to a balanced, positive sequence Δconnected source with a line voltage of 210 V. Calculate the phase currents. Use Vab as a reference. Ex. Practice problem 12.5 In a balanced Δ-Y circuit, Vab = 240∠15o V and ZY = 12 + j15 Ω. Calculate the line currents.