Guided by: Sudhir pandey THREE PHASE CIRCUIT Made by: Vivek pal (13ELEE557) Sher singh (13ELEE564) Gagandeep (13ELEE552) Harsh patel (13ELEE549) Guided by: Sudhir pandey

SINGLE PHASE TWO WIRE

Objectives Explain the differences between single- phase, two-phase and three-phase. Compute and define the Balanced Three- Phase voltages. Determine the phase and line voltages/currents for Three-Phase systems.

SINGLE PHASE SYSTEM A generator connected through a pair of wire to a load – Single Phase Two Wire. Vp is the magnitude of the source voltage, and  is the phase.

SINLGE PHASE THREE WIRE

SINGLE PHASE SYSTEM Most common in practice: two identical sources connected to two loads by two outer wires and the neutral: Single Phase Three Wire. Terminal voltages have same magnitude and the same phase.

POLYPHASE SYSTEM Circuit or system in which AC sources operate at the same frequency but different phases are known as polyphase.

TWO PHASE SYSTEM THREE WIRE

POLYPHASE SYSTEM Two Phase System: A generator consists of two coils placed perpendicular to each other The voltage generated by one lags the other by 90.

POLYPHASE SYSTEM Three Phase System: A generator consists of three coils placed 120 apart. The voltage generated are equal in magnitude but, out of phase by 120. Three phase is the most economical polyphase system.

THREE PHASE FOUR WIRE

IMPORTANCE OF THREE PHASE SYSTEM All electric power is generated and distributed in three phase. One phase, two phase, or more than three phase input can be taken from three phase system rather than generated independently. Melting purposes need 48 phases supply.

IMPORTANCE OF THREE PHASE SYSTEM Uniform power transmission and less vibration of three phase machines. The instantaneous power in a 3 system can be constant (not pulsating). High power motors prefer a steady torque especially one created by a rotating magnetic field.

IMPORTANCE OF THREE PHASE SYSTEM Three phase system is more economical than the single phase. The amount of wire required for a three phase system is less than required for an equivalent single phase system. Conductor: Copper, Aluminum, etc

THREE PHASE GENERATION

FARADAYS LAW Three things must be present in order to produce electrical current: Magnetic field Conductor Relative motion Conductor cuts lines of magnetic flux, a voltage is induced in the conductor Direction and Speed are important

GENERATING A SINGLE PHASE Motion is parallel to the flux. No voltage is induced.

GENERATING A SINGLE PHASE x N S Motion is 45° to flux. Induced voltage is 0.707 of maximum.

GENERATING A SINGLE PHASE x N S Motion is perpendicular to flux. Induced voltage is maximum.

GENERATING A SINGLE PHASE x N S Motion is 45° to flux. Induced voltage is 0.707 of maximum.

GENERATING A SINGLE PHASE Motion is parallel to flux. No voltage is induced.

GENERATING A SINGLE PHASE x N S Motion is 45° to flux. Notice current in the conductor has reversed. Induced voltage is 0.707 of maximum.

GENERATING A SINGLE PHASE x Motion is perpendicular to flux. Induced voltage is maximum.

GENERATING A SINGLE PHASE x Motion is 45° to flux. Induced voltage is 0.707 of maximum.

GENERATING A SINGLE PHASE Motion is parallel to flux. No voltage is induced. Ready to produce another cycle.

THREE PHASE GENERATOR

GENERATOR WORK The generator consists of a rotating magnet (rotor) surrounded by a stationary winding (stator). Three separate windings or coils with terminals a-a’, b-b’, and c-c’ are physically placed 120 apart around the stator.

As the rotor rotates, its magnetic field cuts the flux from the three coils and induces voltages in the coils. The induced voltage have equal magnitude but out of phase by 120.

GENERATION OF THREE-PHASE AC x S

THREE-PHASE WAVEFORM Phase 1 Phase 2 Phase 3 120° 240° Phase 2 lags phase 1 by 120°. Phase 2 leads phase 3 by 120°. Phase 3 lags phase 1 by 240°. Phase 1 leads phase 3 by 240°.

GENERATION OF 3 VOLTAGES x S Phase 1 Phase 2 Phase 3 Phase 1 is ready to go positive. Phase 2 is going more negative. Phase 3 is going less positive.

THREE PHASE QUANTITIES

BALANCED 3 VOLTAGES Balanced three phase voltages: same magnitude (VM ) 120 phase shift

BALANCED 3 CURRENTS Balanced three phase currents: same magnitude (IM ) 120 phase shift

PHASE SEQUENCE NEGATIVE SEQUENCE POSITIVE SEQUENCE

PHASE SEQUENCE

EXAMPLE # 1 Determine the phase sequence of the set voltages:

BALANCED VOLTAGE AND LOAD Balanced Phase Voltage: all phase voltages are equal in magnitude and are out of phase with each other by 120. Balanced Load: the phase impedances are equal in magnitude and in phase.

THREE PHASE CIRCUIT POWER The instantaneous power is constant

THREE PHASE CIRCUIT Three Phase Power,

THREE PHASE QUANTITIES QUANTITY SYMBOL Phase current I Line current IL Phase voltage V Line voltage VL

PHASE VOLTAGES and LINE VOLTAGES Phase voltage is measured between the neutral and any line: line to neutral voltage Line voltage is measured between any two of the three lines: line to line voltage.

PHASE CURRENTS and LINE CURRENTS Line current (IL) is the current in each line of the source or load. Phase current (I) is the current in each phase of the source or load.

THREE PHASE CONNECTION

SOURCE-LOAD CONNECTION Wye Y-Y Delta Y- -  -Y

SOURCE-LOAD CONNECTION Common connection of source: WYE Delta connected sources: the circulating current may result in the delta mesh if the three phase voltages are slightly unbalanced. Common connection of load: DELTA Wye connected load: neutral line may not be accessible, load can not be added or removed easily.

WYE CONNECTION

WYE CONNECTED GENERATOR

WYE CONNECTED LOAD OR

BALANCED Y-Y CONNECTION

PHASE CURRENTS AND LINE CURRENTS In Y-Y system:

PHASE VOLTAGES, V Van Vbn Vcn Phase voltage is measured between the neutral and any line: line to neutral voltage

PHASE VOLTAGES, V

LINE VOLTAGES, VL Vab Vbc Vca Line voltage is measured between any two of the three lines: line to line voltage.

LINE VOLTAGES, VL

PHASE VOLTAGE (V) LINE VOLTAGE (VL)

PHASE DIAGRAM OF VL AND V

PROPERTIES OF PHASE VOLTAGE All phase voltages have the same magnitude, Out of phase with each other by 120 = =

PROPERTIES OF LINE VOLTAGE All line voltages have the same magnitude, Out of phase with each other by 120 = =

RELATIONSHIP BETWEEN V and VL Magnitude Phase - VL LEAD their corresponding V by 30

EXAMPLE 1 Calculate the line currents

DELTA CONNECTION

DELTA CONNECTED SOURCES

DELTA CONNECTED LOAD OR

BALANCED -  CONNECTION

PHASE VOLTAGE AND LINE VOLTAGE In - system, line voltages equal to phase voltages:

PHASE VOLTAGE, V Phase voltages are equal to the voltages across the load impedances.

PHASE CURRENTS, I The phase currents are obtained:

LINE CURRENTS, IL The line currents are obtained from the phase currents by applying KCL at nodes A,B, and C.

LINE CURRENTS, IL

PHASE CURRENTS (I) LINE CURRENTS (IL)

PHASE DIAGRAM OF IL AND I

PROPERTIES OF PHASE CURRENT All phase currents have the same magnitude, Out of phase with each other by 120

PROPERTIES OF LINE CURRENT All line currents have the same magnitude, Out of phase with each other by 120

RELATIONSHIP BETWEEN I and IL Magnitude Phase - IL LAG their corresponding I by 30

EXAMPLE A balanced delta connected load having an impedance 20-j15  is connected to a delta connected, positive sequence generator having Vab = 3300 V. Calculate the phase currents of the load and the line currents.

Given Quantities

Phase Currents

Line Currents

BALANCED WYE-DELTASYSTEM

THREE PHASE POWER MEASUREMENT

Example 1 Unbalanced load In a three-phase four-wire system the line voltage is 400V and non-inductive loads of 5 kW, 8 kW and 10 kW are connected between the three conductors and the neutral. Calculate: (a) the current in each phase (b) the current in the neutral conductor.

Voltage to neutral Current in 10kW resistor Current in 8kW resistor Current in 5kW resistor

IR IY IB IYH IYV IBH IBV INV INH IN Resolve the current components into horizontal and vertical components.

Example 2 A delta –connected load is arranged as in Figure below. The supply voltage is 400V at 50Hz. Calculate: The phase currents; The line currents. (a) I1 is in phase with VRY since there is only resistor in the branch

In branch between YB , there are two components , R2 and X2 In the branch RB , only capacitor in it , so the XC is -90 out of phase.

(b) q=30o q = 71o 34’ -60o= 11o 34’

q = 180-30o-11o 34’ = 138o 34’

Power in three phase Active power per phase = IPVP x power factor Total active power= 3VPIP x power factor If IL and VL are rms values for line current and line voltage respectively. Then for delta () connection: VP = VL and IP = IL/3. therefore: For star connection () : VP = VL/3 and IP = IL. therefore:

Example 3 A three-phase motor operating off a 400V system is developing 20kW at an efficiency of 0.87 p.u and a power factor of 0.82. Calculate: The line current; The phase current if the windings are delta-connected. (a) Since And line current =IL=40.0A (b) For a delta-connected winding

Example 4 Three identical coils, each having a resistance of 20 and an inductance of 0.5 H connected in (a) star and (b) delta to a three phase supply of 400 V; 50 Hz. Calculate the current and the total power absorbed by both method of connections. First of all calculating the impedance of the coils where

Star connection

Example 5 A balanced three phase load connected in star, each phase consists of resistance of 100  paralleled with a capacitance of 31.8 F. The load is connected to a three phase supply of 415 V; 50 Hz. Calculate: (a) the line current; (b) the power absorbed; (c) total kVA; (d) power factor . 415

Admittance of the load where Line current Volt-ampere per phase Active power per phase Total active power

Reactive power per phase (b) Reactive power per phase Total reactive power (c) Total volt-ampere (d) Power Factor = cos = cos 45 = 0.707 (leading)

Example 6 A three phase star-connected system having a phase voltage of 230V and loads consist of non reactive resistance of 4 , 5  and 6 respectively. Calculate: (a) the current in each phase conductor (b) the current in neutral conductor and (c) total power absorbed.

X-component = 46 cos 30 + 38.3 cos 30 - 57.5 = 15.5 A (b) X-component = 46 cos 30 + 38.3 cos 30 - 57.5 = 15.5 A Y-component = 46 sin 30 - 38.3 sin 30 = 3.9 A Therefore (c)