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BASIC ELECTRICAL TECHNOLOGY DET 211/3 Chapter 3 - Three Phase System INTRODUCTION TO THREE PHASE SYSTEM In general, three phase systems are preferred over single phase systems for the transmission of the power system for many reasons, including the following: • Thinner conductors can be used to transmit the same kVA at the same voltage, which reduces the amount of copper required (typically about 25% less) and turn reduces construction and maintenance costs. • The lighter lines are easier to install, and the supporting structures can be less massive and farther apart. • In general, most larger motors are three phase because they are essentially self starting and do not require a special design or additional starting circuitry. Three phase voltages A 3-phase generator basically consists of a rotating magnet (called the rotor) surrounded by a stationary winding (called the stator). Three separate windings or coils with terminals a-a’, b-b’ and c-c’ are physically placed 120o apart around the stator. Generated Voltages The three phase generator can supply power to both single phase and three phase loads The sinusoidal expression for each of the phase voltages v AN Vm( AN ) sin t v BN Vm ( BN ) sin(t 120 o ) vCN Vm(CN ) sin(t 240 o ) Vm (CN ) sin(t 120 o ) The phasor diagram of the phase voltages The effective value of each is determined by V AN VBN Vm( AN ) 2 Vm ( BN ) VCN 2 Vm (CN ) 2 0.707 Vm( AN ) V AN V AN ( m )0 o 0.707 Vm ( BN ) VBN VBN ( m ) 120 o 0.707 Vm (CN ) VCN VCN ( m ) 120 o If the voltage sources have the same amplitude and frequency ω and are out of the phase with each other by 120o, the voltages are said to be balanced. By rearranging the phasors as shown in figure below, so V AN VBN VCN V AN ( m) 0 o VBN ( m) 120 o VCN ( m) 120 o Vm (1.0 0.5 j 0.866 0.5 j 0.866 ) 0 Where | VAN || VBN || VCN | Vm Generator and Load Connections Each generator in a 3-phase system maybe either Yor D-connected and loads may be mixed on a power system. Z Z Z Z Z Z Wye Connected Generator Applying KVL around the indicated loop in figure above, we obtain I L Ig VAB VAN VBN VAN VNB VBC VBN VCN VBN VNC VCA VCN VAN VCN VNA Wye Connected Generator For line-to-line voltage VAB is given by: VAB VA VB V 00 V 120 0 1 3 V V j V 2 2 3 3 V j V 2 2 3 1 3V j 2 2 3V 30 0 Wye Connected Generator VAB V AB30 0 3V AN 30 0 VCA 3VCN 150 0 Phasor Diagram VBC 3VBN 270 0 Wye Connected Generator The relationship between the magnitude of the line-to-line and line-to-neutral (phase) voltage is: VLL 3V The line voltages are shifted 300 with respect to the phase voltages. Phasor diagram of the line and phase voltage for the Y connection is shown below. VCN VAB VCA Rearrange VAN VBN VBC Line-to-line voltages Phase voltages Delta Connected Generator For line-to-line voltage VAB is given by: I A I AB I CA I 00 I 240 0 1 3 I I j I 2 2 V LL V 3 3 I j I 2 2 3 1 3I j 2 2 3 I 30 0 Delta Connected Generator The relationship between the magnitude of the line and phase current is: I L 3I The line currents are shifted 300 relative to the corresponding phase current. Phasor diagram of the line and phase current for the Y connection is shown IC below. ICA IAB IB IA IBC Line-to-line currents Phase currents Phase sequence The phase sequence is the order in which the voltages in the individual phases peak. VC VB VA VB abc phase sequence VA VC acb phase sequence Power relationshipphase quantities The power equations applied to Y-or D load in a balanced 3-phase system are: P 3V I cos P 3I2 Z cos Real power Unit=Watts(W) Q 3V I sin Q 3I2 Z sin Reactive power Unit=Volt-Amps-Reactive (VAR) S 3V I S 3I 2 Z Apparent power Unit=Volt-Amps (VA) - angle between voltage and current in any phase of the load Power relationshipLine quantities The power equations applied to Y-or D load in a balanced 3-phase system are: P 3VLL I L cos Real power Q 3VLL I L sin Reactive power S 3VLL I L Apparent power - angle between phase voltage and phase current in any phase of the load Since both the three-phase source and the threephase load can be either Y- or D- connected, we have 4 possible connections: i) Y-Y connections (i.e: Y-connected source with a Yconnected load). ii) Y-D connection. iii) D-D connection iv) D-Y connection (i) Y connected generator/source with Y connected load Z1 Z 2 Z 3 I g I L I L V E EL 3V (ii) Y-D Connection A balanced Y-D system consists of a balanced Yconnected source feeding a balanced D-connected load Z/3 Z/3 Z Z Z Z/3 ZD ZY 3 D must consists of three equal impedances (iii) ∆-∆ Connection A balanced ∆-D system consists of a balanced ∆connected source feeding a balanced D-connected load Z Z Z Z Z Z (iv) DY Connection A balanced D-Y system consists of a balanced Dconnected source feeding a balanced Y-connected load Z Z Z Z/3 Z/3 Z/3 Example 1 A 208V three-phase power system is shown in Figure 1. It consists of an ideal 208V Y-connected three-phase generator connected to a three-phase transmission line to a Y-connected load. The transmission line has an impedance of 0.06+j0.12W per phase, and the load has an impedance of 12+j9W per phase. For this simple system, find (a) The magnitude of the line current IL (b) The magnitude of the load’s line and phase voltages VLL and VL (c) The real, reactive and apparent powers consumed by the load (d) The power factor of the load (e) The real, reactive and apparent powers consumed by the transmission line (f) The real, reactive and apparent powers supplied by the generator (g) The generator’s power factor Example 1 0.06W + 0.06W i0.12W i0.12W V Z Vcn=120-2400 Z Van=12000 208V Z=12+ i9W Z + Vbn=120-1200 _ 0.06W Figure 1 i0.12W Solution Example 1 (a) The magnitude of the line current IL I line Vline Z line Z load 1200V (0.06 j 0.12)W (12 j 9)W 1200 1200 12.06 j 9.12 15.1237.1 7.94 37.1 A So, the magnitude of the line current is thus 7.94 A Solution Example 1 (b) The magnitude of the load’s line and phase voltages VLL and VL The phase voltage on the load is the voltage across one phase of the load. This voltage is the product of the phase impedance and the phase current of the load: VL I L Z L (7.94 37.1 A)(12 j 9W) (7.94 37.1 A)(1536.9W) 119.1 0.2V Therefore, the magnitude of the load’s phase voltage is VL 119.1V and the magnitude of the load’s line voltage is VLL 3VL 206.3V Solution Example 1 (c) The real power consumed by the load is PLoad 3V I cos 3(119.1V )(7.94 A) cos 36.9 2270W The reactive power consumed by the load is QLoad 3V I sin 3(119.1V )(7.94 A) sin 36.9 1702 var The apparent power consumed by the load is S Load 3V I 3(119.1V )(7.94 A) 2839VA Solution Example 1 PFLoad cos (d) The load power factor is cos 36.9 0.8lagging (e) The current in the transmission line is 7.94 37.1 A and the impedance of the line is (0.06 j 0.12)W or 0.13463.4W per phase. Therefore, the real, reactive and apparent powers consumed in the line are: PLine 3I Z cos QLine 3I Z sin 3(7.94 A) (0.134W) cos 63.4 3(7.94 A) 2 (0.134W) sin 63.4 11.3W 22.7 var 2 2 2 S Line 3I Z 2 3(7.94 A) 2 (0.134W) 25.3VA Solution Example 1 (f) The real and reactive powers supplied by the generator are the sum of the powers consumed by the line and the load: Pgen Pline Pload 11.3W 2270W 2281W Q gen Qline Qload 22.7 var 1702 var 1725 var The apparent power of the generator is the square root of the sum of the squares of the real and reactive powers: S gen Pgen Q gen 2860VA 2 2 Solution Example 1 (g) From the power triangle, the power factor angle is gen tan 1 Q gen Pgen 1725VAR tan 37.1 2281W 1 Therefore, the generator’s power factor is PFgen cos 37.1 0.798lagging Assignment 2 A 208V three-phase power system is shown in Figure 2. It consists of an ideal 208V Y-connected three-phase generator connected to a three-phase transmission line to a D-connected load. The transmission line has an impedance of 0.06+j0.12W per phase, and the load has an impedance of 12+j9W per phase. For this simple system, find (a) The magnitude of the line current IL (b) The magnitude of the load’s line and phase voltages VLL and VL (c) The real, reactive and apparent powers consumed by the load (d) The power factor of the load (e) The real, reactive and apparent powers consumed by the transmission line (f) The real, reactive and apparent powers supplied by the generator (g) The generator’s power factor Assignment 2 Figure 2