Lecture Notes ELE A6



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Transcription:

ecture Notes EE A6 Ramadan El-Shatshat Three Phase circuits 9/12/2006 EE A6 Three-phase Circuits 1

Three-phase Circuits 9/12/2006 EE A6 Three-phase Circuits 2

Advantages of Three-phase Circuits Smooth flow of power (instantaneous power is constant). Constant torque (reduced vibrations). The power delivery capacity tripled (increased by 200%) by increasing the number of conductors from 2 to 3 (increased by 50%). 9/12/2006 EE A6 Three-phase Circuits 3

Three-phase Circuits Wye-Connected System The neutral point is grounded The three-phase voltages have equal magnitude. The phase-shift between the voltages is 120 degrees. an 0 bn 120 cn 240 9/12/2006 EE A6 Three-phase Circuits 4

Three-phase Circuits Wye-Connected System ine-to-line voltages are the difference of the phase voltages ab an - bn 3 30 bc bn - cn 3-90 ca cn - an 3 150 9/12/2006 EE A6 Three-phase Circuits 5

Three-phase Circuits Wye-Connected System Phasor diagram is used to visualize the system voltages Wye system has two type of voltages: ine-to-neutral, and line-to-line. The line-to-neutral voltages are shifted with 120 o The line-to-line voltage leads the line to neutral voltage with 30 o The line-to-line voltage is 3 times the line-to-neutral voltage 9/12/2006 EE A6 Three-phase Circuits 6

Three-phase Circuits Wye-Connected oaded System The load is a balanced load and each one Z Each phase voltage drives current through the load. The phase current expressions are: a an bn, b, c z z z cn 9/12/2006 EE A6 Three-phase Circuits 7

Three-phase Circuits Wye-Connected oaded System Since the load is balanced (Z a Z b Z c ) then: Neutral current 0 an a Z a a This case single phase equivalent circuit can be used (phase a, for instance, only) Phase b and c are eliminated n o 9/12/2006 EE A6 Three-phase Circuits 8

Three-phase Circuits Wye-Connected System with balanced load A single-phase equivalent circuit is used Only phase a is drawn, because the magnitude of currents and voltages are the same in each phase. Only the phase angles are different (-120 o phase shift) The supply voltage is the line to neutral voltage. The single phase loads are connected to neutral or ground. ln oad 9/12/2006 EE A6 Three-phase Circuits 9

Three-phase Circuits Balanced Delta-Connected System The system has only one voltage : the line-to-line voltage ( ) The system has two currents : line current phase current The phase currents are: a z ab 9/12/2006 EE A6 Three-phase Circuits 10

Three-phase Circuits Delta-Connected System The line currents are: a b c ab bc ca ca ab bc n a balanced case the line currents are: line 3 30 phase 9/12/2006 EE A6 Three-phase Circuits 11

Three-phase Circuit Delta-Connected System The phasor diagram is used to visualize the system currents The system has two type of currents: line and phase currents. The delta system has only line-toline voltages, that are shifted by The phase currents lead the line currents by 30 The line current is 3 times the phase current and shifted by 30 degree. 9/12/2006 EE A6 Three-phase Circuits 12

Three-phase Circuit Circuit conversions A delta circuit can be converted to an equivalent wye circuit. The equation for phase a is (will not be used for this course): Z a Z ab Zab Zca + Z + bc Z ca Conversion equation for a balanced system is: Z a Z ab 3 9/12/2006 EE A6 Three-phase Circuits 13

Three-phase Circuit Power Calculation The three phase power is equal the sum of the phase powers P P a + P f the load is balanced: P 3 P b + phase P c 3 phase phase cos ( ) Wye system: P phase N 3 phase phase cos phase ( ) 3 cos ( ) 3 N Delta system: 3 ine phase phase P 3 cos ( ) 3 cos ( ) phase phase 9/12/2006 EE A6 Three-phase Circuits 14

Power measurement Three-phase Circuit n a four-wire system (3 phases and a neutral) the real power is measured using three singlephase watt-meters. n a three-wire system (three phases without neutral) the power is measured using only two single-phase watt-meters. - The watt-meters are supplied by the line current and the line-to-line voltage. 9/12/2006 EE A6 Three-phase Circuits 15

Three-phase Circuit Power measurement oad Watt meter 1 a The total power is the algebraic sum of the two watt-meters reading. W W P T 1 2 W ab cb 1 a c + W cos( θ cos( θ 2 vab vcb 3 Wattmeter 2 θ θ a c ) ) cos( θ ) c b 9/12/2006 EE A6 Three-phase Circuits 16

Three-phase Circuits Example 1 A 345 k, three phase transmission line delivers 500 MA, 0.866 power factor lagging, to a three phase load connected to its receiving end terminals. Assume the load is Y connected and the voltage at the receiving end is 345 k, find: The load impedance per phase. The line and phase currents. The total real and reactive power. 9/12/2006 EE A6 Three-phase Circuits 17

( a) Z 345k 3 0, S 3 500MA 3345k 836.7A 836.7 cos 1 (0.866) 836.7 30 Z 238 30 206 + j119 ( b) 836.7 30 ( c) P Q 3 3 cos( θ ) 433 MW sin( θ ) 249.9 MAR 9/12/2006 EE A6 Three-phase Circuits 18

Three-phase Circuits Example 2 Repeat example 2 assuming the load is Delta connected. 9/12/2006 EE A6 Three-phase Circuits 19

( a) Z S 3 500MA 3*345k 483.1 cos 345k 0, 1 714 30 483.1A (0.866) 618.3 + 483.1 30 j357 ( b) 3 30 836.7 60 ( c) P 3 cos( θ ) 433 MW Q 3 sin( θ ) 249.9 MAR 9/12/2006 EE A6 Three-phase Circuits 20

Power factor correction Power factor (p.f.) correction is the process of making P.f. 1. n order to correct the power factor in any system, a reactive (either inductive or capacitive) will be added to the load. f the load is inductive, then a capacitance is added. Note: Correcting the P.f. W NOT affect the active power, why? 9/12/2006 EE A6 Three-phase Circuits 21

Three-phase Circuits Example 3 A 3-phase load draws 120 kw at a power factor of 0.85 lagging from a 440 bus. n parallel with this load, a three phase capacitor bank that is rated 50 kar is inserted, find: The line current without the capacitor bank. The line current with the capacitor bank. The P.F. without the capacitor bank. The P.F. with the capacitor bank. 9/12/2006 EE A6 Three-phase Circuits 22

( a ) P 3 120 *10 3 440 (0.85 ) 185.25 cos 3 cos( θ ) 185.25 A 1 (0.85 ) 185.2 31.8, ( b ) Q Since we have pure capacitor load sin( Q θ ) 1 capacitor ( capacitor ( capacitor ( new ) capacitor ) ) 3 3 sin( θ ) ( capacitor ) 3 50 *10 65.6 A 3 440 65.6 90 + ( capacitor ) 160.6 11.49 ( c ) P.F. (no capacitor) 0.85 ( d ) P.F. (with capacitor) cos (11.49 ) 0.98 9/12/2006 EE A6 Three-phase Circuits 23

Home Work 9/12/2006 EE A6 Three-phase Circuits 24

Home Work 9/12/2006 EE A6 Three-phase Circuits 25