ECE 420. Review of Three Phase Circuits. Copyright by Chanan Singh, Panida Jirutitijaroen, and Hangtian Lei, For educational use only-not for sale.
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1 ECE 40 Review of Three Phase Circuits
2 Outline Phasor Complex power Power factor Balanced 3Ф circuit Read Appendix A
3 Phasors and in steady state are sinusoidal functions with constant frequency v t cos t v i t max t cos t i max t 0cos 60t 5cos 60t /10 0 1/10 1/60 t
4 Phasor Representation Phasor is an alternative way to represent sinusoidal function, from Euler s identity, Then, e j cos jsin j cos Re e j sin m e v t cos t max max Re Re Thus, and are represented by magnitude, and angles. e jt e j j e e j t max Conventional Phasor Representation
5 Define: Effective Phasor as effective phasor representation in polar form where max Reason will be clear when discuss about complex power. Rectangular representation of phasor is max e j e j cos j sin
6 v + t Let network - p Then, t p nstantaneous Power it t Network v i t max cos t t cos t max = instant power at t absorbed by the v t it 1 max max cos cos t
7 nstantaneous Power Plot p t v t it /10 0 1/10 1/60 t v i t t 5cos 60t 0cos 60t 4
8 Average Power Average power over one period, P 1 T T 0 T 1 T 0 1 Re pt dt 1 1 max max cos T T dt cos t dt max max max max cos cos e j Re * e j => Real Power (Watts) 0 T This is why we use effective phasor representation. 0 max
9 Power Factor Angle v i t t The phase angle differences between and is called power factor angle : 5cos 60t 0cos 60t /10 0 1/10 1/60 t
10 Power Factor Power factor is defined as Terminology: PF cos -Ф +Ф Power factor is lagging. Power factor is leading.
11 Load Description A load of 10 kw or 15 MA Meaning:? A load draws 0 kw at PF lagging Meaning:? A load absorbs 40 kw and 5 kar Meaning:? A load absorbs 30 ka at PF leading Meaning:?
12 Complex Power Define complex power (A- oltampere), S * e j e j cos j S sin * Thus, Q is called reactive power (AR- oltampere reactive). is called apparent power (A). S S P jq Ф S P P Q Re S Q ms
13 Complex Power Calculation From, or We have Or, X j R jx R Z Z S * * Z R P X Q * * * X R X j X R R jx R X R Z Z S X R R P X R X Q Z
14 Power Calculation: Example 1 Calculate instantaneous power and complex power for an inductance, + it Z L jl vt L - No real power dissipated by inductance but instantaneous power keeps on exchanging back and forth between system and an inductance. This requires generation support from system!
15 Power Calculation: Example Calculate instantaneous power and complex j power for a capacitor, + it Z C C vt C -
16 Power Factor Correction Undesirable to have non-zero reactive power, Q, (PF 1). Q causes t-line losses when transmitting power back and forth between load and generation. C Q L P L Reactive power becomes zero when inductive and capacitive reactances cancels. Thus, we can improve PF by adding a capacitor or an inductor. Q C
17 PF Correction: Example A single phase source delivers 00 kw to a 0.9 PF lagging load. What would be reactive power supported by a capacitor in order to increase PF to 0.95 lagging. + - Load, 0.9 PF lagging C Ф S P Q Q C
18 Conservation of Complex Power Theorem: Assume and to be sinusoidal function at the same frequency, sum of complex power supplied by sources equals to sum of complex power absorbed by all other branches in the network.
19 Balanced 3Ф Circuit What is 3-phase and how is it generated? Why 3-phase? More efficient Ruse of equipment and materials: 3 conductors instead of 6. Saving in losses Any pair of voltage sources differ by 10 with equal impedance sequences, positive and negative n practice, phase sequence depends on how we label the wires. Positive sequence, abc Negative sequence, acb cn bn bn cn 1 10 an an
20
21 3Ф oltage Line-line voltage:,, Line-neutral voltage: an, bn, KL: ab 30 an a bn ab bc 3 an ca cn Line-Line 3 Line-neutral + an A + bn ab - n B b C cn ab c - an bn oltage is given as line-line voltage by convention n
22 3Ф Current Phase current: ab, bc, Line current: a, b, KCL: a 30 ab a ca ca c 3 ab c Line 3 Phase ab A ca ca B C c n ab a bc b b bc ca Current is given as line current by convention
23 Δ-Y Load Transformation Z Z Z Y ZY ZY Z Z 3 Usually, load is given as then, Z Y S 3 with power factor, sin 1 P3 S3 p.f. Q S3 sin cos p.f. S3 3
24 Balanced 3Ф Power 3Ф Complex power: S * * * 3 a a b b c c * 3 a a S3 3 Lineneutral Line Line Line Line 3 nstantaneous power: p 3 t v t i t v t i t v t i t a a b b 3 cos 3P Constant power delivery to load Constant torque in motor as contrast with pulsation in 1Ф c c
25 Balanced 3Ф Circuit: Example 1 A 3Ф load of 300 MA, 100 k at 0.85 p.f. lagging. Find and Load P Load S3 300 MW Load Line S LineLine 3 3Line Line 100 k P Load P S 3 3 p.f.
26 ntroduction to Per Phase Analysis 1. All loads and sources are transformed to wye connections. No mutual inductance between phases 3. Circuit reduced to one phase and solved 4. Results extrapolated to the original configuration
27 Balanced 3Ф Circuit: Example Given a one-line diagram, S Source,Δ-Connected 1 j.5 C j5 Load 1, Δ-Connected Capacitor: ZC = -j15 3 Load, Y-Connected nductive: ZL = j10 f the voltage source is Line -Line. Find, 1. Current supplied by source,, and,. Current through a capacitor, C. S 3
28 Equivalent 3 phase diagram, 1a S j.5 a j5 3a ca 1c - + bc ab 1b j.5 c -j15 -j15 -j15 b C j5 3c j10 j10 j10 3b j.5 j5
29 Convert from Δ Y, cn 1c an 1a S bn 1b Z Y an Z j15 j5 3 3 ab j.5 j.5 c -j5 -j5 a -j5 b j5 j5 3c Use voltage source as angle reference j10 j10 3a j10 3b j.5 j5
30 1-phase diagram S j.5 j5 1a a 3a an j5 j10 S Z an eq Z eq j.5 j10 j5 j5 j10 j5 j5 a j. 5 an S We will use this to calculate C
31 1a S j.5 a j5 3a ca 1c - + bc ab 1b j.5 c -j15 -j15 -j15 b C j5 3c j10 j10 j10 3b j.5 j5 10 b a C a j15 b
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