Lecture (5) Power Factor,threephase circuits, and Per Unit Calculations

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1 Lecture (5) Power Factor,threephase circuits, and Per Unit Calculations 5-1

2 Repeating the Example on Power Factor Correction (Given last Class) P? Q? S? Light Motor From source Htz 10kW unity p.f. 0 ka 0.7 p.f. lagging Make this plant take power at 0.95 p.f. lagging 5-

3 Example Solution MOTOR S 0 cos -1 (0.7) ka ka P 0 cos 45.6 kw 14 kw Q 0 sin kar Light S L 10 + j 0 ka 5-3

4 TOTAL P Kw Q 14.8 KAR S KA θ tan p.f. cos lagging This is the total for this particular problem, before Power Factor Correction 5-4

5 When Adding a Capacitor θ new cos Q new P (tan ) 4 x 10 3 x tan kar Q new Q cap Q old Q new kar Need a capacitor that produces 6.38 KAR at 1000 volts 5-5

6 Find the value of capacitance c? -jx 1 c j ω c I * Q Q * - j ω c - jq ω c Q c Q * Q ω c ( ) 16.9 µf 1000 π

7 THREE PHASE CIRCUITS Delta-Connection Y-Connection A B ph L N LN A B C C 5-7

8 THREE PHASE CIRCUITS S 3 ph I ph * phase L 3 Y-Connection ph - ph L AB phase to phase (or line voltage) phase LN AN phase to neutral 5-8

9 Y-connection phase L 3 A B AB BC CA L I L I ph I ph I L N AB AN - BN C 5-9

10 -connection ph L A I A I ph I L 3 S 3 ph I * ph B I B C I C 5-10

11 Power in - and Y Connection If it is not specified in the problem, that the connection is or Y, assume that The connection is a Y-connection Y- Connection S 3 3 L 3 3 I L L * ph I ph * I L * - Connection S 3 I * p h p h 3 I * L L 3 3 L I * L 5-11

12 Example Source 0.4j 0.4j 0.4j ka 0.8 p.f. lagging 3 phase 13.8 k 5-1

13 One Line Diagram Source ~ 100 ka 0.8 p.f. lagging Load Source ~ j Load 5-13

14 Example Solution * I I L L I L θ cos A A 4-14

15 On the other hand, using Y-connection, S 3 p I p * 3 p I L * IL * 100 x 10 3 cos A Α x i.e. I L Α Energy Systems Research Laboratory, Mohammed. FIU All rights reserved

16 Example Solution source j 4 3 source ( )( j 4 source 3 ( ) ( ) ( )( ) ) source S 3 L * I L * I L S 3 L S 3 ph I * ph 5-16

17 PER UNIT SYSTEM P.U. (per unit) I 13.8 k p.f. lagging find eq? I? in p.u. on rated base value Based values for S and 5-17

18 S base 100 ka base 13.8 k S S actual p. u S base actual p. u base the load is operating at 1 p.u. S I 5-18

19 I S S * p. u p. u p. u. I I actual p. u. I base A x

20 actual p. u. S old p. u. new actual base new p. u. new actual S S base new base new p. u. new p u old.. S base old base old S base new base new 5-0

21 Calculate at a Different Base pu actual base new base old base S old base old base S base new base new p. u. new p. u. old S old base old base S base new base new 5-1

22 Example Source Feeder Load ~ p.f. lagging. (use as base) 0. + j0. p.u. on 00 ka at 34.5 k (use as base) Calculate at 100 ka at 13.8 k base. actual P.U. old ( base old ) 5-

23 p. u. new j p.u. new (0.+0.j) p. u. new. +. j p. u. new p.u. 5-3

24 Example on Per Unit R T.L. X L Ref Load Load p.f. 0.8 lagging S 100 MA S b 100 MA 13.8 k S b 13.8 k 5-4

25 Calculating Impedance Using S and S S θ 0 I I θ lagging - leading + I I I θ 5-5

26 S b Base MA b Base oltage S actual S pu S base 1 p. u. pu actual 13 base p. u. 5-6

27 Single Phase I θ ( 3 ) cos actual base 1 p. u. 5-7

28 Power Factor Angle. 1. cos Q cos oltage and Current P 3. I Power Triangle X cos Impedance Triangle R 5-8

29 Another Example on Per Unit Calculations 13.8/40 R T.L. X L Ref Load Base at generator T.L. S b 100 MA 13.8 k X L j0.3 p.u. on 40 k base and 50 MA base at the transmission Line 5-9

30 Calculate (p.u.) on a new base pu actual base S S old base base new new base old base S old base base new old base base new p. u. new p. u. old old base base new S 5-30

31 Example Solution base ( 3) ( T. L Ohms actual p.u. base 0.3 x Ohms p. u. actual base pu 5-31

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