1. BMayer@ChabotCollege.edu ENGR-43_Lec-09-2_Complex_Power.ppt 1 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Bruce Mayer, PE Registered Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Engineering 43 Chp 9 [5-7] Complex Power

2. BMayer@ChabotCollege.edu ENGR-43_Lec-09-2_Complex_Power.ppt 2 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Outline – AC SS Power cont.  Effective or RMS Values Heating Value for Sinusoidal Signals  Power Factor A Measure Of The Angle Between Current And Voltage Phasors within a Load  Power Factor Correction Improve Power Transfer To a Load By “Aligning” Phasors  Single Phase Three-Wire Circuits Typical HouseHold Power Distribution

3. BMayer@ChabotCollege.edu ENGR-43_Lec-09-2_Complex_Power.ppt 3 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Outline – AC Steady State Power  Instantaneous Power Concept For The Special Case Of Steady State Sinusoidal Signals  Average Power Concept Power Absorbed Or Supplied During in Integer Number of Complete Cycles  Maximum Average Power Transfer When The Circuit Is In Sinusoidal Steady State

4. BMayer@ChabotCollege.edu ENGR-43_Lec-09-2_Complex_Power.ppt 4 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Power Factor  Consider A Complex Current Thru a Complex Impedance Load  The Current and Load-Voltage Phasors (Vectors) Can Be Plotted on the Complex Plane  By Ohm & Euler  in the Electrical Power Industry θ Z is the Power Factor Angle, or Simply the Phase Angle

5. BMayer@ChabotCollege.edu ENGR-43_Lec-09-2_Complex_Power.ppt 5 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Power Factor cont  The Phase Angle Can Be Positive or Negative Depending on the Nature of the Load  Typical Industrial Case is the INDUCTIVE Load Large Electric Motors are Essentially Inductors  Now Recall The General Power Eqn  Measuring the Load with an AC DMM yields V rms I rms V is the BaseLine

6. BMayer@ChabotCollege.edu ENGR-43_Lec-09-2_Complex_Power.ppt 6 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Power Factor cont.2  The Product of the DMM Measurements is the APPARENT Power  The Apparent Power is NOT the Actual Power, and is thus NOT stated in Watts. Apparent Power Units = VA or kVA  Now Define the Power Factor for the Load  Some Load Types

7. BMayer@ChabotCollege.edu ENGR-43_Lec-09-2_Complex_Power.ppt 7 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis pf – Why do We Care?  Consider this case V rms = 460 V I rms = 200A pf = 1.5%  Then P apparent = 92kVA P actual =1.4 kW   This Load requires The Same Power as a Hair Dryer  However, Despite the low power levels, The WIRES and CIRCUIT BREAKERS that feed this small Load must be Sized for 200A! The Wires would be nearly an INCH in Diameter

8. BMayer@ChabotCollege.edu ENGR-43_Lec-09-2_Complex_Power.ppt 8 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example  Power Factor  The Local Power Company Services this Large Industrial Load  Find I rms by Pwr Factor  Then the I 2 R Loses in the 100 mΩ line  Improving the pf to 94% Power company I lags V

9. BMayer@ChabotCollege.edu ENGR-43_Lec-09-2_Complex_Power.ppt 9 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example - Power Factor cont  For This Ckt The Effect of the Power Factor on Line Losses

10. BMayer@ChabotCollege.edu ENGR-43_Lec-09-2_Complex_Power.ppt 10 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Complex Power  Consider a general Ckt with an Impedance Ld  Mathematically  For this Situation Define the Complex Power for the Load:  Converting to Rectangular Notation Active Power Reactive Power

11. BMayer@ChabotCollege.edu ENGR-43_Lec-09-2_Complex_Power.ppt 11 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Complex Power cont  Thus S in Shorthand  Alternatively, Reconsider the General Sinusoidal Circuit  S & Q are NOT Actual Power, and Thus all Terms are given Non-Watt Units S→ Volt-Amps (VA) Q → Volt-Amps, Reactive (VAR)  P is Actual Power and hence has Units of W  First: U vs. U rms

12. BMayer@ChabotCollege.edu ENGR-43_Lec-09-2_Complex_Power.ppt 12 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Complex Power cont.2  Now in the General Ckt By Ohm’s Law  In the Last Expression Equate the REAL and Imaginary Parts  And Again by Ohm

13. BMayer@ChabotCollege.edu ENGR-43_Lec-09-2_Complex_Power.ppt 13 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Complex Power cont.3  And by Complex Power Definition  Using the Previous Results for P  Similarly for Q  So Finally the Alternative Expression for S

14. BMayer@ChabotCollege.edu ENGR-43_Lec-09-2_Complex_Power.ppt 14 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Complex Power Triangle  The Expressions for S  Plotting S in the Complex Plane  From The Complex Power “Triangle” Observe  Note also That Complex Power is CONSERVED

15. BMayer@ChabotCollege.edu ENGR-43_Lec-09-2_Complex_Power.ppt 15 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example - Complex Power  For the Circuit At Right Z line =0.09 Ω + j0.3 Ω P load = 20 kW V load = 220  0° pf = 80%, lagging f = 60 Hz → ω = 377s -1  Lagging pf → Inductive  From the Actual Power  Thus inductive capacitive  And Q from Pwr Triangle

16. BMayer@ChabotCollege.edu ENGR-43_Lec-09-2_Complex_Power.ppt 16 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example - Complex Power cont  Then S L  Recall the S Mathematical Definition  Alternatively  Note also that [U*]* = U  In the S Definition, Isolating the Load Current and then Conjugating Both Sides Lagging

17. BMayer@ChabotCollege.edu ENGR-43_Lec-09-2_Complex_Power.ppt 17 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example - Complex Pwr cont.2  Now Determine V S  Then V S  Then The Phase Angle  To find the Src Power Factor, Draw the I & V Phasor Diagram

18. BMayer@ChabotCollege.edu ENGR-43_Lec-09-2_Complex_Power.ppt 18 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example - Complex Power kVAR  For the Circuit At Right, Determine Real And Reactive Power losses in the Ln Real And Reactive Power at the Source  Lagging pf → Inductive  From the Actual Power  Thus inductive capacitive  And by S Definition

19. BMayer@ChabotCollege.edu ENGR-43_Lec-09-2_Complex_Power.ppt 19 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example - Complex kVAR cont.  Also from the S Relation  Now the Power Factor Angle  Then for Line Loses  Quantitatively  pf = cos(θ v − θ i ); hence I Lagging V

20. BMayer@ChabotCollege.edu ENGR-43_Lec-09-2_Complex_Power.ppt 20 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example - Complex kVAR cont.2  Find Power Supplied by Conservation of Complex Power  Then to Summarize the Answer P line = 4.685 kW Q line = 11.713 kVAR P S = 44.685 kW Q S = 37.552 kVAR  In this Case

21. BMayer@ChabotCollege.edu ENGR-43_Lec-09-2_Complex_Power.ppt 21 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Power Factor Correction  As Noted Earlier, Most Industrial Electrical Power Loads are Inductive The Inductive Component is Typically Associated with Motors  The Motor-Related Lagging Power Factor Can Result in Large Line Losses  The Line-Losses can Be Reduced by Power Factor Correction  To Arrive at the Power Factor Correction Strategy Consider A Schematic of a typical Industrial Load

22. BMayer@ChabotCollege.edu ENGR-43_Lec-09-2_Complex_Power.ppt 22 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Power Factor Correction cont.  Prior to The Addition of the Capacitor  For The Capacitive Load  After Addition of the Capacitor

23. BMayer@ChabotCollege.edu ENGR-43_Lec-09-2_Complex_Power.ppt 23 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Power Factor Correction cont.2  Find θ new Cap is a Purely REACTIVE Load  The Vector Plot Below Shows Power Factor Correction Strategy  Use Trig ID to find Q C to give desired θ new QLQL QCQC Q L -Q C P Q new

24. BMayer@ChabotCollege.edu ENGR-43_Lec-09-2_Complex_Power.ppt 24 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Trig ID Digression  Start with the ID  Solve for tan   Recall tanθ new  Or  But: cosθ new = pf new  Substituting

25. BMayer@ChabotCollege.edu ENGR-43_Lec-09-2_Complex_Power.ppt 25 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example – pf Correction  Kayak Centrifugal Injection-Molding Power Analysis Improve Power Factor to 95%  Find S old  Now Q old  Adding A Cap Does NOT Change P Use Trig ID to Find Tan(  new )  And by S Relation Roto-molding process

26. BMayer@ChabotCollege.edu ENGR-43_Lec-09-2_Complex_Power.ppt 26 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example – pf Correction cont  Then the Needed Q C  Recall The Expression for Q C Roto-molding process  Then C from Q C

27. BMayer@ChabotCollege.edu ENGR-43_Lec-09-2_Complex_Power.ppt 27 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis WhiteBoard Work  Let’s Work (w/ 2-changes) Problem 9.81 Determine at the input SOURCE –Voltage & Current –Complex Power –Δθ & pf 60 kVA

28. BMayer@ChabotCollege.edu ENGR-43_Lec-09-2_Complex_Power.ppt 28 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

29. BMayer@ChabotCollege.edu ENGR-43_Lec-09-2_Complex_Power.ppt 29 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

30. BMayer@ChabotCollege.edu ENGR-43_Lec-09-2_Complex_Power.ppt 30 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

31. BMayer@ChabotCollege.edu ENGR-43_Lec-09-2_Complex_Power.ppt 31 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

32. BMayer@ChabotCollege.edu ENGR-43_Lec-09-2_Complex_Power.ppt 32 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

33. BMayer@ChabotCollege.edu ENGR-43_Lec-09-2_Complex_Power.ppt 33 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis