Bruce Mayer, PE Registered Electrical & Mechanical Engineer BMayer@ChabotCollege

1. Engineering 43 Chp 9 [5-7] Complex Power Bruce Mayer, PE Registered Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

2. 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. 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. Consider A Complex Current Thru a Complex Impedance Load Power Factor • By Ohm & Euler • The Current and Load-Voltage Phasors (Vectors) Can Be Plotted on the Complex Plane • in the Electrical Power Industry θZ is the Power Factor Angle, or Simply the Phase Angle

5. The Phase Angle Can Be Positive or Negative Depending on the Nature of the Load V is the BaseLine Power Factor cont • 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 • Vrms • Irms

6. The Product of the DMM Measurements is the APPARENT Power Power Factor cont.2 • Now Define the Power Factor for the Load • The Apparent Power is NOT the Actual Power, and is thus NOT stated in Watts. • Apparent Power Units = VA or kVA • Some Load Types

7. Consider this case Vrms = 460 V Irms = 200A pf = 1.5% Then Papparent = 92kVA Pactual =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 pf – Why do We Care?

8. The Local Power Company Services this Large Industrial Load Example  Power Factor • Then the I2R Loses in the 100 mΩ line • Improving the pf to 94% Power company I lags V • Find Irms by Pwr Factor

9. For This Ckt The Effect of the Power Factor on Line Losses Example - Power Factor cont

10. Consider a general Ckt with an Impedance Ld Active Power Reactive Power Complex Power • Mathematically • Converting to Rectangular Notation • For this Situation Define the Complex Power for the Load:

11. Thus S in Shorthand Complex Power cont • 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. Urms

12. Now in the General Ckt By Ohm’s Law Complex Power cont.2 • And Again by Ohm • In the Last Expression Equate the REAL and Imaginary Parts

13. And by Complex Power Definition Complex Power cont.3 • Similarly for Q • Using the Previous Results for P • So Finally the Alternative Expression for S

14. The Expressions for S Complex Power Triangle • Plotting S in the Complex Plane • From The Complex Power “Triangle” Observe • Note also That Complex Power is CONSERVED

15. For the Circuit At Right Zline =0.09 Ω + j0.3 Ω Pload = 20 kW Vload = 2200° pf = 80%, lagging f = 60 Hz → ω = 377s-1 inductive capacitive Example - Complex Power • From the Actual Power • Lagging pf → Inductive • Thus • And Q from Pwr Triangle

16. Then SL Example - Complex Power cont • Recall the S Mathematical Definition • Note also that [U*]* = U • In the S Definition, Isolating the Load Current and then Conjugating Both Sides • Alternatively Lagging

17. Now Determine VS Example - Complex Pwr cont.2 • Then VS • Then The Phase Angle • To find the Src Power Factor, Draw the I & V Phasor Diagram

18. For the Circuit AtRight, Determine Real And Reactive Power losses in the Ln Real And Reactive Power at the Source inductive capacitive Example - Complex Power kVAR • From the Actual Power • Lagging pf → Inductive • Thus • And by S Definition

19. Also from the SRelation Example - Complex kVAR cont. • Now the PowerFactor Angle • Then for Line Loses I Lagging V • Quantitatively • pf = cos(θv−θi); hence

20. Find Power Supplied by Conservation of Complex Power Example - Complex kVAR cont.2 • In this Case • Then to Summarize the Answer • Pline = 4.685 kW • Qline = 11.713 kVAR • PS = 44.685 kW • QS = 37.552 kVAR

21. 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 Power Factor Correction • To Arrive at the Power Factor Correction Strategy Consider A Schematic of a typical Industrial Load

22. Prior to The Addition of the Capacitor Power Factor Correction cont. • For The Capacitive Load • After Addition of the Capacitor

23. Find θnew Cap is a Purely REACTIVE Load QL QL-QC Qnew QC P Power Factor Correction cont.2 • The Vector Plot Below Shows Power Factor Correction Strategy • Use Trig ID to find QC to give desired θnew

24. Start with the ID Trig ID Digression • Or • Solve for tan • Recall tanθnew • But: cosθnew = pfnew • Substituting

25. Kayak Centrifugal Injection-Molding Power Analysis Improve Power Factor to 95% Find Sold Example – pf Correction Roto-molding process • Adding A Cap DoesNOT Change P • Use Trig ID to Find Tan(new) • Now Qold • And by S Relation

26. Then the Needed QC Example – pf Correction cont Roto-molding process • Recall The Expression for QC • Then C from QC

27. WhiteBoard Work • Let’s Work (w/ 2-changes) Problem 9.81 • Determine at the input SOURCE • Voltage & Current • Complex Power • Δθ & pf 60 kVA