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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 = kW Q line = kVAR P S = kW Q S = kVAR  In this Case

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

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

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

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

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

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

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

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

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

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

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

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

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