EE 221 AC Circuit Power Analysis. Instantaneous and average power RMS value Apparent power and power factor Complex power

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1 EE 1 AC Circuit Power Analysis Instantaneous and average power RMS value Apparent power and power factor Complex power

2 Instantaneous Power Product of time-domain voltage and time-domain current p(t) = v(t) i(t) Determine maximum value Transients Steady-state steady-state As the transient dies out, the circuit returns to steady-state operation. Since the only source remaining in the circuit is dc, the inductor eventually acts as a short circuit absorbing zero power.

3 Power due to Sinusoidal Excitation v(t) = V m cos(ωt+θ) and i(t) = I m cos(ωt +ϕ) p(t) = V m I m cos(ωt+θ) cos(ωt +ϕ) = = ½ V m I m cos(θ - ϕ) + ½ V m I m cos(ωt + θ + ϕ) trigonometric identity p(t) = ½ V m I m cos(θ - ϕ) + ½ V m I m cos(ωt + θ + ϕ) instantaneous two parts average periodic constant independent of t the "average" wanted active or real periodic period is ½T average is zero unwanted

4 Power due to Sinusoidal Excitation p(t) = ½ V m I m cos(θ - ϕ) + ½ V m I m cos(ωt + θ + ϕ) voltage V = 4 0 V, impedance Z = 60 Ω, ω = π/ 6 rad/s I = -60 A p(t) = + 4 cos(π / 3-60 ) W

5 Power due to Sinusoidal Excitation p(t) = ½ V m I m cos(θ - ϕ) + ½ V m I m cos(ωt + θ + ϕ) Example 1: Average power delivered to resistor T P R =½V m I m cos(θ - ϕ) = ½ V m I m cos(0) 1.5 = ½ V m I m = ½ R I m = ½ V m / R = 1 W v(t), i(t), p(t) V = -45 V R = Ω time (s)

6 Power due to Sinusoidal Excitation p(t) = ½ V m I m cos(θ - ϕ) + ½ V m I m cos(ωt + θ + ϕ) Example : Average power delivered to purely reactive elements T P X =½V m I m cos(θ - ϕ) = ½ V m I m cos(±90 ) = v(t), i(t), p(t) V = -45 V X = jω time (s)

7 Power due to Sinusoidal Excitation p(t) = ½ V m I m cos(θ - ϕ) + ½ V m I m cos(ωt + θ + ϕ) Example 3: Voltage across impedance (V = o V and Z = o Ω) Determine active power absorbed. I = V / Z = 5-55 = -30 = j A 5 o V Z = j41.9 Ω I -30 o P = ½ 8.7 = ½ ( ) 8.7 = 57.4 W

8 Power due to Sinusoidal Excitation p(t) = ½ V m I m cos(θ - ϕ) + ½ V m I m cos(ωt + θ + ϕ) Example 4: (Chapter 11, Problem 8.) In the circuit shown in Fig. 11.7, find the average power being (a) dissipated in the 3-Ω resistor; (b) generated by the source. Z R 1 = 3+ = j3= 4+ j3ω 0.1 j0.3 + j5 5 9 Ignore 30 on Vs, IR = 5, IR = 6+ j8 10 (a) (b) P3 Ω = 3 = W 10 (+ j5)(4+ j3) Vs = 5 0 = V 6+ j8 1 Ps, gen = cos = 0.75 W

9 Maximum Power Transfer A simple loop circuit used to illustrate the derivation of the maximum power transfer theorem as it applies to circuits operating in the sinusoidal steady state. Z th = R th + j X th Z L = R L + j X L Z L = Z th * R L = R th and X L = - X th

10 Maximum Power Transfer Example: (Chapter 11, Problem 1.) For the circuit of Fig ; (a) what value of Z L will absorb a maximum average power? (b) What is the value of this maximum power? V Z th th 10j = 10 = V 10 + j5 j10(10 + j15) = = 8 j14ω 10 + j5 (a) (b) Z L = 8 + j14 Ω I L = PL,max = 8 = 180 W 16

11 Effective Values Measure for sinusoidal voltages and currents Power outlets: 60 Hz, "voltage of 115V" Not the mean of T (T/) Not the Amplitude ( 115V) Measure of effectiveness of a source in delivering power to a resistive load Effective value of periodic current is equal to the DC value that delivers the same average power to resistor i(t) R p(t) P R and compare to I DC R

12 Effective Values Mathematical expression P T T 1 R = i Rdt = T T 0 0 i dt = I DC R = I eff R I eff = 1 T T 0 i Rdt (Square) root of the mean of the square current rms value Defined for all periodic signals

13 Effective Values - Sinusoids i(t) = I m cos(ωt +ϕ) with a period of T = π/ω I eff = 1 T T Im cos ( ωt + ϕ) dt 0 = I m π ω ω / π cos(ωt + ϕ ) dt I = eff I m real quantity independent of phase angle equal to the amplitude example: -30 o A delivers the same as I DC = 1A

14 RMS value to compute average power In general P = ½ V m I m cos(θ - ϕ) = V eff I eff cos(θ - ϕ) For resistors P = V eff I eff = V eff / R = I eff R Note: We can use amplitude or rms value Use V and V rms to designate voltages

15 RMS value to compute average power Example: (Chapter 11, Problem 30.) The series combination of a 1-kΩ resistor and a -H inductor must not dissipate more than 50 mw of power at any instant. Assuming a sinusoidal current with ω =500 rad/s, what is the largest rms current that can be tolerated? The peak instantaneous power is 50 mw. The combination of elements yields Z = j1000 Ω = o Ω. o Vm 0 Vm 45 Arbitrarily designate V = V m 0, so that I = = A and V m = 1414 I m. Z 1414 We may write p(t) = ½ V m I m cos φ + ½ V m I m cos (ωt + φ) where φ = the angle of the current (-45 o ). This function has a maximum value of ½ V m I m cos φ + ½ V m I m. Thus, 0.50 = ½ V m I m (1 + cos φ) = ½ (1414) I m (1.707) and I m = ma. In terms of rms current, the largest rms current permitted is 14.39m / = ma rms.

16 Apparent Power We had P = ½ V m I m cos(θ - ϕ) = V eff I eff cos(θ - ϕ) In case of direct current we would use voltage times current: S = V eff I eff This is not the average power Is the "apparent" power (S or AP) Measured in volt-ampere or VA (rather than W to avoid confusion) Magnitude of S is always greater or equal to magnitude of P: S P

17 Power Factor Defined as the ratio of average (real) power to apparent power PF = P / S = P / (V eff I eff ) In the sinusoidal case the power factor is PF = cos(θ - ϕ) θ - ϕ is the angle the voltage leads the current: PF angle note: Purely resistive load has PF = 1 Purely reactive load has PF = 0 PF = 0.5 means a phase angle of ±60 Resolve ambiguity PF leading or lagging for capacitive or inductive load

18 Apparent Power and Power Factor Example: (Chapter 11, Problem 30.) (a) Find the power factor at which the source in the circuit of Fig is operating. (b) Find the average power being supplied by the source. (c) What size capacitor should be placed in parallel with the source to cause its power factor to be unity? (d) Verify your answers with PSpice. (a) 10 Is = = A rms j j 16 PF = cos 6.5 = lag s (b) P = 10 s = 991.7W (c) j48 1 ZL = 4 + = 4 + (19 + j144) 3+ j j5.76 ZL = j5.76 Ω, YL = j5.76 j10π C =, C = 90.09µ F

19 Apparent Power and Power Factor Example: (see also chapter 10) (Chapter 11, Problem 30.) (d) Examine simulation output file (AC sweep at 60Hz). FREQ VM($N_0003,0) VP($N_0003,0) 6.000E E E+00 FREQ IM(V_PRINT1) IP(V_PRINT1) 6.000E E E+01 (a) and (b) are correct Next, add a µF capacitor in parallel with the source: FREQ IM(V_PRINT1) IP(V_PRINT1) 6.000E E E-05 (c) is correct ( degrees is essentially zero, for unity PF).

20 Complex Power Simplifies power calculations We had p(t) = V eff I eff cos(θ - ϕ) + ½ V m I m cos(ωt + θ + ϕ) Where the average (real) power P = V eff I eff cos(θ - ϕ) Using complex nomenclature P = V eff I eff Re{e j(θ - ϕ) } = V eff Re{e j(θ) } I eff Re{e -jϕ) } Hence P = Re{V eff I * eff} phasor voltage complex conjugate of phasor current Define S = V eff I * eff

21 Complex Power S = V eff I * eff can be written as S =V eff I eff e j(θ - ϕ) = P + jq magnitude equals apparent power S = S PF angle average (real) power reactive power Reactive power Q Imaginary part of complex power Dimensions are those of P, S, AP=S ( S ) Avoiding confusion by using volt-ampere-reactive or VAr Q = V eff I eff sin(θ - ϕ) Physical interpretation: Time rate of energy flow back&forth between source and reactive loads Reactive components charge and discharge at ω ( current flows)

22 Complex Power (Example) p(t) = V eff I eff cos(θ - ϕ) + V eff I eff cos(ωt + θ + ϕ) S P Q voltage V = 4 0 o V, impedance Z = 60 o Ω, I = -60 o A, ω = π/ 6 rad/s p(t) = + 4 cos(π / 3-60 ) W voltage V =.83 0 o V rms, I = -60 o A rms S = P + jq = V I * = W + j 3.46VAr = 4 60 o VA

23 Power Triangle Commonly used graphical representation of S = P + jq = V I* = V (V/Z)* = VV* / Z* = V / Z* = I Z Useful relationships: P = S cos(θ), Q = S sin(θ), Θ = power angle = tan -1 (Q/P) Q = P tan(θ) Quadrant means 1st - power factor is lagging, inductive load 4th - power factor is leading, capacitive load Need only two quantities to find third Im S S Q Re P (4 60 o VA = W + j 3.46VAr)

24 Power and Phasors Another interpretation of active and reactive power components Current components In phase with voltage - I eff cos(θ - ϕ) 90 out of phase (quadrature component) - I eff sin(θ - ϕ) Multiplied by V results in P and Q Im θ - ϕ I eff V eff I eff cos(θ - ϕ) I eff sin θ - ϕ Re

25 Power and Phasors Example: (Chapter 11, Problem 4.) Both sources are operating at the same frequency. Find the complex power generated by each source and the complex power absorbed by each passive circuit element. Vx 100 Vx Vx j = 0 6+ j4 j Vx + j = + j0 6+ j4 6+ j4 V = V x I1 = = A 6+ j4 1 S 1. gen = = j 443VA.

26 Power and Phasors Example: (Chapter 11, Problem 4.) Both sources are operating at the same frequency. Find the complex power generated by each source and the complex power absorbed by each passive circuit element. S 1 S 6, abs = = j0 VA 1 = ( j4)9.806 = 0 + j19.3va j I = = , 5 1 S5abs = = j0 VA j4, abs S S, gen j10, abs 1 = ( j 100) = j 39.3VA = ( j 10) = 0 j 14.3VA = VA 10

27 Power Measurement Wattmeter(hours) measures active load Varmeter(hours) measures reactive load Average PF is used to adjust consumer's bill (industry has to pay for unwanted losses) Complex power delivered to individual loads equals their sum no matter how loads are connected S = V I* = V (I 1 + I )* = V (I* 1 + I* ) = VI* 1 + VI*

28 Power Factor Correction Large industrial consumers pay penalty when PF < 0.85 Caused by inductive loads (motors) Why: Causes increased Device ratings Transmission and distribution losses Example: $0./kVAr above 6% of average (real) power demand S = P + jq = P + j0.6p = P (1+j0.6) = P ( ) Targets cos(31.8 ) = 0.85 (pay penalty when PF < 0.85) Use compensation capacitors in parallel with load

29 Power Factor Correction Value of capacitance C = P ( tanθ tanθ ) corrects the PF angle from old to new at the specified frequency and voltage old ωv rms new Derived from Q = V rms /X c = V rms ωc (Q old -Q new ) / V rms = ωc P(tan θ old -tan θ new ) / V rms = ωc

30 Complex Power Example: Source of 115 Vrms supplies two loads. Loads are connected in parallel: 7kW / 3 kvar and 4kVA at 0.85 pf lagging. Find the pf of the equivalent load as seen from the input terminal. S 1 = j 3000 S = 4k [cos(θ) + j sin(θ)] = 4k [ j sin(cos -1 (0.85))] = j 107 S = S 1 + S = j 5107 = θ = tan -1 (Q / P) = tan -1 (5107 / 10400) = 6.15 cos(θ) = cos 6.15 = Z = V / S = 115 / = 1.14Ω Z = 1.14Ω 6.15 = j 0.5 Ω (remember: S = V / Z*)

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