- AC Power 11/17/2015 Reading: Chapter 11 1
Outline Instantaneous power Complex power Average (real) power Reactive power Apparent power Maximum power transfer Power factor correction 2
Power in AC Circuits Consider the situation shown below: A voltage v t = V m cos ωt is applied to an RLC network. The phasor for the voltage source is V = V m 0, and The equivalent impedance of the network is Z = Z θ = R + jx. The phasor current I = V Z = V m 0 = I Z θ m θ, I m = V m Z Resistive (R) load Reactive (L/C) load 3
Review: Effective (=RMS) Value For any periodic function x(t) in general, the rms value is X eff = Xrms = 1 T 0 Tx 2 t dt periodic Same average power dc I eff = 1 T 0 T i 2 t dt V eff = 1 T 0 Tv 2 t dt 4
Example Determine the rms value of the current waveform shown below. Suppose the current is passed through a 2Ω resistor. Find the average power absorbed by the resistor. 5
RMS of a Sinusoidal The RMS value of v(t) = V m cos ωt + φ is Vrms = 1 T t0 t 0 +Tv 2 t dt = 1 T t 0 t 0 +T Vm 2 cos 2 ωt + φ dt = V m 2 Lecture 10 6
AC Power for a General Load: Time Domain cos cos v t V t i t I t m v m i i t Instantaneous power: 1 1 p t V I V I t 2 2 cos cos 2 m m v i m m v i v t Average (or real) power (unit: watts): P 1 T T 0 p t dt P = V mi m 2 cos θ v θ i = V rms I rms cos θ v θ i 7
AC Power for a General Load: Frequency Domain i t v t = V m cos ωt + θ v i t = I m cos ωt + θ i V = V m θ v I = I m θ i We observe that: v t 1 2 VI = 1 2 V mi m θ v θ i = 1 2 V mi m cos θ v θ i + j 1 2 V mi m sin θ v θ i Average (or real) power P = 1 2 Re[VI ] Unit: Watts Reactive power Q = 1 2 Im[VI ] Unit: Volt Amperes Reactive (VARs) 8
Example Find the average power and reactive power absorbed by an impedance Z = 30 j70ω, when a voltage V = 120 0 is applied across it. i t I = V Z = 120 0 30 j70 = 120 0 76.16 66.8 = 1.576 66.8 A v t P = 1 2 V mi m cos θ v θ i = 1 2 120 1.576 cos 0 66.8 = 37.24W Q = 1 2 V mi m sin θ v θ i = 86.91VAR 9
Appliance Rating (Average Power) 10
Average Power -> Apparent Power Average power P = 1 2 V mi m cos θ v θ i = V rms I rms cos θ v θ i Apparent power it seems apparent that the power should be the voltage-current product, by analogy with dc resistive circuits. Unit: volt-amps (VAs) Apparent power = V rms I rms 11
Average Power -> Power Factor The power factor θ v θ i pf = is called power factor angle. >0 means a lagging pf (current lags voltage), an inductive load <0 means a leading pf (current leads voltage), an capacitive load pf ranges from 0 to 1. P Apparent power = cos θ v θ i The reactive factor rf = sin θ v θ i 12
Example A series-connected load draws a current i t = 4cos 100πt + 10 A when the applied voltage is v t = 120cos 100πt 20 Find the apparent power and the power factor of the load. Determine the values that form the series-connected load. 13
Inductive vs. Capacitive Loads Inductive load Capacitive load 14
Instantaneous Power: Resistive Load Z = R, θ = 0 v t = V m cos ωt i t = I m cos ωt p t = v t i t = V m I m cos 2 ωt 15
Average/Reactive Power: Resistive Load T P = 1 T t=0 V m I m cos 2 ωt dt T 1 1 + cos 2ωt = V m I m T 2 t=0 dt Pure resistive load V rms I rms = V mi m 2 pf = P V rms I rms = cos 0 = 1 P = V mi m 2 Q = 0 16
Instantaneous Power: Inductive Load Z = ωl 90, θ = 90 v t = V m cos ωt i t = I m cos ωt 90 = I m sin ωt p t = v t i t = V mi m 2 sin 2ωt 17
Average/Reactive Power: Inductive Load P = 1 T t=0 T Vm I m 2 sin 2ωt dt Pure inductive load = 0 P = 0 Half the time, the energy is delivered to the inductance; The other half time, energy is returned to the source. V Q = 1 2 V mi m sin 90 rms I rms = V mi m 2 P pf = = cos 90 = 0 V rms I rms 18
Instantaneous Power: Capacitive Load Z = 1 ωc 90, θ = 90 v t = V m cos ωt i t = I m cos ωt + 90 = I m sin ωt p t = v t i t = V mi m 2 sin 2ωt 19
Average/Reactive Power: Capacitive Load T P = 1 T t=0 V mi m 2 sin 2ωt dt = 0 P = 0 Pure capacitive load Half the time, the energy is delivered to the capacitance; The other half time, energy is returned to the source. V Q = 1 2 V mi m sin 90 rms I rms = V mi m 2 P pf = = cos 90 = 0 V rms I rms 20
Reactive Power Definition: The peak instantaneous power associated with the energy storage elements contained in a general load. Q = 1 2 Im VI = 1 2 V mi m sin θ v θ i = V rms I rms sin θ v θ i A measure of the energy exchange between the source and the reactive load. Even though no average power is consumed by a pure energystorage element, reactive power is still of concern to powersystem engineers Transmission lines/transformers/fuses et al. must be capable of withstanding the current associated with reactive power. energy-storage elements that draw large currents requiring heavyduty wiring, even though little average power is consumed. 21
Complex Power Complex power is important in power analysis because it contains all the information pertaining to the power absorbed by a given load. Consider the ac load Z = R + jx as shown below V = V m θ v I = I m θ i Complex power, Unit: volt-amps (VAs) S = 1 2 VI = V rms I rms = V rms I rms θ v θ i = P + jq 22
Example The voltage across a load is v t = 60cos ωt 10 V, and the current through the load is i t = 1.5cos ωt + 50. Find The complex and apparent powers. The real and reactive powers. The power factor and the load impedance. 23
Another Way to Calculate Complex Power S = V rms I rms S = V rms I rms = V rms V rms Z = V rms 2 Z = Z I rms I rms = I rms 2 Z = I rms 2 R + jx V rms = I rms Z = P + jq = I rms 2 R + j I rms 2 X = I2 rms R + ji2 rms X = P + jq 24
Example I L = 250 0 40+j30 = 4 j3 = 5 36.87 (rms) V L = I L 39 + j26 = 234 j13 = 234.36 3.18 Find V L and I L. Find the average and reactive power Delivered to the load Delivered to the line Supplied by the source Load: V L I L = 975 + j650 VA Line: P = 5 2 1 = 25 W Q = 5 2 4 = 100 VAR Source: 250 0 I L = 1000 + j750 VA 25
Power Triangle 26
Example Find the power, reactive power and power factor for the source. Phasor current I. 27
Summary of AC Power 28
Maximize Power Transfer in DC circuit Thé venin equivalent circuit V Th + R Th + v L Power absorbed by load resistor: i L R L p i 2 L R L R VTh R Th L 2 R L To find the value of R L for which p is maximum, set to 0: dp drl V Th 2 RTh RL RL 2 RTh RL 4 R R 2 Th 0 R R L Th L dp dr L A resistive load receives maximum power from a circuit if the load resistance equals the Thé venin resistance of the circuit. 29
Maximum Average Power Transfer in AC Circuit Average power delivered to the load: 30
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Combining these one finds that X L =-X Th and R L =R Th satisfy the requirements: The load impedance must be equal to the complex conjugate of the Thevenin impedance. The maximum average power will be: 32
Example Determine the impedance Z L that maximizes the average power transferred to the load Z L. 33
What if Z L is Restricted? Case 1: R L and X L are restricted to a limited range of values. X L X Th R L R 2 Th + X L + X 2 Th Case 2: The phase angle of Z L is restricted, but its magnitude can be varied. Z L = Z Th 34
Example Find Z L that maximizes average power delivered to the load. Find the power number. Assume that R L can be varied between 0 and 4000Ω, and X L can be varied between 0 and -2000Ω. Find the power. X L = 2000Ω R L R L = 3605.55Ω R 2 Th + X L + X 2 Th = 3605.55Ω 35
Example A load impedance having a constant phase angle of 36.87 is connected across the load terminals a and b. the magnitude of Z L is varied until the average power delivered is the most possible under the given restriction. Specify Z L in rectangular form. Calculate the average power delivered to Z L. 36
Power Factor Correction Large currents can flow in energy-storage devices without average power being delivered. Most domestic and industrial loads, such as washing machines, air conditioners, and induction motors are inductive. They have a low, lagging power factor. Energy rates charged to industry (may not for residential customers) depends on the power factor: higher charges for lower power factors. The load cannot be changed, but the power factor can be increased without altering the voltage or current to the original load. This is referred to as power factor correction. 37
Adding a Capacitor To mitigate the inductive aspect of the load, a capacitor is added in parallel with the load. Looking at the phasor diagram, showing before and after adding the capacitor, the power factor has improved. 38
Finding the Capacitance With the same supplied voltage, the current draw is less by adding the capacitor. Overall, the power factor correction benefits the power company and the consumer. By choosing a suitable size for the capacitor, the power factor can be made to be unity. The capacitor needed in order to shift the power factor angle from θ 1 to θ 2 is: C Q V P tan tan C 1 2 2 2 rms Vrms If a load is capacitive in nature, the same treatment with an inductor can be used. 39
Wattmeter (Average) Power consumption in a AC system can be measured using Wattmeter. The meter consists of two coils; the current and voltage coils. The current coil is designed with low impedance and is connected in series with the load. The voltage coil is designed with very large impedance and is connected in parallel with the load. The induced magnetic field from both causes a deflection in the current coil. The physical inertia of the moving coil results in the output being equal to the average power. 40
Further Reading Conservation of AC Power Ch. 11.7 Vampire (standby) Power Refer to [Nilsson, Ch. 10] Wikipedia for AC Power https://en.wikipedia.org/wiki/ac_power 41