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1 Chapter 3 AC Power Analysis. 2 AC Power Analysis Chapter 3 3.1Instantaneous and Average Power 3.2Maximum Average Power Transfer 3.3Effective or RMS.

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Presentation on theme: "1 Chapter 3 AC Power Analysis. 2 AC Power Analysis Chapter 3 3.1Instantaneous and Average Power 3.2Maximum Average Power Transfer 3.3Effective or RMS."— Presentation transcript:

1 1 Chapter 3 AC Power Analysis

2 2 AC Power Analysis Chapter 3 3.1Instantaneous and Average Power 3.2Maximum Average Power Transfer 3.3Effective or RMS Value 3.4Apparent Power and Power Factor 3.5Complex Power 3.6Conservation of AC Power 3.7Power Factor Correction 3.8Power Measurement

3 3 AC Power Analysis Chapter 3 Mostly all electric energy is supplied in the form of sinusoidal voltages and currents After discussion of sinusoidal circuits, now is to consider sinusoidal steady state power calculations. We begin with defining and deriving instantaneous power and average power

4 4 3.1 Instantaneous and Average Power The power at any instant of time is p = vi where V m and I m are the amplitudes and θ v and θ i are the phase angles of the voltage and current, respectively

5 5 3.1 Instantaneous and Average Power Use a zero time corresponding to the instant the current is passing through a positive maximum This reference system requires a shift of both the voltage and current by θ i

6 6 3.1 Instantaneous and Average Power p(t) > 0: power is absorbed by the circuit; p(t) < 0: power is absorbed by the source.

7 7 3.1 Instantaneous and Average Power This shows the instantaneous power has two parts The first part is constant or time independent. Its value depends on the phase difference between the voltage and the current The second part is a sinusoidal function whose frequency is 2, which is twice the frequency of the voltage or current Therefore, the instantaneous power goes through two complete cycles for every cycle of either the voltage or the current Sinusoidal power at 2  t Constant power

8 8 3.1 Instantaneous and Average Power This shows the instantaneous power has two parts Also note that the instantaneous power may be negative for a portion of each cycle, even if the network between the terminal is passive. In a completely passive network, negative power implies that energy stored in the inductors or capacitors is now being extracted Sinusoidal power at 2  t Constant power

9 9 3.1 Instantaneous and Average Power Use the to expand the second right term on the RHS eq. which gives where P = Q =

10 10 3.1 Instantaneous and Average Power P is called the average power, and Q is called the reactive power Average power is sometimes called real power, because it describes the power in a circuit that is transformed from electric to non electric energy The average power associated with sinusoidal signals is the average of the instantaneous power over one period, or, in eq. form, where T is the period of the sinusoidal function The limits imply that we can initiate the integration process at any convenient time t o but must terminate the integration exactly one period later

11 11 3.1 Instantaneous and Average Power We could find the average power by substituting eq. directly into and then performing the integration. But note that the average value of p is given by the first term on the RHS of eq. because the integration of both cos2t and sin2t produces a zero over a cycle

12 12 3.1 Instantaneous and Average Power The average power, P, is the average of the instantaneous power over one period. 1.P is not time dependent. 2.When θ v = θ i, it is a purely resistive load case. 3.When θ v – θ i = ±90o, it is a purely reactive load case. 4.P = 0 means that the circuit absorbs no average power.

13 13 3.1 Instantaneous and Average Power Example 1 Calculate the instantaneous power and average power absorbed by a passive linear network if: Answer:

14 14 Example 2 A current flows through an impedance. Find the average power delivered to the impedance. 3.1 Instantaneous and Average Power Answer: 1854.36W

15 15 3.2 Maximum Average Power Transfer Maximum power is transferred to the load when the load resistance equals the Thevenin resistance as seen from the load (R L = R Th ) We are now extend that result to ac circuit. Consider the circuit below, where an ac circuit is connected to a load Z L and is represented by its Thevenin equivalent. In rectangular form, the Thevenin impedance Z Th and the load impedance Z L are

16 16 3.2 Maximum Average Power Transfer The current through the load is The average power delivered to the load is Our objective is to adjust the load parameters R L and X L so that P is maximum To do this set and equal to zero

17 17 3.2 Maximum Average Power Transfer We obtain Setting to zero gives X L = -X Th and setting to zero results in Our objective is to adjust the load parameters R L and X L so that P is maximum To do this set and equal to zero

18 18 3.2 Maximum Average Power Transfer Combining X L = -X Th and leads to the conclusion that for maximum average power transfer, X L = –X TH and R L = R TH For maximum average power transfer, the load impedance Z L must be equal to the complex conjugate of the Thevenin impedance Z Th If the load is purely real, then

19 19 3.2 Maximum Average Power Transfer Example 3 For the circuit shown below, find the load impedance Z L that absorbs the maximum average power. Calculate that maximum average power. Answer: 3.415 – j0.7317, 1.429W

20 20 3.3 Effective or RMS Value The effective of a periodic current is the dc current that delivers the same average power to a resistor as the periodic current. The idea of effective value arises from the need to measure the effectiveness of a voltage or current source in delivering power to a resistive load The circuit in (a) is ac while that of (b) is dc. To find I eff that will transfer the same power to resistor, R as the sinusoid i. The average power absorbed by the resistor in the ac circuit is

21 21 3.3 Effective or RMS Value While the power absorbed by the resistor in the dc circuit is Equating the eq. and solving for I eff, we obtain The effective value of the voltage is found in the same way as current, that is,

22 22 3.3 Effective or RMS Value This indicates that the effective value is the square root of the mean (or average) of the square of the periodic signal. Thus the effective value is often known as the root-mean- square value, or rms value For any periodic function x(t) in general, the rms value is given by The effective value of a periodic signal is its root mean square (rms) value

23 23 3.3 Effective or RMS Value For the sinusoid i(t)=I m cost, the effective or rms value is Similarly, the average power absorbed by a resistor, R The average power can be written in terms of the rms values: Note: If you express amplitude of a phasor source(s) in rms, then all the answer as a result of this phasor source(s) must also be in rms value.

24 24 3.3 Effective or RMS Value Power engineers conventionally say a voltage and current in terms of its RMS value So if power engineers say the voltage is 240 V, it is a RMS value of 240 V and it means that it peak value is 339.4 V This is only valid for sinusoidal signal For non-sinusoidal waveform, the expression must be modified to include the non-sinusoidal waveform such as by using Fourier series to represent the non-sinusoids.

25 25 3.4 Apparent Power and Power Factor Apparent Power, S, is the product of the r.m.s. values of voltage and current. It is measured in volt-amperes or VA to distinguish it from the average or real power which is measured in watts. Power factor is the cosine of the phase difference between the voltage and current. It is also the cosine of the angle of the load impedance. Apparent Power, SPower Factor, pf

26 26 3.4 Apparent Power and Power Factor Purely resistive load (R) θ v – θ i = 0, Pf = 1 P/S = 1, all power are consumed Power cannot be extracted(all electric energy is dissipated in the form of thermal energy) Purely reactive load (L or C) θ v – θ i = ± 90 o, pf = 0 P (average power)= 0, no real power consumption Resistive and reactive load (R and L/C) θ v – θ i > 0 θ v – θ i < 0 Lagging - inductive load(demand magnetizing vars) Leading - capacitive load(deliver magnetizing vars)

27 27 3.5 Complex Power Complex power S is the product of the voltage and the complex conjugate of the current: The complex sum of real and reactive power

28 28 3.5 Complex Power P: is the average power in watts delivered to a load and it is the only useful power. Q: is the reactive power exchange between the source and the reactive part of the load. It is measured in VAR. Q = 0 for resistive loads (unity pf). Q < 0 for capacitive loads (leading pf). Q > 0 for inductive loads (lagging pf). S = P + j Q

29 29 3.5 Complex Power Apparent Power, S = |S| = Vrms*Irms = Real power, P = Re(S) = S cos(θ v – θ i ) Reactive Power, Q = Im(S) = S sin(θ v – θ i ) Power factor, pf = P/S = cos(θ v – θ i ) S = P + j Q Complex power contains all the information pertaining to the power absorbed by a load

30 30 3.5 Complex Power S = P + j Q Impedance Triangle Power Factor Power Triangle

31 There are 3 types of electric power: – Real power (watts, kilowatts, or kW) – Reactive power (kilovolt-amps reactive or kVAR) – Apparent power (kilovolt-amps or kVA) Real/Active power is said to be the flow of real energy which produces a measureable output such as heat, light or mechanical energy – Household meters measure real power – the power we use at home – Real power does the work – it lights the lights and runs the motors – Example: A light bulb is 100 watts

32 Reactive power provides the magnetic field to make motors operate The magnetic field is needed for real power (MW) to flow through transformers The magnetic field is the invisible force of magnetism – Example: The head on a “teh-tarik” Total Power MVA Real Power MW Reactive Power MVAr

33 Apparent power is the actual/total power the generator must supply to the system – It includes both real power (kilowatts) and reactive power (kVARS) The bulk power system needs both real power and reactive power to be reliable Power Factor (p.f.) is the ratio of real power (kW) to apparent power (kVA) in a circuit

34 34 3.6 Conservation of AC Power The principle applies to both ac and dc circuits The complex real, and reactive powers of the sources equal the respective sums of the complex, real, and reactive powers of the individual loads. The complex power supplied by the source equals the complex powers delivered to loads Z 1 and Z 2

35 35 3.6 Conservation of AC Power The same results can be obtained for a series connection. Whether the loads are connected in series/parallel/general, the total power supplied by the source equals the total power delivered to the load. This means that the total complex power in a network is the sum of the complex powers of the individual elements, BUT not true of apparent power..

36 36 3.7 Power Factor Correction Most all practical loads are inductive (motors, fluorescent lights, etc.) and operate at a lagging power factor. Power factor correction is necessary for economic reason.

37 37 3.7 Power Factor Correction The design of transmission system is very sensitive to the magnitude of the current as determined by the applied load. Increased currents result in increased power losses due to the resistance of the lines Heavier currents also require larger conductors, increasing the amount of copper and obviously the require increased generating capacities Since the voltage is fixed, the apparent power is directly related to the current level

38 38 3.7 Power Factor Correction The smaller the apparent power, the smaller is the current drawn from the supply Minimum current is therefore drawn from a supply when S=P and Q=0. PF angle approaches zero degrees and PF approaches 1, revealing that the network is appearing more and more resistive at the input terminals.

39 39 3.7 Power Factor Correction Power factor correction is the process of increasing the power factor without altering the voltage or current to the original load. Because most loads are inductive, power factor can be corrected by installing a capacitor in parallel with the load. Inductive load with improved pf

40 40 3.7 Power Factor Correction Q 1 = S 1 sin θ 1 = P tan θ 1 Q 2 = P tan θ 2 Q c = Q 1 – Q 2 = P (tan θ 1 - tan θ 2 ) = ωCV 2rms P = S 1 cos θ 1 Phasor diagram of currents relative to V

41 41 3.7 Power Factor Correction When connected to a 120-Vrms, 60-Hz power line, a load absorbs 4 kW at a lagging power factor of 0.8. Find the value of capacitance necessary to raise the pf to 0.95.

42 42 3.8 Power Measurement The wattmeter is the instrument for measuring the average power. Consists of two coils: the current coil and the voltage coil A current coil(very low impedance) is connected in series with the load and responds to the load current The voltage coil(very high impedance) is connected in parallel with the load and respond s to the load voltage The basic structure Equivalent Circuit with load

43 43 3.8 Power Measurement When the two coils are energized, the mechanical inertia of the moving system produces a deflection angle that is proportional to the average value of the product v(t)i(t) The basic structure Equivalent Circuit with load Ifand


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