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ELECTRICAL TECHNOLOGY ET 201 Explain and calculate average power, apparent power, reactive power Calculate the total P, Q and S and sketch the power triangle. 1 Power in AC Circuit • Power is distributed into the resistance and reactance in AC circuit. • The power delivered to a load at any instant is defined by the product of the applied voltage and the resulting current: p vi • Since v and i are sinusoidal quantities v Vm sin t i I m sin t 2 Power in AC Circuit v Vm sin t i I m sin t The chosen v and i include all possibilities because • If the load is purely resistive: = 0° • If the load is purely inductive: = 90° • If the load is purely capacitive: = - 90° • For a network that is primarily inductive, is positive (v leads i or i lags v) • For a network that is primarily capacitive, is negative (i leads v) 3 Power in AC Circuit • Positive power means that power has been distributed from supply into circuit. • Negative power means that the power has been distributed from circuit into supply. • There are three type of power in AC circuit: i) Average power, P ii) Apparent Power, S iii) Reactive Power, Q 4 Power in AC Circuit Average Power, P • The average power (real power) is the power delivered to the load and dissipated by the load. P Vrms I rms cos Where, OR [watt, W] θ : phase angle between Vrms and Irms P I rms R 2 ; Vrms P R 2 [watt, W] 5 14.5 Power Factor (Review) P Vrms I rms cos Average Power, P • For a purely resistive load; T 0 Hence; FP cos 1 P Vrms I rms • For purely inductive or purely capacitive load; T 90 FP cos 0 Hence; P0 6 Power in AC Circuit Apparent Power, S • From analysis of DC networks (and resistive elements above), it would seem apparent that the power delivered to the load is simply determined by P = VI, with no concern for the components of the load. • However, in Chapter 14 (Lecture 10) the power factor (cos θ) of the load has a pronounced effect on the power dissipated, less pronounced for more reactive loads. • Therefore P = VI is called apparent power, S. S Vrms I rms [volt-amperes, VA] 7 Power in AC Circuit Apparent Power, S OR S I rms Z 2 ; S Vrms Z 2 • The average power to the load is; However; Therefore [volt-amperes, VA] P VI cos S VI P S cos • The power factor of a system Fp is P Fp cos S 8 Power in AC Circuit Reactive Power, Q • In general, the reactive power associated with any circuit is defined to be Q Vrms I rms sin [volt-ampere reactive, VAR] Where, θ : phase angle between Vrms and Irms OR Q I rms X • For the resistor, 2 [volt-ampere reactive, VAR] Q 0 VAR 9 Power in AC Circuit Q Vrms I rms sin Reactive Power, Q • For a purely inductive circuit, QL Vrms I rms OR QL I rms X L 2 vL leads iL by 90° ; Vrms QL XL 2 Since the apparent power, S = VI , and the average power for inductor, P = 0 P 0 Fp cos 0 S VI 10 Power in AC Circuit Q Vrms I rms sin Reactive Power, Q • For a purely capacitive circuit, QC Vrms I rms OR QC I rms X C 2 iC leads vC by 90° ; Vrms QC XC 2 Since the apparent power, S = VI , and the average power for capacitor, P = 0 P 0 Fp cos 0 S VI 11 19.7 Power Triangle • The three quantities average power P, apparent power S and reactive power Q can be related in the vector domain by with P P 0 Q L QL 90 QC QC 90 12 19.7 Power Triangle S P 2 Q 2 • For an inductive load, the phasor power S, as it is often called, is defined by S = P + jQL • For a capacitive load, the phasor power S is defined by S = P - jQC 13 19.7 Power Triangle • If a network has both capacitive and inductive elements, the reactive component of the power triangle will be determined by the difference between the reactive power delivered to each. • If QL QC, the resultant power triangle will be similar to the inductive load power diagram. • If QC QL, the resultant power triangle will be similar to the capacitive load power diagram. • That the total reactive power is the difference between the reactive powers of the inductive and capacitive elements. 14 19.7 The Total P, Q, and S The total number of watts PT, volt-amperes reactive QT, and volt-amperes ST, and the power factor Fp of any system can be found using the following procedure: 1. Find the real (average) power and reactive power for each branch of the circuit. 2. The total real power of the system (PT) is the sum of the average power delivered to each branch 3. The total reactive power (QT ) is the difference between the reactive power of the inductive loads and that of the capacitive loads. 4. The total apparent power is ST2 5. The total power factor is PT / ST. = PT2 + QT2. 2 S P2 Q 15 19.7 The Total P, Q, and S There are two important points in the previous slide: • First, the total apparent power, ST must be determined from the total average PT and total reactive powers QT and cannot be determined from the apparent powers of each branch. • Second, and more important, it is not necessary to consider the series-parallel arrangement of branches. In other words, the total real PT , total reactive QT , or total apparent power ST is independent of whether the loads are in series, parallel, or series-parallel. 16 19.7 The Total P, Q, and S Example 19.3 1. Find the total number of watts PT, volt-amperes reactive QT, and volt-amperes ST and draw the power triangle. 2. Find the power factor Fp 3. Find the current in phasor form. 17 19.7 The Total P, Q, and S Example 19.3 – Solution 1. Load 1 S P2 Q2 P1 100 W; S1 100 j 0 Q1 0 VAR 100 VA 18 19.7 The Total P, Q, and S Example 19.3 – solution (cont’d) Load 2 S P2 Q2 P2 200 W; S2 200 j 700 Q2 700 VAR 728 VA 19 19.7 The Total P, Q, and S Example 19.3 – solution (cont’d) Load 3 P3 300 W; Q3 1500 VAR S P2 Q2 S3 300 j1500 1529.71 VA 20 19.7 The Total P, Q, and S Example 19.3 – solution (cont’d) Total PT P1 P2 P3 100 200 300 PT 600 W 21 19.7 The Total P, Q, and S Example 19.3 – solution (cont’d) Total QT Q1 Q2 Q3 0 700 1500 800 VAR QT 800 VAR(C) 22 19.7 The Total P, Q, and S Example 19.3 – solution (cont’d) The power triangle; Total PT 600 W QT 800 VAR(C) ST PT jQT 600 j800 ST 1000 53.13 VA Note: ST is NOT equal to sum of each branch!! 23 19.7 The Total P, Q, and S Example 19.3 – solution (cont’d) 2. The power factor FP PT Fp ST 600 W 1000 VA Fp 0.6 leading (C ) 24 19.7 The Total P, Q, and S Example 19.3 – solution (cont’d) Fp 0.6 leading (C ) 3. The current ST 1000 VA VI 1000 VA 1000 VA I 100 V I 10 A Since Fp is leading I leads E, predominantly capacitive circuit. I 10 A 53.13 25 26