Electric Circuits II AC Circuit Power Analysis Dr. Firas Obeidat 1 Table of Contents 1 1 1 • Conservation of AC power • Power Triangle • Power Factor Correction 2 Dr. Firas Obeidat – Philadelphia University Conservation of AC Power 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 principle of conservation of power applies to ac circuits can be explained by considering AC parallel circuit and AC series circuit. ο AC parallel circuit two load impedances Z1 and Z2 are connected in parallel across an ac source V. KCL gives ππ«π¦π¬ = πππ«π¦π¬ +πππ«π¦π¬ The complex power supplied by the source is π = ππ«π¦π¬ ππ«π¦π¬ ∗ = ππ«π¦π¬ (πππ«π¦π¬ ∗ +πππ«π¦π¬ ∗ ) = ππ«π¦π¬ πππ«π¦π¬ ∗ +ππ«π¦π¬ πππ«π¦π¬ ∗ = ππ + ππ Where S1 and S2 denote the complex powers delivered to loads Z1 and Z2. 3 Dr. Firas Obeidat – Philadelphia University Conservation of AC Power ο AC series circuit If the loads are connected in series with the voltage source, KVL yields ππ«π¦π¬ = πππ«π¦π¬ +πππ«π¦π¬ The complex power supplied by the source is π = ππ«π¦π¬ ππ«π¦π¬ ∗ = (πππ«π¦π¬ +πππ«π¦π¬ )ππ«π¦π¬ ∗ = πππ«π¦π¬ ππ«π¦π¬ ∗ +πππ«π¦π¬ ππ«π¦π¬ ∗ = ππ + ππ Where S1 and S2 denote the complex powers delivered to loads Z1 and Z2. From the above discussion, it can be concluded that whether the loads are connected in series or in parallel (or in general), the total power supplied by the source equals the total power delivered to the load. Thus, in general, for a source connected to N loads, 4 Dr. Firas Obeidat – Philadelphia University Power Triangle The three quantities average power, apparent power, and reactive power can be related in the vector domain by π = ππ«π¦π¬ ππ«π¦π¬ ∗ =P+jQ π°π‘ππ«π π = π∠0, ππ = πππΏ = ππΏ ∠90, ππ = −πππΆ = ππΆ ∠ − 90 For an inductive load, the phasor power S is π=P+jπΈπ³ As in figure (a) For an capacitive load, the phasor power S is π=P-jπΈπͺ As in figure (b) When solving problems involving power, P values can be added to get PT, and Q values to get QT (where Q is positive for inductive elements and negative for capacitive). (a) (b) 5 Dr. Firas Obeidat – Philadelphia University Power Triangle Example: The P and Q values for a circuit are shown in the figure. a. Determine the power triangle. b. Determine the rms current supplied by the source. (a) PT = 700 + 800 + 80 + 120 = 1700 W Q T = 1300 − 600 − 100 − 1200 Q T = −600 VAR ST = PT +jQ T =1700-j600=1803∠-19.4o The net load is capacitive (b) πΌπππ = π ππππ 1803 VA = = 15 π΄πππ 120 V 6 Dr. Firas Obeidat – Philadelphia University Power Triangle Example: The circuit shows a load being fed by a voltage source through a transmission line. The impedance of the line is represented by (4+j2)Ω the impedance and a return path. Find the real power and reactive power absorbed by: (a) the source, (b) the line, and (c) the load. (a) For the source ππ¬ = ππ¬π«π¦π¬ ππ«π¦π¬ ∗ The real power is 2163.5 W and the reactive power is 910.8 VAR (leading). (b) For the line 7 Dr. Firas Obeidat – Philadelphia University Power Triangle ππ₯π’π§π = ππ₯π’π§ππ«π¦π¬ ππ«π¦π¬ ∗ The real power is 455.4 W and the reactive power is 227.76 VAR (lagging). (c) For the load ππ = πππ«π¦π¬ ππ«π¦π¬ ∗ The real power is 1708 W and the reactive power is 1139 VAR (leading). Note that SS=Sline+SL 8 Dr. Firas Obeidat – Philadelphia University Power Triangle Example: A generator (its voltage in rms) supplies power to an electric heater, an inductive element, and a capacitor as in the figure. a. Find P and Q for each load. b. Find total active and reactive power supplied by the generator. c. Draw the power triangle for the combined loads and determine total apparent power. d. Find the current supplied by the generator. (a) The components of power are as follows: (b) 9 Dr. Firas Obeidat – Philadelphia University Power Triangle (c) The apparent power is 3081VA (c) Generator current is πΌ= ππ 3081VA = = 25.7 π΄πππ π 120 V 10 Dr. Firas Obeidat – Philadelphia University Power Triangle Example: In the circuit, Z1=60∠-30oΩ and Z2=40∠45oΩ. Calculate the total: (a) apparent power, (b) real power, (c) reactive power, and (d) pf, supplied by the source and seen by the source. 11 Dr. Firas Obeidat – Philadelphia University Power Triangle 12 Dr. Firas Obeidat – Philadelphia University Power Triangle Example: For the circuit, calculate the complex power for each load then draw the power triangle. 13 Dr. Firas Obeidat – Philadelphia University Power Triangle Example: for the circuit find (the values are given in rms) a. Find the average power, apparent power, reactive power, and pf for each branch. b. Find the total number of watts, volt-amperes reactive, volt-amperes, and the power factor of the system. Sketch the power triangle. (a) βΊ Bulbs βΊ Heating elements π1 = 12 × 60W = 720W π2 = 6.4 kW π1 = 0 VAR π2 = 0 VAR π1 = π1 = 720 VA π2 = π2 = 6.4 kVA ππ1 = 1 ππ2 = 1 14 Dr. Firas Obeidat – Philadelphia University Power Triangle βΊ The motor η= ππ ππ 5 × 746 W → ππ = = = 4548.78 W ππ η 0.82 π3 = ππ = 4548.78W ππ3 = 0.72 πππππππ π3 = π3 πππ θ → π3 = π3 4548.78 W = = 6317.75 VA πππ θ 0.72 π = πππ −1 0.72 = 43.95π π3 = π3 π ππθ = 6317.75 × π ππ43.95π = 6317.75 × 0.694 = 4384.71 VAR (L) βΊ The capacitive load Irms = Vrms 208∠0 208∠0 = = = 13.87∠53.13 Z 9 − j12 15∠ − 53.13 π4 = πΌπππ 2 π = 13.872 × 9 = 1731.39 W π4 = πΌπππ 2 ππΆ = 13.872 × 12 = 2308.52 VAR (C) 15 Dr. Firas Obeidat – Philadelphia University Power Triangle (b) π4 = π4 2 + π4 2 = ππ4 = π4 1731.39 W = = 0.6 πππππππ π4 2885.65 VA 1731.392 + 23.8.522 = 2885.65 VA ππ = π1 +π2 +π3 +π4 ππ = 720 + 6400 + 4548.78 + 1731.39 ππ = 13400.17 W ππ = π1 +π2 +π3 +π4 ππ = 0 + 0 + 4384.71 − 2308.52 ππ = 2076.19 VAR (πΏ) ππ = ππ 2 + ππ 2 = 13400.172 + 2076.192 ππ = 13560.06 VA πππ = ππ 13400.17 W = = 0.988 πππππππ ππ 13560.06 VA π = πππ −1 0.988 = 8.89π 16 Dr. Firas Obeidat – Philadelphia University Power Triangle for Transformers A transformer is a magnetically coupled circuit, that is, a circuit in which the magnetic field produced by time-varying current in one circuit induces voltage in another. Where n : is the turns ratio or transformer ratio. V1: is the voltage across the primary winding. V2: is the voltage across the secondary winding. N1: is the number of turns in the primary. N2: is the number of turns in the secondary. When n=1, the transformer is an isolation transformer. If n>1, the transformer is step-up transformer, as the voltage is increased from primary to secondary (V2>V1). If n<1, the transformer is step-down transformer, since the voltage is decreased from primary to secondary (V2<V1). 17 Dr. Firas Obeidat – Philadelphia University Power Triangle for Transformers Example: A 25 kVA transformer feeds a load with 12 kW. If the pf=0.6 lagging. (1) Find the loading percentage for the maximum load that can be connected to the transformer. (2.a) If a unity pf load is added to the same transformer, find the power which can be added before the transformer becomes fully loaded. (2.b) If the added load has pf=0.866 leading (instead of unity pf load), find the apparent power which can be added so the transformer becomes fully loaded. (1) Transformer apparent power is Load1 apparent power is ππ‘ππππ . = 25πVA πππππ1 = πππππ1 12 πW = = 20 πVA ππ1 0.6 πππππ 20 πVA %πππππππ = × 100% = × 100% = 80% ππ‘ππππ . 25 πVA π1 = πππ −1 0.6 = 53.1π πππππ1 = πππππ1 π ππθ = 20000 × π ππ53.1π = 16 kVAR (2.a) If a unity pf load is added to the transformer means that the load is resistive load, so. πππππ2 = 0 Pf2 =cosπ2 =1 Dr. Firas Obeidat – Philadelphia University 18 Power Triangle for Transformers Total load complex power is πππππ = πππππ1 + πππππ2 = πππππ1 + ππππππ1 + πππππ2 + ππππππ2 πππππ = (πππππ1 + πππππ2 ) + π(πππππ1 + πππππ2 ) Before the transformer becomes fully loaded means that total load complex power equal to transformer complex power 0 ππ‘ππππ . = πππππ = (πππππ1 + πππππ2 ) + π(πππππ1 + πππππ2 ) ππ‘ππππ . = πππππ = πππππ + ππππππ ππ‘ππππ . 2 = πππππ 2 + πππππ 2 = (πππππ1 + πππππ2 )2 +(πππππ1 )2 252 = (12 + πππππ2 )2 +(16)2 625 − 256 = 144 + 24πππππ2 +πππππ2 2 πππππ2 2 + 24πππππ2 − 225 = 0 πππππ2 = 7.2 kW ππ − 31.21 kW Taking the positve value Dr. Firas Obeidat – Philadelphia University ∴ πππππ2 = 7.2 kW 19 Power Triangle for Transformers (2.b) Pf2 =cosπ2 =0.866 π2 = πππ −1 0.866 = 30π πππππ2 = πππππ2 πππ 30 = 0.866πππππ2 πππππ2 = πππππ2 π ππ30 = 0.5πππππ2 πππππ = πππππ1 + πππππ2 = 12 + 0.866πππππ2 πππππ = πππππ1 − πππππ2 = 16 − 0.5πππππ2 πππππ = (πππππ1 + πππππ2 ) + π(πππππ1 + πππππ2 ) πππππ = (12 + 0.866πππππ2 ) + π(16 − 0.5πππππ2 ) Sππππ 2 = πππππ 2 + πππππ 2 252 = (12 + 0.866πππππ2 )2 +(16 − 0.5πππππ2 )2 πππππ2 = 12.8 kVA πππππ2 = 0.866πππππ2 0.866 × 12.8 = 11 ππ πππππ2 = 0.5πππππ2 = 0.5 × 12.8 = 6.4 kVA Dr. Firas Obeidat – Philadelphia University 20 Power Factor Correction Most domestic loads (such as washing machines, air conditioners, and refrigerators) and industrial loads (such as induction motors) are inductive and operate at a low lagging power factor. Although the inductive nature of the load cannot be changed, we can increase its power factor. Power factor correction is the process of increasing the power factor without altering the voltage or current to the original load. ο Most loads are inductive loads as shown in fig(a). A load’s Power factor is improved or corrected by installing a capacitor in parallel with the load, as shown in fig(b). 21 Dr. Firas Obeidat – Philadelphia University Power Factor Correction Perspective (A) The circuit in fig(a) has a power factor of cosθ1, while the circuit in fig(b) has a power factor of cosθ2. It is evident that adding the capacitor has caused the phase angle between the supplied voltage and current to reduce from θ1 to θ2, thereby increasing the power factor. From the magnitudes of the vectors, with the same supplied voltage, the circuit in fig(a) draws larger current IL than the current I drawn by the circuit in fig(b). Perspective (B) If the original inductive load has apparent power We want to increase the power factor from θ1 to θ2 without altering the real power 22 Dr. Firas Obeidat – Philadelphia University Power Factor Correction The new reactive power is The reduction in the reactive power is caused by the shunt capacitor But The value of the required shunt capacitance C is determined as ο If the load is capacitive, an inductor should be connected across the load for power factor correction. The required shunt inductance L can be calculated from 23 Dr. Firas Obeidat – Philadelphia University Power Factor Correction Example: When connected to a 120 Vrms, 60-Hz power line, a load absorbs 4kW at a lagging power factor of 0.8. Find the value of capacitance necessary to raise the pf to 0.95. 24 Dr. Firas Obeidat – Philadelphia University Power Factor Correction Example: Suppose an industrial client is charged a penalty if the plant power factor drops below 0.85. The equivalent plant loads are as in the figure. The frequency is 60 Hz. a. Determine PT and QT. b. Determine what value of capacitance is required to bring the power factor up to 0.85. c. Determine generator current before and after correction. a. π1 = ππ = 132 πΎππ΄π 25 Dr. Firas Obeidat – Philadelphia University Power Factor Correction b. π2 =cos-1(0.85)=31.8o π2 =PT tanπ2 =146tan31.8o=90.5 KVAR ππΆ =Q1-Q2=132-90.5=41.5 KVAR C= QC 41.5×103 = πVrms2 2π×60×6002 = 306 μF c. ST =196.8 KVA I= ST 196.8 πΎππ΄ = π½ 600 V = 328 A S2 =171.8 KVA I= ST 171.8 πΎππ΄ = 600 V π½ = 286 A 26 Dr. Firas Obeidat – Philadelphia University 27