EE1302 Introduction to Electrical Engineering EE1302 Introduction to Electrical Engineering Unit 8: Three Phase Circuits Dr.R.P.S. Chandrasena 1 EE1302 Introduction to Electrical Engineering Generation of three phase voltages ππ = RMS phase voltage 2 EE1302 Introduction to Electrical Engineering 3 Balanced three phase voltages In a balanced three phase supply, the phase voltages are equal in magnitude and are out of phase by 1200 with each other. π£π΄π π£π΅π π£πΆπ EE1302 Introduction to Electrical Engineering 4 abc or positive sequence The phase sequence is the time order in which the voltages pass through their respective maximum values. Hence, the phase sequence of the three phase voltages shown below is abc or positive sequence. π£π΄π π£π΅π π£πΆπ EE1302 Introduction to Electrical Engineering acb or negative sequence The phase sequence of the three phase voltages shown below is acb or negative sequence. π£π΄π π£πΆπ π£π΅π π£π΄π = 2ππ πππ ππ‘ π£πΆπ = 2ππ πππ ππ‘ − 1200 π£π΅π = 2ππ πππ ππ‘ + 1200 5 EE1302 Introduction to Electrical Engineering 6 Phase sequence: Example Determine the phase sequence of the set of voltages given as π£π΄π = 200 πππ ππ‘ + 100 π£πΆπ = 200πππ ππ‘ − 2300 π£π΅π = 200πππ ππ‘ − 1100 . EE1302 Introduction to Electrical Engineering Star (Y) connected three phase source Phase voltages: Line voltages: πππ is 7 3 times πππ and leads by 300. EE1302 Introduction to Electrical Engineering 8 Delta (β) connected three phase source Line voltages = Phase voltages EE1302 Introduction to Electrical Engineering Three Phase Loads Star connected load Delta connected load A balanced load is one in which the phase impedances are equal in magnitude and in phase. Star – Delta transformation can be applied to transform a delta-connected load to an equivalent star-connected load and vise versa. For balanced three phase load 9 EE1302 Introduction to Electrical Engineering Star - Connected Three Phase Loads For a star-connected load, phase currents and line currents are the same, The line voltage is 3 times the phase voltage with 300 phase shift. 10 EE1302 Introduction to Electrical Engineering Delta - Connected Three Phase Loads For a delta-connected load, phase voltages and the line to line (line) voltages are the same. Line currents are 3 times the phase currents with -300 phase shift. 11 Phase currents πΌππ = πΌπ ∠π Line currents πΌπ = 3πΌπ ∠ π − 300 πΌππ = πΌπ ∠ π − 1200 πΌπ = 3πΌπ ∠ π − 1500 πΌππ = πΌπ ∠ π − 2400 πΌπ = 3πΌπ ∠ π + 900 EE1302 Introduction to Electrical Engineering 12 Delta - Connected Three Phase Loads: Example EE1302 Introduction to Electrical Engineering Delta - Connected Three Phase Loads Taking IAB as the reference. For a delta-connected load, phase voltages and the line to line (line) voltages are the same. Line currents are 3 times the phase currents with -300 phase shift. 13 Phase currents πΌππ = πΌπ ∠π Line currents πΌπ = 3πΌπ ∠ π − 300 πΌππ = πΌπ ∠ π − 1200 πΌπ = 3πΌπ ∠ π − 1500 πΌππ = πΌπ ∠ π − 2400 πΌπ = 3πΌπ ∠ π + 900 EE1302 Introduction to Electrical Engineering 14 Source – Loads connection Since both three phase source and three phase load can be either star or delta connected, there are four possible source to load connections. • Y - Y connection • Y - ο connection • ο - ο connection and • ο - Y connection. EE1302 Introduction to Electrical Engineering Balanced Y – Y connection Circuit for phase A (Per phase equivalent circuit) When solving balanced three phase circuits, it is convenient to analyze the per phase circuit and then derive the parameters in other phases using the results obtained through per phase analysis. For a balanced Y – Y connected system, the per phase equivalent circuit can be obtained simply by considering one phase. 15 EE1302 Introduction to Electrical Engineering 16 Balanced Y – Y connection: Example For the given figure, π¬π΄π = 120V∠00 . a) Solve for line currents. b) Solve for the phase voltages at the load. c) Solve for the line voltages at the load. EE1302 Introduction to Electrical Engineering 17 Balanced Y – Y connection: Example For the given figure, π¬π΄π = 120V∠00 . a) Solve for line currents. b) Solve for the phase voltages at the load. c) Solve for the line voltages at the load. EE1302 Introduction to Electrical Engineering 18 Balanced Y – ο or ο - Y connection Transform the ο connected source/load to equivalent Y connected source/load to transform the circuit to Y – Y connected system. Then apply the per phase analysis, considering one phase and used the voltage and current relationships for ο connected system to obtain the required parameters of the original circuit. EE1302 Introduction to Electrical Engineering 19 Balanced Y – ο or ο - Y connection A balanced abc – sequence Y-connected source with πππ = 100∠100 V is connected to a ο-connected balanced load 8 + π4 β¦ per phase. Calculate the phase and line currents. EE1302 Introduction to Electrical Engineering 20 Balanced Y – ο or ο - Y connection A balanced ο-connected load having an impedance 20 – j15 β¦ is connected to a ο-connected, positivesequence generator having πππ = 330∠00 V. Calculate the phase currents of the load and the line currents. EE1302 Introduction to Electrical Engineering 21 Balanced Y – ο or ο - Y connection A balanced Y-connected load with a phase impedance of 40 + j25 β¦ is supplied by a balanced, positive sequence ο-connected source with a line voltage of 210 V. Calculate the phase currents. Use πππ as a reference. EE1302 Introduction to Electrical Engineering 22 Balanced Y – ο or ο - Y connection The line voltage of the generator is 207.8 V. I. Find the line voltage πππ at the load. II. Find the ο currents. EE1302 Introduction to Electrical Engineering 23 Power in a Balanced Three Phase System For a Y-connected three phase load with a per phase impedance π∠π , ππ and πΌπ are rms voltage and rms current. Thus the total instantaneous power, EE1302 Introduction to Electrical Engineering 24 Power in a Balanced Three Phase System Thus the total instantaneous power, Using the trigonometric identity EE1302 Introduction to Electrical Engineering Power in a Balanced Three Phase System Thus the total instantaneous power, π = 3ππ πΌπ πππ π Thus the total instantaneous power in a balanced three-phase system is constant. Hence, average total active and reactive power ππ = 3ππ πΌπ πππ π ππ = 3ππ πΌπ π πππ where ο± is the power factor angle. The average per phase active and reactive power become ππ = ππ πΌπ πππ π ππ = ππ πΌπ π πππ The total real and reactive power, These expressions are valid for both Y and ο connected loads. 25 EE1302 Introduction to Electrical Engineering 26 Power in a Balanced Three Phase System The total complex power πΊπ» = 3ππ πΌπ πππ π + π3ππ πΌπ π πππ πΊ π = ππ + πππ πΊ π = 3π½π π°π ∗ πΊ π = 3 ππΏ πΌπΏ ∠π ππ = 3 ππΏ πΌπΏ 27 EE1302 Introduction to Electrical Engineering Power in a Balanced Three Phase System 28 EE1302 Introduction to Electrical Engineering Power in a Balanced Three Phase System 29 EE1302 Introduction to Electrical Engineering Power in a Balanced Three Phase System EE1302 Introduction to Electrical Engineering 30 Power Factor Correction in Three Phase System In three phase systems, three capacitors are needed and hence Qcap represent the reactive power supplied by all three capacitors. Also note that the voltage applied across the capacitor vary based on the type of connection of the capacitors. EE1302 Introduction to Electrical Engineering 31 Power Factor Correction in Three Phase System Two balanced loads are connected to a 240 kV rms, 60 Hz line, as shown in the figure below. Load 1 draws 30 kW at a power factor of 0.6 lagging, while load 2 draws 45 kVAR at a power factor of 0.8 lagging. Assuming the abc sequence, determine (a) The complex, real, and reactive power absorbed by the combined load (b) The line currents (c) The kVAR rating of the three capacitors β connected in parallel with the load that will raise the power factor to 0.9 lagging and (d) The capacitance of each capacitor.
