ECE430 Power Circuits and Electromechanics Dr Nam Nguyen-Quang Fall 2009 http://www4.hcmut.edu.vn/~nqnam/lecture.php Lecture Three-phase systems  Voltage in each phase differ from the other phases by 1200 In positive (also a-b-c) phase sequence, the three voltages are given by v aa'  Vm cost   vbb '  Vm cos t  120   v cc '  V m cos t  120   Three-phase connections: wye connection and delta connection In wye connection, terminals a’, b’, and c’ are joined and labeled as neutral terminal n a ia, ib, and ic are line currents, which are n  + ib + c in  also equal to the phase currents in is neutral line current ia +  b ic Lecture 2 Three-phase systems (cont.) In delta connection, terminals a’ is connected to b, and b’ to c Because vac’ = v aa’(t) + vbb’(t) + vcc’(t) = 0, as can be verified mathematically, c’ is finally connected to a ia c’ a  Line and phase quantities  Since both supply and load can be wye or + delta connected, there are possible + combinations: wye-wye, wye-delta, delta- b’ a’ +  c  ib b ic wye, and delta-delta (supply-load) • wye-wye connection, balanced condition: Van  V 0 Vbn  V   120 Vcn  V 120 Lecture Three-phase systems (cont.) where V is the rms value of phase-to-neutral voltage The line-to-line voltages are given by V ab  V an  Vbn Vca  Vcn  V an Vbc  Vbn  V cn For example, magnitude of V ab can be calculated as   Vab  2V cos 30  3V From the phasor diagram, it can be seen Vab  3V 30 V cn V ab V ca V an Vbc  3V   90 Vbn Vca  3V 150 Under balanced condition, in = (no neutral current) Lecture Vbc Three-phase systems (cont.) • wye-delta connection, balanced condition: Without losing generality, assuming line-to-line voltages are Vab  VL 00 Vbc  VL   1200 Vca  VL 1200 Phase currents I1, I2, and I3 in the three legs of V ca delta-connected load lag the respective I3 voltages by , and have the same magnitude of I It can be seen from the phasor diagram I a  3I    30   I b  3I    150   I c  3I  90    Wye connection: V L  and I L  I2 V ab I1 Vbc Ia 3V and I L  I  , delta connection: V L  V 3I  Lecture Power in balanced three-phase circuits  Balanced Y-connected load In a balanced system, the magnitudes of voltages in all phases are the same, and so also the current magnitudes Let these be V and I The power per phase is then P  V I  cos  Total power is PT  3P  3V I cos   3VL I L cos  Complex power per phase is S   V I *  V I   And total complex power is S T  3S   3V I    3VL I L  Note that  is the phase angle between the phase voltage and phase current Lecture Power in balanced three-phase circuits (cont.)  Balanced -connected load Similar to the case of balanced Y-connected load, per phase and total power can be calculated using the same formulae It can be seen that for a balanced load, the expression for total complex power is the same both for wye and delta connections, provided line-toline voltages and line currents are used Hence, calculations can be done on a three-phase or per-phase basis  Ex 2.12 and 2.13: see text book Lecture Per-phase equivalents  -Y conversion Given a delta-connected load where impedance of each phase is Z, the equivalent wye circuit has a phase impedance of ZY = Z/3 This can be proved by equating the impedance across arbitrary lines in both cases Instead of analyzing the delta circuit, the per-phase equivalent circuit can be used after doing the -Y conversion  Ex 2.14: Draw the per-phase equivalent circuit for a given circuit Replace -connected capacitor bank by a Y-connected bank with phase impedance of –j15/3 = -j5  The resulting wye-connected circuit can then be simplified to give the per-phase equivalent circuit Lecture Class examples  Ex 2.15: 10 induction motors in parallel, find three-phase kVAR rating of a capacitor bank to improve overall PF to unity? Per-phase real power is 30 x 10 / = 100 kW, at lagging PF = 0.6 Perphase kVA is therefore 100/0.6 Hence, S   S  cos 1 0.6   100  10 0.6  j 0.8 VA  100  j133.33 kVA 0.6 A capacitor bank can be connected in parallel to the load for improving overall PF The capacitor bank needs to supply all the reactive power to bring PF to unity This means per-phase Qcap = 133.33 kVAR, and threephase kVAR required will be 3(133.33) = 400 kVAR Lecture Class examples  Ex 2.16: Suppose in Ex 2.15, the new PF is to be 0.9 lagging, what is the kVAR needed? S  100  j133.33 kVA New PF is 0.9 lagging, therefore new per-phase reactive power is Qnew  P 1 PF    100 1 0.9    48.43 kVAR 2 133.33 kVAR 133.33 + 48.43 = 84.9 kVAR, and three-phase kVAR required will be 3(84.9) = 254.7 kVAR  Ex 2.17: see text book ol d The capacitor bank therefore needs to supply new 48.43 kVAR 100 kW Lecture 10 In class quiz  Problem 2.21: A three-phase load of 15 kVA with a PF of 0.8 lagging is connected in parallel with a three-phase load of 36 kW at 0.6 PF leading The line-to-line voltage is 2000 V a) Find the total complex power and power factor b) How much kVAR is needed to make the PF unity?  Special question: A three-phase wye-connected load with a unity PF is being supplied from a three-phase power system How does the load power change if the load is now rewired in delta configuration? Lecture 11