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EE212 Passive AC Circuits Lecture Notes 5 Three Phase Systems 1 Balanced 3-Phase Systems Example: A balanced 3-phase, line-to-line voltage of 220 V is applied to a balanced 3-phase load. The 3-phase load consists of 3 identical loads (each with an impedance of 6+j8 ) (a) connected in Y (b) connected in D Calculate the line current and the total power consumed for both cases. Draw the complete phasor diagrams for each case. 2 Balanced 3-Phase Systems A 3-phase load in D-connection consumes 3 times more power than in Y- connection. A 3-phase load in D-connection draws 3 times more current than in Y- connection. Y-D motor starting method to reduce starting current 3 Delta-Wye (D-Y) Transformation A A Zab B B Za Zb Zca Zbc Zc C C ZY = Za = (Zab. Zca) / (Zab + Zbc + Zca) Zb = (Zab. Zbc) / (Zab + Zbc + Zca) Zc = (Zca. Zbc) / (Zab + Zbc + Zca) (product of adjacent ZD) sum of all 3 ZD For balanced load, Zab = Zbc = Zca = ZD Therefore, ZY = ZD / 3 4 Wye-Delta (Y- D) Transformation A B Za A Zab B Zb Zbc Zca Zc C C (product of ZY taken in pairs) ZD = Zab = (Za.Zb + Zb.Zc + Zc. Za) / Zc the opposite ZY Zbc = (Za.Zb + Zb.Zc + Zc. Za) / Za For balanced load, Za = Zb = Zc = ZY Zca = (Za.Zb + Zb.Zc + Zc. Za) / Zb Therefore, ZD = 3.ZY 5 Balanced 3-f Source and Load ZL Y-Y System: B b a ~ ~ A ZB ZL n ZA N ZC ~ c Balanced Y source Per phase equivalent circuit: ~ n Line impedance ZL a Vp ZL a C Balanced Y load Ip = IL A ZA N 6 Balanced 3-f Source and Load Y-D System: ZL Ia IAB b a ~ ~ n a ~ ca ZL B Z Z a ZL A Z C 7 ZL Y-D System: Ia IAB b a ~ ZL ~ n a B Z Z a ~ ZL ca A Z C Convert D load to Y load, ZL Ia b a ~ ~ n a ZL N a c Z/3 IAN Z/3 ~ a B A Z/3 ZL C Per phase equivalent circuit: 8 Example: In a balanced Y-D 3-f circuit, the Y-connected source has a Vp = 110 V. Line impedance between the source and the load is ZL = (10 + j5) per phase. Per phase impedance of the D-connected load is ZD = (75 + j225) . Determine the phase currents in the D-connected load. 10+j5 Ia IAB b a ~ Vp = 110 V ~ B n 75+j225 ~ ca A C 9 10+j5 b a ~ Ia IAB ~ n Vp = 110 V A B 75+j225 ~ ca C Convert D load to Y load, 10+j5 Ia b a ~ ~ Vp = 110 V n ~ ca A IAN B N (75+j225)/3 C Per phase equivalent circuit: 10 Multiple Balanced Loads in 3-f Systems a b A1 B1 A2 Z1 N Z1 c B2 Z2 Z1 Z2 Z2 C2 C1 A number of balanced 3-f loads are connected in parallel. Such systems can be evaluated using different methods: 1.Convert all loads to Y, and parallel the impedances 2.Convert all loads to D, and parallel the Z’s 3.Combine the per phase powers 11 Multiple Balanced Loads in 3-f Systems Example: A 3-f, Y-connected motor takes 10 kVA at 0.6 p.f. lagging from a source of 220 volts line-to-line. The motor is in parallel with a balanced D load having 16- resistance and 12- capacitive reactance in series in each phase. Find the total P, line current and p.f. of the combined load. 12 Example: Unbalanced 3-f Load A 3-phase unbalanced load shown below is energized by a balanced 3-phase voltage with 208 volts line-to-line, and phase sequence a-b-c. Determine the total real power delivered to the load. a + 10 208 V _ -j10 10 b + 208 V _ c 13 Example: Unbalanced 3-f Load a + Loop Analysis: Vab = 208/00 V 10 I1 10 -j10 _ b + Vbc = 208/-1200 V _ c I2 14 Example: Unbalanced 3-f Load Ia → a + 10 Vab = 208/00 V -j10 n Nodal Analysis at node n: 10 _ Ib → b + Vbc = 208/-1200 V _ c Ic → The voltage at node n is taken with respect to the neutral node at the source. That is, VnN 15 Watt-meter Connection in Single Phase System Most watt-meters have CC and VC tap settings such that they can be operated close to the rated values (or at high loads) Watt-meter + cc + vc I V + Single phase load cc I Power factor cos vc + V CC: 0 impedance VC: ∞ impedance Power measured by a watt-meter is 16 Watt-meter Connection in 3-f (3-wire) System A Ia + P1 + b a Zb B n Zc + P2 C + Ic Q3f = 3 (P2 – P1) P3f = P1 + P2 P1 = |Vab|.|Ia|. cos f1 where Za f1 = angle between Vab and Ia = 300 - P2 = |Vcb|.|Ic|. cos f2 where c f2 = angle between Vcb and Ic = -300 - 2-Wattmeter Method (connected line – line) Vcb Ic p.f. angle, = angle between Vph and Iph Vca Vcn Vab 2 = angle between Van and Ia (or Vcn and Ic) 300 1 Van Ia P1 + P 2 = VL.IL [cos (300 - ) + cos (-300 - )] = VL.IL [2 cos 300 cos ] = 3.|VL|.|IL|. cos Ib Vbn Vbc 17 Watt-meter Connection in 3-f Balanced System a Pa Ia + b A B + Ib N n c Ic C When neutral point is accessible (3-f, 3-wire system) Three phase power = 3 x Pa 18 Example A balanced Y load having per phase impedance of 10 /450 , is supplied by a 3-f source with a line-to-line voltage of 220 V. Two-wattmeter method is used to measure the power delivered to the load. Determine the reading on each wattmeter, and find the real & reactive power from the wattmeter readings. A Ia + P1 + VL = 220 V b a Z B n Z = 10/450 Z C + P2 + Ic c 19