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EE212 Passive AC Circuits

EE212 Passive AC Circuits. Lecture Notes 5 Three Phase Systems. 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

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EE212 Passive AC Circuits

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  1. EE212 Passive AC Circuits Lecture Notes 5 Three Phase Systems

  2. 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.

  3. 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

  4. Zab A B Za B A Zb Zca Zbc Zc C C Delta-Wye (D-Y) Transformation (product of adjacent ZD) ZY = sum of all 3 ZD Za = (Zab. Zca) / (Zab + Zbc + Zca) Zb = (Zab. Zbc) / (Zab + Zbc + Zca) Zc = (Zca. Zbc) / (Zab + Zbc + Zca) For balanced load, Zab = Zbc = Zca = ZD Therefore, ZY = ZD / 3

  5. A B Za B A Zb Zc C C (product of ZY taken in pairs) ZD = the opposite ZY Wye-Delta (Y- D) Transformation Zab Zbc Zca Zab = (Za.Zb + Zb.Zc + Zc. Za) / Zc Zbc = (Za.Zb + Zb.Zc + Zc. Za) / Za Zca = (Za.Zb + Zb.Zc + Zc. Za) / Zb For balanced load, Za = Zb = Zc = ZY Therefore, ZD = 3.ZY

  6. ZL B b a A ZB ZL ~ ZA n N ZC ~ ZL c C Balanced Y load Balanced Y source Line impedance ~ ZL Ip = IL A a Vp ZA ~a n N Balanced 3-f Source and Load Y-Y System: Per phase equivalent circuit:

  7. ZL Ia IAB b a A B ZL Z ~a n Z Z ~a ZL c C ~a Balanced 3-f Source and Load Y-D System:

  8. ZL Ia IAB b a A B ZL Z ~a n Z Z ~a ZL c C ~a ~a ZL Ia A b a Z/3 ZL IAN B ~a Z/3 n N Z/3 ~a ZL c C Y-D System: Convert D load to Y load, Per phase equivalent circuit:

  9. 10+j5  Ia IAB b a A B ~ ~ n Vp = 110 V 75+j225  ~a c C 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. 10+j5  Ia IAB b a A B ~ ~ n Vp = 110 V 75+j225  ~a c C 10+j5  Ia b A a IAN B ~ ~ n N Vp = 110 V (75+j225)/3  ~a c C Convert D load to Y load, Per phase equivalent circuit:

  11. a b A2 A1 B2 B1 Z1 Z2 Z1 N Z2 Z2 Z1 C2 c C1 Multiple Balanced Loads in 3-f Systems A number of balanced 3-f loads are connected in parallel. Such systems can be evaluated using different methods: • Convert all loads to Y, and parallel the impedances • Convert all loads to D, and parallel the Z’s • Combine the per phase powers

  12. 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.

  13. 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 10  -j10  _ b + 208 V _ c

  14. Example: Unbalanced 3-f Load Loop Analysis: I1 Vbc = 208/-1200 V a + I2 10  Vab = 208/00 V 10  -j10  _ b + _ c

  15. Example: Unbalanced 3-f Load Ia → Nodal Analysis at node n: n Ib → Vbc = 208/-1200 V The voltage at node n is taken with respect to the neutral node at the source. That is, VnN Ic → a + 10  Vab = 208/00 V 10  -j10  _ b + _ c

  16. Watt-meter cc cc I I + + vc + vc Single phase load Power factor cos + V V 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) CC: 0 impedance VC:∞ impedance Power measured by a watt-meter is

  17. + Ia Vcb + A Ic P1 Vca Vcn Vab a b q2 Zb q1 Za 300 B Van q n Ia Ib Zc Vbn + P2 Ic Vbc c C + Watt-meter Connection in 3-f (3-wire) System • Q3f = 3 (P2 – P1) • P3f = P1+ P2 P1 = |Vab|.|Ia|.cos 1where 1= angle between Vab and Ia= 300-  P2 = |Vcb|.|Ic|.cos 2where 2= angle between Vcb and Ic= -300-  2-Wattmeter Method (connected line – line) p.f. angle,  = angle between Vph and Iph = angle between Van and Ia (or Vcn and Ic) P1 + P2 = VL.IL [cos (300- ) + cos (-300- )] =VL.IL [2 cos 300 cos ] = 3.|VL|.|IL|.cos 

  18. + + Pa Ia a A B Ib b N n Ic C c Watt-meter Connection in 3-f Balanced System When neutral point is accessible (3-f, 3-wire system) • Three phase power = 3 xPa

  19. + + Ia A P1 a b Z VL = 220 V B Z = 10/450 n Z + P2 Ic c C + 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.

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