Sequence Network Y-A load Introduction: This experiment solves the problem of symmetrical components using MATLAB This...

 

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Sequence Network Y-A load Introduction: This experiment solves the problem of symmetrical components using MATLAB This experiment solves the sequence networks: balanced-Y and balanced A loads as show in Figure 1. A balanced-Y load is in parallel with a balanced A connected capacitor bank. The Y loa has an impedance Zy = 3+j4 / phase, and its neutral is grounded through an inductiv reactance Xn = 2 Q. The capacitor bank has a reactance Xc =30 n/phase. 1. Draw the sequence networks for this load and calculate the load-sequen impedances. 2. An unbalanced voltage applied to the system, Va=220/0°, Vb-200/-100°, Vc-18 /-200⁰. calculate the sequence voltages and currents of the load. 3. Calculate the total phase currents and neutral current. 4. Repeat 1-3 if the supply voltage has a balanced 220 v and 120° phase shift. 5. Repeat 1-4 if phase a of Y load has the value equals 1+j0.5 0. b 1202 Al Note: First convert Z₁ to Zy as 3+j40 Fig. 1 -1300 1300 Zv2=ZA/3=-j10 Q. Zo-Zy₁+3Z₁ (Zv₂ does not included into Zo because it doesn't connected to the neutral) Z₁=ZYZ//ZYZ Z₂=Z₁ Sequence Network Y-A load Introduction: This experiment solves the problem of symmetrical components using MATLAB This experiment solves the sequence networks: balanced-Y and balanced A loads as show in Figure 1. A balanced-Y load is in parallel with a balanced A connected capacitor bank. The Y loa has an impedance Zy = 3+j4 / phase, and its neutral is grounded through an inductiv reactance Xn = 2 Q. The capacitor bank has a reactance Xc =30 n/phase. 1. Draw the sequence networks for this load and calculate the load-sequen impedances. 2. An unbalanced voltage applied to the system, Va=220/0°, Vb-200/-100°, Vc-18 /-200⁰. calculate the sequence voltages and currents of the load. 3. Calculate the total phase currents and neutral current. 4. Repeat 1-3 if the supply voltage has a balanced 220 v and 120° phase shift. 5. Repeat 1-4 if phase a of Y load has the value equals 1+j0.5 0. b 1202 Al Note: First convert Z₁ to Zy as 3+j40 Fig. 1 -1300 1300 Zv2=ZA/3=-j10 Q. Zo-Zy₁+3Z₁ (Zv₂ does not included into Zo because it doesn't connected to the neutral) Z₁=ZYZ//ZYZ Z₂=Z₁

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Given that The Y load has capacitor bank has a reactance Xc Per phace In positiv... View the full answer