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Solutions -unit4

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  • 1. 1. Determine the fault current, the bus voltages and the line currents during the fault when a 3φ fault occurs with a fault impedance of Zf = j0.16 pu :- Xth = (j0.1 + j0.1) ΙΙ [(j0.4 = j0.4) ΙΙ j0.8 + j0.2 + j0.2] = j0.2 ΙΙ j0.8 = (j0.2 x j0.8 / j1.0) = j0.16 Vth = 1.0(00)
  • 2. Contribution from various generators:- Ig1 = (If x j0.8)/ (j0.8 + j0.2) = 0.8 x (-j3.125) = -j2.5 Ig2 = -j0.625 After the fault, the voltage of the basis will be different:- V1 = Eg1 – (j0.1 + j0.1) Ig1 = 0.1 – j0.2 (-j2.5) = 0.5 pu V2 = Eg2 – (j0.2 + j0.2)Ig2 = 1 – j0.4 (-j0.625) = 0.75 pu Current flowing through bus 3 I3 = Ig2 x 0.8/ (0.8 + 0.8) = Ig2/2 = -j0.3125 V3 = V2 – I3(j0.4) = 0.75 – (-j0.3125) (j0.4) = 0.625 To find the s.c. current in the transmission lines:- I21(F) = (V1-V2)/j0.3 = (0.75 – 0.5)/j0.3 = -j0.3125 I31(F) = (V3-V1)/j0.4 = (0.625-0.5)/j0.4 = -j0.3125 I23(F) = (V2-V3)/j0.4 = (0.75-0.625)/j0.4 = -j0.3125 2. A double line-to-ground fault occurs on lines b and c at point F in the system of given figure. Find the subtransient current in phase c of machine 1 assuming prefault currents to be zero. Both machines are 5%. Each three-phase transformer is rated 1200 kVA, 600 V-Δ/3300 V. The reactances of the neutral grounding reactors are 5% on the kVA base of the machines.
  • 3. X1th = (j0.1 – j0.05) ΙΙ (j0.2 + j0.05 + j0.1) = j0.15 ΙΙ j0.35 = j0.105 X2th = j0.105 X0th = j0.05 ΙΙ j0.45 = j0.045 Ia1 = 1(0)0/ [j0.105 + (j0.105 x j0.045)/ (j0.105 + j0.045)] = 1/j.0.1345 = -j7.326 Ia2 = - [(-j7.326 x j0.045)/ (j0.045 + j0.105)] = +j2.198 Ia0 = j5.128 Ia1 = (-j7.326 x j0.35)/ (j0.15 + j0.35) = -j5.13 Ia2 = (j2.198 x j0.35)/ (j0.15 + j0.35) = j1.5386 IA1 = 5.13(-90)0 x (-90)0 = 5.13(-180)0 IA2 = 1.53 x 6(90)0 x (90)0 = 1.5386(180)0 IC = aIA1 + a2IA2 = 5.13(-180+120)0 + 1.5386(180-120)0 = 5.13(-60)0 + 1.5386(60)0 = 2.565 – j4.44 + 0.7693 + j1.33 = 3.33 – j3.11 = 4.556(-43.04)0
  • 4. ICactual = IC x Ibase = 4.556 x (1.2/√3x0.6) = 5260A 3. An importing area has a total demand of 25MW from an infinite bus via an interconnector. The SSSL is 80MW. Estimate using equal area criteria, the maximum additional area load that could be suddenly switched on without the system losing stability.
  • 5. ; Increase= 4. Three identical resistors are star connected and rated 2500V, 750 kVA. This three-phase unit of resistors is connected to the Y side of a Δ-Y transformer. The following are the voltages at the resistor load: ΙVabΙ =2000V; ΙVbcΙ=2900V; ΙVcaΙ=2500V Choose base as 2500V, 750kVA and determine the line voltages and currents in per unit on the delta side of the transformer. It may be assumed that the load neutral is not connected to the neutral of the transformer secondary. ΙVabΙ=2000/2500=0.8pu ΙVbcΙ=2900/2500=1.16pu ΙVcaΙ=2500/2500=1pu Assuming an angle of 180◦ for Vca , using the law of cosines, the other line voltages can be obtained. Therefore, Vab=0.8(79.397◦) Vbc=1.16(-42.68◦) Vca=1(180◦) Finding the symmetrical components of the line voltages:
  • 6. Vab1=0.975(72.05◦) Vab2=0.2088(-137.27◦) Vab0=0 To find the phase voltages on the Y side: Van1=1/√3* Vab1(-30◦)=0.975(72.05-30)=0.975(42.05◦)_____(on a phase voltage base) Van2=1/√3* Vab2(+30◦)=0.2088(-137.27+30)=0.2088(-107.27◦) (We do not need to divide by √3 as we are taking a phase voltage base here) To find the phase voltages on the delta side: VAN1= VA1=-j Van1=0.975(-47.95◦) VAN2= VA2=+j Van2=0.2088(-17.27◦) VA= VA1+ VA2=1.159(-42.67◦) VB= a2 VA1+a VA2=1(180◦) Vc=a VA1+ a2VA2=0.8(79.39◦) Line voltages on the delta side are: VAB= (VA- VB)/√3=1.16(-22.93◦) VBC=( VB- Vc) /√3=0.8030(-145.57◦) VCA=( VC- VA) /√3=0.995(114.153◦) R=1(0◦)pu Therefore the line currents on the delta side are: IA=VA/1(0◦)=1.16(-42.67◦) IB=VB/1(0◦)=1(180◦) IC=VC/1(0◦)=0.8(79.39◦)