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1. The transfer reactances between a generator and an infinite bus bar operating at
   200KV under various conditions on the interconnector are:
   Prefault:         150ohm per phase
   During Fault: 400ohm/phase
   Post Fault:       200ohm/phase
   If the fault is cleared when the rotor has advanced 600 electrical from its prefault
   position, determine the maximum load that could be transferred without loss of
   stability.                                                    [127.3MW]




   It is nowhere mentioned that the generator is supplying any load and hence we
   have to assume that the voltage is same as that of the infinite bus(=200KV),
   although I have taken it as E and at the end substituted the same.
   A1 = A2
   Integrate to get the above areas and apply the limits as shown in the figure above.
   Pe1 = 1.33Esin(del)
   Pe2 = 0.5Esin(del)
   Pe3 = Esin(del)
   A1 = 0.957Esin(del0) – 0.25Ecos(del0)
   A2 = E(cos(del0 + 600) – cos(pi – del1)) – 1.33E (sin del0)(pi – del1 - del0 - 60)*pi/180
   And P = Esin del1 = 1.33sin(del0)
   So we get a relation between the 2 deltas
   Assume del0 initial value (intelligent guess is < 30 as the increase is 60 and 60+30
   should be less than 90 in most of the cases )
   After 2 or three 3 trials u will get del0 = 280; del1 = 38.630
   P = Pe3 at del1
   Hence P = 200*200/200sin(del1) = 125MW

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Soln unit5

  • 1. 1. The transfer reactances between a generator and an infinite bus bar operating at 200KV under various conditions on the interconnector are: Prefault: 150ohm per phase During Fault: 400ohm/phase Post Fault: 200ohm/phase If the fault is cleared when the rotor has advanced 600 electrical from its prefault position, determine the maximum load that could be transferred without loss of stability. [127.3MW] It is nowhere mentioned that the generator is supplying any load and hence we have to assume that the voltage is same as that of the infinite bus(=200KV), although I have taken it as E and at the end substituted the same. A1 = A2 Integrate to get the above areas and apply the limits as shown in the figure above. Pe1 = 1.33Esin(del) Pe2 = 0.5Esin(del) Pe3 = Esin(del) A1 = 0.957Esin(del0) – 0.25Ecos(del0) A2 = E(cos(del0 + 600) – cos(pi – del1)) – 1.33E (sin del0)(pi – del1 - del0 - 60)*pi/180 And P = Esin del1 = 1.33sin(del0) So we get a relation between the 2 deltas Assume del0 initial value (intelligent guess is < 30 as the increase is 60 and 60+30 should be less than 90 in most of the cases ) After 2 or three 3 trials u will get del0 = 280; del1 = 38.630 P = Pe3 at del1 Hence P = 200*200/200sin(del1) = 125MW