Micro-Scholarship, What it is, How can it help me.pdf
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
![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](https://image.slidesharecdn.com/solnunit5-100506005626-phpapp02/85/Soln-unit5-1-320.jpg)
