The final schedule is calculated to be:
P1 =2.5982 , P2=2.9413 , Q1=1.1795 , Q2 =0.7677
The voltages are V1= 1.0062, V2 = 1.0263, and the phase angular difference d1-d2 = -8 degrees.
The power flow from bus 2 to bus 1 equals 1.4413 pu, and the loss is 0.0395 pu
Comments:
1) |
It is cheaper to import some power from generator 2 for the local load (4 pu) at the bus 1. |
2) |
However, this implies that there will be some loss due to transmission which will increase the cost. |
3) |
Therefore we have to tradeoff : obtain power from a cheaper remote generator and at the same time ensure losses due to transmission are low. |
4) |
Voltages at the 2 ends can be adjusted so as to reduce losses. Obtaining power at voltages above the nominal results in lower losses (due to lesser current) but then voltages have to be restriced within the limits. |
5) |
Voltage at bus 2 is slightly higher than the specified limits. This can be corrected if the penalty function parameter b is increased; however, this may cause divergence in the numerical method used for solving the optimization (try doing it with the program given to you). |
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