Electrical Power Systems

436 Electrical Power Systems
Given that
IC 1 = 4.0 + 0.60 P g1 = 4.0 + 0.6 × 1.024 = 4.6144
IC 2 = 4.0 + 0.60 P g2 = 4.0 + 0.60 × 1.0 = 4.60
Solving eqn (xii), we get,
L 2 =
1
1 - P
P
L 1 is always 1.0, so
L 1 × IC 1 = 4.6144
L
g2
= 1.0203
L 2 × IC 2 = 1.0203 × 4.60 = 4.6933
But L 2 × IC 2 > L 1 × IC 1, therefore we need to decrease P g2.
After few iterations, we have d 23 = 11 0 , d 31 = – 6 0 , P g1 = 1.05 pu, P g2 = 0.963 pu and
L 2 = 1.01792.
\ L 1 × IC 1 = 4.63
L 2 × IC 2 = 4.65
Product L 1IC 1 and L 2IC 2 are quite close to each other and solution has converged.
16.10 TRANSMISSION LOSS ORMULA
Here, main aim is to give a simple expression for loss formula based on certain assumptions.
igure 16.10 shows two generating stations connected through a transmission network to a
number of loads. One of the lines in the power network is designated as branch K.
ig. 16.10: Schematic diagram showing two generating stations connected through a
transmission network to a number of loads.
Now if we assume that the total load current I L is supplied by generating station 1 only, as
shown in ig. 16.11(a)
ig. 16.11(a): Generating Station 2 is off.