Consider the simplified electric power system shown in Figure 6.22 for which the power flow solution can be obtained without resorting to iterative techniques.

(a) Compute the elements of the bus admittance matrix \(\boldsymbol{Y}_{\text {bus }}\)

(b) Calculate the phase angle \(\delta_{2}\) by using the real power equation at bus 2 (voltage-controlled bus).

(c) Determine \(\left|\mathrm{V}_{3}ight|\) and \(\delta_{3}\) by using both the real and reactive power equations at bus 3 (load bus).

(d) Find the real power generated at bus 1 (swing bus).

(e) Evaluate the total real power losses in the system.

Figure 6.22

Transcribed Image Text:

V = 1/0 O Y = 2-j4 Y = 3-j6 P3 = -1.5 p.u. Q3 = +0.8 p.u. P = 1.5 p.u. |Vl = 1.1 p.u. O