Question

6.28 Consider the simplified electric power system shown in Figure 6.17 for which the power- flow solution can be obtained without resorting to iterative techniques. (a) Compute the elements of the bus admittance matrix Ybus. (b) Calculate the phase angle δ, by using the real power equation at bus 2 (voltage-controlled bus). (c) Determine IV and os 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 P2-1.5 p.u ival = 1.1 p.u. FIGURE 6.17 V, 1/0 Problem 6.27 Y=2-j4 Ps1.5 p.u "o,-+0.8 p.u.

Answer 1

(a): - Here we have to find out the bus admittance matrix y_bus. Now consider the elements of y_bus, i.e. y_kk is the sum of admittances connected to bus k. And y_kn is the (-) sum of admittances connected between buses k and n i.e (k notequalto n). The elements are, from given diagram, y_11 = (2 - j4) + (3 - j6) = 2 - j4 + 3 - j6 = 5 - j10 p.u. y_22 = 2 - j4 p.u. and y_33 = 3 - j6 p.u. y_12 = y_21 = -(2 - j4) = -2 + j4 p.u. y_23 = y_32 = 0 y_13 = y_31 = -(3 - j6) = -3 + j6 p.u. Then the bus admittance matrix is becomes as below. y_bus = [5 - j 10 -2 + j4 -3 + j6 -2 + j4 2 - j4 0 -3 + j6 0 3 - j6] = [11.180 angle