![]() ![]() Combining the admittances in parallel, To see this example in PowerWorld Simulator open case Example 1 (see Figure 2). Then, for this regulating transformer in series with line L1, The admittance parameters for line L2 alone are given in part (a) above. For the phase-angle-regulating transformer. From (3.7.5)–(3.7.8), the admittance parameters of the regulating transformer in series with line L1 are For line L2 alone, Combining the above admittances in parallel, b. ![]() For the voltage-magnitude-regulating transformer, c = (1 + ∆V)−1 = (1:05)−1 = 0.9524 per unit. Also, the series resistance and shunt admittance of the lines are neglected.SOLUTION The circuit is shown in Figure 1.a. Assume that the regulating transformer is ideal. Determine the 2 × 2 bus admittance matrix when the regulating transformer (a) provides a 0.05 per-unit increase in voltage magnitude toward bus a′b′c′ and (b) advances the phase 3° toward bus a′b′c′. A regulating transformer is placed in series with line L1 at bus a′b′c′. What value of phase shift minimizes the system losses?EXAMPLE 1 Voltage-regulating and phase-shifting three-phase transformersTwo buses abc and a′b′c′ are connected by two parallel lines L1 and L2 with positive-sequence series reactances XL1 = 0:25 and XL2 = 0:20 per unit. With the LTC tap fixed at 1.05, plot the real power losses as the phase shift angle is varied from −10 to +10 degrees. Since the system is no longer lossless, a field showing the real power losses has also been added to the one-line. By Power System Analysis and Design (5th Edition) Edit edition Solutions for Chapter 3 Problem 60P: PowerWorld Simulator case of this Problem duplicates Example 1 except that a resistance term of 0.06 per unit has been added to the transformer and 0.05 per unit to the transmission line. ![]()
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