Figure 3.32 shows the one-line diagram of a three-phase power system. By selecting a common base of 100 MVA and 22 kV on the generator side, draw an impedance dia- gram showing all impedances including the load impedance in per-unit. The data are given as follows: 3.23 G: 90 MVA 22 kV x = 0.18 per unit T1: 50 MVA 22/220 kV x = 0.10 per unit T2: 40 MVA 220/11 kV x = 0.06 per unit Т3: 40 MVA 22/110 kV x = 0.064 per unit T4: 40 MVA 110/11 kV X = 0.08 per unit M: 66.5 MVA 10.45 kV x = 0.185 per unit Lines 1 and 2 have series reactances of 48.4 and 65.43 N, respectively. At bus 4, the three-phase load absorbs 57 MVA at 10.45 kV and 0.6 power factor lagging. FIGURE 3.32 T, 1 2 4 Line 1 Problem 3.23 220 kV M 6 Line 2 Load 110 kV

Figure 3.32 shows the one-line diagram of a three-phase power system. By selecting a common base of 100 MVA and 22 kV on the generator side, draw an impedance dia- gram showing all impedances including the load impedance in per-unit. The data are given as follows: 3.23 G: 90 MVA 22 kV x = 0.18 per unit T1: 50 MVA 22/220 kV x = 0.10 per unit T2: 40 MVA 220/11 kV x = 0.06 per unit Т3: 40 MVA 22/110 kV x = 0.064 per unit T4: 40 MVA 110/11 kV X = 0.08 per unit M: 66.5 MVA 10.45 kV x = 0.185 per unit Lines 1 and 2 have series reactances of 48.4 and 65.43 N, respectively. At bus 4, the three-phase load absorbs 57 MVA at 10.45 kV and 0.6 power factor lagging. FIGURE 3.32 T, 1 2 4 Line 1 Problem 3.23 220 kV M 6 Line 2 Load 110 kV

Power System Analysis and Design (MindTap Course List)

6th Edition

ISBN:9781305632134

Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Chapter3: Power Transformers

Section: Chapter Questions

Problem 3.23P: Figure 3.32 shows the oneline diagram of a three-phase power system. By selecting a common base of...

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Figure 3.32 shows the oneline diagram of a three-phase power system. By selecting a common base of 100 MVA and 22 kV on the generator side, draw an impedance diagram showing all impedances including the load impedance in per-unit. The data are given a follows: G:90MVA22kVx=0.18perunitT1:50MVA22/220kVx=0.10perunitT2:40MVA220/11kVx=0.06perunitT3:40MVA22/110kVx=0.064perunitT4:40MVA110/11kVx=0.08perunitM:66.5MVA10.45kVx=0.185perunit Lines I and 2 have series reactances of 48.4 and 65.43, respectively. At bus 4, the three-phase load absorbs 57 MVA at 10.45 kV and 0.6 power factor lagging.
Consider a single-phase electric system shown in Figure 3.33. Transformers are rated as follows: XY15MVA,13.8/138kV, leakage reactance 10 YZ15MVA,138/69kV, leakage reactance 8 With the base in circuit Y chosen as 15MVA,138kV determine the per-unit impedance of the 500 resistive load in circuit Z, referred to circuits Z, Y, and X. Neglecting magnetizing currents, transformer resistances, and line impedances, draw the impedance diagram in per unit.
Consider the single-line diagram of the power system shown in Figure 3.38. Equipment ratings are Generator 1: 1000MVA,18kV,X=0.2perunit Generator 2: 1000MVA,18kV,X=0.2p.u. Synchronous motor 3: 1500MVA,20kV,X=0.2p.u. Three-phase -Y transformers T1,T2,T3,T4,: 1000MVA,500kV,Y/20kV,X=0.1p.u. Three-phase YY transformer T5: 1500MVA,500kV,Y/20kVY,X=0.1p.u. Neglecting resistance, transformer phase shift, and magnetizing reactance, draw the equivalent reactance diagram. Use a base of 100 MA and 500 kV for the 50-ohm line. Determine the per-unit reactances.
Determine the positive- and negative-sequence phase shifts for the three- phase transformers shown in Figure 3.36.
Figure 3.39 shows a oneline diagram of a system in which the three-phase generator is rated 300 MVA, 20 kV with a subtransient reactance of 0.2 per unit and with its neutral grounded through a 0.4- reactor. The transmission line is 64km long with a cries reactance of 0.5-/km. The three-phase transformer T1 is rated 350MVA.230/20kV with a leakage reactance of 0.1 per unit. Transformer T2 is composed of three single-phase transformers, each rated 100 MVA, 127/13.2kV with a leakage reactance of 0.1 per unit. Two 13.2kV motors M1 and M2 with a subtransient reactance of 0.2 per unit for each motor represent the load. M1 has a rated input of 200 MVA with its neutral grounded through a 0.4- current-limiting reactor, M2 has a rated input of 100 MVA with its neutral not connected to ground. Neglect phase shifts associated with the transformers. Choose the generator rating as base in the generator circuit and draw the positive-sequence reactance diagram showing all reactances in per unit.
PowerWorid Simulator case Problem 3_60 duplicates Example 3.13 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. Since the system is no longer lossless, a field showing the real power losses has also been added to the oneline. 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. What value of phase shift minimizes the system losses?
Reconsider Problem 3.64 with the change that now Tb includes both a transformer of the same turns ratio as Ta and a regulating transformer with a 4 phase shift. On the base of Ta, the impedance of the two comp onents of Tb is jO.1 per unit. Determine the complex power in per unit transmitted to the load through each transformer. Comment on how the transformers share the real and reactive pors.
With generator conyention, where the current leaves the positive terminal of the circuit element, if P is positive then positive real power is delivered. (a) False (b) True
In developing per-unit equivalent circuits for three-phase transformers. under balanced three-phase operation. (i) A common Sbase is selected for both the H and X terminals. (ii) The ratio of the voltage bases Vbase/VbaseX is selected to be equal to the ratio of the rated line-to-line voltages VratedHLL/VratedXLL. (a) Only one of the above is true. (b) Neither is true. (C) Both statements are true.
In the above figure, a single line model of a three-phase power system is given. Base power 50MVA, base voltage 13.8kV according to its value, a) Per-unit values of power system circuit elements according to the given base power and base voltage values. calculate. b) The perunit you calculate by drawing the equivalent impedance circuit for one phase of the power system given in the figure. show on the values.
An AC three phase distribution system (given in Figure 2), which has end user with balanced three phase (10A) load, is given in the figure. The phase-neutral voltages of the transformers are given as V1=224V, V2=225V, V3=228V and V4=231V respectively. a) Calculate phase-neutral voltage of the bus-bar (B). b) Calculate currents of the transformers (I1, I2, I3 and I4) c) Calculate phase-neutral voltage of the point K while the output of TR3 and TR4 (xy) have been short circuited for a transient time (transformers are still operational) by the line fault.
Draw the per unit impedance diagram for the simple power system in the figure. Catalog of all elementsThe impedances in the information are listed below. The line's actual impedance value and the load'sThe MVA value is also included in this list. Assume a power base value of 100 MVA for all impedances.On the generator side, take the voltage base as 20-kV.