Please enter your answer in the blank's:° p.u.Voltage at load bus, VLoad =• p.u.Motor apparent power, SM=° p.u.Current drawn by the motor, Im=p.u.Current drawn by the load, ILoad=p.uTotal current drawn by load bus, I7=p.u.Equivalent series impedance, ZT =° p.u.Voltage at generator busbar V, =• kVVoltage at generator busbar V1 = The one-line diagram of a three-phase system is shown in Figure 1. By selecting a common based of 90 MVA and 32.75 kV on the generator bus, all impedances including the loadimpedance in per-unit are as follows:XG (new) = j0.1421 p.uXT1 (new) = j0.1219 p.uXT2 (new) = j0.1219 p.uZOH = 0.0016 +j 0.0083 p.uХм пеw) 3D j0.247 р.uZLoad = 2.2759+ j1.2306 p.uT1T290 MVAMotor90 MVA55 MVA33/275 kV275/11kV10.95 kVXT=12%Xr3=12%Xx=15%Generator95 MVAM32.75 kVXg=15%2 x OH Line 138 kmR = 0.01 ohm/kmX = 0.05 ohm/kmLoad35 MVA10.95 kV0.88 pf laggingFigure 1Note:Region 1 = Generator busRegion 2 = Transmission Lines bussesRegion 3 = Load BusIf the motor operates at full load 0.88 power factor lagging at a terminal voltage 10.95 kV, determine the voltage at the generator bus bar.

One line diagram of a three phase system From circuit

Please enter your answer in the blank's: Voltage at load bus, VLoad = ° p.u. Motor apparent power, SM= * p.u. Current drawn by the motor, Im= p.u. Current drawn by the load, ILoad= p.u. Total current drawn by load bus, I7= 'p. Equivalent series impedance, Z7 = * p.u. Voltage at generator busbar V, = p.u. Voltage at generator busbar V, = • kV

Please enter your answer in the blank's: Voltage at load bus, VLoad = ° p.u. Motor apparent power, SM= * p.u. Current drawn by the motor, Im= p.u. Current drawn by the load, ILoad= p.u. Total current drawn by load bus, I7= 'p. Equivalent series impedance, Z7 = * p.u. Voltage at generator busbar V, = p.u. Voltage at generator busbar V, = • 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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Three single-phase two-winding transformers, each rated 25MVA,34.5/13.8kV, are connected to form a three-phase bank. Balanced positive-suence voltages are applied to the high-voltage terminals, and a balanced, resistive Y load connected to the low-voltage terminals absorbs 75 MW at 13.8 kV. If one of the single-phase transformers is removed (resulting in an open connection) and the balanced load is simultaneously reduced to 43.3 MW (57.7 of the original value), determine (a) the load voltages Va,Vb, and Vc; (b) load currents Ia,Ib, and Ic; and (c) the MVA supplied by each of the remaining two transformers. Are balanced voltages still applied to the load? Is the open transformer overloaded?
Determine the positive- and negative-sequence phase shifts for the three- phase transformers shown in Figure 3.36.
Consider the single-Line diagram of a power system shown in Figure 3.42 with equipment ratings given: Generator G1: 50MVA,13.2kV,x=0.15p.u. Generator G2: 20MVA,13.8kV,x=0.15p.u. Three-phase -Y transformer T1: 80MVA,13.2/165YkV,X=0.1p.u. Three-phase Y- transformer T2: 40MVA,165Y/13.8kV,X=0.1p.u. Load: 40MVA,0.8PFlagging,operatingat150kV Choose a base of 100 MVA for the system and 132-kV base in the transmission-line circuit. Let the load be modeled as a parallel combination of resistance and inductance. Neglect transformer phase shifts. Draw a per-phase equivalent circuit of the system showing all impedances in per unit.
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 three ideal single-phase transformers (with a voltage gain of ) put together as three-phase bank as shown in Figure 3.35. Assuming positive-sequence voltages for Va,Vb, and Vc find Va,Vb, and VC. in terms of Va,Vb, and Vc, respectively. (a) Would such relationships hold for the line voltages as well? (b) Looking into the current relationships, express IaIb and Ic in terms of IaIb and Ic respectively. (C) Let S and S be the per-phase complex power output and input. respectively. Find S in terms of S.
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.
With the same transformer banks as in Problem 3.47, Figure 3.41 shows the oneline diagram of a generator, a step-up transformer bank, a transmission line, a stepown transformer bank, and an impedan load. The generator terminal voltage is 15 kV (line-to-line). (a) Draw the per-phase equivalent circuit, aounting for phase shifts for positive-sequence operation. (b) By choosing the line-to-neutral generator terminal voltage as the reference, determine the magnitudes of the generator current, transmiss ion-line current, load current, and line-to-line load voltage. Also, find the three-phase complex power delivered to the load.
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
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.
For a balanced three-phase positive-sequence currents Ia,Ib,Ic, does the equation Ia+Ib+Ic=0 hold good?
Consider the oneline diagram shown in Figure 3.40. The three-phase transformer bank is made up of three identical single-phase transformers, each specified by X1=0.24 (on the low-voltage side), negligible resistance and magnetizing current, and turns ratio =N2/N1=10. The transformer bank is delivering 100 MW at 0.8 p.f. lagging to a substation bus whose voltage is 230 kV. (a) Determine the primary current magnitude, primary voltage (line-to-line) magnitude, and the three-phase complex power supplied by the generator. Choose the line-to-neutral voltage at the bus, Va as the reference Account for the phase shift, and assume positive-sequence operation. (b) Find the phase shift between the primary and secondary voltages.
A Y-connected load bank with a three-phase rating of 500 kVA and 2300 V consists of three identical resistors of 10.58. The load bank has the following applied voltages: Vab=184082.8,Vbc=276041.4, and Vca=2300180V. Determine the symmetrical components of (a) the line-to-line voltages Vab0,Vab1, and Vab2, (b) the line-to-neutral voltages and Van0,Van1, and Van2, (c) and the line currents Ia0,Ia1, and Ia2. (Note that the absence of a neutral connection means that zero-sequence currents are not present.)