The one-line diagram of a three-phase system is shown in Figure 1. By selecting a common based of 100 MVA and 32.75 kVon the generator bus,(a)Calculate all impedances including the load impedance in per-unit.Motor90 MVA33/275 kV90 MVA275/11kV55 MVA10.95 kvXT12%Xrg=12%X15%Generator95 MVA32.75 kvXg+15%2x OH Line 138 kmR= 0.01 ohm/kmX = 0 05 chm/kmLoad35 MVA10.95 kV0.88 pf laggingFigure 1V81V82=V83=Z81 =%3DZ82 =ZB3 =XG (new) =XT1 (new) =XT2(new) =ZOH =Хм (пеw)ZLoad =

Given, Common base power, SB = 100 MVA Base voltage = 32.75 KV

he one-line diagram of a three-phase system is shown in Figure 1. By selecting a common based of 100 MVA and 32.75 kV un the generator bus, ») Calculate all impedances including the load impedance in per-unit. Motor 55 MVA 90 MVA 90 MVA 275/11ky Xry-12% 33/275 kV 10.95 kv X15% Generator 95 MVA 32.75 kv Xg+15% 2x OH Line 138 km Load R=0.01 ohmkm 35 MVA 10.95 kV 0.88 pf laggng X=0 05 chmkm Figure 1 B1 B2= B3= B1 = B2 = B3 = G (new)= T1 (new) = T2 (new) = DH = M (new) = Load =

he one-line diagram of a three-phase system is shown in Figure 1. By selecting a common based of 100 MVA and 32.75 kV un the generator bus, ») Calculate all impedances including the load impedance in per-unit. Motor 55 MVA 90 MVA 90 MVA 275/11ky Xry-12% 33/275 kV 10.95 kv X15% Generator 95 MVA 32.75 kv Xg+15% 2x OH Line 138 km Load R=0.01 ohmkm 35 MVA 10.95 kV 0.88 pf laggng X=0 05 chmkm Figure 1 B1 B2= B3= B1 = B2 = B3 = G (new)= T1 (new) = T2 (new) = DH = M (new) = Load =

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 3.25P: Consider a single-phase electric system shown in Figure 3.33. Transformers are rated as follows:...

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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.
Three single-phase two-winding transformers, each rated 3kVA,220/110volts,60Hz, with a 0.10 per-unit leakage reactance, are connected as a three-phase extended autotransformer bank, as shown in Figure 3.36(c). The low-voltage winding has a 110 volt rating. (a) Draw the positive-sequence phasor diagram and show that the high-voltage winding has a 479.5 volt rating. (b) A three-phase load connected to the low-voltage terminals absorbs 6 kW at 110 volts and at 0.8 power factor lagging. Draw the per-unit impedance diagram and calculate the voltage and current at the high-voltage terminals. Assume positive-sequence operation.
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.
A 30 MVA, 11 kV generator has a reactance of 0.10p.u.on its own base. Determine the per-unit reactance when referred to base kVA of 50,000 kVA and base kV of 33 kV.
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 load impedance in per-unit are as follows: XG (new) = j0.1421 p.u XT1 (new) = j0.1219 p.u XT2 (new) = j0.1219 p.u ZOH = 0.0016 +j 0.0083 p.u Хм пеw) 3D j0.247 р.u ZLoad = 2.2759+ j1.2306 p.u T1 T2 90 MVA Motor 90 MVA 55 MVA 33/275 kV 275/11kV 10.95 kV XT=12% Xr3=12% Xx=15% Generator 95 MVA M 32.75 kV Xg=15% 2 x OH Line 138 km R = 0.01 ohm/km X = 0.05 ohm/km Load 35 MVA 10.95 kV 0.88 pf lagging Figure 1 Note: Region 1 = Generator bus Region 2 = Transmission Lines busses Region 3 = Load Bus If 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.
Connect three single-phase transformers in Y-Y, with subtractive polarity, so that the line-to-line voltages on the primary are 30 degrees delayed to their respective line-to-line voltages on the secondary. It should show each step that is needed to make this connection (example: the phasor diagram) and should show the final connection of the three transformers including the polarity markings. Assume phase A on the primary is terminal "H1".
The one-line diagram of a simple power system is shown in Figure below. The neutral of each generator is grounded through a current-limiting reactor of 0.25/3 per unit on a 100-MVA base. The system data expressed in per unit on a common 100-MVA base is tabulated below. The generators are running on no-load at their rated voltage and rated frequency with their emfs in phase. G Stark Item Base MVA Voltage Rating X' x² 20 kV 20 kV 20/220 kV 20/220 kV 100 0.05 0.15 0.15 0.10 0.10 220 kV 0.125 0.125 0.30 0.15 0.25 025 0.7125 0.15 100 100 0.15 0.05 0.10 0.10 0.10 100 0.10 100 100 Lu La 220 kV 0.15 220 kV 0.35 100 A balanced three-phase fault at bus 3 through a fault impedance Zf= jo.I per unit. The magnitude of the fault current in amperes in phase b for this fault is: Select one: A. 345.3 B. 820.1 C. 312500 3888888 产产

explain why Transformers are necessary in all stages of power distribution systems. Three-phase voltages may be stepped up or down by three-phase transformers
The reactance of a generator is given as 0.25 per-unit based on the generator’s of 18 kV, 500 MVA. Find its per-unit reactance on a base of 20 kV, 100 MVA. a. 0.0505 b. 0.0605 c. 0.0405 d. 0.0606
Three single-phase transformers, each rated 10 MVA , 66.4/12.5 kV, 60 Hz , with an equivalent series reactance of 0.1 per unit divided equally between primary and secondary, are connected in a three-phase bank. The high-voltage windings are Y-connected and their terminals are directly connected to a 115-kV three-phase bus. The secondary terminals are all shorted together. Find the currents entering the high-voltage terminals and leaving the low-voltage terminals if the low-voltage windings are (a) Y-connected and (b) Δ- connected.
Three zones of a single-phase circuit are identified in the figure with a generator voltage specification of 240 V. The zones are connected by the transformers T1 and T2, whose ratings are shown below.Using base values of 30 kVA and 240 volts in zone 1, draw the pu circuit and determine the pu impedance and pu source voltage. Then calculate the load current in pu and in amperes.*Round the final answer to four decimal places, polar form, and label it with the correct units.
L-type equivalent circuit and parameters for a continuous sinusoidal phase of a 5.5 kW, 1480 rpm, line voltage 220 V and frequency 50 Hz star-connected 3-phase ASM, ignoring core (iron) losses given as follows: R1=0.21 Ohm, R2=0.18 Ohm, X1=X2=0.66 Ohm and Xm=9.9 Ohm i, i, R, R, eee R, (1-s) m S How many times the rated current will the inrush current be when the ASM is started directly? O 6.3414 O 1.2558 5.2280 O 3.7473 O 7.1534 4.3763 O 2.6455 8.0815