O e. None of theseClear my choiceThe three phase power and line-line ratings of the below electric power system is given below:Vmut ofMG: 50 MVA11 KVX= 4%T: 50 MVA11/20 KVX= 4%M: 43.2 MVA18 KVX= 4%By choosing 100 MVA and 11 KV on the generator side as base values, the new per unit impedance of the generator isand the new perunit impedance of the motor is4A three-phase, 60-Hz lossless transmission line is 350 Km long. The line has a phase constant B of 0.001266825 rad/Km and a suimpedance Z of 326.09 N. What is the total receiving end reactive power Qr when the sending end voltage is 230 KV and the revoltage is 219 KV and the power angle & is 15 degrees?out of

The given system is, For generator: For motor: And

The three phase power and line-line ratings of the below electric power system is given below: T Vm M G: 50 MVA 11 KV - X= 4% T: 50 MVA 11/20 KV X= 4% M: 43.2 MVA 18 KV X= 4% By choosing 100 MVA and 11 KV on the generator side as base values, the new per unit impedance of the generator is and the new perunit impedance of the motor is

The three phase power and line-line ratings of the below electric power system is given below: T Vm M G: 50 MVA 11 KV - X= 4% T: 50 MVA 11/20 KV X= 4% M: 43.2 MVA 18 KV X= 4% By choosing 100 MVA and 11 KV on the generator side as base values, the new per unit impedance of the generator is and the new perunit impedance of the motor is

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 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.
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.
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.
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".
Equipment ratings and per-unit reactances for the system shown in the below figure are given as follows: Synchronous generators: GI 100 MVA 25 kV X1 = X2 = 0.2 Xo = 0.05 G2 100 MVA 13.8 kV X = X2 = 0.2 Xo =0.05 Transformers: TI 100 MVA 25/230 kV X = X2 = X, = 0.05 T2 100 MVA 13.8/230 kV X = X2 = Xo = 0.05 Transmission lines: TL12 100 MVA 230 kV X = X2 = 0.1 Xo =0.3 TL13 100 MVA 230 kV X = X2 = 0.1 Xo = 0.3 %3D TL23 100 MVA 230 kV X = X2 = 0.1 Xo = 0.3 %3D Using a 100-MVA, 230-kV base for the transmission lines, draw the per-unit sequene networks and reduce them to their Thévenin equivalents, “looking in" at bus 3. Neglect A-Y phase shifts. Compute the fault currents for a bolted three-phase fault at bus 3. 2 T2 b T1 TL12 to G1 G2 TL13 TL23 j0,03 j0,03 3
Problem 1 - Series en Parallel AC networks [19] Look at the circuit in Figure 1 and determine the following: (a) Total Admittance. (b) Total Impedance. (c) Total Current (l:). (d) Current (I1) through impedance Z2. (e) Current (12) through impedance Z3. (f) Current (I3) through impedance Z4. (g) Is this an inductive or capacitive circuit? A. B Zs 220V;50HZ Figure 1 (h) Voltage across Z1. (i) Voltage across A and B. G) Voltage across Zs. Z1 = 3 + j5 ohm Z2 = 10 + jo ohm Z3 = 5 + j15 ohm Z4 = 10 – j30 ohm Zs = 20 – j30 ohm Admittance and Impedance in rectangular notation. All currents and voltage in polar notation. Take voltage as reference.
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.
2A. What is another term for "shunt"? a) in series b) in parallel c) compounded d) none of the above 2B. Which of the following is/are a basic type of DC generator connection? a) series b) shunt c) compound d) all of the above
L-type equivalent circuit and parameters for a continuous sinusoidal phase of a 5.5 kW, 1450 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, jX m १ (1-s) R, S How many times the rated current will the inrush current be when the ASM is started directly? O 4.3763 O 2.6455 O 6.3414 7.1534 O 5.2280 O 8.0815 O 3.7473 O 1.2558
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
7. System Model and Per Unit T1 Line T2 BE 31.25miles 3 Impedance Load R1+jXL-0.1+j0.5 p.u. 30 MVA 30 MVA 13 kV X-0.1 p.u. 14 kV 13/220 kV X-0.2 p.u. 35 MVA * Transformer T2 is composed of "three single-phase" transformers, each rated 40/3 MVA, 132.8:14 kV with a leakage reactance of 0.1 per unit. (a) Fill in the appropriate base values. Gen. G Trans. Ti Line Trans. T: Load New MVA Base New KV Base (b) Calculate the impedance values. Gen G Trans. T1 Line Trans. T2 Load p.u. Impedance
Figure below shows a power system. The rating of generators and transformers are:G1:25MVA, 6.6KV, jO.2 p.u. G2:15MVA, 6.6KV, jo.15 p.u. G3:30MVA, 13.2KV, jo.15 p.u. T1: 30MVA, 6.6A- 115Y KV, jo.1 p.u.T2: 15MVA, 6.6A- 115Y KV, jo.1 p.u. T3:single phase units each rated 10MVA, 69/ 6.9 KV, jo.1 p.u. Find te reactances in p.u, take a base of 30MVA, 6.6 KV in the circuit of G1 .=.XG1 G2} 4 A lubu T2 T1 T3 G1 G3 j120 2 j90 N