A balanced wye-delta system is shown in below figure. Line impedance, %3D %3D per phase Load impedance and generator voltage are Z, = 3434'N, ZA = 302-25°N and Van = 220210'V RMS, respectively. For ABC phase sequence, determine the line currents. 334° 2 I Aa Van I ca 220|10" V^ ZA Ibc I Bh Vhn 334 2 334' 2

A balanced wye-delta system is shown in below figure. Line impedance, %3D %3D per phase Load impedance and generator voltage are Z, = 3434'N, ZA = 302-25°N and Van = 220210'V RMS, respectively. For ABC phase sequence, determine the line currents. 334° 2 I Aa Van I ca 220|10" V^ ZA Ibc I Bh Vhn 334 2 334' 2

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 3.24MCQ: It is stated that (i) balanced three-phase circuits can be solved in per unit on a per-phase basis...

See similar textbooks

Similar questions

A three-phase line with an impedance of (0.2+j1.0)/ phase feeds three balanced three-phase loads connected in parallel. Load 1: Absorbs a total of 150 kW and 120 kvar. Load 2: Delta connected with an impedance of (150j48)/phase. Load 3: 120 kVA at 0.6 PF leading. If the line-to-neutral voltage at the load end of the line is 2000 v (rms), determine the magnitude of the line-to-line voltage at the source end of the line.
Two balanced three-phase loads that are connected in parallel are fed by a three-phase line having a series impedance of (0.4j2.7) per phase. One of the loads absorbs 560 kVA at 0.707 power factor lagging, and the other 132 kW at unity power factor. The line-to-line voltage at the load end of the line is 2203V. Compute (a) the line-to-line voltage at the source end of the line. (b) the total real and reactive power losses in the three-phase line, and (c) the total three-phase real and reactive power supplied at the sending end of the line. Check that the total three-phase complex power delivered by the source equals the total three-phase comp lex power absorbed by the line and loads.
One advantage of balanced three-phase systems over separate singlephase systems is reduced capital and operating costs of transmission and distribution. (a) True (b) False
It is stated that (i) balanced three-phase circuits can be solved in per unit on a per-phase basis after converting - load impedances to equivalent Y impedances. (ii) Base values can be selected either on a per-phase basis or on a three-phase basis. (a) Both statements are true. (b) Neither is true. (c) Only one of the above is true.
Two three-phase generators supply a three-phase load through separate three-phase lines. The load absorbs 30 kW at 0.8 power factor lagging. The line impedance is (1.4+j1.6) per phase between generator G1 and the load, and (0.8+j1) per phase between generator G2 and the load. If generator G1 supplies 15 kW at 0.8 poir factor lagging, with a terminal voltage of 460 V line-to-line, determine (a) the voltage at the load terminals. (b) the voltage at the terminals of generator G2, and (c) the real and reactive power supplied by generator G2. Assume balanced operation.
The one-line diagram of a three-phase power system is as shown in the figure attached. Impedances are marked in per unit on a 100-MVA, 400-kv base. The load at bus 2 is S(sub 2) = 15.93 MW - j33.4 Mvar, and bus 3 is S(sub 3) = 77MW + j14 Mvar. I tis required to hold the voltage at bus 3 at 400 angle 0 degrees kV. Working in per unit, determine the voltage at buses 2 and 1.
1. A 69-kV, three-phase short transmission line is 16 km long. The line has a per phase series impedance of 0.125+j0.4375 Ω per km. Determine the sending end voltage, voltage regulation, the sending end power, and the transmission efficiency when the line delivers (a) 70 MVA, 0.8 lagging power factor at 64 kV. (b) 120 MW, unity power factor at 64 kV. Use lineperf program to verify your results. 2. A three-phase, 765-kV, 60-Hz transposed line is composed of four ACSR, l,431,000-cmil, 45/7 Bobolink conductors per phase with flat horizontal spacing of 14 m. The conductors have a diameter of 3.625 cm and a GMR of 1.439 cm. The bundle spacing is 45 cm. The line is 400 km long, and for the purpose of this problem, a lossless line is assumed. (a) Determine the transmission line surge impedance Zc, phase constant ß, Wavelength, the surge impedance loading SIL, and the ABCD constant. b) The line delivers 2000 MVA at 0.8 lagging power factor at 735 kV. Determine the sending end quantities and…
The line voltage Vab at the terminals of a connected balanced three-phase load at Δ is ??? = 4160∠0° V. The line current IaA is ??? = 69.28∠ − 10° A. a) Calculate the load impedance per phase if the phase sequence is positive (ABC). b) Repeat question (a) for a negative phase sequence (CBA).
3. A balanced delta connected load of "14+J18" ohm - per phase is connected at the end of a three-phase line. The line impedance is "7+J12" ohm - per phase. The line is supplied from a three-phase source with a line-to-line voltage of 207.85 Vrms. Taking phase "a" voltage Va as reference, determine the following: (a) Current in phase a. (b) Total complex power supplied from the source. (c) Magnitude of the line-to-line voltage at the load terminal.
A balanced, positive-sequence, Y-connected voltage source with Eab = 480∠0o volts is applied to abalanced-∆ load with Z∆ = 30 ∠40o Ω. The line impedance between the source and load is ZL = 1 ∠850 Ω.for each phase. Calculate the line currents, the ∆ -load currents, and the voltages at the load terminals.Recall that ZY = Z∆/3
Q1: The parameters of a three-phase line 200 km long are resistance/km-0.15 0, inductivekm -0.5 and capacitive susceptancekm-2x 10° U. The transmission line is delivering 50 MVA at 132 kV with power factor 0.85 lagging. Use nominal T method to determine; 1) sending end voltage 2) sending end current 3) sending end power factor 4) transmission efficiency and 5) voltage (4 points) regulation
Consider a DVR compensated single phase system as shown below. a) Without DVR, find the load terminal voltage (VL) if line resistance (Rs) is negligible with Xs = j0.5 pu , ZL = 0.8 + j0.5 pu and Vs= 1pu with angle equals to 0º.b) With DVR, if Vs=VL=1pu with angle equals to 0º, find the voltage Vtand the DVR voltage injected (Vf).