power factor isThe three phase power and line-line ratings of the below electric power system is given below:VgVmMG: 50 MVA20 KVX= 4%T: 50 MVA20/40 KVX= 4%M: 43.2 MVA36 KVX= 4%By choosing 100 MVA and 20 KV on the generator side as base values, the new per unit impedance of the generator isand the new perunit impedance of the motor is7.5%None of thesea load absorbs 50 MVA at 0.8 pf leading at line to line voltage of 18 KV 10.238%11.43%find the perunit impedance of this load on a base of 100MVA and 208.333%Select one:O a. 5.183-j3.888 puOb. 5.183 +j 3.888 puO c. 1.295 -j0.972 pu1:49 AMEN4/14/20 power factor isThe three phase power and line-line ratings of the below electric power system is given below:T.VmMG: 50 MVA20 KVX= 4%T: 50 MVA20/40 KVX= 4%M: 43.2 MVA36 KVX= 4%By choosing 100 MVA and 20 KV on the generator side as base values, the new per unit impedance of the generator isand the new perunit impedance of the motor is16 %None of these8%VA at 0.8 pf leading at line to line voltage of 18 KV.4%bedance of this load on a base of 100MVA and 20 KV.Select one:O a. 5.183-j3.888 puOb. 5.183 + j 3.888 puOc.1.295-10.972 puEN O1:49

In this question we will find reactance of generator and motor on new base ..

The three phase power and line-line ratings of the below electric power system is given below: Vg Vm M G: 50 MVA 20 KV X= 4% T: 50 MVA 20/40 KV X= 4% M: 43.2 MVA 36 KV X= 4% By choosing 100 MVA and 20 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 7.5% None of these a load absorbs 50 MVA at 0.8 pf leading at line to line voltage of 18 KV 10.238% 11.43% find the perunit impedance of this load on a base of 100MVA and 20 8.333% Select one: O a. 5.183 -j 3.888 pu Ob. 5.183 +j 3.888 pu OC. 1.295-10.972 pu 1:49 EN 4714

The three phase power and line-line ratings of the below electric power system is given below: Vg Vm M G: 50 MVA 20 KV X= 4% T: 50 MVA 20/40 KV X= 4% M: 43.2 MVA 36 KV X= 4% By choosing 100 MVA and 20 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 7.5% None of these a load absorbs 50 MVA at 0.8 pf leading at line to line voltage of 18 KV 10.238% 11.43% find the perunit impedance of this load on a base of 100MVA and 20 8.333% Select one: O a. 5.183 -j 3.888 pu Ob. 5.183 +j 3.888 pu OC. 1.295-10.972 pu 1:49 EN 4714

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

Chapter12: Power System Controls

Section: Chapter Questions

Problem 12.10P

See similar textbooks

Similar questions

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
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.
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.
A balanced three-phase load is connected to a 4.16-kV, three-phase, fourwire, grounded-wye dedicated distribution feeder. The load can be mode led by an impedance of ZL=(4.7+j9)/phase, wye-connected. The impedance of the phase conductors is (0.3+j1). Determine the following by using the phase A to neutral voltage as a reference and assume positive phase sequence: (a) Line currents for phases A, B, and C. (b) Line-to-neutral voltages for all three phases at the load. (c) Apparent. active, and reactive power dissipated per phase, and for all three phases in the load. (d) Active power losses per phase and for all three phases in the phase conductors.
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
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.
The three-phase source line-to-neutral voltages are given by Ean=100,Ebh=10+240, and Ecn=10240volts. Is the source balanced? (a) Yes (b) No
In a balanced three-phase Y-connected system with a positive-sequence source, the line-to-line voltages are 3 times the line-to-neutral voltages and lend by 30. (a) True (b) False
A single-phase source is applied to a two-terminal, passive circuit with equivalent impedance Z=3.045, measured from the terminals. The source current is i(t)=22cos(t)kA. Determine the (a) instantaneous power, (b) real power, (c) reactive power delivered by the source, and (d) source power factor.
A three-phase line, which has an impedance of (2+j4) per phase, feeds two balanced three-phase loads that are connected in parallel. One of the loads is Y-connected with an impedance of (30+j40) per phase, and the other is -connected with an impedance of (60j45) per phase. The line is energized at the sending end from a 60-Hz, three-phase, balanced voltage source of 1203V (rms. line-to-line). Determine (a) the current, real power. and reactive power delivered by the sending-end source: (b) the line-to-line voltage at the load: (C) the current per phase in each load: and (d) the total three-phase real and reactive powers absorbed by each load and by the line. Check that the total three- phase complex power delivered by the source equals the total three-phase power absorbed by the line and loads.
Three single-phase transformers, each rated 10MVA,66.4/12.5kV,60Hz, 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 V-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) - connected.
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.