2. Three parallel three-phase loads are supplied from a 207.85-Vrms, 60-Hz three-phase supply. The loads are as follows: Load 1: "B" kW with 0.6 lagging power factor. Load 2: Resistive load of "B" kW. Load 3: capacitor load of “C" kvar. (a) What is the total system kW, kvar, power factor, and the supply current per phase?

2. Three parallel three-phase loads are supplied from a 207.85-Vrms, 60-Hz three-phase supply. The loads are as follows: Load 1: "B" kW with 0.6 lagging power factor. Load 2: Resistive load of "B" kW. Load 3: capacitor load of “C" kvar. (a) What is the total system kW, kvar, power factor, and the supply current per phase?

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

Chapter2: Fundamentals

Section: Chapter Questions

Problem 2.44P: Two balanced three-phase loads that are connected in parallel are fed by a three-phase line having a...

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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.
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.
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.
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.
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 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.
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.
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 Y-connected loads, one drawing 10 kW at 0.8 power factor lagging and the other IS kW at 0.9 power factor leading, are connected in parallel and supplied by a balanced three-phase Y-connected, 480-V source. (a) Determine the source current. (b) If the load neutrals are connected to the source neutral by a zero-ohm neutral wire through an ammeter, what will the ammeter read?
In a balanced system, the phasor sum of the line-to-line voltages and the phasor sum of the line-to-neutral voltages are always equal to zero. (a) False (b) True
Figure 2.33 gives the general -Y transformation. (a) Show that the general transformation reduces to that given in Figure 2.16 for a balanced three-phase load. (b) Determine the impedances of the equivalent Y for the following impedances: ZAB=j10,ZBC=j20, and ZCA=j25. ZAB=ZAZB+ZBAC+ZCZAZCZA=ZABZCAZAB+ZBC+ZCAZBC=ZAZB+ZBAC+ZCZAZAZB=ZABZBCZAB+ZBC+ZCAZCA=ZAZB+ZBAC+ZCZAZBZA=ZCAZBCZAB+ZBC+ZCA
For the circuit element of Problem 2.3, calculate (a) the instantaneous power absorbed, (b) the real power (state whether it is delivered or absorbed). (c) the reactive power (state whether delivered or absorbed). (d) the power factor (state whether lagging or leading). [Note: By convention the power factor cos() is positive. If | | is greater than 90, then the reference direction for current may be reversed, resulting in a positive value of cos() ].