1). A SINGLE PHASE 400 KVA TRANSFORMERANE EFFICIENCY OF 99.13% AT HAIF LOAD UNITYPf WHEREAS IHASIT EFFICIENCY IS 98.77 % ATFULL-LOAD 0.8 of LAGGING. CALCULATE THE FULLLOAD COPPER LOSSIN KW.

Efficiency = outputinput×100 Input = output + copper losses + iron losses. Output = apparent power…

Answered: :). A SINGLE PHASE 400 KVA TRANSFORMER…

:). A SINGLE PHASE 400 KVA TRANSFORMER ANG EFFICIENCY OF 99.13% AT HAIF LOAD UNITY Pf WHEREAS I EFFICIENCY IS 98.77% AT FULL-LOAD 0.8 of LAGGING. CALCULATE THE FULL LOAD COPPER LOSS IN KW. HAS

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

Chapter8: Symmetrical Components

Section: Chapter Questions

Problem 8.39P

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An infinite bus, which is a constant voltage source, is connected to the primary of the three-winding transformer of Problem 3.53. A 7.5-MVA,13.2-kV synchronous motor with a sub transient reactance of 0.2 per unit is connected to the transformer secondary. A5-MW,2.3-kV three-phase resistive load is connected to the tertiary Choosing a base of 66 kV and 15 MVA in the primary, draw the impedance diagram of the system showing per-unit impedances. Neglect transformer exciting current, phase shifts, and all resistances except the resistive load.
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 single-phase 100-kVA,2400/240-volt,60-Hz distribution transformer is used as a step-down transformer. The load, which is connected to the 240-volt secondary winding, absorbs 60 kVA at 0.8 power factor lagging and is at 230 volts. Assuming an ideal transformer, calculate the following: (a) primary voltage, (b) load impedance, (c) load impedance referred to the primary, and (d) the real and reactive power supplied to the primary winding.
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.
Consider the single-line diagram of the power system shown in Figure 3.38. Equipment ratings are Generator 1: 1000MVA,18kV,X=0.2perunit Generator 2: 1000MVA,18kV,X=0.2p.u. Synchronous motor 3: 1500MVA,20kV,X=0.2p.u. Three-phase -Y transformers T1,T2,T3,T4,: 1000MVA,500kV,Y/20kV,X=0.1p.u. Three-phase YY transformer T5: 1500MVA,500kV,Y/20kVY,X=0.1p.u. Neglecting resistance, transformer phase shift, and magnetizing reactance, draw the equivalent reactance diagram. Use a base of 100 MA and 500 kV for the 50-ohm line. Determine the per-unit reactances.
In developing per-unit equivalent circuits for three-phase transformers. under balanced three-phase operation. (i) A common Sbase is selected for both the H and X terminals. (ii) The ratio of the voltage bases Vbase/VbaseX is selected to be equal to the ratio of the rated line-to-line voltages VratedHLL/VratedXLL. (a) Only one of the above is true. (b) Neither is true. (C) Both statements are true.
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.
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() ].
Consider the single-Line diagram of a power system shown in Figure 3.42 with equipment ratings given: Generator G1: 50MVA,13.2kV,x=0.15p.u. Generator G2: 20MVA,13.8kV,x=0.15p.u. Three-phase -Y transformer T1: 80MVA,13.2/165YkV,X=0.1p.u. Three-phase Y- transformer T2: 40MVA,165Y/13.8kV,X=0.1p.u. Load: 40MVA,0.8PFlagging,operatingat150kV Choose a base of 100 MVA for the system and 132-kV base in the transmission-line circuit. Let the load be modeled as a parallel combination of resistance and inductance. Neglect transformer phase shifts. Draw a per-phase equivalent circuit of the system showing all impedances in per unit.
PowerWorid Simulator case Problem 3_60 duplicates Example 3.13 except that a resistance term of 0.06 per unit has been added to the transformer and 0.05 per unit to the transmission line. Since the system is no longer lossless, a field showing the real power losses has also been added to the oneline. With the LTC tap fixed at 1.05, plot the real power losses as the phase shift angle is varied from 10 to +10 degrees. What value of phase shift minimizes the system losses?
Three single-phase two-winding transformers, each rated 25MVA,34.5/13.8kV, are connected to form a three-phase bank. Balanced positive-suence voltages are applied to the high-voltage terminals, and a balanced, resistive Y load connected to the low-voltage terminals absorbs 75 MW at 13.8 kV. If one of the single-phase transformers is removed (resulting in an open connection) and the balanced load is simultaneously reduced to 43.3 MW (57.7 of the original value), determine (a) the load voltages Va,Vb, and Vc; (b) load currents Ia,Ib, and Ic; and (c) the MVA supplied by each of the remaining two transformers. Are balanced voltages still applied to the load? Is the open transformer overloaded?
Consider Figure 3.4. For an ideal phase-shifting transformer, the imda nce is unchanged when it is referred from one side to the other. (a) True (b) False