Xline 26.4Xline 0.1213 ln12

0.052

40

2 60 2 10 7 1609 0.1213Zb 132.25Zb1152

100

XT2 0.5XT2 0.1010020

T2

XT1 0.48XT1 0.1210025

T1

Xq 3.2Xq 0.810025

Xd 4.8Xd 1.210025

Gen

Reactance data given in per unit or percent on rating basis: Generator: 15 kV, 25 MVA, Xd = 1.2 pu, Xq = 0.8 pu Transformer T1: 15 kV: 115 kV, 25 MVA, X = 12% Line: 115 kV, Deq = 12 ft, Ds = 0.0520 ft, length = 40 miles Transformer T2: 115 kV: 12.5 kV, 20 MVA, X = 10%

1. Set up the system circuit diagram in per unit on the system base of 100 MVA and rated voltages. Hint: all reactances must be converted to per unit on the system base.

ELCT 751 Practice Exam 1

The exam is based on the small power system shown in the one-line diagram below. This is a balanced three-phase 60-Hz system. The generator maintains its terminal voltage at 1.00 per unit and its phase angle is taken to be zero. Instrumentation at the generating plant indicates that the generator output is P = 20.0 MW and Q = 5.15 MVAr. Resistances and shunt admittances may be neglected. Use a system base of 100 MVA for system per-unit calculations. Solve the problem in per unit, then convert all answers to SI units such as kV and MW.

Load

T2 T1

Generator Transmission Line

arg Ig 14.44 degIbG

100 1063 15 103

A IbG 3.849 kA Ig IbG 794.9 A

IbH100 1063 115 103

A IbH 502 A Ig IbH 103.7 A

I Ig same current for T1, Line, T2, and load since all in series

3. Calculate the voltage and current at the high-voltage side of the step-up transformer T1.

VT1 Vg j XT1 I j XT1 I 0.025 0.096j

VT1 0.975 0.096j VT1 0.98 arg VT1 5.62 deg115

3kV VT1 65.07 kV L N arg VT1 5.62 deg

115 kV VT1 112.7 kV L L arg VT1 30 deg 24.38 deg

XlineXlineZb

Xline 0.200

2. Calculate the generator power factor and current.

Vg 1.0 j 0.0 Pg20100

Pg 0.2 Qg5.15100

Qg 0.052

Sg Pg2 Qg

2 Sg 0.207 pfgPgSg

pfg 0.968 lag

acos pfg 14.44 deg Ig SgVg e j Ig 0.207 arg Ig 14.44 deg

or IgPg j Qg

Vg Ig 0.207

115 kV VL 111.4 kV L L arg VL 30 deg 15.9deg6. Calculate the generator excitation voltage Ea = Va + jXd Iad + jXq Iaq. (Remember to use the phasor diagram construction method in the notes to locate q and d axes)

Vg 1 arg Vg 0 Ig 0.207 arg Ig 14.44 deglocate the q axis: Ex Vg j Xq Ig j Xq Ig 0.165 0.64jEx 1.165 0.64j arg Ex 28.79deg

gives the angle by which q leads Vg

43.23deg angle by which the q axis leads Ig

Iq Ig cos ( ) Iq 0.15 convert to phasor Iq Iq ej

4. Calculate the line inductive reactance, and the voltage and current at the high-voltage side of the step-down transformer T2.

Xline 0.2 VT2 VT1 j Xline I j Xline I 0.01 0.04jVT2 0.965 0.136j VT2 0.975 arg VT2 8.02 deg

115

3kV VT2 64.7 kV L N arg VT2 8.02 deg

115 kV VT2 112.1 kV L L arg VT2 30 deg 21.98deg5. Calculate the voltage and current at the low-voltage side of transformerT2. Calculate the power delivered to the load (both P and Q).

VL VT2 j XT2 I j XT2 I 0.026 0.1jVL 0.939 0.236j VL 0.968 arg VL 14.1 deg

115

3kV VL 64.3 kV L N arg VL 14.1 deg

arg Ea 28.79degEa 1.555Ea 1.363 0.749jj Xd Id j Xq Iq 0.363 0.749j

j Xq Iq 0.232 0.422jj Xd Id 0.595 0.327jarg Id 61.21 degId 0.141arg Iq 28.79degIq 0.15

check

arg Ea 28.79degEa 1.555Ea Vg j Xd Id j Xq IqId Id e

j 2

convert to phasorId 0.141Id Ig sin ( )