Distribution system and power quality

Cal Poly Pomona ECE 5750 Distribution system and power quality - Instructor Dr. Ha Thu Le

1

California State Polytechnic University Pomona Department of Electrical and Computer Engineering

ECE 5750 Distribution system and power quality – Spring 2020

Short Answer for Homework 4 Distribution feeder analysis

Problem 1

tol = 0.0001; % tolerance level

number of iterations: 4

V1 is the voltage at the primary of the substation

Q1 TRF forward and backward sweep matrices Ztabc = 0.6475 + 3.2375i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.6475 + 3.2375i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.6475 + 3.2375i Specified line-neutral source voltage and load VLN_ABC = 1.0e+04 * 3.4500 - 1.9919i -3.4500 - 1.9919i 0.0000 + 3.9837i S3 = 1.0e+05 * 5.5250 + 3.4241i 4.5000 + 2.1794i 9.0250 + 2.9664i V3 = 1.0e+03 * 7.9674 + 0.0000i -3.9837 - 6.9000i -3.9837 + 6.9000i

Cal Poly Pomona ECE 5750 Distribution system and power quality - Instructor Dr. Ha Thu Le

2

Final result when system converged V1_mag = 1.0e+04 * 3.9837 3.9847 3.9840

V1_ang = -30.0081 -149.9951 90.0010

V2_mag = 1.0e+03 * 7.7754 7.8380 7.7583

V2_ang = -61.4511 178.7891 57.4620

V3_mag = 1.0e+03 * 7.7452 7.8130 7.7042

V3_ang = -61.6420 178.7369 57.0198

Final voltages in pu & volt on a 120-V base V1_pu = 1.0000 1.0002 1.0001 V2_pu = 0.9759 0.9838 0.9737

Cal Poly Pomona ECE 5750 Distribution system and power quality - Instructor Dr. Ha Thu Le

3

V3_pu = 0.9721 0.9806 0.9670 V3_120V = 116.6528 117.6745 116.0355

Problem 2

Q 1 n 2 Line impedance - Compensator R and X setting

Zline_aver = 0.2360 + 0.4399i ohm

Npt = 66.3953 CTp = 209.1849 CTs = 5

Zcomp_volt = 0.4489 + 1.4227i

0.9520 + 0.8521i

0.8293 + 1.8829i

Zcomp_ohm = 0.0898 + 0.2845i

0.1904 + 0.1704i

0.1659 + 0.3766i

Q3 Compensator voltage and current input - Load new

Icomp_mag = 2.0060

1.5296

2.9474 A

Icomp_ang = -93.4303

152.8949

38.8249 deg.

Vrelay_mag = 116.6528

117.6745

116.0355 V

Cal Poly Pomona ECE 5750 Distribution system and power quality - Instructor Dr. Ha Thu Le

4

Vrelay_ang = -61.6420

178.7369

57.0198

Q4 Final tap position, V=121V, bandwidth=2V

Tap = [5; 4; 6]

Q5 With the regulators taps set, compute the actual load voltages in volts and in per unit.

tol = 0.0001; % tolerance level

number of iterations: 4

V2r is the input voltage to regulator

V2 is the output voltage from regulator

V1 is the substation primary voltage

Regulator matrices (Self)

Final result when system converged

V1_mag = 1.0e+04 *

3.9837

3.9847

3.9840

V1_ang = -30.0079

-149.9956

90.0009

V2_mag = 1.0e+03 *

8.0263

8.0390

8.0609

V2_ang = -61.4512

178.7894

57.4622

V2r_mag = 1.0e+03 *

7.7754

7.8380

7.7586

Cal Poly Pomona ECE 5750 Distribution system and power quality - Instructor Dr. Ha Thu Le

5

V2r_ang = -61.4512

178.7894

57.4622

V3_mag = 1.0e+03 *

7.9968

8.0146

8.0091

V3_ang = -61.6288

178.7365

57.0535

V3_pu = 1.0037

1.0059

1.0052

V3_120V = 120.4417

120.7107

120.6276