Distribution system and power quality

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

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California State Polytechnic University Pomona Department of Electrical and Computer Engineering

ECE 5750 Distribution system and power quality – Spring 2020

Solution for Homework 4 Distribution feeder analysis

Problem 1 Consider the feeder in the following figure.

The substation transformer is connected to an infinite bus. The infinite bus voltages (i.e. the

substation primary voltages) are balanced and being held at 69 kV for all power-flow problems. The

substation transformer ratings are:

5000 kVA, 69 kV delta – 13.8 kV grounded Y, Z = 1.7 + j8.5 %

The phase impedance matrix for a four-wire wye line between Node 2-3 is

The four-wire wye feeder between Node 2-3 is 0.75 miles long. An unbalanced wye-connected load

is located at node 3 and has the following values:

Phase a: 650 kVA at 0.85 lagging power factor

Phase b: 500 kVA at 0.90 lagging power factor

Phase c: 950 kVA at 0.95 lagging power factor

Assuming that the regulators are in the neutral position:

1) Determine the forward and backward sweep matrices for the substation transformer and the line segment.

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

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2) Use the modified ladder technique to determine the line-to-ground voltages at all nodes. Use a tolerance of 0.0001 per unit.

3) Calculate all node voltages in actual values in volts and on a 120-V base. Solution: tol = 0.0001; % tolerance level

number of iterations: 4

V1 is the voltage at the primary of the substation

sweep matrices for line segment a1 = 1 0 0 0 1 0 0 0 1 b1 = 0.3432 + 0.8085i 0.1170 + 0.3763i 0.1151 + 0.2887i 0.1170 + 0.3763i 0.3499 + 0.7862i 0.1185 + 0.3177i 0.1151 + 0.2887i 0.1185 + 0.3177i 0.3461 + 0.7988i d1 = 1 0 0 0 1 0 0 0 1 A1 = 1 0 0 0 1 0 0 0 1 B1 = 0.3432 + 0.8085i 0.1170 + 0.3763i 0.1151 + 0.2887i 0.1170 + 0.3763i 0.3499 + 0.7862i 0.1185 + 0.3177i 0.1151 + 0.2887i 0.1185 + 0.3177i 0.3461 + 0.7988i 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 at = 0 -5.7735 -2.8868 -2.8868 0 -5.7735 -5.7735 -2.8868 0

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

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bt = 0.0000 + 0.0000i -3.7383 -18.6916i -1.8692 - 9.3458i -1.8692 - 9.3458i 0.0000 + 0.0000i -3.7383 -18.6916i -3.7383 -18.6916i -1.8692 - 9.3458i 0.0000 + 0.0000i dt = 0.1155 -0.1155 0 0 0.1155 -0.1155 -0.1155 0 0.1155 At = 0.1155 0 -0.1155 -0.1155 0.1155 0 0 -0.1155 0.1155 Bt = 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 VLN_ABC_mag = 1.0e+04 * 3.9837 3.9837 3.9837

VLN_ABC_ang = -30.0000 -150.0000 90.0000

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 V3_mag = 1.0e+03 * 7.9674 7.9674 7.9674

V3_ang = 0 -120.0000 120.0000

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

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End of first iteration V1_mag = 1.0e+04 * 4.1254 4.0597 4.0785

V1_ang = 30.8211 -87.0387 151.8043

V2_mag = 1.0e+03 * 7.9967 7.9930 8.0190

V2_ang = 0.1723 -119.9450 120.4171

V3_mag = 1.0e+03 * 7.9674 7.9674 7.9674

V3_ang = 0 -120.0000 120.0000

I3_mag = 81.5821 62.7555 119.2354

I3_ang = -31.7883 -145.8419 101.8051

End of second iteration V1_mag = 1.0e+04 * 4.1021 4.2111 4.1460

V1_ang = -31.3711 -149.5510 89.7250

V2_mag = 1.0e+03 * 8.0873 8.0690 8.2524

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

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V2_ang = -61.9272 178.5134 57.5030

V3_mag = 1.0e+03 * 8.0574 8.0447 8.2025

V3_ang = -62.0986 178.4547 57.1146

I3_mag = 80.6716 62.1530 115.8188

I3_ang = -93.8869 152.6127 38.9197

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

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

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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

V3_pu = 0.9721 0.9806 0.9670

V3_120V = 116.6528 117.6745 116.0355

Error history

er = 0.0356 0.0191 0.0238 er = 0.0297 0.0571 0.0407 er = 0.0019 0.0011 0.0002 er = 1.0e-03 * 0.0062 0.2419 0.0591

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

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Problem 2

Three type B step-voltage regulators are installed in a wye connection at the substation in order to

hold the load voltages (node 3) at a voltage level of 121 V and a bandwidth of 2 V.

1) Compute the actual equivalent line impedance between nodes 2 and 3.

2) Determine a potential transformer ratio and current transformer ratio given that the

compensator circuit ratings are 120V and 5A. Determine the R and X compensator settings

calibrated in volts and Ohms. The settings must be the same for all three regulators.

3) For the load conditions of Problem 1 with the regulators in the neutral position, compute the

voltages across the voltage relays in the compensator circuits.

4) Determine the appropriate tap settings for the three regulators to hold the node 3 voltages

at 121 V in a bandwidth of 2 V.

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

Solution:

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

Zline_ohm =

0.1425 + 0.4516i

0.3022 + 0.2705i

0.2632 + 0.5976i

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.

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

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Vrelay_mag = 116.6528

117.6745

116.0355 V

Vrelay_ang = -61.6420

178.7369

57.0198

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

tap_req =

4.4630

3.1007

5.2860

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

a_reg = 0.9688 0 0

0 0.9750 0

0 0 0.9625

d_reg = 1.0323 0 0

0 1.0256 0

0 0 1.0390

Areg = 1.0323 0 0

0 1.0256 0

0 0 1.0390

First iteration

I3_mag =

81.5821

62.7555

119.2354

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

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I3_ang =

-31.7883

-145.8419

101.8051

V1_mag = 1.0e+04 *

3.9837

3.9837

3.9837

V1_ang =

-30.0000

-150.0000

90.0000

V2r_mag = 1.0e+03 *

8.0527

8.0523

8.1826

V2r_ang =

-61.8932

178.6064

57.5315

V2_mag = 1.0e+03 *

8.3124

8.2588

8.5014

V2_ang = -61.8932

178.6064

57.5315

V3_mag = 1.0e+03 *

8.3200

8.2533

8.5297

V3_ang = -62.1487

178.4249

57.0442

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

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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

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