Homework5Sol

pdf

School

University of British Columbia *

*We aren’t endorsed by this school

Course

453

Subject

Electrical Engineering

Date

Dec 6, 2023

Type

pdf

Pages

Uploaded by MateRose22143

ELEC 453 – Power System Analysis Fall 2023 Homework 5 Solution Due Date: Thursday, November 9, 2023 Reading: Chapters 3.1–3.5 in 7th edition textbook Problem 1. The following data are obtained when open-circuit and short-circuit tests are performed on a single-phase, 50-kVA, 2400/240-volt, 60-Hz distribution transformer Measurement on low-voltage side with high-voltage winding open. Voltage: 240 V. Current: 5.97 A. Power: 213 W. Measurements on high-voltage side with low-voltage winding shorted. Voltage: 60 V. Current: 20.8 A. Power: 750 W. 1. Neglecting the series impedance, determine the exciting admittance referred to the high-voltage side. The open circuit test data can be used to find the exciting admittance by neglecting the series impedance. First, we compute the turns ratio as a = V 1 ,rated V 2 ,rated = 2400 240 = 10 . G c , Y m , and B m are then determined as follows: G c = P 2 V 2 1 = 213 2400 2 = 3 . 698 × 10 − 5 S , | Y m | = I 1 V 1 = 1 a I 2 V 1 = 0 . 1 · 5 . 97 2400 = 2 . 488 × 10 − 4 S , B m = p | Y m | 2 − G 2 c = p (2 . 488 × 10 − 4 ) 2 − (3 . 698 × 10 − 5 ) 2 = 2 . 460 × 10 − 4 S , Y m = G c − jB m = 3 . 698 × 10 − 5 − j 2 . 460 × 10 − 4 = 2 . 488 × 10 − 4 ∠ − 81 . 45 ◦ S . 1

2. Neglecting the exciting admittance, determine the equivalent series impedance referred to the high- voltage side. The short circuit test data can be used to find the equivalent series impedance referred to the high- voltage side. The rated current for the high-voltage side is I 1 ,rated = S rated V 1 ,rated = 50000 2400 = 20 . 83 A . R eq 1 , Z eq 1 , and X eq 1 are then determined as follows: R eq 1 = P 1 I 2 1 ,rated = 750 20 . 83 2 = 1 . 728 Ω | Z eq 1 | = V 1 I 1 ,rated = 60 20 . 83 = 2 . 880 Ω X eq 1 = q | Z eq 1 | 2 − R 2 eq 1 = p 2 . 880 2 − 1 . 728 2 = 2 . 305 Ω Z eq 1 = R eq 1 + jX eq 1 = 1 . 728 + j 2 . 305 = 2 . 880 ∠ 53 . 14 ◦ Ω 3. Assuming equal series impedances for the primary and referred secondary, obtain an equivalent T- circuit referred to the high-voltage side.

Problem 2. A single-phase 50-kVA, 2400/240-volt, 60-Hz distribution transformer is used as a step-down transformer at the load end of a 2400-volt feeder whose series impedance is (1.0 + j2.0) Ω. The equivalent series impedance of the transformer is (1.0 + j2.5) Ω referred to the high-voltage (primary) side. The transformer is delivering rated load at 0.8 power factor lagging and at rated secondary voltage. Neglecting the transformer exciting current, determine: 1. The voltage at the transformer primary terminals. A circuit representation of the feeder and transformer is as follows: The voltage at the transformer primary terminals is denoted by ¯ V 1 . Let ¯ V 2 denote the voltage at the secondary terminals and set ¯ V 2 with reference angle 0. Then, the secondary-side current is ¯ I 2 = ¯ S rated ¯ V 2 ∗ = 50000 ∠ cos − 1 0 . 8 240 ∠ 0 ◦ ∗ = 208 . 3 ∠ − 36 . 87 ◦ A . The current referred to the primary side is ¯ I 1 = ¯ I 2 a = 208 . 3 ∠ − 36 . 87 ◦ 10 = 20 . 83 ∠ − 36 . 87 ◦ A . The voltage at the primary side of the ideal transformer is ¯ E 1 = a ¯ V 2 = 10 · 240 ∠ 0 ◦ = 2400 ∠ 0 ◦ V. Then, we get ¯ V 1 = ¯ E 1 + ¯ Z eq ¯ I 1 = 2400 ∠ 0 ◦ + (1 . 0 + j 2 . 5)(20 . 83 ∠ − 36 . 87 ◦ ) = 2448 ∠ 0 . 683 ◦ V . 2. The voltage at the sending end of the feeder. ¯ V s = ¯ V 1 + ¯ Z feed ¯ I 1 = 2448 ∠ 0 . 683 ◦ + (1 . 0 + j 2 . 0)(20 . 83 ∠ − 36 . 87 ◦ ) = 2490 ∠ 1 . 150 ◦ V 3. The real and reactive power delivered to the sending end of the feeder. The complex power at the sending end of the feeder is ¯ S s = ¯ V s ¯ I ∗ 1 = (2490 ∠ 1 . 150 ◦ )(20 . 83 ∠ − 36 . 87 ◦ ) ∗ = 40 . 861 + j 31 . 948 kVA . Therefore, the real and reactive power delivered at the sending end are P s = 40 . 861 kW , Q s = 31 . 948 kVar .

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Problem 3. A three-phase generator rated 300 MVA, 23 kV, is supplying a system load of 240 MVA and 0.9 power factor lagging at 230 kV through a 330 MVA, 23 kV Delta-230 kV Wye step-up transformer with a leakage reactance of 0.11 p.u. Use ¯ V A = 1 . 0 ∠ 0 ◦ as reference. 1. Neglecting the exciting current and choosing base values at the load of 100 MVA and 230 kV, find the phasor currents I A , I B , and I C supplied to the load in per unit (magnitude and angle). The current drawn by each phase of the load is ¯ I A = ¯ S load 1 ϕ ¯ V A = (240 / 3) × 10 6 ∠ cos − 1 0 . 9 (230 / √ 3) × 10 3 ∠ 0 ◦ ∗ = 602 . 45 ∠ − 25 . 84 ◦ A , ¯ I B = ¯ S load 1 ϕ ¯ V B = (240 / 3) × 10 6 ∠ cos − 1 0 . 9 (230 / √ 3) × 10 3 ∠ − 120 ◦ ∗ = 602 . 45 ∠ − 145 . 84 ◦ A , ¯ I C = ¯ S load 1 ϕ ¯ V C = (240 / 3) × 10 6 ∠ cos − 1 0 . 9 (230 / √ 3) × 10 3 ∠ 120 ◦ ∗ = 602 . 45 ∠ 94 . 16 ◦ A , The base quantities are S base, 3 ϕ = 100 MVA , V base,ll = 230 kV , I base = S base 1 ϕ V base,ln = 100 / 3 × 10 6 230 / √ 3 × 10 3 = 251 . 02 A . Therefore, the current drawn by each phase of the load, in per-unit, is ¯ I A,pu = ¯ I A I base = 602 . 45 ∠ − 25 . 84 ◦ 251 . 02 = 2 . 4 ∠ − 25 . 84 ◦ p.u. , ¯ I B,pu = ¯ I B I base = 602 . 45 ∠ − 145 . 84 ◦ 251 . 02 = 2 . 4 ∠ − 145 . 84 ◦ p.u. , ¯ I C,pu = ¯ I C I base = 602 . 45 ∠ 94 . 16 ◦ 251 . 02 = 2 . 4 ∠ 94 . 16 ◦ p.u. . 2. Draw the per phase equivalent circuit and compute the phasor currents I a , I b , and I c , from the generator in per unit. (Note: Take into account the phase shift of the transformer.)

The high-voltage side leads low-voltage side by 30 ◦ . So on the low-voltage side, in per-unit, ¯ I a,pu = 2 . 4 ∠ − 55 . 84 ◦ p.u. , ¯ I b,pu = 2 . 4 ∠ − 175 . 84 ◦ p.u. , ¯ I c,pu = 2 . 4 ∠ 64 . 16 ◦ p.u. . 3. Find the generator terminal voltage magnitude in kV and the total three-phase real power supplied by generator in MW. Even though the leakage reactance X l is given in per unit, this is per unit with respect to the low-voltage side transformer. So we convert this to per-unit value in the system base. X l,p.u. = X l Z base,sys Z base,low = X l    V 2 base,low S base,low V 2 base,sys S base,sys    = 0 . 11 23 2 330 23 2 100 ! = 0 . 11 100 330 = 0 . 0333 p.u. The high-voltage side leads low-voltage side by 30 ◦ . So on the low-voltage side, ¯ V a,pu = 1 ∠ − 30 ◦ p.u. . Therefore, ¯ V G,pu = ¯ I a,pu ( jX l,pu ) + ¯ V a,pu = (2 . 4 ∠ − 55 . 84 ◦ )( j 0 . 033) + 1 ∠ − 30 ◦ = 1 . 037 ∠ − 26 . 02 ◦ p.u. Then, | ¯ V G | = V base,low | ¯ V G,pu | = 23(1 . 037) = 23 . 86 kV . The complex power supplied by the generator is ¯ S G,pu = ¯ V G,pu ¯ I ∗ a,pu = (1 . 037 ∠ − 26 . 02 ◦ )(2 . 4 ∠ − 55 . 84 ◦ ) ∗ = 2 . 16 + j 1 . 24 p.u. , which corresponds to 2.16 p.u. or 216 MW real power supplied. This matches the real power absorbed (240 × 0 . 9 = 216 MW) since there are no I 2 R losses. 4. By omitting the transformer phase shift altogether, check to see whether you get the same magnitude of generator terminal voltage and real power delivered by the generator (must show work). By omitting the transformer phase shift, the high-voltage side and low-voltage side have the same phase shift. So on the low-voltage side, ¯ V a,pu = 1 ∠ 0 ◦ p.u. . Therefore, ¯ V G,pu = ¯ I a,pu ( jX l,pu ) + ¯ V a,pu = (2 . 4 ∠ − 25 . 84 ◦ )( j 0 . 033) + 1 ∠ 0 ◦ = 1 . 037 ∠ 3 . 98 ◦ p.u. Then, | ¯ V G | = V base,low | ¯ V G,pu | = 23(1 . 037) = 23 . 86 kV . ¯ S G,pu = ¯ V G,pu ¯ I ∗ a,pu = (1 . 037 ∠ 3 . 98 ◦ )(2 . 4 ∠ − 25 . 84 ◦ ) ∗ = 2 . 16 + j 1 . 24 p.u.

Problem 4. Consider the system below. Select a base of 100 MVA and 161 kV in Zone 2. The 3 ϕ line-to-neutral rated values are given: G : 150 MVA, 14.2 kV, X ′ d = 0 . 3 Ω T 1 : 250 MVA, 13.8 - 161 kV, X T 1 = 0 . 15 Ω referred to the low-voltage side T 2 : 150 MVA, 161 - 13.2 kV, X T 2 = 0 . 2 Ω referred to the low-voltage side Line: ¯ z line = j 100 Ω 1. Fill in the table below. Zone 1 Zone 2 Zone 3 Base Voltage 13 . 8 kV 161 kV 13 . 2 kV Base Power 100 MVA 100 MVA 100 MVA Base Impedance ( 13 . 8 × 10 3 ) 2 100 × 10 6 = 1 . 9044 Ω ( 161 × 10 3 ) 2 100 × 10 6 = 259 . 21 Ω ( 13 . 2 × 10 3 ) 2 100 × 10 6 = 1 . 7424 Ω 2. Fill in per-unit values for: 1) V G = 14 . 2 13 . 8 = 1 . 029 p.u. 2) X ′ d = 0 . 3 1 . 9044 = 0 . 1575 p.u. 3) X T 1 = 0 . 15 1 . 9044 = 0 . 0788 p.u. 4) ¯ z line = j 100 259 . 21 = j 0 . 3858 p.u. 5) X T 2 = 0 . 2 1 . 7424 = 0 . 1148 p.u.

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

3. Draw the per-phase diagram using per-unit quantities. Label figure with quantities from (2). φ + - φ 4. The generator is delivering 90 MW at 0.9 p.f. lagging with terminal voltage 14.2 kV. In per-unit quantities, compute generator current ¯ I G . ¯ S Gϕ = 0 . 3 + j 0 . 3 × tan ( cos − 1 0 . 9 ) = 0 . 3 + j 0 . 145 p.u. ¯ I Gϕ = ¯ S Gϕ ¯ V Gϕ ∗ = 0 . 3 + j 0 . 145 1 . 029 ∠ 0 ◦ ∗ = 0 . 3238 ∠ − 25 . 8 ◦ p.u. 5. In actual quantities, determine the line-to-neutral voltage at the load terminal. ¯ V LNϕ = ¯ V Gϕ − j ¯ I Gϕ ( X T 1 + X line + X T 2 ) V base = [1 . 029 ∠ 0 ◦ − j (0 . 3238 ∠ − 25 . 8 ◦ ) (0 . 0788 + 0 . 3858 + 0 . 1148)] × 13 . 2 = (0 . 9623 ∠ − 10 . 11 ◦ ) × 13 . 2 = 12 . 70 ∠ − 10 . 11 ◦ kV 6. In per-unit quantities and using the ZIP model, describe a possible load that fits this circuit. All types of loads are possible: constant power load constant impedance load constant current load