Electrical machinery and power system fundamentals

Instructor’s Manual to accompany Chapman Electric Machinery and Power System Fundamentals First Edition Stephen J Chapman BAE SYSTEMS Australia i Instructor’s Manual to accompany Electric Machinery and Power System Fundamentals, First Edition Copyright  2001 McGraw-Hill, Inc All rights reserved Printed in the United States of America No part of this book may be used or reproduced in any manner whatsoever without written permission, with the following exception: homework solutions may be copied for classroom use ISBN: ??? ii TABLE OF CONTENTS 10 11 12 13 Preface Mechanical and Electromagnetic Fundamentals iv Three-Phase Circuits Transformers 23 27 AC Machine Fundamentals Synchronous Machines Parallel Operation of Synchronous Generators 66 69 103 Induction Motors DC Motors Transmission Lines 114 148 178 Power System Representation and Equations 193 205 256 285 Introduction to Power-Flow Studies Symmetrical Faults Unsymmetrical Faults iii PREFACE TO THE INSTRUCTOR This Instructor’s Manual is intended to accompany the third edition of Electric Machinery and Power System Fundamentals To make this manual easier to use, it has been made self-contained Both the original problem statement and the problem solution are given for each problem in the book This structure should make it easier to copy pages from the manual for posting after problems have been assigned Many of the problems in Chapters 2, 5, 6, and require that a student read one or more values from a magnetization curve The required curves are given within the textbook, but they are shown with relatively few vertical and horizontal lines so that they will not appear too cluttered Electronic copies of the corresponding open-circuit characteristics, short-circuit characteristics, and magnetization curves as also supplied with the book They are supplied in two forms, as MATLAB MAT-files and as ASCII text files Students can use these files for electronic solutions to homework problems The ASCII files are supplied so that the information can be used with non-MATLAB software The solutions in this manual have been checked carefully, but inevitably some errors will have slipped through If you locate errors which you would like to see corrected, please feel free to contact me at the address shown below, or at my email address schapman@tpgi.com.au I greatly appreciate your input! My physical and email addresses may change from time to time, but my contact details will always be available at the book’s Web site, which is http://www.mhhe.com/engcs/electrical/chapman/ Thank you Stephen J Chapman Melbourne, Australia August 16, 2001 Stephen J Chapman 276 Orrong Road Caulfield North, VIC 3161 Australia Phone +61-3-9527-9372 iv Chapter 1: Mechanical and Electromagnetic Fundamentals 1-1 A motor’s shaft is spinning at a speed of 1800 r/min What is the shaft speed in radians per second? SOLUTION The speed in radians per second is   2π rad    = 188.5 rad/s  60 s  r  ω = (1800 r/min) 1-2 A flywheel with a moment of inertia of kg ⋅ m2 is initially at rest If a torque of N ⋅ m (counterclockwise) is suddenly applied to the flywheel, what will be the speed of the flywheel after s? Express that speed in both radians per second and revolutions per minute SOLUTION The speed in radians per second is: N ⋅m τ  (5 s ) = 25 rad/s t = kg ⋅ m J ω =α t =  The speed in revolutions per minute is:  r   60 s  n = ( 6.25 rad/s)  = 59.7 r/min  2π rad    1-3 A force of N is applied to a cylinder, as shown in Figure P1-1 What are the magnitude and direction of the torque produced on the cylinder? What is the angular acceleration α of the cylinder? SOLUTION The magnitude and the direction of the torque on this cylinder is: τ ind = rF sin θ , CCW τ ind = (0.5 m )(5 kg ⋅ m ) sin 40° = 1.607 N ⋅ m, CCW 1-4 A motor is supplying 70 N ⋅ m of torque to its load If the motor’s shaft is turning at 1500 r/min, what is the mechanical power supplied to the load in watts? In horsepower? SOLUTION The mechanical power supplied to the load is   2π rad  P = τω = (70 N ⋅ m )(1500 r/min)   = 11,000 W  60 s  r   hp  P = (11,000 W )  = 14.7 hp  746 W  1-5 A ferromagnetic core is shown in Figure P1-2 The depth of the core is cm The other dimensions of the core are as shown in the figure Find the value of the current that will produce a flux of 0.003 Wb With this current, what is the flux density at the top of the core? What is the flux density at the right side of the core? Assume that the relative permeability of the core is 1000 SOLUTION There are three regions in this core The top and bottom form one region, the left side forms a second region, and the right side forms a third region If we assume that the mean path length of the flux is in the center of each leg of the core, and if we ignore spreading at the corners of the core, then the path lengths are l1 = 2(27.5 cm) = 55 cm, l = 30 cm, and l3 = 30 cm The reluctances of these regions are: 0.55 m l l = = = 58.36 kA ⋅ t/Wb −7 µA µ r µ o A (1000)(4π × 10 H/m )(0.05 m )(0.15 m ) 0.30 m l l R2 = = = = 47.75 kA ⋅ t/Wb −7 µA µ r µ o A (1000)(4π × 10 H/m )(0.05 m )(0.10 m ) 0.30 m l l R3 = = = = 95.49 kA ⋅ t/Wb −7 µA µ r µ o A (1000)(4π × 10 H/m )(0.05 m )(0.05 m ) R1 = The total reluctance is thus RTOT = R1 + R2 + R3 = 58.36 + 47.75 + 95.49 = 201.6 kA ⋅ t/Wb and the magnetomotive force required to produce a flux of 0.003 Wb is F = φ R = ( 0.003 Wb )( 201.6 kA ⋅ t/Wb ) = 605 A ⋅ t and the required current is i= F 605 A ⋅ t = = 1.21 A N 500 t The flux density on the top of the core is B= φ A = 0.003 Wb = 0.4 T (0.15 m )(0.05 m ) The flux density on the right side of the core is B= 1-6 φ A = 0.003 Wb = 1.2 T (0.05 m )(0.05 m ) A ferromagnetic core with a relative permeability of 2000 is shown in Figure P1-3 The dimensions are as shown in the diagram, and the depth of the core is cm The air gaps on the left and right sides of the core are 0.050 and 0.070 cm, respectively Because of fringing effects, the effective area of the air gaps is percent larger than their physical size If there are 300 turns in the coil wrapped around the center leg of the core and if the current in the coil is 1.0 A, what is the flux in each of the left, center, and right legs of the core? What is the flux density in each air gap? SOLUTION This core can be divided up into five regions Let R1 be the reluctance of the left-hand portion of the core, R2 be the reluctance of the left-hand air gap, R3 be the reluctance of the right-hand portion of the core, R4 be the reluctance of the right-hand air gap, and R5 be the reluctance of the center leg of the core Then the total reluctance of the core is RTOT = R5 + (R1 + R2 )(R3 + R4 ) R1 + R2 + R3 + R4 l1 1.11 m R1 = = = 90.1 kA ⋅ t/Wb −7 µ r µ A1 (2000)(4π × 10 H/m )(0.07 m )(0.07 m ) l 0.0005 m R2 = = = 77.3 kA ⋅ t/Wb −7 µ A2 (4π × 10 H/m )(0.07 m )(0.07 m )(1.05) l3 1.11 m R3 = = = 90.1 kA ⋅ t/Wb −7 µ r µ A3 (2000)(4π × 10 H/m )(0.07 m )(0.07 m ) l 0.0007 m R4 = = = 108.3 kA ⋅ t/Wb −7 µ A4 (4π × 10 H/m )(0.07 m )(0.07 m )(1.05) l5 0.37 m R5 = = = 30.0 kA ⋅ t/Wb −7 µ r µ A5 (2000)(4π × 10 H/m )(0.07 m )(0.07 m ) The total reluctance is RTOT = R5 + (R1 + R2 )(R3 + R4 ) = 30.0 + (90.1 + 77.3)(90.1 + 108.3) = 120.8 kA ⋅ t/Wb R1 + R2 + R3 + R4 90.1 + 77.3 + 90.1 + 108.3 The total flux in the core is equal to the flux in the center leg: φ center = φ TOT = (300 t )(1.0 A ) = 0.00248 Wb F = RTOT 120.8 kA ⋅ t/Wb The fluxes in the left and right legs can be found by the “flux divider rule”, which is analogous to the current divider rule φ left = (R3 + R4 ) R1 + R2 + R3 + R4 φ TOT = (R1 + R2 ) φ right = R1 + R2 + R3 + R4 (90.1 + 108.3) 90.1 + 77.3 + 90.1 + 108.3 φ TOT = (0.00248 Wb) = 0.00135 Wb (90.1 + 77.3) 90.1 + 77.3 + 90.1 + 108.3 (0.00248 Wb) = 0.00113 Wb The flux density in the air gaps can be determined from the equation φ = BA : Bleft = φ left Aeff Bright = 1-7 = φ right Aeff 0.00135 Wb = 0.262 T (0.07 cm )(0.07 cm )(1.05) = 0.00113 Wb = 0.220 T (0.07 cm )(0.07 cm )(1.05) A two-legged core is shown in Figure P1-4 The winding on the left leg of the core (N1) has 600 turns, and the winding on the right (N2) has 200 turns The coils are wound in the directions shown in the figure If the dimensions are as shown, then what flux would be produced by currents i1 = 0.5 A and i2 = 1.00 A? Assume µr = 1000 and constant SOLUTION The two coils on this core are would so that their magnetomotive forces are additive, so the total magnetomotive force on this core is FTOT = N 1i1 + N i2 = (600 t )(0.5 A ) + (200 t )(1.0 A ) = 500 A ⋅ t The total reluctance in the core is RTOT = l µ r µ0 A = 2.60 m = 92.0 kA ⋅ t/Wb (1000)(4π × 10 H/m )(0.15 m )(0.15 m ) −7 and the flux in the core is: φ= FTOT 500 A ⋅ t = = 0.0054 Wb RTOT 92.0 kA ⋅ t/Wb 1-8 A core with three legs is shown in Figure P1-5 Its depth is cm, and there are 200 turns on the leftmost leg The relative permeability of the core can be assumed to be 1500 and constant What flux exists in each of the three legs of the core? What is the flux density in each of the legs? Assume a 4% increase in the effective area of the air gap due to fringing effects SOLUTION This core can be divided up into four regions Let R1 be the reluctance of the left-hand portion of the core, R2 be the reluctance of the center leg of the core, R3 be the reluctance of the center air gap, and R4 be the reluctance of the right-hand portion of the core Then the total reluctance of the core is RTOT = R1 + l1 (R2 + R3 )R4 R + R3 + R 1.08 m = 127.3 kA ⋅ t/Wb µ r µ A1 (1500)(4π × 10 H/m )(0.09 m )(0.05 m ) l2 0.34 m R2 = = = 24.0 kA ⋅ t/Wb −7 µ r µ A2 (1500)(4π × 10 H/m )(0.15 m )(0.05 m ) l 0.0004 m R3 = = = 40.8 kA ⋅ t/Wb −7 µ A3 (4π × 10 H/m )(0.15 m )(0.05 m )(1.04 ) l4 1.08 m R4 = = = 127.3 kA ⋅ t/Wb −7 µ r µ A4 (1500)(4π × 10 H/m )(0.09 m )(0.05 m ) R1 = = −7 The total reluctance is RTOT = R1 + (R2 + R3 )R4 R + R3 + R = 127.3 + (24.0 + 40.8)127.3 24.0 + 40.8 + 127.3 = 170.2 kA ⋅ t/Wb The total flux in the core is equal to the flux in the left leg: φ left = φ TOT = (200 t )(2.0 A ) = 0.00235 Wb F = RTOT 170.2 kA ⋅ t/Wb The fluxes in the center and right legs can be found by the “flux divider rule”, which is analogous to the current divider rule φ center = φ right = R4 127.3 (0.00235 Wb) = 0.00156 Wb φ TOT = R2 + R3 + R4 24.0 + 40.8 + 127.3 R2 + R3 24.0 + 40.8 (0.00235 Wb) = 0.00079 Wb φ TOT = 24.0 + 40.8 + 127.3 R + R3 + R The flux density in the legs can be determined from the equation φ = BA : Bleft = φ left A Bcenter = Bright = 1-9 = 0.00235 Wb = 0.522 T (0.09 cm )(0.05 cm ) φ center A φ left A = = 0.00156 Wb = 0.208 T (0.15 cm )(0.05 cm ) 0.00079 Wb = 0.176 T (0.09 cm )(0.05 cm ) A wire is shown in Figure P1-6 which is carrying 2.0 A in the presence of a magnetic field Calculate the magnitude and direction of the force induced on the wire SOLUTION The force on this wire can be calculated from the equation F = i (l × B ) = ilB = (2 A )(1 m )(0.35 T ) = 0.7 N, into the page 1-10 The wire shown in Figure P1-7 is moving in the presence of a magnetic field With the information given in the figure, determine the magnitude and direction of the induced voltage in the wire SOLUTION The induced voltage on this wire can be calculated from the equation shown below The voltage on the wire is positive downward because the vector quantity v × B points downward eind = ( v × B ) ⋅ l = vBl cos 45° = ( m/s )( 0.2 T )( 0.75 m ) cos 45° = 0.636 V, positive down c 0.911/ 118.97 0.000/ 0.00 One Three Three a b c 0.000/ 14.04 0.769/-111.04 0.670/ 110.64 Four Four a b c 0.433/ 12.04 0.728/-109.02 0.626/ 107.24 1.180/ 105.69 0.400/ -41.57 0.476/-107.97 1.180/ -74.31 0.400/ 138.43 0.476/ 72.03 a b c a b c 1.180/ 105.69 0.400/ -41.57 0.476/-107.97 2.137/ 114.64 1.088/ -78.21 1.103/ -52.69 4.737/ -68.56 0.000/ 0.00 0.000/ 0.00 Two a b c a b c 0.000/ 0.000/ 0.000/ 0.00 0.00 0.00 Three a 2.137/ -65.36 b 1.088/ 101.79 c 1.103/ 127.31 |==============================================================================| The subtransient, transient, and steady-state fault currents are given in per-unit above The actual fault currents are found by multiplying by the base current in Region 2: I f , A = I f , A,pu I base,2 = ( 4.737∠ − 68.6°)( 524.9 A ) = 2487∠ − 68.6° A I f ,B = I f ,B ,pu I base,2 = 0∠0.0° A I f ,C = I f ,C ,pu I base,2 = 0∠0.0° A The subtransient generator current will be the same as the current flowing from Bus to Bus This current is given in per-unit above The actual generator current is found by multiplying by the base current in Region 1: I12, A = I12, A,pu I base,2 = (1.180∠ − 74.3°)( 4184 A ) = 4937∠ − 74.3° A I12,B = I12,B ,pu I base,2 = ( 0.400∠138.4°)( 4184 A ) = 1674∠ − 138.4° A I12,C = I12,C ,pu I base,2 = ( 0.476∠72.0°)( 4184 A ) = 1992∠72.0° A The subtransient motor current will be the same as the current flowing from Bus to Bus This current is given in per-unit above The actual motor current is found by multiplying by the base current in Region 3: I 43, A = I 43, A,pu I base,2 = ( 2.137∠ − 65.4°)( 4374 A ) = 9347∠ − 65.4° A I 43,B = I 43,B ,pu I base,2 = (1.088∠101.8°)( 4374 A ) = 4759∠101.8° A I 43,C = I 43,C ,pu I base,2 = (1.103∠127.3°)( 4374 A ) = 4825∠127.3° A 13-9 The Ozzie Outback Electric Power (OOEP) Company maintains the power system shown in Figure P133 This power system contains five busses, with generators attached to two of them and loads attached to the remaining ones The power system has six transmission lines connecting the busses together, with the characteristics shown in Table 13-1 There are generators at busses Bunya and Mulga, and loads at all other busses The characteristics of the two generators are shown in Table 13-2 Note that the neutrals of 319 the two generators are both grounded through an inductive reactance of 0.60 pu Note that all values are in per-unit on a 100 MVA base (a) Assume that a single-line-to-ground fault occurs at the Mallee bus What is the per-unit fault current during the subtransient period? during the transient period? (b) Assume that the neutrals of the generators are now solidly grounded, and that a single-line-toground fault occurs at the Mallee bus What is the per-unit fault current during the subtransient period now? during the transient period? (c) Assume that a line-to-line fault occurs at the Mallee bus What is the per-unit fault current during the subtransient period? during the transient period? (d) How much does the per-unit fault current change for a line-to-line fault if the generators are either solidly grounded or grounded through an impedance? From Bunya Mulga Bunya Myall Myall Mallee Table 13-1: Transmission lines in the OOEP system To RSE , pu X , pu X , pu Mulga Satinay Myall Satinay Mallee Satinay Name Bus R (pu) G1 G2 Bunya Mulga 0.02 0.01 0.011 0.007 0.007 0.007 0.011 0.011 0.051 0.035 0.035 0.035 0.051 0.051 0.051 0.035 0.035 0.035 0.051 0.051 Table 13-2: Generators in the OOEP system XS (pu) X' (pu) X" (pu) X2 (pu) 1.5 1.0 0.35 0.25 0.20 0.12 0.20 0.12 X , pu 0.090 0.090 0.060 0.060 0.110 0.090 Xg0 (pu) XN (pu) 0.10 0.05 0.60 0.60 G2 Mulga G1 Bunya Satinay Myall Mallee Figure P13-3 The Ozzie Outback Electric Power Company system SOLUTION (a) Note that the impedance to ground for the two generators will be equally to Z g + 3Z N The input file required for program faults is: 320 % File describing a possible fault at bus Mallee on the % Ozzie Outback Electric Power system % % System data has the form: %SYSTEM name baseMVA SYSTEM OOEP_p13a 100 % % Bus data has the form: %BUS name volts BUS Bunya 1.00 BUS Mulga 1.00 BUS Mallee 1.00 BUS Myall 1.00 BUS Satinay 1.00 % % Transmission line data has the form: %LINE from to Rse Xse Gsh Bsh X0 Vis LINE Bunya Mulga 0.011 0.051 0.000 0.000 0.090 LINE Mulga Satinay 0.007 0.035 0.000 0.000 0.090 LINE Bunya Myall 0.007 0.035 0.000 0.000 0.060 LINE Myall Satinay 0.007 0.035 0.000 0.000 0.060 LINE Myall Mallee 0.011 0.051 0.000 0.000 0.110 LINE Mallee Satinay 0.011 0.051 0.000 0.000 0.090 % % Generator data has the form: %GENERATOR bus R Xs Xp Xpp X2 X0 GENERATOR Bunya 0.02 1.5 0.35 0.20 0.20 1.90 GENERATOR Mulga 0.01 1.0 0.25 0.12 0.00 1.85 % % type data has the form: %FAULT bus Calc Type Calc_time (0=all;1=sub;2=trans;3=ss) FAULT Mallee SLG The resulting fault currents are shown below >> faults prob_13_9a_fault Input summary statistics: 32 lines in system file SYSTEM lines BUS lines LINE lines GENERATOR lines MOTOR lines TYPE lines Results for Case OOEP_p13a Single Line-to-Ground Fault at Bus Mallee Calculating Subtransient Currents |=====================Bus Information=============|======Line Information======| Bus P Volts / angle Amps / angle | To | P Amps / angle | no Name h (pu) (deg) (pu) (deg) | Bus | h (pu) (deg) | |==============================================================================| Bunya a 0.137/ -6.64 0.000/ 0.00 b 1.515/-145.69 0.000/ 0.00 c 1.471/ 147.65 0.000/ 0.00 Mulga a 0.333/ 97.32 321 Myall Mulga a b c 0.156/ -6.31 1.514/-146.04 1.471/ 147.96 0.000/ 0.000/ 0.000/ Satinay Mallee a b c 0.000/-176.42 1.546/-146.58 1.502/ 148.59 Satinay Myall a b c 0.085/ -6.71 1.527/-146.09 1.483/ 148.07 0.000/ 0.000/ 0.000/ Satinay Mallee Satinay a b c 0.086/ -7.01 1.530/-146.28 1.487/ 148.25 0.333/ 0.188/ 0.176/ 1.377/ 0.142/ 0.134/ a b c a b c 1.193/ 92.18 0.021/ 0.45 0.020/-157.29 1.319/ 92.72 0.021/-179.55 0.020/ 22.71 a b c a b c a b c 1.136/ 92.12 0.142/ 63.86 0.134/ 133.15 0.057/ 93.44 0.134/-108.17 0.129/ -55.24 1.193/ -87.82 0.021/-179.55 0.020/ 22.71 -82.68 45.98 154.31 -87.25 63.86 133.15 0.00 0.00 0.00 Bunya a b c a b c 2.512/ -87.54 0.000/ 180.00 0.000/ 180.00 Myall 0.188/-134.02 0.176/ -25.69 1.136/ -87.88 0.142/-116.14 0.134/ -46.85 0.00 0.00 0.00 Bunya b c a b c 0.000/ 0.000/ 0.000/ 0.00 0.00 0.00 Mulga a 1.377/ 92.75 b 0.142/-116.14 c 0.134/ -46.85 Myall a 0.057/ -86.56 b 0.134/ 71.83 c 0.129/ 124.76 Mallee a 1.319/ -87.28 b 0.021/ 0.45 c 0.020/-157.29 |==============================================================================| Results for Case OOEP_p13a Single Line-to-Ground Fault at Bus Mallee Calculating Transient Currents |=====================Bus Information=============|======Line Information======| Bus P Volts / angle Amps / angle | To | P Amps / angle | 322 no Name h (pu) (deg) (pu) (deg) | Bus | h (pu) (deg) | |==============================================================================| Bunya a 0.130/ -6.78 0.000/ 0.00 b 1.431/-145.82 0.000/ 0.00 c 1.390/ 147.50 0.000/ 0.00 Mulga a 0.297/ 97.37 b 0.184/-139.40 c 0.175/ -20.80 Myall a 1.082/ -88.00 b 0.134/-120.31 c 0.129/ -42.84 Mulga a 0.147/ -6.43 0.000/ 0.00 b 1.429/-146.18 0.000/ 0.00 c 1.389/ 147.84 0.000/ 0.00 Bunya a 0.297/ -82.63 b 0.184/ 40.60 c 0.175/ 159.20 Satinay a 1.290/ -87.39 b 0.134/ 59.69 c 0.129/ 137.16 Mallee a 0.000/ -90.00 2.372/ -87.67 b 1.460/-146.71 0.000/ 0.00 c 1.418/ 148.46 0.000/ 0.00 Myall a 1.129/ 92.05 b 0.022/ -2.48 c 0.021/-156.13 Satinay a 1.243/ 92.58 b 0.022/ 177.52 c 0.021/ 23.87 Myall a 0.081/ -6.84 0.000/ 0.00 b 1.442/-146.23 0.000/ 0.00 c 1.400/ 147.93 0.000/ 0.00 Bunya a 1.082/ 92.00 b 0.134/ 59.69 c 0.129/ 137.16 Satinay a 0.047/ 93.32 b 0.125/-111.48 c 0.122/ -52.07 Mallee a 1.129/ -87.95 b 0.022/ 177.52 c 0.021/ 23.87 Satinay a 0.081/ -7.14 0.000/ 0.00 b 1.445/-146.41 0.000/ 0.00 c 1.404/ 148.12 0.000/ 0.00 Mulga a 1.290/ 92.61 b 0.134/-120.31 c 0.129/ -42.84 Myall a 0.047/ -86.68 b 0.125/ 68.52 c 0.122/ 127.93 Mallee a 1.243/ -87.42 b 0.022/ -2.48 c 0.021/-156.13 |==============================================================================| 323 Results for Case OOEP_p13a Single Line-to-Ground Fault at Bus Mallee Calculating Steady-State Currents |=====================Bus Information=============|======Line Information======| Bus P Volts / angle Amps / angle | To | P Amps / angle | no Name h (pu) (deg) (pu) (deg) | Bus | h (pu) (deg) | |==============================================================================| Bunya a 0.095/ -7.41 0.000/ 0.00 b 1.053/-146.44 0.000/ 0.00 c 1.023/ 146.89 0.000/ 0.00 Mulga a 0.226/ 96.85 b 0.132/-137.19 c 0.126/ -24.42 Myall a 0.792/ -88.66 b 0.098/-118.63 c 0.094/ -45.86 Mulga a 0.108/ -7.04 0.000/ 0.00 b 1.052/-146.79 0.000/ 0.00 c 1.022/ 147.22 0.000/ 0.00 Bunya a 0.226/ -83.15 b 0.132/ 42.81 c 0.126/ 155.58 Satinay a 0.953/ -87.98 b 0.098/ 61.37 c 0.094/ 134.14 Mallee a 0.000/ 180.00 1.746/ -88.29 b 1.074/-147.33 0.000/ 0.00 c 1.043/ 147.84 0.000/ 0.00 Myall a 0.830/ 91.42 b 0.015/ -1.95 c 0.015/-157.88 Satinay a 0.916/ 91.97 b 0.015/ 178.05 c 0.015/ 22.12 Myall a 0.059/ -7.47 0.000/ 0.00 b 1.061/-146.85 0.000/ 0.00 c 1.030/ 147.31 0.000/ 0.00 Bunya a 0.792/ 91.34 b 0.098/ 61.37 c 0.094/ 134.14 Satinay a 0.037/ 93.19 b 0.092/-110.25 c 0.090/ -54.59 Mallee a 0.830/ -88.58 b 0.015/ 178.05 c 0.015/ 22.12 Satinay a 0.060/ -7.75 0.000/ 0.00 b 1.063/-147.03 0.000/ 0.00 c 1.033/ 147.50 0.000/ 0.00 Mulga a 0.953/ 92.02 b 0.098/-118.63 c 0.094/ -45.86 Myall a 0.037/ -86.81 b 0.092/ 69.75 c 0.090/ 125.41 324 Mallee a 0.916/ -88.03 b 0.015/ -1.95 c 0.015/-157.88 |==============================================================================| The subtransient fault current at the Mallee bus for a single-line-to-ground fault is I′′f = 2.512∠ − 87.5 pu , and the transient fault current at the Mallee bus for a single-line-to-ground fault is I′f = 2.372∠ − 87.7 pu (b) If the generators are solidly grounded, their zero sequence impedances will become just Z g This smaller zero sequence impedance will result in a much higher fault current for a single-line-to-ground fault The input file required for program faults is: % File describing a possible fault at bus Mallee on the % Ozzie Outback Electric Power system % % System data has the form: %SYSTEM name baseMVA SYSTEM OOEP_p13b 100 % % Bus data has the form: %BUS name volts BUS Bunya 1.00 BUS Mulga 1.00 BUS Mallee 1.00 BUS Myall 1.00 BUS Satinay 1.00 % % Transmission line data has the form: %LINE from to Rse Xse Gsh Bsh X0 Vis LINE Bunya Mulga 0.011 0.051 0.000 0.000 0.090 LINE Mulga Satinay 0.007 0.035 0.000 0.000 0.090 LINE Bunya Myall 0.007 0.035 0.000 0.000 0.060 LINE Myall Satinay 0.007 0.035 0.000 0.000 0.060 LINE Myall Mallee 0.011 0.051 0.000 0.000 0.110 LINE Mallee Satinay 0.011 0.051 0.000 0.000 0.090 % % Generator data has the form: %GENERATOR bus R Xs Xp Xpp X2 X0 GENERATOR Bunya 0.02 1.5 0.35 0.20 0.20 0.10 GENERATOR Mulga 0.01 1.0 0.25 0.12 0.00 0.05 % % type data has the form: %FAULT bus Calc Type Calc_time (0=all;1=sub;2=trans;3=ss) FAULT Mallee SLG The resulting fault currents are shown below >> faults prob_13_9b_fault Input summary statistics: 32 lines in system file SYSTEM lines BUS lines LINE lines GENERATOR lines MOTOR lines 325 TYPE lines Results for Case OOEP_p13b Single Line-to-Ground Fault at Bus Mallee Calculating Subtransient Currents |=====================Bus Information=============|======Line Information======| Bus P Volts / angle Amps / angle | To | P Amps / angle | no Name h (pu) (deg) (pu) (deg) | Bus | h (pu) (deg) | |==============================================================================| Bunya a 0.544/ 0.80 0.000/ 0.00 b 0.863/-127.07 0.000/ 0.00 c 0.813/ 122.57 0.000/ 0.00 Mulga a 1.565/ 103.75 b 0.621/-140.86 c 0.645/ -0.41 Myall a 4.489/ -80.38 b 0.483/-113.77 c 0.476/ -30.88 Mulga a 0.641/ 1.13 0.000/ 0.00 b 0.838/-128.38 0.000/ 0.00 c 0.784/ 123.31 0.000/ 0.00 Bunya a 1.565/ -76.25 b 0.621/ 39.14 c 0.645/ 179.59 Satinay a 5.664/ -79.89 b 0.483/ 66.23 c 0.476/ 149.12 Mallee a 0.000/-172.87 10.153/ -80.11 b 0.966/-133.70 0.000/ 0.00 c 0.868/ 131.99 0.000/ 0.00 Myall a 4.797/ 99.63 b 0.086/ -7.81 c 0.093/-136.64 Satinay a 5.356/ 100.13 b 0.086/ 172.19 c 0.093/ 43.36 Myall a 0.342/ 0.72 0.000/ 0.00 b 0.897/-130.03 0.000/ 0.00 c 0.827/ 126.50 0.000/ 0.00 Bunya a 4.489/ 99.62 b 0.483/ 66.23 c 0.476/ 149.12 Satinay a 0.309/ 99.72 b 0.467/-103.54 c 0.459/ -42.12 Mallee a 4.797/ -80.37 b 0.086/ 172.19 c 0.093/ 43.36 Satinay a 0.349/ 0.41 0.000/ 0.00 b 0.905/-131.17 0.000/ 0.00 c 0.828/ 127.90 0.000/ 0.00 Mulga a 5.664/ 100.11 b 0.483/-113.77 c 0.476/ -30.88 Myall a 0.309/ -80.28 326 b 0.467/ 76.46 c 0.459/ 137.88 Mallee a 5.356/ -79.87 b 0.086/ -7.81 c 0.093/-136.64 |==============================================================================| Results for Case OOEP_p13b Single Line-to-Ground Fault at Bus Mallee Calculating Transient Currents |=====================Bus Information=============|======Line Information======| Bus P Volts / angle Amps / angle | To | P Amps / angle | no Name h (pu) (deg) (pu) (deg) | Bus | h (pu) (deg) | |==============================================================================| Bunya a 0.442/ -1.09 0.000/ 0.00 b 0.700/-128.94 0.000/ 0.00 c 0.660/ 120.68 0.000/ 0.00 Mulga a 1.205/ 101.96 b 0.538/-148.25 c 0.569/ 1.86 Myall a 3.664/ -82.25 b 0.395/-120.39 c 0.396/ -28.27 Mulga a 0.517/ -0.76 0.000/ 0.00 b 0.677/-130.28 0.000/ 0.00 c 0.633/ 121.43 0.000/ 0.00 Bunya a 1.205/ -78.04 b 0.538/ 31.75 c 0.569/-178.14 Satinay a 4.550/ -81.79 b 0.395/ 59.61 c 0.396/ 151.73 Mallee a 0.000/ 180.00 8.214/ -82.00 b 0.781/-135.58 0.000/ 180.00 c 0.702/ 130.11 0.000/ 180.00 Myall a 3.889/ 97.74 b 0.078/ -10.81 c 0.084/-138.69 Satinay a 4.324/ 98.24 b 0.078/ 169.19 c 0.084/ 41.31 Myall a 0.277/ -1.16 0.000/ 0.00 b 0.726/-131.91 0.000/ 0.00 c 0.670/ 124.60 0.000/ 0.00 Bunya a 3.664/ 97.75 b 0.395/ 59.61 c 0.396/ 151.73 Satinay a 0.225/ 97.72 b 0.376/-109.12 c 0.375/ -40.32 Mallee a 3.889/ -82.26 b 0.078/ 169.19 c 0.084/ 41.31 Satinay a 0.282/ -1.47 0.000/ 0.00 327 b c 0.732/-133.07 0.670/ 126.02 0.000/ 0.000/ 0.00 0.00 Mulga a 4.550/ 98.21 b 0.395/-120.39 c 0.396/ -28.27 Myall a 0.225/ -82.28 b 0.376/ 70.88 c 0.375/ 139.68 Mallee a 4.324/ -81.76 b 0.078/ -10.81 c 0.084/-138.69 |==============================================================================| Results for Case OOEP_p13b Single Line-to-Ground Fault at Bus Mallee Calculating Steady-State Currents |=====================Bus Information=============|======Line Information======| Bus P Volts / angle Amps / angle | To | P Amps / angle | no Name h (pu) (deg) (pu) (deg) | Bus | h (pu) (deg) | |==============================================================================| Bunya a 0.197/ -5.54 0.000/ 0.00 b 0.312/-133.40 0.000/ 0.00 c 0.294/ 116.23 0.000/ 0.00 Mulga a 0.554/ 97.64 b 0.230/-149.89 c 0.243/ -5.19 Myall a 1.629/ -86.74 b 0.174/-122.18 c 0.175/ -35.35 Mulga a 0.231/ -5.19 0.000/ 0.00 b 0.303/-134.71 0.000/ 0.00 c 0.283/ 116.99 0.000/ 0.00 Bunya a 0.554/ -82.36 b 0.230/ 30.11 c 0.243/ 174.81 Satinay a 2.041/ -86.20 b 0.174/ 57.82 c 0.175/ 144.65 Mallee a 0.000/ -90.00 3.670/ -86.44 b 0.349/-140.03 0.000/ 0.00 c 0.314/ 125.66 0.000/ 0.00 Myall a 1.736/ 93.29 b 0.033/ -15.05 c 0.035/-143.43 Satinay a 1.934/ 93.81 b 0.033/ 164.95 c 0.035/ 36.57 Myall a 0.124/ -5.61 0.000/ 0.00 b 0.324/-136.36 0.000/ 0.00 c 0.299/ 120.16 0.000/ 0.00 Bunya a 1.629/ 93.26 b 0.174/ 57.82 c 0.175/ 144.65 Satinay a 0.107/ 93.71 328 Mallee Satinay a b c 0.126/ -5.91 0.327/-137.50 0.299/ 121.58 0.000/ 0.000/ 0.000/ b c a b c 0.168/-111.43 0.167/ -46.90 1.736/ -86.71 0.033/ 164.95 0.035/ 36.57 0.00 0.00 0.00 Mulga a 2.041/ 93.80 b 0.174/-122.18 c 0.175/ -35.35 Myall a 0.107/ -86.29 b 0.168/ 68.57 c 0.167/ 133.10 Mallee a 1.934/ -86.19 b 0.033/ -15.05 c 0.035/-143.43 |==============================================================================| The subtransient fault current at the Mallee bus for a single-line-to-ground fault is I′′f = 10.153∠ − 80.1 pu , and the transient fault current at the Mallee bus for a single-line-to-ground fault is I′f = 8.214∠ − 82.0 pu These numbers are about times larger than before Clearly, the reactances in the neutrals of the generators help to reduce the fault current (c) The input file required for a line-to-line fault is: % File describing a possible fault at bus % Ozzie Outback Electric Power system % % System data has the form: %SYSTEM name baseMVA SYSTEM OOEP_p13c 100 % % Bus data has the form: %BUS name volts BUS Bunya 1.00 BUS Mulga 1.00 BUS Mallee 1.00 BUS Myall 1.00 BUS Satinay 1.00 % % Transmission line data has the form: %LINE from to Rse Xse LINE Bunya Mulga 0.011 0.051 LINE Mulga Satinay 0.007 0.035 LINE Bunya Myall 0.007 0.035 LINE Myall Satinay 0.007 0.035 LINE Myall Mallee 0.011 0.051 LINE Mallee Satinay 0.011 0.051 % % Generator data has the form: %GENERATOR bus R Xs Xp GENERATOR Bunya 0.02 1.5 0.35 GENERATOR Mulga 0.01 1.0 0.25 % % type data has the form: 329 Mallee on the Gsh 0.000 0.000 0.000 0.000 0.000 0.000 Bsh 0.000 0.000 0.000 0.000 0.000 0.000 X0 Vis 0.090 0.090 0.060 0.060 0.110 0.090 Xpp 0.20 0.12 X2 0.20 0.00 X0 1.90 1.85 %FAULT FAULT bus Mallee Calc Type LL Calc_time (0=all;1=sub;2=trans;3=ss) The resulting fault currents are shown below >> faults prob_13_9c_fault Input summary statistics: 32 lines in system file SYSTEM lines BUS lines LINE lines GENERATOR lines MOTOR lines TYPE lines Results for Case OOEP_p13c Line-to-Line Fault at Bus Mallee Calculating Subtransient Currents |=====================Bus Information=============|======Line Information======| Bus P Volts / angle Amps / angle | To | P Amps / angle | no Name h (pu) (deg) (pu) (deg) | Bus | h (pu) (deg) | |==============================================================================| Bunya a 0.656/ -8.08 0.000/ 0.00 b 0.480/-133.36 0.000/ 0.00 c 0.545/ 125.92 0.000/ 0.00 Mulga a 1.178/ -69.30 b 1.811/ 35.43 c 1.893/ 178.44 Myall a 0.627/ -70.10 b 4.146/-173.60 c 4.046/ 15.07 Mulga a 0.597/ -9.77 0.000/ 0.00 b 0.525/-123.85 0.000/ 0.00 c 0.613/ 118.87 0.000/ 0.00 Bunya a 1.178/ 110.70 b 1.811/-144.57 c 1.893/ -1.56 Satinay a 0.627/ 109.90 b 5.916/-164.72 c 5.999/ 9.29 Mallee a 0.626/ -8.88 0.000/ 0.00 b 0.313/ 171.12 10.032/-168.38 c 0.313/ 171.12 10.032/ 11.62 Myall a 0.160/ 110.54 b 4.792/ 10.44 c 4.767/-167.67 Satinay a 0.160/ -69.46 b 5.242/ 12.70 c 5.267/-169.02 Myall a 0.634/ -8.66 0.000/ 0.00 b 0.376/-147.54 0.000/ 0.00 c 0.429/ 136.23 0.000/ 0.00 Bunya a 0.627/ 109.90 b 4.146/ 6.40 c 4.046/-164.93 Satinay a 0.467/ -70.32 330 Mallee Satinay a b c 0.618/ -9.11 0.381/-143.73 0.443/ 133.22 0.000/ 0.000/ 0.000/ b c a b c 0.719/ 34.41 0.751/ 177.42 0.160/ -69.46 4.792/-169.56 4.767/ 12.33 0.00 0.00 0.00 Mulga a 0.627/ -70.10 b 5.916/ 15.28 c 5.999/-170.71 Myall a 0.467/ 109.68 b 0.719/-145.59 c 0.751/ -2.58 Mallee a 0.160/ 110.54 b 5.242/-167.30 c 5.267/ 10.98 |==============================================================================| Results for Case OOEP_p13c Line-to-Line Fault at Bus Mallee Calculating Transient Currents |=====================Bus Information=============|======Line Information======| Bus P Volts / angle Amps / angle | To | P Amps / angle | no Name h (pu) (deg) (pu) (deg) | Bus | h (pu) (deg) | |==============================================================================| Bunya a 0.470/ -11.37 0.000/ 0.00 b 0.344/-136.60 0.000/ 0.00 c 0.391/ 122.55 0.000/ 0.00 Mulga a 0.931/ -73.35 b 1.236/ 35.47 c 1.286/ 172.17 Myall a 0.496/ -74.15 b 3.000/-177.32 c 2.927/ 12.18 Mulga a 0.423/ -13.17 0.000/ 0.00 b 0.372/-127.22 0.000/ 0.00 c 0.435/ 115.51 0.000/ 0.00 Bunya a 0.931/ 106.65 b 1.236/-144.53 c 1.286/ -7.83 Satinay a 0.496/ 105.85 b 4.175/-167.70 c 4.235/ 5.58 Mallee a 0.446/ -12.23 0.000/ 0.00 b 0.223/ 167.77 7.151/-171.72 c 0.223/ 167.77 7.151/ 8.28 Myall a 0.126/ 106.49 b 3.427/ 7.00 c 3.408/-170.90 Satinay a 0.126/ -73.51 b 3.726/ 9.45 c 3.743/-172.47 Myall a 0.453/ -11.99 0.000/ 0.00 331 b c 0.268/-150.82 0.307/ 132.85 0.000/ 0.000/ 0.00 0.00 Bunya Satinay Mallee Satinay a b c 0.440/ -12.47 0.271/-147.13 0.315/ 129.91 0.000/ 0.000/ 0.000/ a b c a b c a b c 0.496/ 105.85 3.000/ 2.68 2.927/-167.82 0.370/ -74.37 0.490/ 34.44 0.510/ 171.15 0.126/ -73.51 3.427/-173.00 3.408/ 9.10 0.00 0.00 0.00 Mulga a 0.496/ -74.15 b 4.175/ 12.30 c 4.235/-174.42 Myall a 0.370/ 105.63 b 0.490/-145.56 c 0.510/ -8.85 Mallee a 0.126/ 106.49 b 3.726/-170.55 c 3.743/ 7.53 |==============================================================================| Results for Case OOEP_p13c Line-to-Line Fault at Bus Mallee Calculating Steady-State Currents |=====================Bus Information=============|======Line Information======| Bus P Volts / angle Amps / angle | To | P Amps / angle | no Name h (pu) (deg) (pu) (deg) | Bus | h (pu) (deg) | |==============================================================================| Bunya a 0.163/ -16.83 0.000/ 0.00 b 0.120/-142.09 0.000/ 0.00 c 0.136/ 117.13 0.000/ 0.00 Mulga a 0.306/ -78.74 b 0.443/ 28.25 c 0.459/ 168.64 Myall a 0.163/ -79.55 b 1.036/ 177.44 c 1.012/ 6.46 Mulga a 0.148/ -18.52 0.000/ 0.00 b 0.130/-132.59 0.000/ 0.00 c 0.152/ 110.13 0.000/ 0.00 Bunya a 0.306/ 101.26 b 0.443/-151.75 c 0.459/ -11.36 Satinay a 0.163/ 100.45 b 1.464/-173.28 c 1.484/ 0.42 Mallee a 0.156/ -17.63 0.000/ 0.00 b 0.078/ 162.37 2.493/-177.13 c 0.078/ 162.37 2.493/ 2.87 Myall a 0.042/ 101.10 332 Satinay Myall a b c 0.158/ -17.41 0.093/-156.27 0.107/ 127.46 0.000/ 0.000/ 0.000/ Satinay Mallee Satinay a b c 0.154/ -17.86 0.094/-152.50 0.110/ 124.49 0.000/ 0.000/ 0.000/ 1.192/ 1.64 1.186/-176.38 0.042/ -78.90 1.301/ 4.00 1.307/-177.81 a b c a b c a b c 0.163/ 100.45 1.036/ -2.56 1.012/-173.54 0.121/ -79.77 0.176/ 27.23 0.182/ 167.62 0.042/ -78.90 1.192/-178.36 1.186/ 3.62 0.00 0.00 0.00 Bunya b c a b c 0.00 0.00 0.00 Mulga a 0.163/ -79.55 b 1.464/ 6.72 c 1.484/-179.58 Myall a 0.121/ 100.23 b 0.176/-152.77 c 0.182/ -12.38 Mallee a 0.042/ 101.10 b 1.301/-176.00 c 1.307/ 2.19 |==============================================================================| The subtransient fault current at the Mallee bus for a line-to-line fault is I′′f = 10.032∠ − 168.4 pu , and the transient fault current at the Mallee bus for a single-line-to-ground fault is I′f = 7.151∠ − 171.7 pu (d) The line-to-line fault current is not affected by whether or not the neutrals of the generators are solidly grounded, because there are no zero-phase components in a line-to-line fault 333 [...]... the following questions about this power system 16 (a) Assume that the switch shown in the figure is open, and calculate the current I, the power factor, and the real, reactive, and apparent power being supplied by the source (b) Assume that the switch shown in the figure is closed, and calculate the current I, the power factor, and the real, reactive, and apparent power being supplied by the source... 3 ( 480 V ) cos ( 30.45°) θ = tan −1 22 P , where θ = tan −1 QG PG 3 VL cos θ (e) The real and reactive power supplied by the generator when the switch is closed is just the sum of the real and reactive powers of Loads 1, 2, and 3 The powers of Loads 1 and 2 have already been calculated The real and reactive power of Load 3 are: P3 = 80 kW ( ) Q3 = P tan θ = P tan cos−1 PF = ( 80 kW )  tan ( −31.79°)... containing a single 480 V generator and three loads Assume that the transmission lines in this power system are lossless, and answer the following questions (a) Assume that Load 1 is Y-connected What are the phase voltage and currents in that load? (b) Assume that Load 2 is ∆-connected What are the phase voltage and currents in that load? (c) What real, reactive, and apparent power does the generator supply... transmission lines? (c) Find the real and reactive powers supplied to each load (d) Find the real and reactive power losses in the transmission line (e) Find the real power, reactive power, and power factor supplied by the generator SOLUTION To solve this problem, first convert the two deltas to equivalent wyes, and get the per-phase equivalent circuit 0.090 Ω j0.16 Ω + Line 277∠0° V Zφ 2 Zφ 1 + - Vφ ,load... small 480-V distribution system Assume that the lines in the system have zero impedance (a) If the switch shown is open, find the real, reactive, and apparent powers in the system Find the total current supplied to the distribution system by the utility (b) Repeat part (a) with the switch closed What happened to the total current supplied? Why? SOLUTION (a) With the switch open, the power supplied to each... = 208∠ − 30° V and the current flowing through the load is I = 5∠15° A (a) Calculate the complex power S consumed by this load (b) Is this load inductive or capacitive? (c) Calculate the power factor of this load? (d) Calculate the reactive power consumed or supplied by this load Does the load consume reactive power from the source or supply it to the source? SOLUTION (a) The complex power S consumed... three-phase power system with two loads The ∆-connected generator is producing a line voltage of 480 V, and the line impedance is 0.09 + j0.16 Ω Load 1 is Y-connected, with a phase impedance of 2.5∠36.87° Ω and load 2 is ∆-connected, with a phase impedance of 5∠-20° Ω 19 (a) What is the line voltage of the two loads? (b) What is the voltage drop on the transmission lines? (c) Find the real and reactive powers... W + 32.0 W A single-phase power system is shown in Figure P3-1 The power source feeds a 100-kVA 14/2.4-kV transformer through a feeder impedance of 38.2 + j140 Ω The transformer’s equivalent series impedance referred to its low-voltage side is 0.12 + j0.5 Ω The load on the transformer is 90 kW at 0.85 PF lagging and 2300 V (a) What is the voltage at the power source of the system? (b) What is the voltage... 480 V The phase current can be derived from the equation S = 3Vφ I φ as follows: Iφ 2 = S 80 kVA = = 55.56 A 3Vφ 3 ( 480 V ) (c) The real and reactive power supplied by the generator when the switch is open is just the sum of the real and reactive powers of Loads 1 and 2 P1 = 100 kW ( ) Q1 = P tan θ = P tan cos −1 PF = (100 kW )( tan 25.84°) = 48.4 kvar P2 = S cos θ = ( 80 kVA )( 0.8) = 64 kW Q2 = S... cos 2ω t ) + VI sin θ sin 2ω t 18 Chapter 2: Three-Phase Circuits 2-1 Three impedances of 4 + j3 Ω are ∆-connected and tied to a three-phase 208-V power line Find I φ , I L , P, Q, S, and the power factor of this load SOLUTION IL + 240 V Iφ Zφ - Zφ = 3 + j4 Ω Zφ Zφ Here, VL = Vφ = 208 V , and Z φ = 4 + j 3 Ω = 5∠36.87° Ω , so Iφ = Vφ Zφ = 208 V = 41.6 A 5Ω I L = 3I φ = 3 ( 41.6 A ) = 72.05 A P=3 Vφ 2 ... Find the real and reactive powers supplied to each load (d) Find the real and reactive power losses in the transmission line (e) Find the real power, reactive power, and power factor supplied by... cos θ (e) The real and reactive power supplied by the generator when the switch is closed is just the sum of the real and reactive powers of Loads 1, 2, and The powers of Loads and have already... third edition of Electric Machinery and Power System Fundamentals To make this manual easier to use, it has been made self-contained Both the original problem statement and the problem solution