INtroductor solution manunual to accomapy power system analysis and design 5th

INSTRUCTOR'S SOLUTIONS MANUAL     TO ACCOMPANY            POWER SYSTEM  ANALYSIS AND DESIGN        FIFTH EDITION                      J. DUNCAN GLOVER    MULUKUTLA S. SARMA    THOMAS J. OVERBYE      Contents      Chapter 2                  1  Chapter 3    Chapter 4    Chapter 5    Chapter 6    Chapter 7    Chapter 8    Chapter 9    Chapter 10    Chapter 11                  27                  71                  95                   137                           175                          195                             231                           303                  323                  339                  353                  379      Chapter 12    Chapter 13    Chapter 14                Chapter Fundamentals ANSWERS TO MULTIPLE-CHOICE TYPE QUESTIONS 2.1 b 2.19 a 2.2 a 2.20 A c 2.3 c B a 2.4 a C b 2.5 b 2.21 a 2.6 c 2.22 a 2.7 a 2.23 b 2.8 c 2.24 a 2.9 a 2.25 a 2.10 c 2.26 b 2.11 a 2.27 a 2.12 b 2.28 b 2.13 b 2.29 a 2.14 c 2.30 (i) c (ii) b 2.15 a (iii) a 2.16 b (iv) d 2.17 A a 2.31 a B b 2.32 a C a 2.18 c © 2012 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 2.1 (a) A1 = 5∠30° = [ cos30° + j sin 30°] = 4.33 + j 2.5 = ∠126.87° = 5e j126.87° −3 (c) A3 = ( 4.33 + j 2.5 ) + ( −3 + j ) = 1.33 + j 6.5 = 6.635∠78.44° (b) A2 = −3 + j = + 16 ∠ tan −1 (d) A4 = ( 5∠30° )( ∠126.87° ) = 25 ∠156.87° = −22.99 + j 9.821 (e) A5 = ( 5∠30° ) / ( 5∠ − 126.87° ) = 1∠156.87° = e j156.87° 2.2 (a) I = 400∠ − 30° = 346.4 − j 200 (b) i(t ) = 5sin (ω t + 15° ) = 5cos (ω t + 15° − 90° ) = 5cos (ω t − 75° ) ( ) ∠ − 75° = 3.536∠ − 75° = 0.9151− j3.415 (c) I = ( ) ∠ − 30° + 5∠ − 75° = ( 2.449 − j1.414 ) + (1.294 − j 4.83 ) I = = 3.743 − j 6.244 = 7.28∠ − 59.06° 2.3 (a) Vmax = 359.3V; I max = 100 A (b) V = 359.3 = 254.1V; I = 100 = 70.71A (c) V = 254.1∠15° V; I = 70.71 ∠ − 85° A 2.4 (a) I1 = 10∠0° − j6 6∠ − 90° = 10 = 7.5∠ − 90° A + j6 − j6 I = I − I1 = 10∠0° − 7.3∠ − 90° = 10 + j 7.5 = 12.5∠36.87° A V = I ( − j ) = (12.5∠36.87° ) ( 6∠ − 90° ) = 75∠ − 53.13° V (b) 2.5 (a) υ (t ) = 277 cos (ω t + 30° ) = 391.7cos (ω t + 30° ) V (b) I = V / 20 = 13.85∠30° A i(t ) = 19.58cos (ω t + 30° ) A © 2012 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part (c) Z = jω L = j ( 2π 60 ) (10 × 10 −3 ) = 3.771∠90° Ω I = V Z = ( 277 ∠30° ) ( 3.771 ∠90° ) = 73.46 ∠ − 60° A i(t ) = 73.46 cos (ω t − 60° ) =103.9cos (ω t − 60° ) A (d) Z = − j 25 Ω I = V Z = ( 277∠30° ) ( 25∠ − 90° ) = 11.08∠120° A i(t ) = 11.08 cos (ω t + 120° ) = 15.67cos (ω t + 120° ) A 2.6 ( (a) V = 100 ) ∠ − 30°= 70.7∠ − 30° ; ω does not appear in the answer (b) υ (t ) = 100 cos (ω t + 20° ) ; with ω = 377, υ (t ) = 141.4 cos ( 377t + 20° ) (c) A = A∠α ; B = B∠β ; C = A + B c(t ) = a(t ) + b(t ) = Re Ce jωt  The resultant has the same frequency ω 2.7 (a) The circuit diagram is shown below: (b) Z = + j8 − j = + j = 5∠53.1° Ω (c) I = (100∠0° ) ( 5∠53.1° ) = 20∠ − 53.1° A The current lags the source voltage by 53.1° Power Factor = cos53.1° = 0.6 Lagging 2.8 Z LT = j ( 377 ) ( 30.6 × 10 −6 ) = j11.536 m Ω Z LL = j ( 377 ) ( × 10 −3 ) = j1.885 Ω ZC = − j V= = − j 2.88 Ω ( 377 ) ( 921 × 10−6 ) 120 2 ∠ − 30° V © 2012 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part The circuit transformed to phasor domain is shown below: 2.9 KVL : 120∠0° = ( 60∠0° )( 0.1 + j 0.5 ) + VLOAD ∴ VLOAD = 120∠0° − ( 60∠0° )( 0.1 + j 0.5 ) = 114.1 − j 30.0 = 117.9∠ − 14.7° V ← 2.10 (a) p(t ) = υ (t )i(t ) = 359.3cos (ω t + 15° )  100 cos (ω t − 85° )  ( 359.3)(100 ) cos100° + cos ( 2ω t − 70°)  = −3120 + 1.797 × 10 cos ( 2ω t − 70° ) W = (b) P = VI cos (δ − β ) = ( 254.1)( 70.71) cos (15° + 85° ) = −3120 W Absorbed = +3120 W Delivered (c) Q = VI sin (δ − β ) = ( 254.1)( 70.71) sin100° = 17.69 kVAR Absorbed (d) The phasor current ( − I ) = 70.71∠ − 85° + 180° = 70.71 ∠ 95° A leaves the positive terminal of the generator The generator power factor is then cos (15° − 95° ) = 0.1736 leading 2.11 (a) p(t ) = υ (t )i(t ) = 391.7 × 19.58cos2 (ω t + 30° ) 1 = 0.7669 × 10   1 + cos ( 2ω t + 60° )  2 = 3.834 × 103 + 3.834 × 103 cos ( 2ω t + 60° ) W P = VI cos (δ − β ) = 277 × 13.85cos0° = 3.836 kW Q = VI sin (δ − β ) = VAR Source Power Factor = cos (δ − β ) = cos ( 30° − 30° ) = 1.0 (b) p(t ) = υ (t )i(t ) = 391.7 × 103.9cos (ω t + 30° ) cos (ω t − 60° ) 1 = 4.07 × 10    cos90° + cos ( 2ω t − 30° )  2 = 2.035 × 10 cos ( 2ω t − 30° ) W P = VI cos (δ − β ) = 277 × 73.46 cos ( 30° + 60° ) = W © 2012 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Q = VI sin (δ − β ) = 277 × 73.46 sin 90° = 20.35 kVAR pf = cos (δ − β ) = Lagging (c) p(t ) = υ (t )i(t ) = 391.7 × 15.67 cos (ω t + 30° ) cos (ω t + 120° ) 1 = 6.138 × 103   cos ( −90° ) + cos ( 2ω t + 150° )  = 3.069 × 103 cos ( 2ω t + 150° ) W 2 P = VI cos (δ − β ) = 277 × 11.08cos ( 30° − 120° ) = W Q = VI sin (δ − β ) = 277 × 11.08sin ( −90° ) = −3.069 kVAR Absorbed = +3.069 kVAR Delivered pf = cos (δ − β ) = cos ( −90° ) = Leading 2.12 (a) pR (t ) = ( 359.3cos ω t )( 35.93cos ω t ) = 6455 + 6455cos2ω t W (b) px (t ) = ( 359.3cos ω t ) 14.37cos (ω t + 90° )  = 2582 cos ( 2cot + 90° ) = −2582sin 2ω t W 2) ( X = ( 359.3 ) (c) P = V R = 359.3 (d) Q = V 2 10 = 6455 W Absorbed 25 = 2582 VAR S Delivered (e) ( β − δ ) = tan −1 ( Q / P ) = tan −1 ( 2582 6455 ) = 21.8° Power factor = cos (δ − β ) = cos ( 21.8° ) = 0.9285 Leading 2.13 Z = R − jxc = 10 − j 25 = 26.93 ∠ − 68.2° Ω i(t ) = ( 359.3 / 26.93 ) cos (ω t + 68.2° ) = 13.34 cos (ω t + 68.2° ) A (a) pR (t ) = 13.34 cos (ω t + 68.2° )  133.4 cos (ω t + 68.2° )  = 889.8 + 889.8cos 2 (ω t + 68.2° )  W (b) px (t ) = 13.34 cos (ω t + 68.2° )  333.5cos (ω t + 68.2° − 90° )  = 2224sin  (ω t + 68.2° )  W ) 10 = 889.8 W ( (d) Q = I X = (13.34 ) 25 = 2224 VAR S (c) P = I R = 13.34 2 (e) pf = cos  tan −1 ( Q / P )  = cos  tan −1 (2224 / 889.8) = 0.3714 Leading © 2012 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 2.14 (a) I = 4∠0° kA V = Z I = ( 2∠ − 45° )( 4∠0° ) = 8∠ − 45° kV υ (t ) = cos (ω t − 45° ) kV p(t ) = υ (t )i(t ) = 8 cos (ω t − 45° )   cos ω t     1 = 64    cos ( −45° ) + cos ( 2ω t − 45° )  2 = 22.63 + 32 cos ( 2ω t − 45° ) MW (b) P = VI cos (δ − β ) = × cos ( −45° − 0° ) = 22.63MW Delivered (c) Q = VI sin (δ − β ) = × 4sin ( −45° − 0° ) = −22.63 MVAR Delivered = + 22.63MVAR Absorbed (d) pf = cos (δ − β ) = cos ( −45° − 0° ) = 0.707 Leading ( 2.15 (a) I =   ) ∠60°  ( 2∠30°) = ∠30° A i(t ) = cos (ω t + 30° ) A with ω = 377 rad/s p(t ) = υ (t )i(t ) = cos30° + cos ( 2ω t + 90° )  = 3.46 + cos ( 2ω t + 90° ) W (b) υ(t), i(t), and p(t) are plotted below: (See next page) (c) The instantaneous power has an average value of 3.46 W, and the frequency is twice that of the voltage or current © 2012 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 2.16 (a) Z = 10 + j 120 π × 0.04 = 10 + j15.1 = 18.1∠56.4° Ω pf = cos56.4° = 0.553 Lagging (b) V = 120 ∠0° V The current supplied by the source is I = (120 ∠0° ) (18.1∠56.4° ) = 6.63∠ − 56.4° A The real power absorbed by the load is given by P = 120 × 6.63 × cos56.4° = 440 W which can be checked by I R = ( 6.63 ) 10 = 440 W The reactive power absorbed by the load is Q = 120 × 6.63 × sin 36.4° = 663VAR (c) Peak Magnetic Energy = W = LI = 0.04 ( 6.63 ) = 1.76 J Q = ωW = 377 × 1.76 = 663VAR is satisfied 2.17 (a) S = V I * = Z I I * = Z I = jω LI Q = Im[ S ] = ω LI ← (b) υ (t ) = L di = − 2ω L I sin (ω t + θ ) dt p(t ) = υ (t ) ⋅ i(t ) = −2ω L I sin (ω t + θ ) cos (ω t + θ ) = −ω L I sin (ω t + θ ) ← = − Q sin (ω t + θ ) ← Average real power P supplied to the inductor = ← Instantaneous power supplied (to sustain the changing energy in the magnetic field) has a maximum value of Q ← 2.18 (a) S = V I * = Z I I * = Re  Z I  + j Im  Z I  = P + jQ ∴P = Z I cos ∠Z ; Q = Z I sin ∠Z ← (b) Choosing i(t ) = I cos ω t , Then υ (t ) = Z I cos (ω t + ∠Z ) ∴ p(t ) = υ (t ) ⋅ i(t ) = Z I cos (ω t + ∠Z ) ⋅ cos ω t = Z I cos ∠Z + cos ( 2ω t + ∠Z )  = Z I [ cos ∠Z + cos2ω t cos ∠Z − sin 2ω t sin ∠Z ] = P (1 + cos2ω t ) − Q sin 2ω t ← © 2012 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part jωC (c) Z = R + jω L + From part (a), P = RI and Q = QL + QC I ωC which are the reactive powers into L and C, respectively Thus p(t ) = P (1 + cos2ω t ) − QL sin 2ω t − QC sin 2ω t ← where QL = ω LI and QC = − If ω LC = 1,   ← p(t ) = P (1 + cos2ω t )  QL + QC = Q = Then *  150   ∠10°  ∠ − 50°  = 375 ∠60° 2.19 (a) S = V I =     = 187.5 + j 324.8 * P = Re S = 187.5 W Absorbed Q = Im S = 324.8 VAR SAbsorbed (b) pf = cos ( 60° ) = 0.5 Lagging (c) QS = P tan QS = 187.5 tan cos −1 0.9  = 90.81VAR S QC = QL − QS = 324.8 − 90.81 = 234 VAR S 2.20 Y1 = 1 = = 0.05∠ − 30° = ( 0.0433 − j 0.025 ) S = G1 − jB1 Z1 20∠30° Y2 = 1 = = 0.04∠ − 60° = ( 0.02 − j 0.03464 ) S = G2 + jB2 Z 25∠60° P1 = V G1 = (100 ) 0.0433 = 433 W Absorbed Q1 = V B1 = (100 ) 0.025 = 250 VAR S Absorbed P2 = V G2 = (100 ) 0.02 = 200 W Absorbed Q2 = V B2 = (100 ) 0.03464 = 346.4 VAR SAbsorbed © 2012 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Current sources:   (12.4.9) I m ( t ) = I k ( t − 500 ) +   Vk ( t )  400    (12.4.10) I k ( t ) = I m ( t − 500 ) −   Vm ( t )  400  (12.4.14) I L ( t ) = I L ( t − 100 ) + Vm ( t ) − VC ( t )  1000    (12.4.18) IC ( t ) = − IC ( t − 100 ) +   VC ( t )  25  t Vk Vm VC Im Ik IL IC μs kV kV kV kA kA kA kA 50.0 0 0.25 0 100 50.0 0 0.25 0 200 50.0 0 0.25 0 300 50.0 0 0.25 0 400 50.0 0 0.25 0 500 50.0 83.68 2.034 0.25 0.2398 0.0816 0.0814 600 50.0 57.69 0.25 374 © 2012 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 13.21 E = 100 kV Z G = Z c = 299.73 Ω RR = 150 Ω  = 25 Ω Δt = 50.μ s ↑= 200.μ s  Δt   2C R  Writing nodal equations:    t    299.73 + 299.73       Vk ( t )  =  2.9973 − I k ( t − 200 )      1   Vm ( t )    + +  I m ( t − 200 ) + I C ( t − 50 )     299.73 150 25      Solving: Vk ( t ) = 50.t − 149.865 I k ( t − 200 ) Vm ( t ) = 19.999  I m ( t − 200 ) + I C ( t − 50 )  Current sources: (12.4.9) (12.4.10) (12.4.18)   I m ( t ) = I k ( t − 200 ) +   Vk ( t )  299.73    I k ( t ) = I m ( t − 200 ) −   Vm ( t )  299.73    I C ( t ) = − I C ( t − 50 ) +   Vm ( t )  12.5  t Vk Vm Im Ik IC μs kV kV kA kA kA 0.0 0 0 50 0.0025 1.66 × 10 −5 0 100 0.0050 3.33 × 10 −5 0 150 0.0075 5.0 × 10 −5 0 200 0.0100 6.67 × 10 −5 0 250 0.0125 3.32 × 10 −4 8.34 × 10 −5 1.43 × 10 −5 2.66 × 10 −5 300 0.0150 1.20 × 10 −3 10.0 × 10 −5 2.53 × 10 −5 6.94 × 10 −5 375 © 2012 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 13.22 Nodal Equations: −1 10 V1 ( t )   0.1767 −0.1667      =  −0.1667 0.1767    V2 ( t )     − I ( t − 0.2 )  −1 V3 ( t )   0.1767 −0.1667   I ( t − 0.2 )     =  V4 ( t )   −0.1667 0.1677   − I L ( t − 0.02 )  (a ) (b) (c ) (d ) Dependent current sources: Eq (13.4.10)   I ( t ) = I ( t − 0.2 ) −   V3 ( t )  100  ( e) Eq (13.4.9)   I ( t ) = I ( t − 0.2 ) +   V2 ( t )  100  (f) Eq (13.4.14) I L ( t ) = I L ( t − 0.02 ) + V4 ( t ) 500 (g) Equations (a) – (g) can be solved iteratively for t = 0, Δt ,2 Δt  where Δt = 0.02ms I ( and I ( ) ) on the right hand side of Eqs (a) – (g) are zero for the first 10 iterations 13.23 (a) The maximum 60-Hz voltage operating voltage under normal operating conditions is 1.08 115/ = 71.7 kV From Table 13.2, select a station-class surge arrester with 84kV MCOV This is the station-class arrester with the lowest MCOV that exceeds 71.7kV, providing the greatest protective margin and economy (Note: where additional economy is required, an intermediate-class surge arrester with an 84-kV MCOV may be selected.) ( ) 376 © 2012 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part (b) From Table 13.2 for the selected station-class arrester, the maximum discharge voltage (also called Front-of-Wave Protective Level) for a 10-kA impulse current cresting in 0.5 μs ranges from 2.19 to 2.39 in per unit of MCOV, or 184 to 201 kV, depending on arrester manufacturer Therefore, the protective margin varies from ( 450 − 201) = 249kV to ( 450 − 184 ) = 266kV Note From Table of the Case Study for Chapter 13, select a variSTAR Type AZE station-class surge arrester, manufactured by Cooper Power Systems, rated at 108kV with an 84-kV MCOV From Table for the selected arrester, the Front-of-Wave Protective Level is 313kV, and the protective margin is therefore (450–313) = 137kV or 137/84 = 1.63 per unit of MCOV 13.24 The maximum 60-Hz line-to-neutral voltage under normal operating conditions on the HV side of the transformer is 1.1 345 / = 219.1kV From Table of the Case Study for Chapter 13, select a VariSTAR Type AZE station-class surge arrester, manufactured by Cooper Power Systems, with a 276-kV rating and a 220-kV MCOV This is the type AZE station-class arrester with the lowest MCOV that exceeds 219.1 kV, providing the greatest protective margin and economy For this arrester, the maximum discharge voltage (also called Front-of-Wave Protective Level) for a 10-kA impulse current cresting in 0.5 μs is 720kV The protective margin is (1300 – 720) = 580 kV = 580/220 = 2.64 per unit of MCOV ( ) 377 © 2012 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 14 Power Distribution 14.1 See Figure 14.2 Yes, laterals on primary radial systems are typically protected from short circuits Fuses are typically used for short-circuit protection on the laterals 14.2 The three-phase, four-wire multigrounded primary system is the most widely used primary distribution configuration The fourth wire in these Y-connected systems is used as a neutral for the primaries, or as a common neutral when both primaries and secondaries are present Usually the windings of distribution substation transformers are Y-connected on the primary distribution side, with the neutral point grounded and connected to the common neutral wire The neutral is also grounded at frequent intervals along the primary, at distribution transformers, and at customers’ service entrances Sometimes distribution substation transformers are grounded through an impedance (approximately one ohm) to limit short circuit currents and improve coordination of protective devices 14.3 See Table 14.1 for a list of typical primary distribution voltages in the United States [1–9] The most common primary distribution voltage is 15-kV Class, which includes 12.47, 13.2, and 13.8 kV 14.4 Reclosers are typically used on (a) overhead primary radial systems (see Figure 14.2) and on (c) overhead primary loop systems (see Figure 14.7) Studies have shown that the large majority of faults on overhead primaries are temporary, caused by lightning flashover of line insulators, momentary contact of two conductors, momentary bird or animal contact, or momentary tree limb contact As such, reclosers are used on overhead primary systems to reduce the duration of interruptions for these temporary faults Reclosers are not used on primary systems that are primarily underground, because faults on underground systems are usually permanent 14.5 See Table 14.2 Typical secondary distribution voltages in the United States are 120/240 V, single-phase, three-wire for Residential applications; 208Y/120V, three-phase, four-wire for Residential/Commercial applications; and 480Y/277V, three-phase, four-wire for Commercial/Industrial/High-Rise applications 14.6 Secondary Network Advantages Secondary networks provide a high degree of service reliability and operating flexibility They can be used to supply high-density load areas in downtown sections of cities In secondary network systems, a forced or scheduled outage of a primary feeder does not result in customer outages Because the secondary mains provide parallel paths to customer loads, secondary cable failures usually not result in customer outages Also, each secondary network is designed to share the load equally among transformers and to handle large motor starting and other abrupt load changes without severe voltage drops 379 © 2012 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Secondary Network Disadvantage Secondary networks are expensive They are typically used in high-density load areas where revenues justify grid costs 14.7 As stated in Section 14.3, more than 260 cities in the United States have secondary networks If one Googles “secondary electric power distribution networks United States,” one of the web sites that appears in the Google list is: Interconnection of Distributed Energy Resources in Secondary File Format: PDF/Adobe Acrobat Electric Power Research Institute and EPRI are registered service marks of the Electric SECONDARY DISTRIBUTION NETWORK OVERVIEW Power System Design and Operation Printed on recycled paper in the United States of America mydocs.epri.com/docs/public/000000000001012922.pdf The following is stated on Page 2-5 of the above-referenced publication: “Major cities, such as New York, Seattle, and Chicago have extensive distribution network systems However, even smaller cities, such as Albany or Syracuse, New York, or Knoxville, Tennessee, have small spot or grid networks in downtown areas.” 14.8 (a) At the OA rating of 40 MVA, I OA,L = 40 (13.8 × √ 3) = 1.673 kA per phase Similarly, I FA, L = 50 (13.8 × √ 3) I FOA L = 65 = 2.092 kA per phase (13.8 × √ 3) = 2.719 kA per phase (b) The transformer impedance is 8% or 0.08 per unit based on the OA rating of 40 MVA Using (3.3.11), the transformer per unit impedance on a 100 MVA system base is:  100  Z transformerPUSystem Base = 0.08   = 0.20 per unit  40  (c) For a three-phase bolted fault, using the transformer OA ratings as the base quantities, Isc3ϕ = VF Z transformerPU = 1.0 = 12.5 per unit = (12.5 )(1.673 ) ( 0.08 ) = 20.91 kA/phase Note that in (c) above, the OA rating is used to calculate the short-circuit current, because the transformer manufacturer gives the per unit transformer impedance using the OA rating as the base quantity 380 © 2012 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 14.9 (a) During normal operations, all four transformers are in service Using a 5% reduction to account for unequal transformer loadings, the summer normal substation rating is 1.20 × (30 + 33.3 + 33.3 + 33.3) × 0.95 = 148 MVA With all four transformers in service, the substation can operate as high as 148 MVA without exceeding the summer normal rating of 120% of each transformer rating (b) The summer allowable substation rating, based on the single-contingency loss of one of the 33.3 MVA transformers, would be 1.5 × (30 + 33.3 + 33.3) × 0.95 = 137 MVA Each of the three transformers that remain in service is each allowed to operate at 150% of its nameplate rating for two hours (reduced by 5% for unequal transformer loadings), which gives time to perform switching operations to reduce the transformer loading to its 30-day summer emergency rating However, from the solution to (c), the 30-day summer emergency rating of the substation is 119.3 MVA Since it is assumed that a maximum reduction of 10% in the total substation load can be achieved through switching operations, the summer allowable substation rating is limited to 119.3/0.9 = 132.6 MVA Note that, even though the normal summer substation rating is 148 MVA, it is only allowed to operate up to 132.6 MVA, so that in case one transformer has a permanent outage: (1) the remaining in-service transformers not exceed their two-hour emergency ratings; and (2) switching can be performed to reduce the total substation load to its 30-day emergency rating (c) Based on the permanent loss of one 33.3-MVA transformer, the 30-day summer emergency rating of the substation is 1.3 × (30 + 33.3 + 33.3) × 0.95 = 119.3 MVA When one transformer has a permanent failure, each of the other three transformers can operate at 130% of their rating for 30 days (reduced by 5% for unequal transformer loadings), which gives time to replace the failed transformer with a spare that is in stock 14.10 Based on a maximum continuous current of 2.0 kA per phase, the maximum power through each circuit breaker at 12.5 kV is 12.5 × 2.0 × √3 = 43.3 MVA The summer allowable substation rating under the single-contingency loss of one transformer, based on not exceeding the maximum continuous current of the circuit breakers, is 43.3 × × 0.95 = 123.4 MVA Comparing this result with the summer allowable substation rating of 132.6 MVA determined in Problem 14.9(b), we conclude that the circuit breakers limit the summer allowable substation rating to 123.4 MVA –1 14.11 (a) The power factor angle of the load is θL = cos (0.85) = 31.79° Using Figure 2.5, the load reactive power is: QL = SLsin(θL) = 10 × sin (31.79°) = 5.268 Mvar absorbed Also, the real power absorbed by the load is SL × p.f = 10 × 0.85 = 8.5 MW Similarly, the power factor angle of the source is θS = cos–1(0.90) = 25.84° The real power delivered by the source to the load is 8.5 MW, which is unchanged by the shunt capacitors Using Figure 2.5, the source reactive power is: P   8.5  QS = SS sin (θS ) =  S  sin (θ S ) =   sin (25.84°)  0.90   p.f  = 4.117 Mvar delivered 381 © 2012 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part The reactive power delivered by the shunt capacitors is the load reactive power minus the source reactive power: QC = QL − QS = 5.268 − 4.117 = 1.151 Mvar –1 (b) The power factor angle of the load is θL = cos (0.90) = 25.84° Using Figure 2.5, the load reactive power is: QL = SL sin (θ L ) = 10 × sin ( 25.84° ) = 4.359 Mvar absorbed Also, the real power absorbed by the load is SL × p.f = 10 × 0.90 = 9.0 MW Similarly, the power factor angle of the source is θS = cos–1(0.95) = 18.19° The real power delivered by the source to the load is 9.0 MW, which is unchanged by the shunt capacitors Using Figure 2.5, the source reactive power is: P   9.0  QS = SS sin (θ S ) =  S  × sin (θS ) =   sin(18.19°)  0.95   p.f  = 2.958 Mvar delivered The reactive power delivered by the shunt capacitors is the load reactive power minus the source reactive power: QC = QL − QS = 4.359 − 2.958 = 1.401 Mvar (c) Comparing the results of (a) which requires 1.151 Mvar to increase the power factor from 0.85 to 0.90 lagging; and (b) which requires 1.401 Mvar (22% higher than 1.151 Mvar) to increase the power factor from 0.90 to 0.95 lagging; it is concluded that improving the high power-factor load requires more reactive power 14.12 (a) Without the capacitor bank, the total impedance seen by the source is: Z TOTAL = RLINE + jX LINE + RLOAD Z TOTAL = + j + Z TOTAL = + j + + jX LOAD 1 + 40 j 60 33.333 = + j6 + 0.03 33.69° −33.69° Z TOTAL = + j + 27.74 + j18.49 = 30.74 + j 24.49 = 39.30 Ω/phase 38.54° (a1) The line current is: I LINE = VSLN Z TOTAL  13.8  1.05   /0° 0.2129 √3   = = kA/phase −38.54°  39.30   38.54°    382 © 2012 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part (a2) The voltage drop across the line is: Z  0.2129  VDROP = LINE = (3 + j 6)   I LINE  −38.54°   6.708  0.2129  =    63.43°  −38.54°  1.428 kV 24.89° | VDROP | = 1.428 kV (a3) The load voltage is: VLOAD    13.8  1.428  − = VSLN − Z LINE I LINE = 1.05   √  24.89°    0°  = 8.366 − (1.295 + j 0.60010 ) = 7.071 − j 0.6010 = 7.096 kVLN −4.86° |VLOAD | = 7.096 √ = 12.29 kVLL (a4) The real and reactive power delivered to the three-phase load is: 3(VLOADLN )2 3(7.096)2 PLOAD3ϕ = = = 3.776 MW RLOAD 40 QLOAD3ϕ = (VLOADLN )2 3(7.096)2 = = 2.518 Mvar X LOAD 60 (a5) The load power factor is:    Q   2.518   p.f = cos  tan −1    = cos  tan −1    = 0.83 lagging P    3.776     (a6) The real and reactive line losses are: PLINELOSS3ϕ = 3I LINE RLINE = 3(0.2129)2 (3) = 0.408 MW QLINELOSS3ϕ = I LINE X LINE = 3(0.2129)2 (6) = 0.816 Mvar (a7) The real power, reactive power, and apparent power delivered by the distribution substation are: PSOURCE3ϕ = PLOAD3ϕ + PLINELOSS3ϕ = 3.776 + 0.408 = 4.184 MW QSOURCE3ϕ = QLOAD3ϕ + QLINELOSS3ϕ = 2.518 + 0.816 = 3.334 Mvar SSOURCE3ϕ = √ (4.1842 + 3.3342 ) = 5.350 MVA 383 © 2012 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part (b) With the capacitor bank in service, the total impedance seen by the source is: Z TOTAL = RLINE + jX LINE + 1 + − RLOAD jX LOAD jXC 1 1 + − 40 j60 j60 43.42 Z TOTAL = + j6 + = 43 + j6 = Ω /phase 0.025 7.94° (b1) The line current is:  13.8  1.05   0° VSLN 0.1927 √3   = = I LINE = kA/phase − 7.94° Z TOTAL 43.42 7.94 ° (b2) The voltage drop across the line is: Z TOTAL = + j6 + 1.293  6.708   0.1927  VDROP = Z LINE I LINE =  = kV     63.43°   − 7.94°  55.49° | VDROP | = 1.293 kV (b3) The load voltage is: 1.293  13.8  VLOAD = VSLN − Z LINE I LINE = 1.05   /0° − 55.49° √   = 8.366 − (0.7325 + j1.065) = 7.633 − j1.065 7.707 = kVLN −7.94° |VLOAD | = 7.707 √ = 13.35 kVLL (b4) The real and reactive power delivered to the three-phase load is: 3(VLOADLN )2 3(7.707)2 = = 4.455 MW PLOAD3ϕ = 40 RLOAD QLOAD3ϕ = 3(VLOADLN )2 3(7.707)2 = = 2.970 Mvar 60 X LOAD (b5) The load power factor is:    Q   2.970  p.f = cos tan −   = cos tan −1   = 0.83 lagging  P   4.455    (b6) The real and reactive line losses are: PLINELOSS3ϕ = 3I LINE RLINE = 3(0.1927)2 (3) = 0.3342 MW QLINELOSS3ϕ = I LINE X LINE = 3(0.1927)2 (6) = 0.6684 Mvar (b7) The reactive power delivered by the shunt capacitor bank is: 3(VLOADLN )2 3(7.707)2 QC = = = 2.970 Mvar XC 60 384 © 2012 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part (b8) The real power, reactive power, and apparent power delivered by the distribution substation are: PSOURCE3ϕ = PLOAD3ϕ + PLINELOSS3ϕ = 4.455 + 0.3342 = 4.789 MW QSOURCE3ϕ = QLOAD3ϕ + QLINELOSS3ϕ − QC = 2.970 + 0.6684 − 2.97 = 0.6684 Mvar SSOURCE3ϕ = √ (4.7892 + 0.66842) = 4.835 MVA (c) Comparing the results of (a) and (b), with the shunt capacitor bank in service, the real power delivered to the load increases by 18% (from 3.776 to 4.445 MW) while at the same time: • The line current decreases from 0.2129 to 0.1927 kA/phase • The real line losses decrease from 0.408 to 0.334 MW • The reactive line losses decrease from 0.816 to 0.668 Mvar • The voltage drop across the line decreases from 1.428 to 1.293 kV • The reactive power delivered by the source decreases from 3.334 to 0.6684 Mvar • The load voltage increases from 12.29 to 13.35 kVLL The above benefits are achieved by having the shunt capacitor bank (instead of the distribution substation) deliver reactive power to the load 14.13 (a) Using all the outage data from the Table in (14.7.1) – (14.7.4): 342 + 950 + 125 + 15 + 2200 + 4000 + 370 4500 = 1.7782 interruptions/year SAIFI = (14.4 × 342) + (151.2 × 950) + (89.8 × 125) + (654.6 × 15) + (32.7 × 2200) + (10053 × 4000) + (40 × 370) SAIDI = 4500 SAIDI = 8992.966 minutes/year = 149.88 hours/year CAIDI = SAIDI 8992.966 = = 5057.34 minutes/year = 84.289 hours/year SAIFI 1.7782 8760 × 4500 − [(14.4 × 342) + (151.2 × 950) + (89.8 × 125) + (654.6 × 15) + (32.7 × 2200) + (10053 × 4000) + (40 × 370)]/60 ASAI = 8760 × 4500 ASAI = 0.98289 = 98.289% (b) Omitting the major event on 11/04 2010 and using the remaining outage data from the table in (14.7.1) – (14.7.4): 342 + 950 + 125 + 15 + 2200 + 370 SAIFI = = 0.889 interruptions/year 4500 (14.4 × 342) + (151.2 × 950) + (89.8 × 125) + (654.6 × 15) + (32.7 × 2200) + (40 × 370) SAIDI = 4500 SAIDI = 56.966 minutes/year 385 © 2012 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part CAIDI = SAIDI 56.966 = = 64.079 minutes/year SAIFI 0.889 8760 × 4500 − [(14.4 × 342) + (151.2 × 950) + (89.8 × 125) + (654.6 × 15) + (32.7 × 2200) + (40 × 370)]/60 8760 × 4500 ASAI = 0.99989 = 99.989% ASAI = Comparing the results of (a) and (b), the CAIDI is 84.289 hours/year including the major event versus 64.079 minutes/year or 1.068 hours/year excluding the major event 14.14 Using the outage data from the two tables, and omitting the major event: 200 + 600 + 25 + 90 + 700 + 1500 + 100 + 342 + 950 + 125 + 15 + 2200 + 370 2000 + 4500 SAIFI = 1.110 interruptions/year SAIFI = SAIDI = [(8.17 × 200) + (71.3 × 600) + (30.3 × 25) + (267.2 × 90) + (120 × 700) + (10 × 1500) + (40 × 100) + (14.4 × 342) + (151.2 × 950) + (89.8 × 125) + (654.6 × 15) + (32.7 × 2200) + (40 × 370)]/[2000 + 4500] SAIDI = 428,568.3/6500 = 65.934 minutes/year SAIDI 65.934 CAIDI = = = 59.400 minutes/year SAIFI 1.110 ASAI = [8760 × 6500 − 428,568.3/60]/[8760 × 6500] = 0.99987 = 99.987% 14.15 The optimal solution space is rather flat around a value of 0.066 MW for total system losses The below system solution represents an optimal (or near optimal) solution 1_TransmissionBus 138 kV 1.050 tap 3.3 MW -0.4 Mvar A MVA 13.8 kV 1.050 pu 4com Residential: 0.60 Commercial: 0.55 Industrial: 0.80 MVA 1.048 pu A MVA 5ind 1.1 Mvar 6com MVA 1.046 pu A 0.0 Mvar 1.044 pu A MVA 9com 0.55 MW 1.041 pu 10res 0.27 Mvar A MVA 1.038 pu 0.0 Mvar 0.80 MW 0.48 Mvar A 0.55 MW 0.27 Mvar 1.045 pu MVA 0.55 MW 0.22 Mvar 1.047 pu MVA Case Losses: 0.066 MW A 1.00 pu 1.050 tap 3.2 MW 0.3 Mvar A 0.60 MW 0.18 Mvar 1.1 Mvar 0.0 Mvar A MVA 11res 1.037 pu 0.60 MW 0.18 Mvar A 7res 1.035 pu 1.046 pu MVA 1.036 pu 1.1 Mvar A A MVA 0.60 MW 0.18 Mvar MVA 8ind 13ind 0.80 MW 0.48 Mvar 12res 0.80 MW 0.48 Mvar 0.60 MW 0.18 Mvar 386 © 2012 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 14.16 Initial losses are 0.082 MW When the bus tie breaker is closed the losses increase to 0.124 MW This increase is due to circulating reactive power between the two transformers The losses can be reduced by balancing the taps; opening some of the capacitors can also help reduce losses The below system solution has reduced the losses to 0.053 MW Again the solution space is flat so there are a number of near optimal solutions that would be acceptable 1_TransmissionBus 138 kV 1.025 tap 2.9 MW -0.2 Mvar A MVA 13.8 kV 1.024 pu 4com 1.00 pu 1.025 tap 2.8 MW -0.2 Mvar A A MVA MVA 1.024 pu Case Losses: 0.053 MW A A MVA Residential: 0.60 Commercial: 0.55 Industrial: 0.80 MVA 1.022 pu 0.55 MW 0.22 Mvar A 1.021 pu 5ind MVA 1.0 Mvar 6com MVA 0.55 MW 0.27 Mvar 7res 1.018 pu MVA 0.55 MW 0.27 Mvar A MVA 1.022 pu 0.0 Mvar 0.60 MW 0.18 Mvar 0.0 Mvar 1.019 pu A 10res 0.80 MW 0.48 Mvar A 9com 1.022 pu 1.1 Mvar A MVA 11res 1.022 pu 0.60 MW 0.18 Mvar A 1.025 pu 1.017 pu 1.0 Mvar MVA 1.023 pu 0.0 Mvar A A MVA MVA 8ind 0.60 MW 0.18 Mvar 13ind 12res 0.60 MW 0.00 MW 0.00 Mvar 0.80 MW 0.48 Mvar 0.18 Mvar 14.17 Because of the load voltage dependence, the total load plus losses are minimized by reducing the voltages to close to the minimum constraint of 0.97 per unit Again, the solution space is quite flat, with an optimal (or near optimal) solution shown below with total load + losses of 9.882 MW (versus a starting value of 10.644 MW) 1_TransmissionBus 138 kV 0.981 tap 4.9 MW -0.2 Mvar A MVA 13.8 kV 0.979 pu 4com Residential: 1.00 Commercial: 1.00 Industrial: 1.00 0.975 pu A MVA 5ind 0.9 Mvar 6com MVA 0.971 pu A MVA A MVA 9com 0.98 MW 0.975 pu 10res 0.49 Mvar A MVA 0.972 pu 0.0 Mvar 0.97 MW 0.58 Mvar A 0.97 MW 0.49 Mvar 0.9 Mvar 0.980 pu MVA 0.971 pu MVA Total Load+Losses: 9.882 MW A 0.97 MW 0.39 Mvar 0.972 pu 1.00 pu 0.987 tap 4.9 MW 0.4 Mvar A 0.97 MW 0.29 Mvar 0.9 Mvar 0.9 Mvar A MVA 11res 0.971 pu 0.97 MW 0.29 Mvar A 7res 0.971 pu 0.971 pu A MVA 0.971 pu 0.9 Mvar A MVA MVA 0.97 MW 0.29 Mvar 8ind 13ind 0.97 MW 0.58 Mvar 12res 0.97 MW 0.58 Mvar 0.97 MW 0.29 Mvar 387 © 2012 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part ... V1 leads V2 , δ = δ1 − δ is positive and the real power flows from node to node If V1 Lags V2 , δ is negative and power flows from node to node (b) Maximum power transfer occurs when δ = 90° =... = 15 + j STOTAL = S1 + S2 + S3 = 60 + j80 = 100∠53.13° kVA = P + jQ ∴ PTOTAL = 60 kW; QTOTAL = 80 kVAR; kVA TOTAL = STOTAL = 100 kVA ← Supply pf = cos ( 53.13° ) = 0.6 Lagging ← (b) ITOTAL = S... = −ω L I sin (ω t + θ ) ← = − Q sin (ω t + θ ) ← Average real power P supplied to the inductor = ← Instantaneous power supplied (to sustain the changing energy in the magnetic field) has a maximum