Power System Analysis And Design 5th Edition Glover Solutions Manual by edithcj93

INSTRUCTOR'S SOLUTIONS MANUAL TO ACCOMPANY POWER SYSTEM ANALYSIS AND DESIGN FIFTH EDITION J. DUNCAN GLOVER MULUKUTLA S. SARMA THOMAS J. OVERBYE Power System Analysis And Design 5th Edition Glover Solutions Manual Full Download: http://testbanktip.com/download/power-system-analysis-and-design-5th-edition-glover-solutions-manual/ Download all pages and all chapters at: TestBankTip.com

Contents Chapter 2 1 Chapter 3 27 Chapter 4 71 Chapter 5 95 Chapter 6 137 Chapter 7 175 Chapter 8 195 Chapter 9 231 Chapter 10 303 Chapter 11 323 Chapter 12 339 Chapter 13 353 Chapter 14 379

Chapter 2 Fundamentals

ANSWERS TO MULTIPLE-CHOICE TYPE QUESTIONS

2.1 b 2.2 a 2.3 c 2.4 a 2.5 b 2.6 c 2.7 a 2.8 c 2.9 a 2.10 c 2.11 a 2.12 b 2.13 b 2.14 c 2.15 a 2.16 b 2.17 A. a B. b C. a 2.18 c 2.19 a 2.20 A. c B. a C. b 2.21 a 2.22 a 2.23 b 2.24 a 2.25 a 2.26 b 2.27 a 2.28 b 2.29 a 2.30 (i) c (ii) b (iii) a (iv) d 2.31 a 2.32 a

2.1 (a) [] 1 5305cos30sin304.33 2.5 Ajj =∠°=°+°=+

(b) 1126 87 2 4 34916tan5126.875 3 j Aj e =−+=+∠=∠°=−°

(d) ()() 4 5305126.8725156.8722.999.821Aj =∠°∠°=∠°=−+

(e) ()() 156.87 5 530/5126.871156.871 j Ae ° =∠°∠−°=∠°=

2.2 (a) 40030346.4200Ij =∠−°=−

(b) ()()() ()5sin155cos15905cos75 itttt ωωω =+°=+°−°=−°

52753.536750.91513.415Ij =∠−°=∠−°=−

2.3 (a) max max 359.3V;100AVI==

(b) 359.32254.1V;100270.71AVI ====

(a) ()() ()2772cos30391.7cos30V ttt υωω

(b) () /2013.8530A ()19.58cos30A IV it tω ==∠° =+°

2 ©

2012 Cengage Learning.

()

2.4 (a) ()()() 1 21 2 6690 100107.590A 8668 1007.390107.512.536.87A 612.536.876907553.13V j I jj IIIj VIj −∠−° =∠°==∠−° +− =−=∠°−∠−°=+=∠° =−=∠°∠−°=∠−° (b) 2.5

=+° =+°

3 © 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) () () ()() ()() 3 26010103.77190 277303.7719073.4660A ()73.462cos60103.9cos60A ZjLj IVZ ittt ωπ ωω ==×=∠°Ω ==∠°∠°=∠−° =−°=−° (d) ()() ()() 25 27730259011.08120A ()11.082cos12015.67cos120A Zj IVZ ittt ωω =−Ω ==∠°∠−°=∠° =+°=+° 2.6 (a) ()

;

()

()()()2Re



 The

ω 2.7 (a)

The

2.8 () () () () () () 6 3 6 37730.61011.536m 3775101.885 1 2.88 37792110 1202 30V 2 LT LL C Zj j Zj j Zj j V =×=Ω =×=Ω =−=−Ω × =∠−°

10023070.730 V =∠−°=∠−°

ω does not appear in the answer. (b) ()()1002cos20 tt υω=+° ; with ω = 377,

()141.4cos37720 tt υ =+ ° (c) ;; AABBCAB αβ =∠=∠=+

j t ctatbtCe ω

=+=

resultant has the same frequency

The circuit diagram is shown below: (b) 38434553.1Zjjj=+−=+=∠°Ω

1000553.12053.1A I =∠°∠°=∠−°

current lags the source voltage by 53.1° Power Factorcos53.10.6 Lagging =°=

The

circuit

2.9 ()() ()() :12006000.10.5 12006000.10.5 114.130.0117.914.7V LOAD LOAD KVL jV Vj j ∠°=∠°++ ∴=∠°−∠°+ =−=∠−°← 2.10 (a) ()() ()()() () 4 ()()()359.3cos15100cos85 1 359.3100cos100cos270 2 31201.79710cos270W pttittt t t υωω ω ω  ==+°−°   =° + °  =−+× −° (b) ()()()()cos254.170.71 cos1585 3120WAbsorbed +3120WDelivered PVI δβ =−= °+° =− = (c) ()()() sin254.170.71sin100 17.69 kVAR Absorbed QVI δβ =−= ° = (d) The phasor current () 70.718518070.7195 I −=∠−°+°=∠° A leaves the positive terminal of the generator. The generator power factor is then () cos15950.1736 °−°= leading 2.11 (a) () () () () () 2 4 33 ()()()391.719.58cos30 1 0.7669101cos260 2 3.834103.83410cos260W cos27713.85cos03.836kW sin0VAR pttitt t t PVI QVI υω ω ω δβ δβ ==×+°   =×++°    =×+×+° =−=×°= =−= ()() Source Power Factorcoscos30301.0 δβ =−=°−°= (b) ()() () () ()() υω ω ω ω δβ ==×+°−°   =×°+−°    =×−° =−=×°+°= 4 4 ()()()391.7103.9cos30cos60 1 4.0710cos90cos230 2 2.03510cos230W cos27773.46cos30600W pttittt t t PVI

transformed to phasor domain is shown below:

(d) ()2 2 13.342252224VARS QIX

5 © 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. () () δβ δβ =−=×°= =−= sin27773.46sin9020.35kVAR cos0 Lagging QVI pf (c) ()() ()()()391.715.67cos30cos120 pttittt υω ω ==×+°+° ()()() ()() ()() ()() 3 3 1 6.13810cos90cos21503.06910cos2150W 2 cos27711.08cos301200W sin27711.08sin90 3.069kVAR Absorbed3.069kVAR Delivered coscos900 Leading tt PVI QVI pf ωω δβ δβ δβ   =×−°++°=×+°    =−=×°−°= =−=×−° =−=+ =−=−°= 2.12 (a) ()()()359.3cos35.93cos 64556455cos2W R p ttt t ωω ω = =+ (b) ()() () ()359.3cos14.37cos90 2582cos2cot90 2582sin2W x pt t t t ωω ω =+ °  =+° =− (c) ()2 2 359.32106455W Absorbed PVR=== (d) () 2 2 359.32252582VARS Delivered QVX=== (e) ()() () ()() 11 tan/tan2582645521.8 Power factorcoscos21.80.9285 Leading QP βδ δβ −== =° =−=°= 2.13 ()() () 102526.9368.2 ()359.3/26.93cos68.2 13.34cos68.2A c ZRjxj itt t ω ω =−=−=∠−°Ω =+° =+° (a)

()

R pt

ωω ω   =+°+°   =++°

()

x pt

ω  =+°+° −°   =+°

()()

()13.34cos68.2133.4cos68.2 889.8889.8cos268.2W

(b) ()()

()13.34cos68.2333.5cos68.290 2224sin268.2W

t t t ωω

===

0.3714

pfQP ==      =

=== (e) () 11 costan/costan(2224/889.8)

Leading

(b) ()() cos84cos45022.63MW Delivered PVI δβ =−=×−°−°=

Delivered22.63MVAR Absorbed

(d) ()() coscos4500.707 Leading pf δβ =−=−°−°=

(b) υ(t), i(t), and p(t) are plotted below: (See next page)

(a) 40kA I =∠° ()() () () ()() () 24540845kV ()82cos45kV ()()()82cos4542cos 1 64cos45cos245 2 22.6332cos245MW VZI tt p ttitt t t t υω υωω ω ω ==∠−°∠°=∠−° =−° ==−°   =−°+−°    =+−°

2.14

22.63MVAR

QVI

=−=×−°−° =−=+

δβ

2.15 (a) () () 4260230230A I  =∠°∠°=∠°  () () () ()2cos30A with 377rad/s ()()()4cos30cos290 3.464cos290W itt pttitt t ωω υω ω =+°=  ==°++°  =++°

2.16 (a) 101200.041015.118.156.4Zjj π =+×=+=∠°Ω cos56.40.553 Lagging pf =°=

(b) 1200V V =∠°

The current supplied by the source is ()() 120018.156.46.6356.4A I =∠°∠°=∠−°

The real power absorbed by the load is given by 1206.63cos56.4440W P =××°=

which can be checked by () 2 2 6.6310440W IR ==

The reactive power absorbed by the load is 1206.63sin36.4663VAR Q =××°=

(c)

(b)

Average real power supplied to the inductor0

Instantaneous power supplied (to sustain the changing energy in the magnetic field) has a maximum value of Q. ←

QW ω ==×=

(a) 2

ω ==== 2 Im[] QSLI ω ==←

()

di tLLIt dt υωω θ ==−+ ()() () () 2 2 ()()()2sincos sin2 sin2 pttitLItt LIt Qt υωωθωθ ωωθ ωθ =⋅=−++ =− +← =−+←

()2 2 Peak Magnetic Energy 0.046.631.76J WLI ==== 3771.76663VAR is satisfied.

2.17

**2 SVIZIIZIjLI

(b)

()2sin

P =←

2.18 (a)

SVIZIIZIjZI===+ 22 cos;sin PjQ PZIZQZIZ =+ ∴=∠=∠←

**22 ReIm

∠ () () [] 2 2 2 ()()()coscos coscos2 coscos2cossin2sin (1cos2)sin2 pttitZItZt ZIZtZ ZIZtZtZ PtQt υωω ω ωω ωω ∴=⋅=+∠⋅ =∠++∠ =∠+∠−∠ =+−←

Choosing ()2cos, itIt ω = Then () ()2cos tZItZυω=+

8 © 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) 1 ZRjL j C ω ω =++ From part (a), 2 PRI = and L QQQC =+ where 2 L QLI ω = and 2 1 C QI Cω =− which are the reactive powers into L and C, respectively. Thus () ()1cos2sin2sin2 LC ptPtQtQt ωωω =+−−← () 2 If 1, 0 Then ()1cos2 LC LCQQQ ptPt ω ω =+==   ←  =+   2.19 (a) * * 1505 105037560 22 187.5324.8 SVI j  ==∠°∠−°=∠°   =+ Re187.5WAbsorbed Im324.8VARSAbsorbed PS QS == == (b) () cos600.5 Lagging pf =°= (c) 1 tan187.5tancos0.990.81VARS 324.890.81234VARS SS CLS QPQ QQQ ===   =−=−= 2.20 () 1 11 1 11 0.05300.04330.025S 2030 YjGjB Z ===∠−°=−=− ∠° () () () () () 2 22 2 2 2 11 2 2 11 2 2 22 2 2 22 11 0.04600.020.03464S 2560 1000.0433433WAbsorbed 1000.025250VARSAbsorbed 1000.02200WAbsorbed 1000.03464346.4VARSAbsorbed YjGjB Z PVG QVB PVG QVB ===∠−°=−=+ ∠° === === === ===

9 © 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. 2.21 (a) 1 cos0.653.13 tan500tan53.13666.7 kVAR cos0.925.84 L LL S QP φ φ φ ==°= ==° tan500tan25.84242.2kVAR 666.7242.2424.5kVAR 424.5kVA SS CLS CC QP QQQ SQ φ ==°= =−=−= == (b) The ()5000.746 Synchronous motor absorbs414.4kW 0.9 m P == and 0kVAR Qm = Source () 1 PFcostan666.7914.40.808 Lagging  ==  2.22 (a) () () 1 1 111 0.253.13 34553.13 0.120.16S Y Zj j ====∠−° +∠° =− () () () () 2 2 2 12 12 2 2 11 2 2 22 11 0.1S 10 1100 70.71 V 0.120.1 70.710.12600W 70.710.1500W Y Z P PVGGV GG PVG PVG === =+  === ++ === === (b) () 12 0.120.160.10.220.16 0.27236.03S eq YYYj j =+=−+=− =∠−° () 70.710.27219.23A IVYSeq===

power delivered by the source is

The complex power delivered by the source is equal to the total complex power absorbed by the load.

2.26 (a) The problem is modeled as shown in figure below:

10 © 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. 2.23 ()() * 12001030120030 1039.2600 Re1039.2WDelivered Im600VARSDelivered600VARSAbsorbed SVI j PS QS ==∠°∠−°=∠−° =− == ==− =+ 2.24 1 111 2 100;10cos0.994.359SPjQjSj =+=+=∠=+ 1 3 123 100.746 cos0.959.23818.198.7762.885 0.850.95 27.781.47427.823.04 Re()27.78kW Im()1.474kVAR 27.82kVA S SS SS SS Sj SSSSj PS QS SS × =∠−=∠−°=− × =++=+=∠° == == == 2.25 **22 **22 **22 (20)312000 ()8(20)03200 ()4(20)01600 RR LLLL CCCC SVIRIIIR j SVIjXIIjXIjj SVIjIXIjXIjj =====+ =====+ ==−=−=−=− Complex

load 200053.1 LOADRLC SSSS=++=∠°

Triangle: Complex

SOURCE SVI==∠°∠−°=∠°

power absorbed by the total

Power

()()*

10002053.1200053.1

1 120kW 0.85Lagging cos0.8531.79 L L L P pf θ = = ==°

The input voltage is given by /682.4V(rms)

141.1831.79kVA /141,180/480294.13A LLL L SPjQ ISV =+=∠° === ()tan31.79 74.364kVAR LL QP=° =

Reactive

()2 2 294.131 QIXLINELINE== 86.512kVAR = () 12074.36486.512200.753.28kVA SSS SPjQj ∴=+=++=∠°

Power triangle for the load:

Real power loss in the line is zero.

power loss in the line is

SS VSI==

S Vj=∠°+∠−° () 635250682.421.5V(rms) ()cos21.531.790.6 Lagging S j pf =+=∠° =°+°=

() 1 50kW;cos0.836.87;36.87;tan 37.5kVAR old OLD PQoldoldold P ==°=°=θθ = 50,00037,500 old Sj ∴=+ () 1 cos0.9518.19;50,00050,000tan18.19 50,00016,430 new new Sj j θ ==°=+° =+ Hence 21,070VA capnewold SSSj =−=− ()() 2 21,070 1155F 377220 C μ ∴==←

The power factor at the input is cos53.280.6 Lagging °= (b) Applying KVL, () 48001.0294.1331.79

2.27 The circuit diagram is shown below:

12 © 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. 2.28 () () () 1 1 2 3 123 126.667 40.964sincos0.963.841.12 150 30.845.547kVA TOTAL Sj Sjj Sj SSSSj =+  =−=−  =+ =++=+ (i) Let Z be the impedance of a series combination of R and X Since * 2 * * VV SVIV ZZ  ===   , it follows that () () () 2 2 * 3 240 (1.8090.3254) 30.845.54710 1.8090.3254 V Zj Sj Zj == =−Ω + ∴=+Ω← (ii) Let Z be the impedance of a parallel combination of R and X Then () () () () () 2 3 2 3 240 1.8677 30.8410 240 10.3838 5.54710 1.867710.3838 R X Zj ==Ω ==Ω ∴=Ω← 2.29 Since complex powers satisfy KCL at each bus, it follows that ()()() 13 * 3113 11110.40.20.41.8 0.41.8 Sjjjj SSj =+−−−+=−+← =−=+← Similarly, ()()() 23 0.50.5110.40.20.10.7 Sjjjj =+−+−−+=−−← * 3223 0.10.7SSj=−=−← At Bus 3, ()() 33132 0.41.80.10.70.51.1 G SSSjjj =+=++−=+← 2.30 (a) For load 1: 1 1 cos(0.28)73.74 Lagging θ ==° 1 2 3 12573.7435120 1040 150 Sj Sj Sj =∠°=+ =− =+ () 123 608010053.13kVA 60kW;80kVAR;kVA100kVA. Supplycos53.130.6 Lagging TOTAL TOTALTOTALTOTALTOTAL SSSSj PjQ PQ S pf =++=+=∠°=+ ∴====← =°=← (b) *3 * 1001053.13 10053.13A 10000 TOTAL S I V ×∠−° ===∠−° ∠°

At the new pf of 0.8 lagging, PTOTAL of 60kW results in the new reactive power Q ′ , such that

∴

(b) Maximum power transfer occurs when 12 90

2.32 4 Mvar minimizes the real power line losses, while 4.5 Mvar minimizes the MVA power flow into the feeder.

() 1 cos0.836.87 θ′ ==°

() 60tan36.8745kVAR Q ′ =°=

and

The

capacitor’s kVAR is 804535kVAR QC =−=←

follows then () 2 2 * 1000 28.57 35000 C C V Xj Sj ===−Ω and ()() 6 10 92.85F 26028.57 C μ π ==← The new current is * * 60,00045,000 6045 10000 Sj Ij V ′ ′ ===− ∠° 7536.87A=∠−° The supply current, in magnitude, is reduced from 100A to 75A ← 2.31 (a) 11221 2 12 1 2 90 90 90 VVVV I XXX δδ δδ ∠−∠ ==∠ ° ∠ °  ∠°  Complex power 12 * 121121112 2 112 12 9090 9090 VV SVIV XX VVV XX δδδ δδ   ==∠∠°−−∠°−     =∠°−∠°+−

The real and reactive power at the sending end are () () 2 112 1212 12 12 cos90cos90 sin VVV P XX VV X δδ δδ =°−°+− =−← () () 2 112 1212 1 1212 sin90sin90 cos VVV Q XX V VV X δδ δδ =°−°+−  =−−← 

required

∴

2V , 12 δδδ =−

2. If 1V Lags 2V , δ

and power

Note: If

leads

is positive and the real power flows from node 1 to node

is negative

flows from node 2 to node 1.

12 MAX VV P X =←

δδδ=°=−←

Power System Analysis And Design 5th Edition Glover Solutions Manual Full Download: http://testbanktip.com/download/power-system-analysis-and-design-5th-edition-glover-solutions-manual/ Download all pages and all chapters at: TestBankTip.com