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Solution Manual For Power System Analysis and Design 7th Edition by J. Duncan Glover, Mulukutla S. Sarma, Thomas Overbye, Adam Birchfield R327,67   Add to cart
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Solution Manual For Power System Analysis and Design 7th Edition by J. Duncan Glover, Mulukutla S. Sarma, Thomas Overbye, Adam Birchfield

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Solution Manual For Power System Analysis and Design 7th Edition by J. Duncan Glover, Mulukutla S. Sarma, Thomas Overbye, Adam Birchfield

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  • April 30, 2024
  • 402
  • 2023/2024
  • Exam (elaborations)
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  • solution manual for power system analysis and desi

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AN INSTRUCTOR’S SOLUT IONS MANUAL TO ACCOMPANY POWER SYST EM ANALYSIS AND DESIGN SEVENTH EDITION J. DUNCAN GLOVER THOMAS J. OVERBYE MULUKUT LA S. SARMA Adam B. Birchfield Contents Chapter 2........................................ ................................................... ................................. . 1 Chapter 3........................................ ................................................... ............................... . 24 Chapter 4........................................ ................................................... ............................... . 63 Chapter 5........................................ ................................................... ............................... . 85 Chapter 6........................................ ................................................... ............................. . 125 Chapter 7........................................ ................................................... ............................. . 160 Chapter 8........................................ ................................................... ............................. . 166 Chapter 9........................................ ................................................... ............................. .185 Chapter 10....................................... ................................................... ............................ . 219 Chapter 11....................................... ................................................... ............................ . 293 Chapter 12....................................... ................................................... ............................ . 314 Chapter 13....................................... ................................................... ............................ . 333 Chapter 14....................................... ................................................... ............................ . 346 Chapter 15....................................... ................................................... ............................ . 369 1 © 20 23 Cengage Learning®. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 2 Fundamentals 2.1 (a) 𝐴̅1=9∠30°=9[𝑐𝑜𝑠 30°+𝑗 𝑠𝑖𝑛 30°]=7.8+𝑗4.5 (b) 𝐴̄2=4+𝑗5=√16+25 ∠𝑡𝑎𝑛−15
4=6.40∠128.66°=6.40𝑒𝑗128.66° (c) 𝐴̅3=(7.8+𝑗4.5)+(−4+𝑗5)=(11.8+𝑗9.5)=15.15∠38.8° (d) 𝐴4̅̅̅=(9∠30°)(6.4∠51.34°)=57.6∠81.34°=8.673+𝑗56.9 (e) 𝐴5̅̅̅=9∠30°
6.4∠−52.34°=1.41∠81.34°=1.41𝑒𝑗81.34° 2.2 (a) 𝐼̄=500∠−30°=433.01−𝑗250 (b) 𝑖(𝑡)=4𝑠𝑖𝑛(𝜔𝑡+15°)=4𝑐𝑜𝑠(𝜔𝑡+15°−90°)=4cos(𝜔𝑡−75°) 𝐼̅=4
√2∠−75°=2.83∠−75°=0.73−𝑗2.73 (c) 𝐼̄=(5/√2)∠−15°+4∠−60°=(3.42−𝑗0.92)+(2−𝑗3.46) =5.42−𝑗4.38=6.964∠−38.94° 2.3 𝑉̅2=13
√2∠(−(125×10−6)(2𝜋60)) kV=9.19∠−2.7° kV 2.4 (a) ()( )()1
21
26 6 9010 0 10 7.5 90 A8 6 6 8
10 0 7.3 90 10 7.5 12.5 36.87 A
6 12.5 36.87 6 90 75 53.13 VjIjj
I I I j
V I j− − =   = = − +−
= − =  − − = + =  
= − =   −  = −  (b) 2 © 20 23 Cengage Learning®. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. 2.5 (a) () () ( ) 277 2 cos 30 391.7cos 30 Vt t t  = +  = +  (b) 𝐼̅=𝑉̅
45=6.155∠30° A 𝑖(𝑡)=8.7 cos (𝜔𝑡+30°) A (c) ()()
()( )
()()32 60 10 10 3.771 90
277 30 3.771 90 73.46 60 A
( ) 73.46 2 cos 60 103.9cos 60 AZ j L j
I V Z
i t t t
−= =  =  
= =     = − 
= −  = −  (d) ()()
() ()25
277 30 25 90 11.08 120 A
( ) 11.08 2 cos 120 15.67cos 120 AZj
I V Z
i t t t =− 
= =   −  =  
= +  = +  2.6 (a) 𝑉̄=(75
√2)∠−15°=53.03∠−15°;  does not appear in the answer. (b) 𝜐(𝑡)=50√2𝑐𝑜𝑠(𝜔𝑡+10°); with  = 377, 𝜐(𝑡)=70.71𝑐𝑜𝑠(377𝑡+10°) (c) 𝐴̄=𝐴∠𝛼;𝐵̄=𝐵∠𝛽;𝐶̄=𝐴̄+𝐵̄ ( ) ( ) ( ) 2 Rejtc t a t b t Ce = + = The resultant has the same frequency . 2.7 (a) The circuit diagram is shown below: (b) 𝑍̶̄=3+𝑗8−𝑗4=3+𝑗4=5∠53.1° 𝛺 (c) ()() 100 0 5 53.1 20 53.1 AI=     = −  The current lags the source voltage by 53.1  Power Factor cos53.1 0.6 Lagging = =

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