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Lecture 1 electric machinery fundamental

(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 1 pot

(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 1 pot

... 0.933 0.966 1. 033 1. 067 1. 1 1. 133 1. 167 1. 2 1. 233 1. 267 1. 3 1. 333 1. 367 1. 4 1. 433 1. 466 1. 5 10 4.4 11 8.86 13 2.86 14 6.46 15 9.78 17 2 .18 18 3.98 19 5.04 205 .18 214 .52 223.06 2 31. 2 238 244 .14 249.74 255.08 ... R3 + R4 ) R1 + R2 + R3 + R4 ( R1 + R2 ) φ right = φTOT = R1 + R2 + R3 + R4 (90 .1 + 77.3) 90 .1 + 10 8.3 + 90 .1 + 77.3 φTOT = (90 .1 + 10 8.3) (0.0033 Wb) = 0.0 019 3 Wb 90 .1 + 10 8.3 + 90 .1 + 77.3 (0.0033 ... RTOT = R5 + R1 = R2 = R3 = R4 = R5 = l1 µ r µ0 A1 ( R1 + R2 ) ( R3 + R4 ) R1 + R2 + R3 + R4 = 1. 11 m = 90 .1 kA ⋅ t/Wb (2000) 4π × 10 H/m (0.07 m )(0.07 m ) ( −7 ) l2 0.0007 m = = 10 8.3 kA ⋅ t/Wb...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 2 ppt

(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 2 ppt

... 11 6.3 2. 28° V 27 VR = (3) 11 6.3 -11 5 × 10 0% = 1. 1% 11 5 0.8 PF Leading: VP ′ = VS + Z EQ IS = 11 5∠0° V + ( 0 .14 0 + j 0.5 32 Ω )(8.7∠36.87° A ) VP ′ = 11 3.3 2. 24° V 11 3.3 -11 5 VR = × 10 0% = 1. 5% 11 5 ... 11 5∠0° V + ( 0 .14 0 + j 0.5 32 Ω )(8.7∠ − 36.87° A ) VP ′ = 11 8.8 1. 4° V 11 8.8 -11 5 VR = × 10 0% = 3.3% 11 5 (2) 1. 0 PF: VP ′ = VS + Z EQ I S = 11 5∠0° V + ( 0 .14 0 + j0.5 32 Ω )(8.7∠0° A ) VP ′ = 11 6.3 2. 28° ... TEST: 19 .1 V Z EQ = REQ + jX EQ = = 2. 2 Ω 8.7 A P 42. 3 W θ = cos 1 SC = cos 1 = 75.3° VSC I SC 19 .1 V )(8.7 A ) ( Z EQ = REQ + jX EQ = 2. 20∠75.3° Ω = 0.558 + j 2 . 12 8 Ω REQ = 0.558 Ω X EQ = j 2 . 12 8...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 3 ppsx

(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 3 ppsx

... 19 .9 kV 7.97 kV 200 kVA 2.50 :1 19.9 kV 13 .8 kV 200 kVA 1. 44 :1 Y-∆ 34 .5 kV 7.97 kV 200 kVA 4 .33 :1 ∆-Y 34 .5 kV 13 .8 kV 200 kVA 2.50 :1 ∆-∆ 34 .5 kV 13 .8 kV 34 6 kVA 2.50 :1 open-∆ 19 .9 kV 13 .8 kV 34 6 ... (low-voltage) side is Vbase (15 kV ) = 1. 125 Ω = S base 200 MVA Z base = so REQ = ( 0. 012 ) (1. 125 Ω ) = 0. 0 13 5 Ω X EQ = ( 0.05) (1. 125 Ω ) = 0.05 63 36 X M = (10 0 ) (1. 125 Ω ) = 11 2.5 Ω The equivalent ... VSC = 16 00 V , I φ ,SC = I SC / = 1. 1547 A , and Pφ ,SC = PSC / = 38 3 W Z EQ = Vφ ,SC 16 00 V = = 13 85 Ω I φ ,SC 1. 155 A θ = cos 1 Pφ ,SC Vφ ,SC I φ ,SC = cos 1 38 3 W = 78.0° (16 00 V ) (1. 155 A...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 4 pot

(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 4 pot

... Load = (0 .47 65∠ − 41 . 6°) (1. 513 + j1 .13 4 ) = 0.9 01 4. 7° VLoad = VLoad,puVbase3 = (0.9 01) (48 0 V ) = 43 2 V The power supplied to the load is PLoad,pu = I RLoad = ( 0 .47 65) (1. 513 ) = 0. 344 PLoad ... j0. 040 + 0.00723 + j0. 048 2 + 0. 040 + j 0 .17 0 + 1. 513 + j1 .13 4 Z EQ = 1. 5702 + j1.3922 = 2.099∠ 41 . 6° The resulting current is I= 1 0° = 0 .47 65∠ − 41 . 6° 2.099∠ 41 . 6° The load voltage under these ... 0. 010 j0. 040 0.00723 T1 1 0° j0. 048 2 0. 040 Line + - j0 .17 0 T2 1. 513 j1 .13 4 (b) L1 With the switch opened, the equivalent impedance of this circuit is Z EQ = 0. 010 + j0. 040 + 0.00723 + j0. 048 2...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 5 ppsx

(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 5 ppsx

... Phase − T / 12 T / 12 3T / 12 5T / 12 7T / 12 9T / 12 11 T / 12 T / 12 3T / 12 5T / 12 7T / 12 9T / 12 11 T / 12 T / 12 c a a b b c c b b c c a a b 88 Conducting SCR (Positive) SCR3 SCR1 SCR1 SCR2 SCR2 ... from R1, the time at which iD(t) reaches IH is t2 = − R2C ln I H R2 (0.00 05 A ) ( 15 00 Ω ) = 5. 5 ms = − ( 0.0 0 15 ) ln 30 V VBO Therefore, the period of the relaxation oscillator is T = 17 8 ms + 5. 5 ... of R1 and R2, so the time constant for the discharge is τ2 = (50 0 kΩ) (1. 5 kΩ) 1. 0 µ F = 0.0 0 15 s R1R2 C= ( ) R1 + R2 50 0 kΩ + 1. 5 kΩ The equation for the voltage on the capacitor during the discharge...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 6 pps

(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 6 pps

... below: 11 2 0 .15 Ω j1 .1 Ω IA + EA + - Vφ Z 6. 667 ∠30° - The magnitude of the phase current flowing in this generator is IA = EA 13 77 V 13 77 V = = = 18 6 A RA + jX S + Z 0 .15 + j1 .1 + 6. 667 ∠30° 1. 829 ... A = 12 40∠0° + ( 0 .15 Ω ) (18 6 − 30° A ) + j (1. 1 Ω) (18 6 − 30° A ) E A = 13 77 6. 8° V The resulting phasor diagram is shown below (not to scale): E = 13 77 6. 8° V A θ V = 12 40∠0° V φ IA = 18 6 -30° ... cos θ = (12 40 V ) (18 6 A )( 0.8) = 554 kW PCU = I A2 RA = (18 6 A ) ( 0 .15 Ω ) = 15 .6 kW PF&W = 24 kW Pcore = 18 kW Pstray = (assumed 0) PIN = POUT + PCU + PF&W + Pcore + Pstray = 61 2 kW 11 3 η= (d)...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 7 doc

(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 7 doc

... MW = (1. 5) 61. 0 − f sys + (1. 676 ) 61. 5 − f sys + (1. 9 61) 60.5 − f sys ) MW = 91. 5 − 1. 5f sys + 10 3. 07 1. 676 f sys + 11 8.64 − 1. 961f sys 5 .13 7 fsys = 306.2 f sys = 59. 61 Hz The power supplied by ... SD B 3.0 +1 +1 100 10 0 f nl,C 60.5 Hz f fl,C = = = 58. 97 Hz SDC 2.6 +1 +1 100 10 0 and the slopes of the power-frequency curves are: MW S PA = = 1. 5 MW/Hz Hz MW S PB = = 1. 676 MW/Hz 1. 79 Hz MW ... will be ( (f (f ) ) = (1. 676 MW/Hz )( 61. 5 Hz − 59. 61 Hz ) = 3 . 17 MW ) = (1. 9 61 MW/Hz )(60.5 Hz − 59. 61 Hz ) = 1. 74 MW PA = s PA f nlA − fsys = (1. 5 MW/Hz )( 61. 0 Hz − 59. 61 Hz ) = 2.09 MW PB =...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 8 ppt

(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 8 ppt

... = 1. 15 E A1 = 1. 15 ( 384 V ) = 4 41. 6 V 15 0 δ = sin 1 E A1 384 V sin 1 = sin 1 sin ( −36.4° ) = − 31. 1° E A2 4 41. 6 V The new armature current is I A2 = Vφ − E A2 jX S = 480 ∠0° V − 4 41. 6∠ − 31. 1° ... = sin 1 E A1 13 ,230 V sin δ = sin 1 sin 27.9° = 31. 3° E A2 11 ,907 V Therefore, the new armature current is IA = E A2 − Vφ jX S = 11 ,907∠ 31. 3° − 7044∠0° = 84 8∠ − 26 .8 A j8 . 18 With a 15 % decrease, ... A Line voltage, kV Extrapolated air-gap voltage, kV 320 13 .0 15 .4 365 13 .8 17 .5 380 14 .1 18. 3 475 15 .2 22 .8 570 16 .0 27.4 475 15 50 570 18 85 The short-circuit test was then performed with the following...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 9 pot

(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 9 pot

... PIN = (c) Pout POUT 10 0 kW = = 11 0 kW 0. 91 η The mechanical speed is nm = 15 00 r/min (d) The armature current is IA = IL = P 11 0 kW = = 15 6 A VT PF ( 480 V )(0.85) I A = 15 6∠ 31. 8° A Therefore, ... English units is τ load = POUT ωm = ( 210 00 hp)(746 W/hp) (12 00 r/min ) 2π rad 60 s τ load = 6 -13 = 12 4,700 N ⋅ m 1r 5252 P 5252 ( 210 00 hp ) = = 91 , 91 0 lb ⋅ ft nm (12 00 r/min ) A 440-V three-phase ... nfl 17 90 17 20 × 10 0% = × 10 0% = 4 .1% 17 20 nfl A 208-V, two-pole, 60-Hz Y-connected wound-rotor induction motor is rated at 15 hp Its equivalent circuit components are 17 3 R1 = 0.200 Ω R2 = 0 .12 0...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 10 pps

(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 10 pps

... = 0. 211 Ω and X = 0. 317 Ω Therefore, X M = 5.455 Ω − 0. 211 Ω = 5.244 Ω The resulting equivalent circuit is shown below: 19 2 IA R1 + Vφ jX1 0 .10 5 Ω jX2 j0. 211 Ω j0. 317 Ω j5.244 Ω R2 0.0 71 Ω jXM ... M ( R1 + jX ) ( j15 Ω )( 0.20 Ω + j 0. 41 Ω ) = = 0 .18 95 + j0.4 016 Ω = 0.444∠64.7° Ω R1 + j ( X + X M ) 0.20 Ω + j ( 0. 41 Ω + 15 Ω ) VTH = jX M ( j15 Ω ) Vφ = (12 0∠0° V ) = 11 6.8∠0.7° V R1 + j ... the DC test, R1 = 13 .5 V 64 A ⇒ R1 = 0 .10 5 Ω IDC + R1 VDC R1 R1 - In the no-load test, the line voltage is 208 V, so the phase voltage is 12 0 V Therefore, X1 + X M = Vφ I A,nl = 12 0 V = 5.455...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 2 Part 1 pptx

(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 2 Part 1 pptx

... (18 8.5 rad/s) (0. 3 21 + 0 .1 72 /1. 9 62) 2 + (0. 418 + 0. 420 )2 ( 26 2 V ) (0.0877 ) τ ind = (18 8.5 rad/s) (0. 3 21 + 0.0877 )2 + ( 0. 418 + 0. 420 )2 τ ind = 11 0 N ⋅ m, opposite the direction of motion 20 3 ... Motor Torque-Speed Characteristic 450 R2 = 0.0059 ohms R2 = 0 .1 72 ohms 400 350 300 τ ind 25 0 20 0 15 0 10 0 50 16 00 1 620 16 40 16 60 16 80 17 00 n 1 720 17 40 17 60 17 80 18 00 m (b) The slip at pullout torque ... nsync = (1 − 0 .1 92 ) (18 00 r/min ) = 14 54 r/min and the pullout torque of the motor is τ max = τ max = 3VTH 2 sync RTH + RTH + ( X TH + X ) ( 26 2 V ) (18 8.5 rad/s) 0. 3 21 Ω + (0. 3 21)2 + (0. 418 Ω...
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Electric Machinery Fundamentals (Solutions Manual) Part 1 pdf

Electric Machinery Fundamentals (Solutions Manual) Part 1 pdf

... below: 11 2 0 .15 & j1 .1 & IA + + - E A 6.667 30° Z V ⎞ - The magnitude of the phase current flowing in this generator is EA I =A + S + jX R A 13 77 V 13 77 V = 0 .15 1. 1 + Z + 6.667 30 1. 82 j ° 18 6 ... = ) 18 A AR P = W re 12 40 V 18 6 A 0.8 ( = 24 kW = 18 kW )( ) = 6A ( 0 .15 & ) = 15 .6 kW 554 kW 18 6 30 A) P ray = (assumed 0) st IN = P+ OUT P+ P + P+ P CU F&W core P = 612 kW stray 11 3 OUT 10 0% ... )( 18 6 ° EA = 12 40 + (0 .1 & ° + j 30 A) & EA 13 77 6.8 V = ° The resulting phasor diagram is shown below (not to scale): A ⎞ A R A S A (1. 1 ° E )( = 13 77 6.8° V A ⎝ V = ⎞ IA = 18 6 -30° (c) 12 40...
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Tài liệu Electric Machinery Fundamentals docx

Tài liệu Electric Machinery Fundamentals docx

... Manual to accompany Chapman Electric Machinery Fundamentals Fourth Edition Stephen J Chapman BAE SYSTEMS Australia i Instructor’s Manual to accompany Electric Machinery Fundamentals, Fourth Edition ... APPENDIX D: ERRATA FOR ELECTRIC MACHINERY FUNDAMENTALS 4/E 301 iii PREFACE TO THE INSTRUCTOR This Instructor’s Manual is intended to accompany the fourth edition of Electric Machinery Fundamentals To ... CONTENTS CHAPTER 1: INTRODUCTION TO MACHINERY PRINCIPLES CHAPTER 2: TRANSFORMERS 23 CHAPTER 3: INTRODUCTION TO POWER ELECTRONICS 63 CHAPTER 4: AC MACHINERY FUNDAMENTALS 103 CHAPTER 5: SYNCHRONOUS...
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Electric Machinery Fundamentals Power & Energy_2 potx

Electric Machinery Fundamentals Power & Energy_2 potx

... The overall efficiency of the power system will be the ratio of the output power to the input power The output power supplied to the load is POUT = 90 kW The input power supplied by the source ... questions about this power system (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 ... A + 24∠ − 45° A = 47.59∠ − 37.5° A The power factor supplied by the source is PF = cosθ = cos ( −37.5° ) = 0.793 lagging The real, reactive, and apparent power supplied by the source are P = VI...
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Electric Machinery Fundamentals Power & Energy_3 docx

Electric Machinery Fundamentals Power & Energy_3 docx

... equivalent circuit of this power system 59 (b) With the switch opened, find the real power P, reactive power Q, and apparent power S supplied by the generator What is the power factor of the generator? ... generator? (c) With the switch closed, find the real power P, reactive power Q, and apparent power S supplied by the generator What is the power factor of the generator? (d) What are the transmission ... transformers is equal to the power consumed by the loads (within roundoff error), while the total reactive power Q A + QB supplied by the transformers is equal to the reactive power consumed by the...
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