... Usage Usage designates rules of a less basic and binding sort, concerning how we should use the language in certain situations These sentences, for instance, violate formal usage: She dresses beautiful ... papers or business letters Which is not to dismiss such writing as easy Being clear and concise is never easy (To say nothing of being interesting!) But at least the writing process is stru...
Ngày tải lên: 24/07/2014, 12:22
... 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 +...
Ngày tải lên: 05/08/2014, 20:22
(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...
Ngày tải lên: 05/08/2014, 20:22
(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 Ω...
Ngày tải lên: 05/08/2014, 20:22
(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....
Ngày tải lên: 05/08/2014, 20:22
(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...
Ngày tải lên: 05/08/2014, 20:22
(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...
Ngày tải lên: 05/08/2014, 20:22
(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...
Ngày tải lên: 05/08/2014, 20:22
(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...
Ngày tải lên: 05/08/2014, 20:22
(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/...
Ngày tải lên: 05/08/2014, 20:22
(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...
Ngày tải lên: 05/08/2014, 20:22
(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 Ω /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...
Ngày tải lên: 05/08/2014, 20:22
McGraw.Hill PIC Robotics A Beginners Guide to Robotics Projects Using the PIC Micro eBook-LiB Part 1 pdf
... Mounting the Servomotors Leg Positioning Linkage vii 88 89 90 91 97 99 10 0 10 1 10 4 10 7 10 9 11 2 11 4 11 5 12 1 12 1 12 1 12 1 12 1 12 3 12 3 12 5 12 6 12 6 12 8 12 9 13 1 13 3 13 7 13 9 13 9 14 1 14 1 14 3 14 3 14 3 14 4 14 4 14 5 ... Little Feedback Servomotors 15 4 15 5 15 8 15 9 16 4 16 5 16 7 16 7 16 7 16 7 16 8 16 8 16 8 16 9 16 9 17...
Ngày tải lên: 10/08/2014, 04:22
McGraw-Hill - Robot Mechanisms and Mechanical Devices Illustrated Part 1 pot
... Ten Torque-Limiters One Time Use Torque Limiting 73 73 73 75 76 77 79 80 80 82 82 83 84 85 87 88 88 90 90 93 95 96 99 10 0 10 2 10 2 10 4 10 7 10 9 11 0 11 4 11 4 11 5 11 7 11 8 11 8 12 0 12 1 12 1 12 5 vii viii ... Six-Wheeled Layouts Eight-Wheeled Layouts Chapter Steering History Steering Basics The Next Step Up Chapter 13 0 13 0 13 1 13 4 13 6 14 1 14 4 15 0 15 5...
Ngày tải lên: 10/08/2014, 04:22
McGraw-Hill - The Robot Builder''''s Bonanza Part 1 ppt
... without the prior written permission of the publisher 0-0 7 -1 3867 1- 8 The material in this eBook also appears in the print version of this title: 0-0 7 -1 3629 6-7 All trademarks are trademarks of their ... Plastic Robot Platform 10 1 Chapter Building a Basic Wooden Platform 11 5 Chapter 10 Building a Metal Platform 12 3 Chapter 11 Constructing High Tech Robots from...
Ngày tải lên: 10/08/2014, 04:22