Solution manual elements of forecasting 4th edition diebold

... for the dramatists of the Restoration The first would be the right form in a paragraph on the dramatists of the Restoration; the second, in a paragraph on the tastes of modern readers The need of ... one of the Italian mariners whom the decline of their own republics had put at the service of the world and of adventure, seeking for Spain a westward pas...

Ngày tải lên: 15/03/2014, 09:20

... the course was a slim volume called The Elements of Style, whose author was the professor himself The year was 1919 The book was known on the campus in those days as "the little book," with the ... for the dramatists of the Restoration The first would be the preferred form in a paragraph on the dramatists of the Restoration, the second in a paragraph on...

Ngày tải lên: 09/05/2014, 13:08

... 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

... 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

... 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

... 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

... 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

... 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

... 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

... = 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

... 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

... = 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

... (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