Fundamentals of Electrical Drivess - Chapter 1 pptx

... were open %%%%%%define variables A_c =10 e-3 *10 e-3; % cross-section of core and armature 10 mm x10 mm A_a =10 e-3 *10 e-3; % assumed cross-section of airgap g =10 e-3; % distance (airgap) between armature ... ring of width ‘h’ placed on the middle of the airgap at radius r + g/2. An observation of figure 1. 17 learns that the flux crossing the airgap has a larger cross-sectional...

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... 16 6 FUNDAMENTALS OF ELECTRICAL DRIVES SCOPE2 SCOPE1 u 1 i 2S U 1 @ 2D i 2D 0 i 2S i 1 @ 2S 0 @ 2S i 1 i 2D @ 1 @ 2D @ 1 3 81. 0 51 17 .14 8 220.000 782.597m 9.9 01 17 .14 8 9.088 4 51. 833m ... results NOTE1 PI @ 1 i m u 1 1.25 @ 1 @ s U 2 u 2 u 1 U 1 i 2 i 1 I1 I 2 P 1 Q 1 @ s @ s @ 2 1. 250 833.333m 392.699 1. 250 314 .15 9 45.3...

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... Heidelberg Printed on acid-free paper SPIN: 11 7604 81 62/techbooks 543 210 1- 4 02 0-5 50 4-8 ebook 97 8 -1 -4 02 0-5 50 4 -1 ebook Springer2007 ISBN -1 0 () ( ) (HB) (HB) ISBN -1 0 1- 4 02 0-5 503-X ISBN -1 3 97 8 -1 -4 02 0-5 50 3-4 ... Pulle Fundamentals of Electrical Drives ABC With 288 Figures and Rik W. De Doncker x FUNDAMENTALS OF ELECTRICAL DRIVE...

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... ‘vector of input values:’ set to tanh( [-5 :0 .1: 5]), and ‘vector 38 FUNDAMENTALS OF ELECTRICAL DRIVES 0 0 .1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.5 1 1.5 (a) time (s) voltage (V) 0 0 .1 0.2 0.3 ... as Simple Electro-Magnetic Circuits 41 0 0.005 0. 01 0. 015 0.02 0.025 0.03 0.035 0.04 −200 10 0 0 10 0 200 (a) time (s) voltage (V) 0 0.005 0. 01 0. 015 0.02 0.025 0.03 0...

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... COPE1GAI 0.0998 GAI 0.02 i 1 i RM i 2 i M @' 2 i 11 u 1/ 50 u 2 /10 u 1 @ 2 @ 1 i' 2 -1 2.282 -4 2.855m -3 3.689 11 .18 2m 11 8.755m -1 2.239 -1 8.668 -3 3.706 -9 33.3 81 43 .18 4m 1. 257m -1 2.2 51 Figure 3.28. Caspoc model: transformer ... is set to 10 0ms, solver type ‘ode4’ and step size 1 s. i _11 u2 i _1 u1 −K− attent datout To...

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... to Three-phase Circuits 11 9 0 0.002 0.004 0.006 0.008 0. 01 0. 012 0. 014 0. 016 0. 018 0.02 −500 0 500 time (s) u S1 (V) 0 0.002 0.004 0.006 0.008 0. 01 0. 012 0. 014 0. 016 0. 018 0.02 −20 10 0 10 20 time ... current 10 6 FUNDAMENTALS OF ELECTRICAL DRIVES 0 0.002 0.004 0.006 0.008 0. 01 0. 012 0. 014 0. 016 0. 018 0.02 −400 −200 0 200 400 time (s) u Sα (V) 0 0.002 0.004 0....

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... 0. 01 0. 012 0. 014 0. 016 0. 018 0.02 1 0 1 time (s) p L VA 0 0.002 0.004 0.006 0.008 0. 01 0. 012 0. 014 0. 016 0. 018 0.02 1 0 1 time (s) p e re VA 0 0.002 0.004 0.006 0.008 0. 01 0. 012 0. 014 0. 016 ... SCOPE5 SCOPE4 SCOPE1 SCOPE3 B_S + - i p c u c i p L u L i u c u L u s i p s u s i u c u L 10 ground -9 89. 810 u -1 5 81. 294n 15 97.573u -9 89. 810 u -9 89...

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... Machines 18 9 0 1 2 3 4 5 6 15 10 −5 0 5 10 15 rotor angle radials stator current and torque i α i β Torque Figure 7 .19 . Simulink results of IRTF simulation a function of time, where the duration of ... the maximum value of 12 .5Nm. Four cases are considered which reflect the basic operation of electrical ma- chines. Run your simulation for a ‘run’ time of 1s and tim...

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... 373.06 -5 3 .13 0.75 0.42 2 41. 73 -3 2.23 1. 0 0.30 13 0.96 -2 3.58 1. 25 0.23 25. 81 -1 8.66 1. 50 0.26 -7 6.84 -1 5.47 1. 75 0.35 -1 78 .16 -1 3. 21 2.00 0.47 -2 78.66 -1 1. 54 2.25 0.60 -3 78.62 -1 0.24 Build an m-file ... no-load→load T e (Nm) ρ m ( ◦ ) P s (W) Q s (VA) I s (A) 0.0 0 0 98 .14 0 .14 0 .12 5 -5 .74 39.27 10 0 .14 0 .16 0.250 -1...

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... 2859.34]; Te3= [15 .15 17 .17 18 .83 19 .68 19 .87 19 . 81 17.39 13 .93 8. 31 4.54]; Is3= [16 .68 15 .94 14 .94 13 . 91 13.29 12 .40 9.32 6.95 3.93 2 .11 ]; Ps3=[4760.36 5393 .19 5 916 . 81 618 2.37 6243. 31 6224.90 5464.44 ... E M (V) 10 00 369.29 15 .15 16 .68 4760.36 10 299.87 83. 41 200 878.93 17 .17 15 .94 5393 .19 9408.48 79.72 10 0 13 01. 94 18 .83 14 .94 5 91...

Ngày tải lên: 09/08/2014, 06:22