Fundamentals of Electrical Drivess - Chapter 4 doc

... its transfer matrix is of the form given by equation (4. 49).   u Dα u Dβ   = √ 3   cos γ 4 sin γ 4 −sin γ 4 cos γ 4     u α u β   (4. 49) 98 FUNDAMENTALS OF ELECTRICAL DRIVES 0 0.05 ... u T (4. 14a) u R = −u D2 + u S (4. 14b) Adding equations (4. 14a), (4. 14b) gives 2u R = u D1 − u D2 +(u T + u S )    −u R +3u S0 (4. 15) where the term (u T + u S ) can ac...

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... Heidelberg Printed on acid-free paper SPIN: 1176 048 1 62/techbooks 543 210 1 -4 02 0-5 50 4- 8 ebook 97 8-1 -4 0 2 0-5 50 4- 1 ebook Springer2007 ISBN-10 () ( ) (HB) (HB) ISBN-10 1 -4 02 0-5 503-X ISBN-13 97 8-1 -4 0 2 0-5 50 3 -4 ... VA p -number of pole pairs PI -proportional-integral PWM -pulse width modulation Q -reactive power VA R -resistance Ω R -reluctance At/Wb...

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... variables A_c=10e-3*10e-3; % cross-section of core and armature 10mm x10 mm A_a=10e-3*10e-3; % assumed cross-section of airgap g=10e-3; % distance (airgap) between armature and C core mu_c=0. 64/ 79.58; ... parameters depth=20e-3; % thickness E and I core g=10e-3; % gap w_A =45 e-3; % E core center leg width w_B=22.5e-3; % E core side leg width n=500; % number of turns coil i_coil=20; %...

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... 40 FUNDAMENTALS OF ELECTRICAL DRIVES i Psi u i(psi) dat To Workspace Sine Wave 100 R 1 s Clock Figure 2.15. Simulink model of non-linear inductance with sinusoidal excitation function of ... gradient of the flux-linkage/current curve becomes zero for currents in excess of ±3 A. In reality the gradient will be non-zero when saturation occurs. 4 −3 −2 −1 0 1 2 3 4 −1 −0.8 −0...

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... MATLAB 74 FUNDAMENTALS OF ELECTRICAL DRIVES S 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 11.182m 118.755m -1 2.239 -1 8.668 ... model of figure 3.15 is given by equation (3. 24) . u 1 = dψ 1 dt (3.24a) u 2 = dψ 2 dt (3.24b) ψ 1 = i 1 L σ + ψ  2 (3.24c) ψ  2 = i M L M (3.24d) i M = i 1 −...

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... input power of the circuit is equal to the real power, hence P = p in (5.33) 144 FUNDAMENTALS OF ELECTRICAL DRIVES m-file Tutorial 3, chapter 5 %Tutorial 3, chapter 5 %we set E=100; eta =-7 *pi/6 assume ... example of an m-file implementation for this tutorial is given below: m-file Tutorial 1, chapter 5 %Tutorial 1, chapter 5 close all L=100e-3; C=101.32e-6; Ub=10; T=pi*sqr...

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... 3 14. 159 45 . 345 78. 540 392.699 226.725 1 54. 418 25. 642 18.132 89.153 9.538k 7.818k 3 14. 159 3 14. 159 250.000m Figure 6.15. Caspoc simulation: simplified transformer model 6 .4. 3 ... period of 0.4s, using a ‘step-size’ of 10 −5 s and ‘solver’ ode4. Outputs of the simulation should be the vectors  i 1 ,  i 2D ,  ψ 1 and  ψ 2D . An example of a possible...

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... 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 time step of 10 4 s, using the ode4 solver. ... corre- sponds to machine configurations discussed in this section. The diagram shows 1 74 FUNDAMENTALS OF ELECTRICAL DRIVES (a) IRTF-flux (b) IRTF-current Figure 7.5. Generic repres...

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... (deg) rhoM=[0 -5 . 74 -1 1. 54 -1 7 .46 -2 3.58 -3 0.00 -3 6.87 -4 4. 43 -5 3.13 -6 4. 16 -8 9.29]; %measured, from display, real power (W) PM=[0 39.27 78. 54 117.81 157.08 196.35 235.62 2 74. 89 3 14. 16 353 .43 392.67]; %measured, ... 373.06 -5 3.13 0.75 0 .42 241 .73 -3 2.23 1.0 0.30 130.96 -2 3.58 1.25 0.23 25.81 -1 8.66 1.50 0.26 -7 6. 84...

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... 66 .46 50 1888.55 19.81 12 .40 62 24. 90 5690 .40 62.00 30 22 84. 33 17.39 9.32 546 4 .44 3216 .46 46 .61 20 2503 .43 13.93 6.95 43 75.32 1786.92 34. 74 10 27 34. 06 8.31 3.93 2609.29 570.73 19.63 5 2859. 34 4. 54 ... 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= [47 60.36 5393.19 5916.81 6182.37 6 243 .31 62 24. 90 546 4 .44 43 75.32 2609.29 1...

Ngày tải lên: 09/08/2014, 06:22