Fundamentals of Electrical Drivess - Chapter 8 potx

... 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 -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 which will display the ... (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, fro...

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... 15.15 16. 68 4760.36 10299 .87 83 .41 200 87 8.93 17.17 15.94 5393.19 94 08. 48 79.72 100 1301.94 18. 83 14.94 5916 .81 82 63.42 74.71 70 1590.66 19. 68 13.91 6 182 .47 7166.30 69.57 60 1726.55 19 .87 13.29 ... inductance nm2=[ 184 4.66 2100 .86 2 288 .64 2556.27 2744.29]; Te2=[ 18. 90 14.71 11.64 7.26 4. 18] ; Is2= [8. 73 6 .80 5. 38 3.35 1.93]; Ps2=[59 38. 94 4621.94 3656.71 2 280 .9...

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... Heidelberg Printed on acid-free paper SPIN: 11760 481 62/techbooks 543210 1-4 02 0-5 50 4 -8 ebook 97 8- 1 -4 02 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 02 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 r...

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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; % ... located between 28 FUNDAMENTALS OF ELECTRICAL DRIVES two-dimensional finite element package which was also used to calculate the flux distribution shown in figure 1. 18. Observat...

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... following entries under: ‘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 ... concept of ‘incremental flux-linkage’ ∆ψ = ψ(t)−ψ(0) which is fundamental to the control of electrical drives. The equation basically states that a flux-linkage variation c...

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... MATLAB 74 FUNDAMENTALS OF ELECTRICAL DRIVES S COPE1GAI 0.09 98 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 .85 5m -3 3. 689 11. 182 m 1 18. 755m -1 2.239 -1 8. 6 68 ... -3 3. 689 11. 182 m 1 18. 755m -1 2.239 -1 8. 6 68 -3 3.706 -9 33. 381 43. 184 m 1.257m -1 2.251 Figure 3. 28. Caspoc model...

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... An example of this phasor calculation together with the commands needed to display the results of the simulation are shown in the m-file below. m-file Tutorial 8, chapter 4 %tutorial 8, chapter 4 close ... 40]) grid legend(’i_{D\alpha}’,’i_{D\beta}’) subplot(3,1,3) plot(datoutn(:,6),datoutn(:,5),’g’) xlabel(’time (s)’) ylabel(’i_R (A)’) grid axis([0 20e-3 -6 0 60]) 4 .8. 8 Tutorial...

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... 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 .81 0u -1 581 .294n 1597.573u -9 89 .81 0u -9 89 6.523u 9.9 98 -9 89 .81 0u 1597.573u 9.9 98 10.000 -9 89 .81 0u -9 89 8.102u 10.000 -9 89 .81 0u 1597.573u 9.9 98 0 [Volt] Figure 5.16. ... input power of the circuit is equal to the real...

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... 166 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 @ 1 381 .051 17.1 48 220.000 782 .597m 9.901 17.1 48 9. 088 451 .83 3m ... 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 83 3.333m 392.699 1.250 314.159 45.345 78. 540 392.699 226.725 154.4...

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... directly 182 FUNDAMENTALS OF ELECTRICAL DRIVES example, in the two-pole machine we can rotate the resultant magnetic field by an angle of π 2 radians by way of a voltage excitation sequence of the ... physically Chapter 7 INTRODUCTION TO ELECTRICAL MACHINES 7.1 Introduction This chapter considers the basic working principles of the so-called ‘clas- sical’ set of machines...

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