Fundamentals of Electrical Drivess - Chapter 5 ppt

... 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 89.810u -1 58 1.294n 159 7 .57 3u -9 89.810u -9 896 .52 3u 9.998 -9 89.810u 159 7 .57 3u 9.998 10.000 -9 89.810u -9 898.102u 10.000 -9 89.810u 159 7 .57 3u 9.998 0 [Volt] Figure 5. 16. Caspoc ... (5. 39), (5. 40) leads to P R = i (i) ∗ R (5. 45) Q...

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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=45e-3; % E core center leg width w_B=22.5e-3; % E core side leg width n =50 0; % number of turns coil i_coil=20;...

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... as Simple Electro-Magnetic Circuits 41 0 0.0 05 0.01 0.0 15 0.02 0.0 25 0.03 0.0 35 0.04 −200 −100 0 100 200 (a) time (s) voltage (V) 0 0.0 05 0.01 0.0 15 0.02 0.0 25 0.03 0.0 35 0.04 −1 −0 .5 0 0 .5 1 (b) time ... 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)...

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... 833.333m 392.699 1. 250 314. 159 45. 3 45 78 .54 0 392.699 226.7 25 154 .418 25. 642 18.132 89. 153 9 .53 8k 7.818k 314. 159 314. 159 250 .000m Figure 6. 15. Caspoc simulation: simplified ... Tutorial 3, chapter 6 %Tutorial 3, chapter 6 psi1_hat=1. 25; % primary flux amplitude psi1=psi1_hat; % primary flux phasor k =5; % n1/n2 =5 winding ratio Lm=1 .5; % magne...

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... Heidelberg Printed on acid-free paper SPIN: 11760481 62/techbooks 54 3210 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 50 3-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 A...

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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. 855 m -3 3.689 11.182m 118. 755 m -1 2.239 -1 8.668 ... u2(t) t=[0:0.1e-3:60e-3]; u2_t=u2_hat*sin(w*t+u2_rho); plot(t,u2_t /5, ’g’); %plot u2(t) /5 legend(’u2 /5 ,’i_1’,’u2 /5 phasor’) xlabel(’time (s)’) ylabel(’...

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... (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 0.1 0. 15 0.2 0. 25 0.3 0. 35 0.4 −2000 − 150 0 −1000 50 0 0 50 0 1000 150 0 2000 time → [s] phase currents → [A] 0 0. 05 0.1 0. 15 0.2 0. 25 0.3 0. 35 0.4 −2000 − 150 0 −1000 50 0 0 50 0 1000 150 0 2000 time ... (s)’) ylabel(’u_{...

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... corre- sponds to machine configurations discussed in this section. The diagram shows 174 FUNDAMENTALS OF ELECTRICAL DRIVES (a) IRTF-flux (b) IRTF-current Figure 7 .5. Generic representations of IRTF ... 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. Th...

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... 0.30 0.6 25 -3 0.00 196. 35 150 .79 0.36 0. 750 -3 6.87 2 35. 62 176.71 0.43 0.8 75 -4 4.43 274.89 210.43 0 .51 1.00 -5 3.13 314.16 255 . 25 0.60 1.1 25 -6 4.16 353 .43 319.43 0.70 1. 250 -8 9.29 392.67 4 85. 98 0.92 Build ... i F ⊂ [0 .5, , 2. 25] A i F (A) I s (A) Q s (VA) ρ m ◦ 0 .5 0.60 373.06 -5 3.13 0. 75 0.42 241.73 -3 2.23 1.0 0.30 130.96 -2 3 .58 1....

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... Machines 255 0 50 0 1000 150 0 2000 250 0 3000 0 10 20 torque (Nm) 0 50 0 1000 150 0 2000 250 0 3000 0 10 20 RMS stat. cur. (A) 0 50 0 1000 150 0 2000 250 0 3000 0 50 00 10000 Real power (W) 0 50 0 1000 150 0 ... DRIVES 150 0 2000 250 0 3000 0 10 20 30 torque (Nm) 150 0 2000 250 0 3000 0 5 10 15 RMS stat. cur. (A) 150 0 2000 250 0 3000 0 50 00 10000 Real power (W) shaft speed (r...

Ngày tải lên: 09/08/2014, 06:22