华科大辜承林主编《电机学》课后习题答案

33202?0.6667 kq3??3203sin2sinkN3?ky3kq3??0.3333

E3?4.44fNkN3?3?4.4450360.33330.13?799.11(V)

E??E12?E32?5462(V)

El?3E1?9363(V)

注:因为星接,故线电势中无三次谐波

4.21 JO2-82-4三相感应电动机,PN=40kW,UN=38V,IN=75A,定子绕组采用三角形联接,双层叠绕组,4极,48槽,y1=10槽,每极导体数为22,a=2,试求:

(1)计算脉振磁动势基波和3、5、7等次谐波的振幅,并写出各相基波脉振磁动势的表达式;

(2)当B相电流为最大值时,写出各相基波磁动势的表达式;

(3)计算三相合成磁动势基波及5、7、11次谐波的幅值,并说明各次谐波的转向、极对数和转速;

(4)写出三相合成磁动势的基波及5、7、11次谐波的表达式;

(5)分析基波和5、7、11次谐波的绕组系数值,说明采用短距和分布绕组对磁动势波形有什么影响。 (1)I??IN75??43.3(A) 33

?1?q?p3602360??15? z4848?443

48???124ky1?sin(sin1090?)?sin(90?)?0.9659 ?12y1q?1415sin2?2?0.9577kq1??15qsin14sin22kN1?ky1kq1?0.925 N?2pq44Nc?11?88a2

F?1?0.9kN3NkN1880.925I??0.943.3?1586.1P2?ky3kq3??0.4619

kN5?ky5kq5?0.05314kN7?ky7kq7??0.04077688(?0.4619)43.3??26423880.05314F?5?0.943.3?18.245288(?0.040776)F?7?0.943.3??9.9972F?3?0.9(2)

FA1?1586.1coswtcos? FB1?1586.1cos(wt?120?)cos(??120?)

FC1?1586.1cos(wt?240?)cos(??240?)NkN13I??1586.1?2379.2 P260f6050 基波:正转,n1???1500rpm,p?2

P23 5次谐波:F5?F?5?1.518.24?27.36

2n 反转,n5???300rpm,pv?vp?10对极

53 7次谐波:F7?F?7?1.5(?9.99)??15

21500 正转,n7??214.3rpm,pv?vp?14

7 (3)F1?1.35sin(1130?)?0.5?0.9659??0.12184sin(117.5)3.965888(?0.1218)F?0.943.3??18.98 ?11 1123F11?F?11?1.5(?18.98)??28.472kN11?sin1175? 反转,p11?112?22对极 n11?1500?136.36rpm 11(4)f1(t,?)?F1cos(wt??)?2379.2cos(wt??)

f5(t,?)?27.36cos(wt?5?) f7(t,?)??15cos(wt?7?)

f11(t,?)??28.47cos(wt?11?)

(15)kN1:kN5:kN7:kN11?1:0.0574:0.0441:0.1317

采用短距分布后,5,7次谐波幅大为减少

4.22 一台50000 kW的2极汽轮发电机,50Hz,三相,UN=10.5 kV星形联接, cosфN=0.85,定子为双层叠绕组,Z=72槽,每个线圈一匝,y1=7τ/9,a=2,试求当定子电

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