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1、13章12周电机学作业13-1 解： E =U2+(Ix)2=2=1.414 0*s*DU%=0*UE-U*100%=41.4% Ixd=arctan*s*=45o U*I=0.9 *q=31.79o E=U+jIx0*s*=1+j0.9-31.79o =1.474+j0.765=1.6627.43oE0*=1.66 d=27.43o DU%=1.66-1100%=66% 1I=0.9 *q=31.79o E=U+jIx0*s*=1+j0.931.79o =0.526+j0.765=0.92855.49oE0*=0.928 d=55.49o DU%=0.928-1100%=-7.2% 113-
2、2 解： o设U=10 则I=10o *N*UN*+jIx=1.0+j0.6 *q*0.6=30.96o 1d=arctanj=q+d=0o+30.96o=30.96o Id*=Isinj=0.514 Iq*=Icosj=0.857 E=Ucosd+Ix0*Nd*d*=1cos30.96o+0.514 =1.3721.372-1DU%=100%=37.2% 1U2(x-x)*d*q*P=0*sind+sin2d M*x2xxd*d*q*=1.372sind+0.333sin2dEUo=10 则I=0.9 设UN*I=0.765-j0.474=0.931.79o *UN*+jIx=1.284+j
3、0.459 *q*0.459=19.67o 1.284d=arctanj=q+d=31.79o+19.67o=51.46o Id*=Isinj=0.704 Iq*=Icosj=0.561 E=Ucosd+Ix0*Nd*d*=1cos19.67o+0.704 =1.6461.646-1DU%=100%=64.6% 1U2(x-x)*d*q*P=0*sind+sin2d M*x2xxd*d*q*=1.646sind+0.333sin2dEUo设U=10 则I=0.9 N*I=0.765+j0.474=0.9-31.79o *UN*+jIx=0.716+j0.459 *q*0.459=32.66o
4、0.716d=arctanj=q+d=32.66o-31.79o=0.87o Id*=Isinj=0.014 E=Ucosd+Ix0*Nd*d*=1cos32.66o+0.014 =0.8560.856-1DU%=100%=-14.4% 1U2(x-x)*d*q*P=0*sind+sin2d M*x2xxd*d*q*=0.856sind+0.333sin2dEU13-3 解： Ud=UN=105000=6.06KV 33P3UN=24000310.5=1.32KA I=I=dl故Z=Ud=6.06=4.6W I1.32dxx5q2.76d所以x=1.09 x=0.6 d*q*Z4.6Z4.6U
5、2(x-x)*d*q*0*P=sind+sin2d *x2xxd*d*q*=1.376sind+0.375sin2dEU20MKVA=0.833 *24MKVA解方程 1.376sind+0.375sin2d=0.833 得 d=23.8 oP= -Ucosd1.5-1cos23.8o*I=0.537 d*x1.09d*Usind1sin23.8o I=*=0.673 q*x0.6q*=0*EI=I2+I2=0.86 *d*q*j=arctand*=38.57o Iq*Iq=j-d=14.77o 所以Q=SUIsinq=24MVA10.86sin14.77=5.256MVA *odP*=1.3
6、76cosd+0.75cos2d=0 dd因为1cosd1，解这个方程得cosd=0.3842 即 d=67.4 所以sind=0.923 sin2d=0.71 故Pomax*=1.3760.923+0.3750.71=1.54 P=1.5424MW=36.96MW max 13-4 解： /E=U+jIx ，U=jIx+E0* *s*0*s*/所以E0*=2U-E *0*o令U=10，P=0.5 *U0*sind=0.5d=24.62o 则P=*xs*=U+jIx=1.224.62o 所以E0*s*/E0*=2U-E=2-1.224.62o=1.038-28.81 *0*EEUP=0*sin
7、d=0.5 *xs*E0*/=1.1时E0*=1.038不变 1)=61.24o 1.11.04j=arcsin(2Ix=E2+E/2-2EE/cos61.24o=1.09 *s*0*0*0*0*/2U=0.52E2+E0*-(2Ix)2)=0.92 0*0*s*13-6 解： o=0.8+j0.6 设U=10 ID*I=-0.8-j0.6 *U+jIx=1.36-j0.48=1.44-19.44o *q*故d=-19.4 oq=arccos0.8=36.9o 所以Id*=Isin(q-d)=0.832 *E=Ucosd+Ix=1cos(-19.4o)+0.8321=1.77 0*d*d*所以该电动机在过励状态下运行。
