Tespe 18602.13×60 =592N·m To= TemN-T 2x×3000 P。=T09=5.1× =1602.2 Pin=PeN+ P+ P Pemv+TaRa+TR 18602.13+87.6882×0.0896+1.2122×181.5 19557.7w P×100%=869% 220×3000 (4) no =3111.2r/m Cp、E 212.14 (5)因为调速前后T不变,所以l。不变 E。=Ux-(Rn+R)=220-87688×(00896+0.15)=199 3000 ×199=2814.2r/ E E 212.14 1-581-58解:Ea=Ux-IR=220-40×0.5=200V Ea200 C.、= 0.2 1000 )4,C.p。=40×C.=0 所以 E。Ux-lR220-20×05 =2100r/ C 0.1 0.1 Tm=CMpI=(955301)×20 (2)串励且转矩保持不变,所以L不变,C不变 Ⅰ。=259.2 2 18602.13 60 =   = =  n P T N N emN emN  N·m （3） 59.2 54.1 5.1 0 T =T −T = − = emN N N·m 1602.2 60 2 3000 5.1 0 0 =  =  =   P T w P N PemN Pcua Pcuf = + + 1 PemN I aRa If Rf 2 2 = + + ＝18602.13+87.6882×0.0896+1.2122×181.5 ＝19557.7 w 100% 86.9% 1 =  = P P N N  N （4） r m E U n C U n aN N N e N N 3111.2 / 212.14 220 3000 0 =  = = =  （5）因为调速前后 T em 不变，所以 I a 不变 R R V Ea UN I a a ( ) 220 87.688 (0.0896 0.15) 199 ' = − + = −  + = E r m E n n a aN N 199 2814.2 / 212.14 ' ' 3000 =  =  = 1－58 1－58 解： = − R = 220 − 400.5 = 200 EaN UN I N a V 0.2 1000 200 = = = n E C N aN e N （1） A I a 20 ' = ， 0.1 40 ' 20  =   = Ce N Ce N 所以 r m U I R C E n N a a e a 2100 / 0.1 220 20 0.5 0.1 ' ' ' ' = −  = − = =  (9.55 0.1) 20 19.1 ' ' ' = =   = Tem C M  I a Ｎ·m （2）串励且转矩保持不变，所以 I a 不变，  ' Ce 不变 I a ＝20 A