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Torque- Curve 9.55P 9.55|ER |2 9.55| ER | 2 T r g2  g 2 E 2 2 ns  n 2 R2 g n R s 2 R2 ns s, PIr  |1 | , I 1  Z s n n s R 1 s Z1 (ns n ) s Z  2 s n n 1 s s

Stable operating Region (n, T)

no-load

• The curve is nearly linear between no-load and full-load because s is small and

R2/s is big (Z1 is ignored) 2 2 9.55 |Eg | 9.55 | E g | I1 sEg / R 2 T2 (ns  n ) s R2 nsR 2 n s 33 Two practical squirrel-cage induction motors

34 The 5 hp induction motor Torque-Speed Curve

Eg=440/1.73 V R2=1.2  Z1=1.5+6j  9.55|E |2 R T  g 2 n =1800 r/min 2 s R2 ns Z1 (ns n ) ns n

Torque-Slip Curve Current-Speed Curve

E 9.55|E |2 R I  g T  g 2 1 R n R 2 Z  2 s Z 2 s n 1 n  n 1 s s s

35 Torque-Speed Curve: 5 hp motor

s I1 Pr T n

• When the motor is stalled, i.e. locked-rotor condition, the current is 5-6 the full-load current, making I2R losses 25-36 times higher than normal, so the rotor must never remain locked for more than a few

• Small motors (15 hp and less) develop their breakdown torque at about 80% of ns 36 Torque-Speed Curve: 5000 hp motor

Big motors (>1500 hp) : • Relatively low starting (locked-rotor) torque • Breakdown torque at

about 98% of ns • Rated n is close to ns

37 Effect of rotor resistance (5 hp induction motor) 9.55|E |2 R E g 2 I  g T  2 1 R 2 n s R2 ns Z 1  Z1 (ns n ) n  n n n s R2 s

Rated Torque =15Nm Current at the rated torque

2.5R2

Rated torque =15Nm Current at the rated torque

10R2 Rated torque =15Nm Current at the rated torque

38 Effects of rotor resistance

• When R2 , starting (locked-rotor) torque TLR , starting current I1,LR , and breakdown torque Tb remains the same 2 2 2 9.55|Eg | 9.55|Eg |R2 9.55| E g | Eg T  T  R 2 b LR 2 2 2 I1,LR  Pjr |I1 | R2 2  |Z1 R2 | ns ||Z 1 n s Z  R 4n | Z | cos 1 2 s 1 2

• Pros & Cons with a high rotor resistance R2

– It produces a high starting torque TLR and a relatively low starting current I1,LR – However, because the torque-speed curve becomes flat it produces a rapid fall- off in speed with increasing load around the rated torque, and the motor has high copper losses and low efficiency and tends to overheat • Solution – For a squirrel-cage induction motor, design the rotor bars in a special way so that the rotor resistance R2 is high at starting and low under normal operations – If the rotor resistance needs to be varied over a wide range, a wound-rotor motor needs to be used.

39 P P Asynchronous generator js f Pjr Connect the 5 hp, 1800 r/min, 60Hz motor Pe P r Pm to a 440 V, 3-phase line and drive it at a speed of 1845 r/min s=(ns-n)/ns=(1800-1845)/1800= -0.025 <0

R2/s=1.2/(-0.025)= - 48 <0 The negative resistance indicates the actual flow from the rotor to the stator Power flow from the rotor to the stator: |E|= 440/1.73=254 V

|I1|=|E|/ |-48+1.5+ j6|=254/46.88= 5.42A 2 Pr=|I1| R2/s= -1410 W (in fact, rotor  stator) Mech. power & torque inputs to the shaft: Reactive power absorbed from the line: P =|I |2R =35.2 W 2 2 jr 1 2 Q3=(|I1| x+|E| /Xm)3=(176+586)3=2286 var Pm=Pr+Pjr=1410 + 35.2=1445W Complex power delivered to the line: T=39.55 P /n=22.3 Nm m S3=P3 - Q3=3882 - j2286 VA Total active power delivered to the line: cos=86.2% 2 2 Pjs=|I1| r1=44.1 W, Pf =|E| /Rm=71.1 W Efficiency of this asynchronous generator P =P –P –P = 1410-44.1-71.7=1294 W e r js f =Pe/Pm=1294/1445=89.5% P3=3Pe=3882W 40