tailieunhanh - (McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 2 Part 5

Tham khảo tài liệu '(mcgraw-hill) (instructors manual) electric machinery fundamentals 4th edition episode 2 part 5', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | The resulting torque-speed characteristic is shown below 10-5. A 220-V 50-Hz two-pole capacitor-start induction motor has the following main-winding impedances R Q X1 Q XM 105 Q R2 Q X2 Q At a slip of the motor s rotational losses are 291 W. The rotational losses may be assumed constant over the normal operating range of the motor. Find the following quantities for this motor at 5 percent slip a Stator current b Stator power factor c Input power d Pag e Pconv f Pout g Tnd h Tload i Efficiency Solution The equivalent circuit of the motor is shown below 275 a 11 jX V 220Z0 V _ Rị s jX2 jXM ZF R2 s jX 2 jXM Z A 0 Q F 30 j100 Rị 2 s jX 2 jXM R2 2 s jX2 jXM j100 __ ZB j Q B j j100 ZB The input stator current is V I1 I1 R jX 1 220Z0 V s 1. Forward Rị 2 s Reverse A b The stator power factor is PF cos 27 lagging c The input power is PIN VIcos0 220 v A cos 27 2548 W d The air-gap power is PAG F I12 A 2 Q 2246 W PAG B I12 A 2 Q W PAG PAG F PAG B 2246 W W 2184 W 276 e The power converted from electrical to mechanical form is PconvF 1 - 5 Pag f 1 - 2246 W 2134 W Pconv B 1 - 5 Pag b 1 - W 59 W Pconv PconvF - PonvB 2134 W - 59 W 2075 W f The output power is Pout Pconv - Prot 2134 W - 291 W 1843 W g The induced torque is Tind --------2I84 W N m rn 2n rad 1 min sync 3000 r min v 7 1 r 60 s h The load torque is 1843 W POUT load . 2nrad 1 min 6-18 F 3000 r min Ỵ - - i The overall efficiency is P 1843 W n Z X100 1843 W X 100 PIN 2548 W 10-6. Find the induced torque in the motor in Problem 10-5 if it is operating at 5 percent slip and its terminal voltage is a 190 V b 208 V c 230 V. Z R2 5 jX 2 jX M F R2 5 jX 2 jXM Z ijji j Q F 30 j100 Z R2 2 5 jX 2 jXM B R2 2 - 5 jX2 jXM j100 __ ZB Q B .