上海交通大学：《Design & Manufacturing II and Project》课程教学资源（作业）02_Motor Solution

Motor Homework Solutions Problem 1: Statement: Which type(s)of electric motor would you specify a.To drive a load with large inertia. b.To minimize variation of speed with load variation. c.To maintain accurate constant speed regardless of load variations. Solution: See Mathcad file P0216. Motors with high starting torque are suited to drive large inertia loads.Those with this characteristic series-wound.compound-wound,and shunt-wound DC motors,and capacitor-start AC motors. b. Motors with flat torque-speed curves (in the operating range)will minimize variation of speed with load variation.Those with this characteristic include shunt-wound DC motors,and synchronous and capacitor-start AC motors. c. Speed-controlled DC motors will maintain accurate constant speed regardless of load variations. Problem 2: Refer to the AC motor performance curve in the following figure. a)What type of motor is the curve likely to represent? b)If the motor is a six-pole type,rated at 0.75 hp,how much torque can it exert at the rated load? c)How much torque can the motor develop to start a load? d What is the breakdown torque for the motor? 100 80 60 40 20 0 0 100 200 300 400 Torque (percentage of full-load torque)

Motor Homework Solutions Problem 1: Problem 2: Refer to the AC motor performance curve in the following figure. a) What type of motor is the curve likely to represent? b) If the motor is a six-pole type, rated at 0.75 hp, how much torque can it exert at the rated load? c) How much torque can the motor develop to start a load? d) What is the breakdown torque for the motor?

Sol. a) Single phase,split-phase AC motor b) 6-pole No-load Speed 1200 rpm (Norton Table 2-6) Full-load Speed 1140 rpm 63000H T= n T= 63000H63000×0.75 1140 =41.4bin n c) Tsart =1.5Trated Ta=1.5Tad=1.5×41.4lb-in=62.2lb·in d)Threak =345%Trated T6reak=345%T,aed=345%×41.4=142.81b-in Problem 3: Three examples of motors,with their parameters are listed in the Table below.For each case pick the best motor. a)A rotary load with a moment of inertia of 0.02 Kg m2 is driven by a gear reducer. Select the motor that produces the minimum power dissipation and the motor that results in minimum temperature rise.Find the optimal gear ratio in either case. b)A mass of M=0.75 Kg must be driven by a pulley.Select the motor that can drive it under minimum temperature rise and find the optimal radius size

Sol. a) Single phase, split-phase AC motor b) 6-pole No-load Speed 1200 rpm (Norton Table 2-6) Full-load Speed 1140 rpm c) d) Tbreak  345%Trated Problem 3: Three examples of motors, with their parameters are listed in the Table below. For each case pick the best motor. a) A rotary load with a moment of inertia of 0.02 Kg m2 is driven by a gear reducer. Select the motor that produces the minimum power dissipation and the motor that results in minimum temperature rise. Find the optimal gear ratio in either case. b) A mass of M = 0.75 Kg must be driven by a pulley. Select the motor that can drive it under minimum temperature rise and find the optima1 radius size. 63000H T n  63000 63000 0.75 41.4 1140 H T lb in n      1.5 Tstart  Trated 1.5 1.5 41.4 62.2 Tstart Trated    lbin  lbin 345% 345% 41.4 142.8 Tbreak Trated     lbin

Table ll-Motor Parameters Parameter Motor #1 Motor #2 Motor #3 Motor Parameters Torque constant K [N-/A] 0.1 0.2 0.25 Moment of inertiIm [K.m2] 2105 3-105 1-103 Armature【esistance r (2] 2 g 1.25 Continuous torque Ta [Nl 0.5 2.2 6.2 Derived p2r2me设rs Electromechanical time 0.004 0.001 0.002 constant Tm n s Power rate Tc2/lm 12,000 31,000 38,000 Sol. a)rotary load I =.02kgm'with gear reducer Select motor with min power dissipation and min temperature rise. Find optimal gear ratio 1)Min Power dissipation motor with min electromechanical time constant. Select motor #2 N-(y=(02gm '-25.8 3E-5kgm 2.Min Temperature rise motor withlargest power rate. Select motor #3 N=y= .02kgm 1E-3kgm )=45

Sol. m Tc 2 / Im m

b)mass M=.75kg Select motor with min temperature rise and find optimal radius size. Min temperature rise motor with largest power rate. Select motor #3 Optimal pulley radius inertia matching 12=m2=J 入滑 =.037m =3.7cm Problem 4: A rotating antenna has a moment of inertia(I)of 0.8 lb ft s2.It is driven through a 10: 1 gearbox by a DC motor(motor runs faster).The antenna must be able to rotate 180 degrees from rest in 1 s.The motor characteristics are given below.Find the motor voltage required. 500 400 10V 300 -12V 200 15V 5V 100 6 0 50 1001118150 Torgue(in.·oz) 0 0.5 1.0+ Torque (N-m)

Problem 4: A rotating antenna has a moment of inertia (I) of 0.8 lb ft s 2 . It is driven through a 10: 1 gearbox by a DC motor (motor runs faster). The antenna must be able to rotate 180 degrees from rest in 1 s. The motor characteristics are given below. Find the motor voltage required

3 Rotating antenna I.=0.8lbfis Gear Box 10:1>N=10 6(t=1sec)=180=万rad Find motor voltage Start up Torque Problem 1)Find required load acceleration 6t0-号f (from rest) 9.280 4-是-2md1g au =Na =10*2m=20xrad/s 2)Find inertia I=1(negligable)+ -0.8bt-0.08bm 100 3)Find startup torque T=I*a=0.008lbfis**20xrad /s2 =0.51bft =96.5in-oz 4)Find voltage T =96.5in-oz 0=0 Startup speed is zero 500 400 10V 300 -12V 200 15V 5V 100 0 50 T004150 T:9a.5 Torque (in.-oz) 0.5 1.0 Torque (N.m) V is about 10 voltage