2015 /1/19 اإلجابة النموذجية لمادة نظم التشغيل الكهربية ك 563 د شوقي حامد عرفه ابراهيم يوم االثنين الموافق Benha University Benha Faculty of Engineering Subject: Electrical drives (E563) Time: 3hours Fifth Year 2014/2015 Elect.Eng.Dept. Solve & draw as much as you can (questions in two pages) Question (1) [15] Points a- Write about: Electrical drive- Speed control of a separately excited DC motor? b- A [15 hp, 220V, 2000 rpm] separately excited DC motor drives a load requiring a torque of 50 Nm at a speed of 1200 rpm. R a = 0.25Ω, R f = 147Ω, KΦ = 1.5Nm A, V f = 220V. a- Find the armature current and the field current? b- Find the armature voltage required? c- Find the motor speed if it is supplied by fully controlled single phase rectifier with 30 degrees firing angle and 240 V, 60Hz AC supply and assume constant armature current? d- Draw the wave forms of voltages and currents and the power circuit for c? Question (2) [10] Points a- Explain how to control the speed of the 3-phase induction motor? b- A three phase 460V, 60Hz, 4 poles, wye-connected induction motor has a stator impedance of (0.4+j0.8) Ω/phase and (0.2+j0.8)Ω/phase of the rotor winding referred to the stator side. The exciting branch impedance viewed from the stator side is (0 Ω j25 Ω). The no load loss= 100 watt and may be assumed constant. The rated rotor speed is 1700rpm. i-draw the equivalent circuit? ii-find the ω syn, S, I s, I r, P gap, P copper, T dev? Question (3) [10] Points a-explain the AC/AC converter used to control the speed of the 3-phase induction motor? b-if the motor in question 2 is connected to a 3-phase full wave AC/AC converter. Find the firing angle required to run the motor with speed range of 1000 rpm to 1700 rpm. P.T.O.
Question (3) [10] Points a- A three phase 460V, 60Hz, 8 poles, wye-connected cylindrical rotor synchronous motor has a synchronous reactance of 2 Ω/phase. R s is negligible and I s =20A/phase and unity p.f. i-draw the equivalent circuit? ii-find the rotor speed and the torque angle? iii-find the P out and the maximum torque? b-for a speed control of a separately excited DC motor using the closed loop control system. Draw the steady state block diagram and prove that ω r V r = K 1 Kφ ВR a +Kφ K 1 K 2 +Kφ ω r Ƭ w = R a ВR a +Kφ K 1 K 2 +Kφ Question (4) [15] Points a- A 40 Kw, 240V,1150 rpm separately DC motor is to be used in a speed control system. The field current is held constant at a value for which kφ = 1.95V.s/rad. Armature resistance and viscous friction factor are R a = 0.089Ω, В = 0.275 N. m. s rad The tachometer delivers10v/1000 rpm and the amplification of the controller and power modulator is 200. i- Find V r required to drive the motor with no load? ii- Find the motor speed if V r is not changed and the motor supplied rated torque? b- If the motor in question 1 is connected to class A chopper with DC of 400 V and duty cycle is equal to 50% assume constant armature current? i-find the armature current?? ii-find the motor speed? iii-draw the wave forms of voltages and currents and the power circuit?
Answer Question (1) [15] Points a- Write about: Electrical drive- Speed control of a separately excited DC motor? The electrical Drive is a system converts the electrical Energy to the mechanical energy with electrical control. Drive types are: 1-line shaft drive 2-single motor single load drive 3-multimotor drives.
b- A [15 hp, 220V, 2000 rpm] separately excited DC motor drives a load requiring a torque of 50 Nm at a speed of 1200 rpm. R a = 0.25Ω, R f = 147Ω, KΦ = 1.5Nm A, V f = 220V. a-find the armature current and the field current? b-find the armature voltage required? c-find the motor speed if it is supplied by fully controlled single phase rectifier with 30 degrees firing angle and 240 V, 60Hz AC supply and assume constant armature current? d-draw the wave forms of voltages and currents and the power circuit for c? I f =220/147=1.5A, I a =50/1.5=33.3A, E a =1200*2pi*1.5/60=188.5V Va=E a +IaRa =188.5+33.3*0.25=196.82V V a =2V max sinα=(2*240*1.414sin30)/pi=108v ω r =(108-8.325)/1.5=66.5rad/s=66.5*30/pi=635rpm
Question (2) [10] Points a-explain how to control the speed of the 3-phase induction motor? Using three tools are 1-inverter 2-AC/AC converter 3-cycloconverter b-a three phase 460V, 60Hz, 4 poles, wye-connected induction motor has a stator impedance of (0.4+j0.8) Ω/phase and (0.2+j0.8)Ω/phase of the rotor winding referred to the stator side. The exciting branch impedance viewed from the stator side is (0 Ω j25 Ω). The no load loss= 100 watt and may be assumed constant. The rated rotor speed is 1700rpm. i-draw the equivalent circuit? ii-find the ω syn, S, I s, I r, P gap, P copper, T dev? Figure 6.9 Single-phase equivalent circuit for a three-phase induction motor. Equivalent circuits with the core-loss resistance Rc neglected
n s =120*60/4=1800rpm, ω s =1800*pi/30=188.5rad/s S1=(n s -n r )/ n s =0.056, ω r =1700*pi/30=178.02rad/s, S2=(n s -n r )/ n s =0.44 R 2 S1 = 0.2 0.056 = 3.6, Z 2 = 3.6 + j0.8 = 3.7 12.53 Ω, j25 Z 2 = 3.54 20.5 = 3.32 + j1.24 Ω (j25 Z 2)+ Z 1 = 3.72 + j2.04 Ω = 4.24 29Ω, 460/ 3=265.6 0V I s1 =(265.6 0/4.24 29)=62.64-29A I' r = Copper losses= 3I L 2 R eq =W Output power= Pgap-losses= η=p out /p in = p out /(p out +losses)= P mech = = W, T mech = = Nm P shaft = W, T shaft = Nm Question (3) [10] Points a-explain the AC/AC converter used to control the speed of the 3-phase induction motor?
b-if the motor in question 2 is connected to a 3-phase full wave AC/AC converter. Find the firing angle required to run the motor with speed range of 1000 rpm to 1700 rpm. n s =120*60/4=1800rpm, ω s =1800*pi/30=188.5rad/s S1=(n s -n r )/ n s =0.056, ω r =1700*pi/30=178.02rad/s, S2=(n s -n r )/ n s =0.44 I a1 =(265.6 0/4.24 29)=62.64-29A, I a2 =, ( I a /I a rat )^2= [S(1-S)^2]/ [S rat (1-S rat )^2]=( ST/S rat T s rat ) I N =I s2 /I s1 = From table or chart Question (3) [10] Points
a- A three phase 460V, 60Hz, 8 poles, wye-connected cylindrical rotor synchronous motor has a synchronous reactance of 2 Ω/phase. R s is negligible and I s =20A/phase and unity p.f. i-draw the equivalent circuit? ii-find the rotor speed and the torque angle? iii-find the P out and the maximum torque? n r =n s =120*60/8=900rpm, ω r =ω s =900*pi/30=94.25rad/s V f = V t ji a X s =460/ 3=265.6 0-j20*2=265.6-j40=268.6-8.6V Torque angle=δ=-8.6 o, P out = 3V f V t sin δ 3 268.6 265.6 sin 8.6 = = 15936.3W X s 2 T max = 3V f V t 3 268.6 265.6 = = 1135.4Nm ωsx s 94.25 2 b-for a speed control of a separately excited DC motor using the closed loop control system. Draw the steady state block diagram and prove that ω r V r = K 1 Kφ ВR a +Kφ K 1 K 2 +Kφ ω r Ƭ w = R a ВR a +Kφ K 1 K 2 +Kφ
Question (4) [15] Points a- A 40 Kw, 240V,1150 rpm separately DC motor is to be used in a speed control system. The field current is held constant at a value for which kφ = 1.95V.s/rad. Armature resistance and viscous friction factor are R a = 0.089Ω, В = 0.275 N. m. s rad The tachometer delivers10v/1000 rpm and the amplification of the controller and power modulator is 200. i- Find V r required to drive the motor with no load? ω nl =(240)/1.95=123.1rad/s=1175.5rpm, T rat =40000*30/(1150*pi)=332.15Nm ω/v r =300/30.922=9.7, V r =9.7*123.1=12.7V ii- Find the motor speed if V r is not changed and the motor supplied rated torque? -ω/t w =0.089/30.922=0.0029, -ω =0.0029*332.15=0.96rad/s ω =123.1-0.96=122.14rad/s=1166.4rpm b- If the motor in question 1 is connected to class A chopper with DC of 400 V and duty cycle is equal to 50% assume constant armature current? i-find the armature current?? ii-find the motor speed? iii-draw the wave forms of voltages and currents and the power circuit?
I a rat =(40000/240)=166.7A, V ave =0.5*400=200V, ω=(200-166.7*0.089)/1.95=94.96rad/s=906.8rpm