1.
1. A 220 V, 500 A, 600 rpm separately excited motor has armature and field resistance ofA 220 V, 500 A, 600 rpm separately excited motor has armature and field resistance of 0.02 and 10
0.02 and 10 respectively. The load torque respectively. The load torque is given the expressionis given the expression T
TLL = 2000 – 2 = 2000 – 2 N N ,,
where
where N N is the speed in rpm. Speeds below the rated are obtained by armature voltage is the speed in rpm. Speeds below the rated are obtained by armature voltage
control and speeds above the rated are obtained by field control. control and speeds above the rated are obtained by field control.
i) Calculate motor terminal voltage and armature current when the speed is 450 rpm. i) Calculate motor terminal voltage and armature current when the speed is 450 rpm. ii) Calculate field winding voltage and armature current when the speed is 750 rpm. ii) Calculate field winding voltage and armature current when the speed is 750 rpm. Assume the rated field voltage is the same as the rated armature voltage.
Assume the rated field voltage is the same as the rated armature voltage. V =220V , I
V =220V , Ia1a1 = 500 A, N = 500 A, N11 = 600 rpm, R = 600 rpm, R aa = 0.02 ohm, R = 0.02 ohm, R f f = 10 ohm, = 10 ohm,
T
TLL = 2000 _ = 2000 _ 2N 2N Nm ( Nm ( separately excited mseparately excited motor)otor)
(i) V =? I (i) V =? Ia2a2 = = ? ? (450 rpm(450 rpm)) T T22 = 2000_ 2 (450) = = 2000_ 2 (450) = 1100 Nm1100 Nm E E11 = V _ I = V _ Ia1a1R R aa = 220 _ 500 (0.02) = 210 V = 220 _ 500 (0.02) = 210 V T T11 = P / W = P / Wmm= (E= (Ea1a1 IIa1a1) / W) / Wmm = 210 x 500 / ( 600 x 2 = 210 x 500 / ( 600 x 2 / 60)/ 60) = 1671 Nm = 1671 Nm T T22 / T / T11 = I = Ia2a2/ I/ Ia1a1 IIa2a2 = (T= (T22 / T / T11) x I) x Ia1a1 = (1100 / 1671 ) x 210 = 138.24 A = (1100 / 1671 ) x 210 = 138.24 A E E22= V _ I= V _ Ia2a2R R aa = 220 _ (138.24 = 220 _ (138.24 x 0.02) = 217.2 Vx 0.02) = 217.2 V 600 rpm ---E = 217.2 V 600 rpm ---E = 217.2 V 450 rpm ---E = ? 450 rpm ---E = ? E= (450 / E= (450 / 600) 600) x 217.2 x 217.2 = 163 = 163 VV V = 163 + (138.24 x 0.02) = 165.8 V V = 163 + (138.24 x 0.02) = 165.8 V (ii) V (ii) Vf f = ?, I = ?, Ia2a2= ?, N = 750 rpm= ?, N = 750 rpm T T22 = 2000_ 2 (750) = = 2000_ 2 (750) = 500 Nm500 Nm T T22 = T = T22 K K eeø Iø Ia2a2 = 500 = 500 K K eeø =(500) / Iø =(500) / Ia2a2 V = E V = E22 + I + Ia2a2 R R aa 220 = K 220 = K ee ø N ø N22 (2 (2 /60) + 0.02I/60) + 0.02Ia2a2 220 =(500 / I 220 =(500 / Ia2a2 ) x 750 x (2 /60) + 0.02I ) x 750 x (2 /60) + 0.02Ia2a2 0 = 0.02I
0 = 0.02I22a2a2 _ 220I _ 220Ia2a2 + 39270+ 39270
IIa2a2= 10818.5 A (or) 181.5 A= 10818.5 A (or) 181.5 A K K ee ø ø11 = E = E11 / W / Wm1m1 = 210 / (600 x 2 /60) = 3.342 = 210 / (600 x 2 /60) = 3.342 K K ee ø ø22 = 500 / I = 500 / Ia2a2 = 2.75 = 2.75 K K ee ø ø11 = = 3.342---3.342---V--Vff= 220 V= 220 V K K ee ø ø22 =2.75 =2.75--- V--- Vff=?=?
=(2.75 / 3.342) x 220 = 181.03 V Vf=181.03 V
2. A 220 V, 1500 rpm, 10 A separately excited D.C. motor has an armature resistance of 1 Ohm. It is fed from a single phase fully controlled bridge converter with an AC source voltage of 230 V, 50 Hz. Assuming continuous load current. Calculate (i) Motor speed at the firing angle of 300 and torque of 5 N-m, (ii) Developed torque at the firing angle of 450 and speed of 1000rpm.
3. The chopper used for on-off control of a D.C. separately excited motor has supply voltage of 230 V DC, an on time of 10 ms and off time of 15 ms. Neglecting armature inductance and assuming continuous conduction of motor current. Calculate the average load current when the motor speed is 1500 rpm and has voltage constant of 0.4 V/rad/sec. The armature resistance is 2 ohms.
(ii) The speed of motor for α=45o and Ia=20A.
5. A 230V, 960rpm and 200A separately excited dc motor has an armature resistance of 0.02 ohms. The motor is fed from a chopper which provides both motoring and braking operations. Assume continuous conduction.
i) Calculate duty ratio of chopper for motoring operation at rated torque and 350rpm ii) Calculate duty ratio of chopper for braking operation at rated torque and 350rpm iii) If maximum duty ratio of chopper is limited to 0.95 and maximum permissible motor
current rating is twice the rated, calculate the maximum permissible motor speed obtainable without field weakening and power fed to the ac source
torque?
4. A three phase full wave bridge converter is connected to the armature of a dc motor. The motor runs at 1000 rpm at no load, when connected to a 220V dc source with R a=1 ohm.
Find
(i) Line to line voltage of the co nverter so that maximum output voltage of the co nverter is equal to the rated voltage of the motor.
6. A 230 V, 1200 rpm, 15 A separately excited D.C. motor has an armature resistance of 1.2 ohms. Motor is operated under dynamic braking with chopper control. Braking resistance has a value of 20 ohms.
(i) Calculate duty ratio of chopper for motor speed of 1000 rpm and braking torque equal to 1.5 times rated motor torque.
7. A 220V, 970 rpm, 100A dc separately excited motor has armature resistance of 0.05 ohms. It is braked by plugging from an initial speed of 1000 rpm. Calculate
i) Resistance to be placed in armature circuit to limit the braking current to twice the full load
ii) Braking torque
iii) Torque when speed has fallen zero
ii) firing angle for rated motor torque and -500 rpm (3) iii) motor speed for firing angle is 160 degrees & rated torque (2+2)
