Texts: Answer 8, 9, 10
A starter is to be designed for a 10 kW, 250 V shunt motor. The armature resistance is 0.15 ohms. This motor is to be started with a resistance in the armature circuit so that during the starting period the armature current does not exceed 200% of the rated value or fall below the rated value. That is, the machine is to start with 200% of armature current and, as soon as the current falls to the rated value, sufficient series resistance is to be cut out to restore the current to 200% (or less in the last step). The process is to be repeated until all the resistance is cut out.
a) Calculate the total resistance of the starter.
b) Also calculate the resistance to be cut out in each step in the starting operation.
A 200 V, 14.92 kW DC shunt motor, when tested by the Swinburne method, gave the following results:
- Running light: armature current was 6.5 A and field current was 2.2 A.
- With the armature locked, the current was 70 A when a potential difference of 3 V was applied to the brushes.
Estimate the efficiency of the motor when working under full-load conditions.
0) A shunt-wound motor runs at 600 r.p.m. from a 230 V supply when taking a line current of 50 A. Its armature and field resistances are 0.4 ohms and 104.5 ohms, respectively. Neglecting the effects of armature reaction and allowing a 2 V brush drop, calculate:
a) The no-load speed if the no-load line current is 5 A.
b) The resistance to be placed in the armature circuit in order to reduce the speed to 500 r.p.m. when the motor is taking a line current of 50 A.
c) The percentage reduction in the flux per pole in order that the speed may be 750 r.p.m. when the armature current is 30 A with no added resistance in the armature circuit.
