Torque and Speed of a D.C. Motor
For any motor, the torque and speed are very important factors. When the torque increases, the speed of a motor increases and vice-versa. We have seen that for a d.c. motor;
N = K (V- IaRa)/ Ф = K Eb/ Ф…………………………………………….(i)
Ta α ФIa…………………………………………………………………………(ii)
If the flux decreases, from Eq.(i), the motor speed increases but from Eq.(ii) the motor torque decreases. This is not possible because the increase in motor speed must be the result of increased torque. Indeed, it is so in this case. When the flux decreases slightly, the armature current increases to a large value. As a result, in spite of the weakened field, the torque is momentarily increased to a high value and will exceed considerably the value corresponding to the load. The surplus torque available causes the motor to accelerate and back e.m.f (Ea=PФZN/60A) to rise. Steady conditions of speed will ultimately be achieved when back e.m.f. has risen to such a value that armature current[Ia = (V- Ea)/ Ra]develops torque just sufficient to drive the load.
Illustration
Let us illustrate the above point with a numerical example. Suppose a 400 V
shunt motor is running at 600 r.p.m., taking an armature current of 50 A. The armature resistance is 0.28 Ω. Let us see the effect of sudden reduction of flux by 5% on the motor.
Let us illustrate the above point with a numerical example. Suppose a 400 V
shunt motor is running at 600 r.p.m., taking an armature current of 50 A. The armature resistance is 0.28 Ω. Let us see the effect of sudden reduction of flux by 5% on the motor.
Initially (prior to weakening of field), we have,
Ea = V-IaRa= 400 – 50 × 0.28 = 386 volts
We know that Eb α Ф N. If the flux is reduced suddenly, Eb α Ф because inertia
of heavy armature prevents any rapid change in speed. It follows that when the flux is reduced by 5%, the generated e.m.f. must follow suit. Thus at the instant of reduction of flux, E’b = 0.95 × 386 = 366.7 volts.
of heavy armature prevents any rapid change in speed. It follows that when the flux is reduced by 5%, the generated e.m.f. must follow suit. Thus at the instant of reduction of flux, E’b = 0.95 × 386 = 366.7 volts.
Instantaneous armature current is
I’a=(V- E’b)/ Ra =(400-366.7)/0.28=118.9A
Note that a sudden reduction of 5% in the flux has caused the armature current to increase about 2.5 times the initial value. This will result in the production of high value of torque. However, soon the steady conditions will prevail. This will depend on the system inertia; the more rapidly the motor can alter the speed, the sooner the e.m.f. rises and the armature current falls.
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