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Rheostatic Control

  • This method consists of obtaining reduced speeds by the insertion of external series resistance in the armature circuit. It can be used with series, shunt and compound motors; for the last two types, the series resistor must be connected between the shunt field and the armature, not between line and the motor.
  • It is common method of speed control for series motors and is generally analogous in action; to wound-rotor induction-motor control by series rotor resistance.
    • This method is used when speeds below the no-load speed is required.

Advantages:

1.The ability to achieve speeds below the basic speed.

2.Simplicity and ease of connection.

3.The possibility of combining the functions of motor starting with speed control.

Disadvantages:

  1. The relatively high cost of large, continuously rated, variable resistors capable of dissipating large amounts of power (particularly at higher power ratings).
  2. Poor speed regulation for any given no-load speed setting.
  3. Low efficiency resulting in high operating cost.
  4. Difficulty in obtaining stepless control of speed in higher power ratings.

Shunt motors:

  • ·In armature or rheostatic control method of speed the voltage across the armature (which in normally constant) is varied by inserting a variable rheostat or resistance, called  controller resistance, in series with the armature circuit. As the controller resistance is increased, the potential difference across the armature is decreased thereby decreasing the armature speed. For a load of constant torque, speed is approximately proportional to the potential difference across the armature

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From the speed/armature current characteristic it is seen that greater the resistance in armature, greater is the fall in speed.

There is a particular load current for which the speed would be zero. This is the maximum current and is known as ‘stalling current’.

  • This method is very wasteful, expensive and unsuitable for rapidly changing load, because for a given value of R, the speed will change with load. A more stable operation can be obtained by using a divertor across the armature (Fig. 88) in addition to armature control resistance. Now, the changes in armature current due to changes in the load torque will not be so effective in changing the potential difference across the armature and hence the
    speed of the armature.

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By increasing the resistance in series with the armature the voltage applied across the armature terminals can be decreased. With the reduced voltage across the armature, the speed is reduced.

Since full motor current passes through the resistance, the loss of power is considerable. Although terminal-voltage control by means of a variable voltage supply would effectively control the speed of a D.C. series motor, the high cost of the control equipment is seldom warranted.

Series-parallel control:

  • This system is widely used in electric traction. Here two or more similar mechanically-coupled motors are employed.
  • At low speeds the motors are joined ill series as shown in Fig. 91 (a). The additional resistance is gradually cut out by controller as the motors attain the speed, and finally the resistance is totally removed, then each motor has half of line voltage. In this arrangement, for any given value of armature current, each motor will run at half of its normal speed. As there is no external resistance in the circuit, therefore there is no waste of energy and. so motors operate at all efficiency nearly equal to that obtainable with full line voltage across the terminals of each motor.

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Example 40. The armature and shunt field resistances of a 230 V shunt motor are 0.1 ohm and 230 ohms respectively. It takes a current of 61 A at 1000 r.p.m. If the current taken remains unaltered find the resistance to be included in series with the armature circuit to reduce the speed to 750r.p.m.

Solution. Supply voltage, V = 230 Volts

Armature resistance, Ra = 0.1 ohm

Shunt field resistance, Rsh = 230 ohms

Load current, I= 61 A

Speed, N1 = 1000 r.p.m

Speed, N2 = 750 r.p.m,

Additional resistance required, R :

Shunt field current,

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Example 41. The armature resistance of a 230 V D.C. shunt motor is 0.2 ohm. It takes 15 A at rated voltage and runs at 800 r.p.m. Calculate the value of additional resistance required in the armature circuit to reduce the speed to 600 r.p.m. when the load torque is independent of speed.

Ignore the field current.

Solution. Supply voltage,             V = 230 Volts

Armature resistance, Ra = 0.2 ohm.

Armature current, I1 = Ia1 = 15 A

Speed, N1 = 800 r.p.m,

Speed, N2 = 600 r.p.m.

Additional resistance required, R :

Back e.m.f., Eb1 = V – Ia1Ra

= 230 – 15 × 0.2 = 227 V

Since as per given data load torque is independent of speed and flux is constant

Ia1 = Ia2 = 15 A

ɸ1 = ɸ2

Back e.m.f., Eb2 = V -Ia2(Ra + R)

= 230 – 15(0.2 + R)

= 230 – 3 – 15R = 227 – 15R

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