Question

(a) List the factors that determine the starting torque of the three- phase induction motor. How does this torque generally compare with the value of the rated torque. (6)

(b) The low-voltage release of an a.c. motor-starter consists of a solenoid into which an iron plunger is drawn against a spring. The resistance of the solenoid is 35 ohm. When connected to a 220 V, 50 Hz, a.c. supply the current taken is at first 2 A, and when the plunger is drawn into ths "full-in" position the current falls to 0.7 A. Calculate the inductance of the solenoid for both positions of the plunger and the maximum value of flux-linkages in weber-turns for the "full-in" position of the plunger.

Answer 1

Part (a)

Factors that determine the starting torque of the three-phase induction motor:

Starting torque is proportional to the square of the supply voltage. A reduced supply voltage will significantly reduce the starting torque.
Higher rotor resistance leads to higher starting torque but also lower efficiency and higher rotor heating at running speeds. This is because higher resistance allows for greater current flow in the rotor at standstill.
Increasing rotor resistance enhances starting torque by improving the power factor, but it also leads to higher losses and reduced efficiency during normal operation.
Changes in frequency affect the motor's magnetic field and slip, impacting starting torque. Generally, lower frequency results in higher torque.
The design of the stator windings, including the number of turns, winding configuration (star or delta), and the shape and placement of the stator poles, significantly affects the strength of the magnetic field produced, thus affecting torque.
A smaller air gap between the stator and rotor generally leads to higher starting torque due to increased magnetic coupling.

Comparison with Rated Torque:

Typically, the starting torque of a standard three-phase induction motor is designed to be equal to or slightly higher than the rated (full-load) torque. This design ensures that the motor can overcome initial inertia and start the load effectively. However, the exact ratio varies based on motor design and application requirements. For instance, some motors may have a starting torque that is 1.25 times the full-load torque, while others, especially those designed for high starting torque applications, may have even higher ratios.

While higher starting torque is beneficial for initiating motion, it often comes with an increased starting current, which can impact the electrical supply system. Therefore, motor designs aim to balance starting torque and current to suit specific application needs.

Part (b)

Given:

$$Resistance \space of \space solenoid \space = \space 35Ω$$

$$Supply \space voltage \space = 220V, \space 50Hz$$

$$Initial \space current \space = \space 2A$$

$$Current \space in \space "Full-in" \space position \space = \space 0.7A$$

When plunger is out (initial stage),

$$Impedance \space (Z) \space = \space {{220} \over 2} \space = \space 110Ω$$

$$Inductive \space reactance \space (X) \space = \space \sqrt{Z^2 - R^2}$$

$$X \space = \space \sqrt{110^2 - 35^2} \space = \space 104.28Ω$$

$$Also, \space X \space = \space 2 \pi fl$$

$$l \space = \space {{X} \over 2 \pi f} \space = \space {{104.28} \over 2 \times 3.14 \times 50} $$

$$l \space = \space 0.33H$$

Inductance of solenoid when plunger is out = 0.33H

When "Full-in",

$$Impedance \space = \space {{V} \over I} \space = \space {{220}\over 0.7} \space = \space 314.286Ω$$

$$∴ \space X \space = \space \sqrt{314.286^2 - 35^2} \space = \space 312.33Ω$$

$$l \space = \space {{312.33} \over 2 \times 3.14 \times 50 } \space = \space 0.994H $$

$$Also, L \space = \space {{N \phi} \over I_{peak}}$$

$$∴ \space N \phi \space = \space L \space I_{peak}$$

$$∴ \space N \phi \space = \space 0.994 \times \sqrt{2} \times 0.7$$

$$∴ \space N \phi \space = \space 0.984 \space weber-turn$$

✓ Verified Model Answer (Text Solution)

Factors that determine the starting torque of the three-phase induction motor:

Given: