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Starting of Induction Motors

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Induction motors comes with various range of capacity. For example, Small induction motors which are upto 2KW do not require a starting device as it can switch to the supply main without any rushing current. Induction motors with high capacity on the other requires the starting device also commonly called as starters.

The functions of these starters restrict the initial rush in the current which is supposed to be 5 times of the full load current has the following upsetting effects on the motor:

  • The distributing network suffers the drop in the large voltage
  • Machines which are already running on the supply chains are also bought to stop

Such effects have got the Electrical undertaking authorities into a serious thought of forbidding the large capacity induction motors usage without external starting devices.

Following are the principle elements of the starting devices:

  • Push the voltage to a lower level when starting the motor
  • Include external resistance if the motor in use is slip ring to maintain the low value of initial rotor current

Direct-on-line starting of induction motors

This method is restricted to the small motors of 2kW implying the switch of the motor under the absence of any external device. The starting period of such motors only stands for few seconds stating that the starting torque is twice the value of the full load torque.

As we know,

Rotor input = 2ΠNsT = kT

Also, rotor copper loss = s × rotor input

3I22R2 = skT

Most common used starting devices in order to start the induction motors are as follow:

For squirrel-Cage motors:

  • Stator rheostat starter
  • Auto-transformers (auto-starters)
  • Star-delta starter

For Slip Ring Motors:

  • Rotor rheostat

20.1. Squirrel-Cage Motors.

 

Stator rheostat starter. The connection diagram is shown in Fig. 30.

Fig. 30. Stator rheostat starter

Due to the resistance R of the rheostat on each stator phase, reduced voltage is captivated

If the machines are switched directly on to supply chains, the initial current will be lesser on the supply chains.

If x = fraction of voltage (V) reduced by the stator resistors

Then Ist = xIsc and Tst = x2Tsc

Tst/Tf = (Ist/If)2sf = (xIsc/If)2sf = x2 (Isc/If)2sf

  • Please note that such a method is successful for starting only the small machines.

Advantages:

  • While you start the squirrel cage motor, it provides high power factor
  • It gives suave acceleration
  • These are more reasonable than the auto transformer starter under the selection of lower output ratings
  • It provides closed transition starting

Disadvantages

  1. Under the role of resistors, heat is given of T
  2. Efficiency of torque is low
  3. The starting time of the motor is more than 5 seconds and thus require expensive resistors

Auto-transformers. It depicts connection diagram for autotransformer starting of squirrel-cage induction motors.

This method takes tapings from the three phase auto transformer at suitable points in order to reduce the voltage. The adjustments at the particular voltage are made for proper starting torque requirements noting that all to-transformers are tapped at 50, 60 and 80 percent of the points.

Auto-transformers may be either normally or magnetically operated

Relationship between starting and full load torque

Let us assume that the transformation ratio is k of the auto transformer of the motor. Then,

Motor input current ISC =KISC

Where t is the starting current at the applied normal voltage and KV is the applied voltage to stator wilding.

Supply current = Primary current of auto-transformer

= K2Isc

Tst/Tf = (Ist/If)2 X  Sf  =  K2 (Isc/If)2  X  Sf

 

Advantages

  1. The starting or the transformer is closed transition.
  2. As compared to resistor starting the current and power drawn from the supply mains are also reduced because of the voltage reduction by transformation and not by resistor dropping.
  3. One can notice the highest torque as per ampere of supply current.
  4. The supply current is smaller than the motor current.
  5. This method is best suited for long starting periods.
  6. One can adjust the starting voltage under this method by selecting proper tabs on auto transformer.

Disadvantages

  1. No power factor
  2. It is expensive in case of lower output rating motor

Star delta starter

As the name suggests this switching method is based on the principle of 3 windings connected in a star. In such a method each winding comes across the voltage value at 1 which states that 57.7% of the voltage value is equal to 9-9 voltage, on the other hand the same windings will encounter the voltage across each of the full line to  line.

Of the starting instance, the star delta connects the stator winding in the form of star which is reconnected in delta across the same supply voltage when the motor attains a specific speed.

As you can see in the figure, you will realize that the actual stator is incorporated under voltage and overload coins. It is important to handle the run position and to prevent the same the stator provides the mechanical interlocking device which are employed for initiating the three phase squirrel cage induction motor ranging from 4KW to 15KW.

The torque becomes one-third of the reduced factor when the start connected reduces the applied voltage over every motor phase. Please note that this torque could have also been realized when motor were directly connected in delta in reducing the line current to one-third.

Relationship between Tst and Tf:

Where Isc is the current per phase that delta connected motor would have acquired if switched on to the supply directly.

  1. This method helps in reducing the starting line current one-third but also reduces the starting torque with the same.
  2. This method is applied to such tools that do not require high starting torque, e.g., pounds and machine tools
  3. For the motor exceeding 3000 V of voltage, this method is not suitable.
  4. As mentioned before, this method uses starters for 3 phase squirrel cage induction motor which is rating between 4 and 20KW

Precaution with star delta starting

It is important for the motor to attain 90% of synchronous speed while changing over from star to delta connection. As mentioned before, 57.7% of the short-circuit current is initially flowing when the motor starts together with transient in each phase. If 90% of synchronous speed is not maintained, one can notice current surge situation which will be greater than full load current and sometimes greater than stand still current along with star connections.

20.2. Slip-ring Induction Motors-Starting of

 

Rotor Rheostat:

Generally the slip ring induction motors starts with full line voltage crosswise the terminals of stator. Talking about the resistance in rotor circuit, the variable element of the same adjusts the value of the starting current in the motor. While the induction motor catches more speed, the controlling resistance which is present in the form of star connected rheostat starts vanishing or cutting out of the rotor circuit.

When talking about torque in the induction motor, one can notice the increase in the value of torque because of the improved power factor when the rotor resistance is also increased followed by reduced rotor current.

Rheostat is examined as hand operated as well as automatic depending on the need of the same which is either a stud or contractor form.

In fact under the presence of external resistance added in the rotor circuit, it develops high starting torque under the shelter of moderate starting current. Such dynamics states that motors like slip-ring induction motor can also initiate under load because while starting such motors on normal condition, the rings are short circuited resulting into lifted brushes.

Example:

Example 20:

A small 3-phase induction motor has short-circuit current which is 3 times the full-load currents. Calculate starting torque percentage of full-load torque if full-load slip is 4 per cent.

A small 3-phase induction motor has a short-circuit current which is 3 times the full-load currents.

Solution:

Short-circuit current, Ist = 3If

Full-load slip, sf = 4% or 0.04

Answer: Starting torque is 36% of full-load torque

 

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