Types of Starters for Three Phase Induction Motor

Types of Starters for Three Phase Induction Motor AU : May-05, 07, 08, 11, 12, 13, 14, 16,18, Dec.-04, 05, 10, 11, 12, 14, 15,16,17

From the expression of rotor current it can be seen that the current at start can be controlled by reducing E₂ which is possible by supplying reduced voltage at start or by increasing the rotor resistance R₂ at start. The second method is possible only in case of slip ring induction motors. The various types of starters based on the above two methods of reducing the starting current are,

1. Stator resistance starter

2. Autotransformer starter

3. Star-delta starter

4. Rotor resistance starter

5. Direct on line starter

Let us study the details of each of these starters.

 

1. Stator Resistance Starter

In order to apply the reduced voltage to the stator of the induction motor, three resistances are added in series with each phase of the stator winding. Initially the resistances are kept maximum in the circuit. Due to this, large voltage gets dropped across the resistances. Hence a reduced voltage gets applied to the stator which reduces the high starting current. The schematic diagram showing stator resistances is shown in the Fig. 7.3.1.

When the motor starts running, the resistances are gradually cut-off from the stator circuit. When the resistances are entirely removed from the stator circuit i.e. rheostats in 

RUN position then rated voltage gets applied to the stator. Motor runs with normal speed.

The starter is simple in construction and cheap. It can be used for both star and delta connected stator. But there are large power losses due to resistances. Also the starting torque of the motor reduces due to reduced voltage applied to the stator.

a. Relation between T_st and T_{F. L .}

We know, P₂ = T × w_s

 where T is torque produced and P₂ is the rotor input at N_s.

 

But rotor current I2r and stator current are related to each other through transformer action.

When stator resistance starter is used, the factor by which stator voltage reduces is say x< 1. The starting current is proportional to this factor x. So if I_sc is the normal current drawn under full rated voltage condition at start then,

Key point As x < 1, it can be seen that the starting torque reduces fey the fraction x² due to the stator resistance starter.

 

2. Autotransformer Starter

A three phase star connected autotransformer can be used to reduce the voltage applied to the stator. Such a starter is called an autotransformer starter. The schematic diagram of autotransformer starter is shown in the Fig. 7.3.2.

It consists of a suitable change over switch.                 

When the switch is in the start position, start the stator winding is supplied with reduced voltage. This can be controlled by tappings provided   with autotransformer.

The reduction in applied voltage by the fractional percentage tappings x, used for an autotransformer is shown in the Fig. 7.3.3. 

When motor gathers 80 % of the normal speed, the change over switch is thrown into run position.

Due to this, rated voltage gets applied to stator winding. The motor starts rotating with normal speed. Changing of switch is done automatically by using relays. The power loss is much less in this type of starting. It can be used for both star and delta connected motors. But it is expensive than stator resistance starter.

a. Relation between T_st and TF.L.

Let x be the fractional percentage tappings used for an autotransformer to apply reduced voltage to    the stator.

So if, I_sc =   Starting motor current at         rated voltage

and    I_st =    Starting motor current with starter

then   I_st =    x I_sc ... Motor side … (7.3.5)

But there exists a fixed ratio between starting current drawn from supply I_st (supply) and starting motor current Ist (motor) due to autotransformer, as shown in the Fig. 7.3.4.

Key Point Thus starting torque reduces by x2 where x is the transformation ratio.

 

Example 7.3.1 A squirrel cage induction motor has a full load slip of 5 %. The motor starting current at rated voltage is 6 times its full load current. Find the tapping on the autotransformer starter which would give full load torque at start. What would then be the supply starting current.

Solution :

Starting current at rated voltage = I_sc

Let x = Tapping on autotransformer

T_F.L. = T_st

Thus 74.53 % tapping is required

Now  I_st (supply) = x I_st (motor) = x [x I_sc] = x² I_sc

= x² x 6 I_F.L = 3.33 I_F.L

Thus supply starting current is 3.33 times the full load current.

 

3. Star-Delta Starter

This is the cheapest starter of all and hence used very commonly for the induction motors. It uses Tripple Pole Double Throw (TPDT) switch. The switch connects the stator winding in star at start. Hence per phase voltage gets reduced by the factor 1 /√3. Due to this reduced voltage, the starting current is limited.

When the switch is thrown on other side, the winding gets connected in delta, across the supply. So it gets normal rated voltage. The windings are connected in delta when motor gathers sufficient speed.

The arrangement of star-delta starter is shown in the Fig. 7.3.5.

The operation of the switch can be automatic by using relays which ensures that motor will not start with the switch in Rim position. The cheapest of all and maintenance free operation are the two important advantages of this starter. While its limitations are, it is suitable for normal delta connected motors and the factor by which voltage changes is 1 / √3 which can not be changed.

a. Ratio of Tst to TF.L.

We have seen in case of autotransformer that if x is the factor by which the voltage is reduced then,

where I_sc = Starting phase current when delta connection with rated voltage.

I_F.L.  = Full load phase current when delta connection.

 

Example 7.3.2 A three phase induction motor has a ratio of maximum torque to full load torque as 2.5 : 1. Determine the ratio of starting torque to full load torque if star-delta starter is used. The rotor resistance and standstill reactance per phase are 0.4 Ω and 4 Ω respectively. 

Solution :

 

4. Rotor Resistance Starter

To limit the rotor current which consequently reduces the current drawn by the motor from the supply, the resistance can be inserted in the rotor circuit at start. This addition of the resistance in rotor is in the form of 3 phase star connected rheostat. The arrangement is shown in the Fig. 7.3.6.

The external resistance is inserted in each phase of the rotor winding through slip ring and brush assembly. Initially maximum resistance is in the circuit. As motor gathers speed, the resistance is gradually cutoff. The operation may be manual or automatic. 

We have seen that the starting torque is proportional to the rotor resistance. Hence important advantage of this method is not only the starting current is limited but starting torque of the motor also gets improved. The only limitation of the starter is that it can be used only for slip ring induction motors as in squirrel cage motors, the rotor is parmanently short circuited.

 

5. Direct On Line Starter (D.O.L.)

In case of small capacity motors having rating less than 5 h.p., the starting current is not very high and such motors can withstand such starting current without any starter. Thus there is no need to reduce applied voltage, to control the starting current. Such motors use a type of starter which is used to connect stator directly to the supply lines without any reduction in voltage. Hence the starter is known as direct on line starter.

Though this starter does not reduce the applied voltage, it is used because it protects the motor from various severe abnormal conditions like over loading, low voltage, single phasing etc.

The Fig. 7.3.7 shows the arrangement of various components in direct on line starter.

The NO contact is normally open and NC is normally closed. At start, NO is pushed for fraction of second due to which coil gets energised and attracts the contactor. So stator directly gets supply. The additional contact provided, ensures that as long as supply is ON, the coil gets supply and keeps contactor in ON position. When NC is pressed, the coil circuit gets opened due to which coil gets de-energised and motor gets switched OFF from the supply. 

Under over load condition, current drawn by the motor increases due to which there is an excessive heat produced, which increases temperature beyond limit. Thermal relays get opened due to high temperature, protecting the motor from overload conditions.

 

6. Comparison of Various Starters

 

Example 7.3.3 A 3-phase, 6-pole, 50 Hz induction motor takes 60 A at full load speed of 940 rpm develops a torque of 150 N-m. The starting current at rated voltage is 300 A. What is starting torque ? If a star/delta starter is used determine the starting torque and starting current.

Solution :

The given values are, I_F.L. = 60 A, P = 6 pole

N = 940 r.p.m., T_FL = 150 N-m, I_sc = 300 A

 

Example 7.3.4 A 3-phase squirrel cage induction motor has a maximum torque equal to thrice the full load torque. Determine the ratio of starting torque to full load torque if started by i) DOL starter ii) Star delta starter. The maximum torque occurs at 0.1 slip.

Solution :  

i) D.O.L. starter

ii) With star-delta starter, the starting torque is one third of that obtained by D.O.L. starter.

 

Example 7.3.5 A 3-phase, 400 V, distribution circuit is designed to supply 1200 A. Assuming that three phase squirrel cage induction motor has full load efficiency of 0.85 power factor 0.8 starting current is 5 times the rated current what is the maximum possible kW of motor if it is designed to use star-delta starter ? What is the maximum possible kW rating of the motor if it is to be started using an autotransformer stepping down the voltage to 80 % ? AU : Dec.-14, Marks 4

Solution :

Maximum possible permissible current that induction motor can take from the distribution circuit is 1200 A at the time of starting.

If it is designed to have star-delta starter, then star-delta starter is equivalent to autotransformer with 57.8 % tapping. i.e. ratio (1/√3)

 

Example 7.3.6 A 3-phase, delta-connected cage type induction motor when connected directly to 400 V, 50 Hz supply takes a starting current of 100 A in each stator phase. Calculate :

i) The line current for direct-on-line starting.

ii) Line and phase starting currents for star-delta starting.

iii) Line and phase starting currents for a 70 % tapping on auto-transformer starting.

Solution :

i) In D.O.L. starter,

This will produce a phase current of,

I_ph = 100 / √3 = 57.735 A

This is because voltage is reduced by 1/√3, hence current will also reduce by 1/√3.

Starting phase current = 57.735 A

In start connection, I_L = I_ph .

Starting line current = 57.735 A

iii) 70 % tapping on autotransformer starting

The autotransformer is star connected. So phase voltage of autotransformer is 400/√3 = 230.94 V.

The line voltage for delta connected stator is 0.7 × 400 = 280 V .

The phase voltage for delta connected stator is 280 V.

 The 400 V circulates 100 A phase current hence 280 V will circulate a phase current of 280 / 400 × 100 = 70 A

Starting phase current of motor = 70 A

Starting line current of motor = √3 × 70 = 121.243 A

Supply line current = 0.7 × 121.243 = 84.87 A

 

Example 7.3.7 The ratio of maximum torque to full-load torque in a three phase squirrel cage induction motor is 2.2:1 Determine the ratio of actual starting torque of full load torque for Direct starting, Star-delta starting and Auto transformer starting with tapping of 70 %. The rotor resistance and standstill reactance per phase are 0.5 Ω and 5 Ω. respectively.

Solution :

c) Auto transformer starting,

Stator voltage per phase = 0.7 times the voltage on direct on line.

i.e. Rotor voltage at starting = 0.7 E₂

 

Example 7.3.8 A small squirrel-cage induction motor has a starting current of six times the full load current and a full-load slip of 0.05. Find in pu of full-load values, the current (line) and starting torque with the following methods of starting ((a) to (d)). a) Direct switching, b) Stator-resistance starting with motor current limited to 2 p.u., c) Auto-transformer starting with motor current limited to 2 p.u. and d) Y-delta starting, e) What auto trnasformer ratio would give 1 pu starting torque ?

Solution :

Let F.L. current = 1 p.u., Ist = 6I_FL,  s_fl; = 0.05

a) Direct switching

T_st  = s_fl I²_st = 0.05 × 6² × (1 p.u.) = 1.8 p.u.

b) Stator resistance starting

I_st =2 p.u. (limited)

T_st = 0.05 × (2)² = 0.2 p.u.

c) Autotransformer starting

I_st (motor) = 2 p.u.

Let × be the fraction by which the voltage applied to the motor is reduced.

I_st = x I_sc      ... I_sc is starting current for full voltage. 

 

Example 7.3.9 A three phase induction motor takes a starting current which is 5 times full-load current at normal voltage. Its full-load slip is 4 percent. What auto-tarnsformer ratio would enable the motor to be started with not more than twice the full load current drawn from the supply ? What would be the starting torque under this condition ? AU : May-14, Marks 8

Solution :

I_sc = 5I_FL, s_f = 4% = 0.04

For autotransformer starting, x be the ratio.

Ist = Starting motor current with starter = × I_sc

 

Examples for Practice

Example 7.3.10 A cage type induction motor when started by means of a star-delta starter takes 180 % of full load line current and develops 35 % of full load torque at starting. Calculate the starting torque and current in terms of full load values, if an autotransformer with 80 % tappings were employed. UPTU : 2005-06

[Ans.: Tst = 3.456 IfL, Tst = 0.6718 TfL]

Example 7.3.11 A cage induction motor when started by means of a star-delta starter takes 180 % of full-load line current and develops 35 % of full-load torque at starting. Calculate the starting torque and current in terms of full-laod values, if an auto-transformer with 75 % tapping were employed. UPTU : 2007-08

[Ans.: I_st: 3.0375 I_fz, T_st = 59.05 % T_fl]

Example 7.3.12 A 3o squirrel cage IM (SCIM) has maximum torque equal to twice the full load torque. Determine the ratio of motor torque to its full load torque, if it is started by:

i) D.O.L. starter

ii) Auto-transformer starter with 70 % tapping.

iii) Star-delta starter.

The per phase rotor resistance and per phase standstill reactance referred to stator are 0.2 Ω and 2 Ω respectively. Neglect stator impedance. UPTU : 2011-12

[Ans.: i) 0.396, ii) 0.194, iii) 0.132]

Example 7.3.13 A 3-phase induction motor in a short circuit current equal to 4 times the full load current. Determine the starting torque as a percentage of full load torque if full load slip is 2.5 %, for direct starting and star-delta starting.

[Ans. ; 40 % of T_FI, 13.33 % of T_FI]

Review Questions

1. Explain the various starting methods used for three phase induction motor. Analyse and compare them.

2. Which are the various starters used for three phase induction motors ? Explain star-delta starter in detail.

3. Explain the autotransformer starter used for three phase induction motors. AU : May-08, Dec.-ll, 16 Marks 8

4. Describe a starter suitable for a three phase slip-ring induction motor.

5. Explain the working of rotor resistance starter with the help of neat circuit diagram.

6. Explain the working of direct on line starter with the help of neat circuit diagram.

7. Compare the performance of various types of starters used for squirrel cage induction motors. AU : May-18, Marks 13