11 January 2018

Starting Torque in the Three Phase Induction Motor

Starting torque
  • The torque in the DC Motor is related by the product of field flux and armature current whereas the torque in the induction motor is product of rotor flux, rotor current and rotor power factor.

Ts a F I2 Cos F2
    = KF I2 Cos F2      ( As E2 a F2 )
    = KF [ E2 / Z2 ] [ R2 / Z2 ]
    = KF E2 R2 / Z22
    = KF E2 R2 / [ R22 + X22 ]……………… ( 1 )
Where
K = Constant
F = Stator flux in weber
E2 = Rotor emf at standstill condition
R2 = Rotor resistance in ohm
R2 = Rotor resistance in ohm
X2 = Rotor reactance in ohm
Z2 = Rotor impedance in ohm
Condition for maximum starting torque
dTs / dR2 = 0
              = d { [ KF E2 / [ R22 + X22 ] }  / dR= 0
              = [ R22 + X22 ] KFE2 – KFE2 [ 2R2 ] / [ R22 + X22 ]2 = 0
              = KFE2 { R22 + X22 – 2R22 } = 0
The rotor induced emf and flux should not be zero therefore
{ R22 + X22 – 2R22 } = 0
{ X22 – R22 } = 0
( X2 – R2 )( X2 + R2 ) = 0
If
( X2 + R2 ) = 0,  X2 = – R2 is not possible
Therefore ( X2 – R2 ) = 0 resulting X2 = R2………..( 2 )
When the rotor resistance is equal to rotor reactance, the starting torque becomes maximum.
Maximum starting torque
Ts = KFE2 R2 / [ R22 + X22 ]
Putting R2 = X2 in the torque equation
Ts = KFE2 X2 / [ X22 + X22 ]
Ts ( MAX ) = KFE2  / 2X2………….( 3 )
Conclusion on Starting torque
  • The maximum starting torque does not depend upon rotor resistance.
  • The maximum starting torque is inversely proportional to rotor reactance
  • Higher the rotor reactance lesser the maximum starting torque and vice versa.

        Ts ( MAX ) a ( 1 / X2 )
Ratio of starting torque to maximum torque
Ts = KF E2 R2 / [ R22 + X22 ] and
Ts ( MAX ) = KFE2  / 2X2
Ts / Ts ( MAX ) = ( KF E2 R2 ) ( 2X2 ) / ( KFE2 ) ( R22 + X22 )
                    = 2 R2 X2 / ( R22 + X22 )
Multiplying and dividing equation by X22
Ts / Ts ( MAX ) = { 2 R2 / X2 } / ( R22 / X22  + 1 )
Putting R2 / X2 = a
Ts / Ts ( MAX ) = 2a / ( a2 + 1 )
Effect of Supply voltage on the starting torque
The starting torque Ts = KFE2 R2 / [ R22 + X22 ]
  • As the rotor resistance R2 and rotor reactance X2 and stator flux F are constant, the starting torque is directly proportional to supply voltage.

        Ts a V2   ( As flux F a V and E2 a V  ) 
( a ) If the supply voltage is increased by 10%, the starting torque increases by 20%
 Ts a V2  
[ Ts2 / Ts1 ]  = [ V2 / V1 ]2 
Ts2 = [ V2 / V1 ]2 × Ts1
Ts2 = [ 1.1V1 / V1 ]2 × Ts1         [ As V2 = 1.1V1 ]
Ts2 = [ 1.21] Ts1   
[ Ts2 / Ts1 ]  = [ 1.21 / 1 ]  
[ Ts2 – Ts1 / Ts1 ] = [ 1.21 – 1 / 1 ]  × 100 %
                           = 21 %
( b ) If the supply voltage is increased by 25%, the starting torque increases by  56.25%
Ts a V2  
[ Ts2 / Ts1 ]  = [ V2 / V1 ]2 
Ts2 = [ V2 / V1 ]2 × Ts1
Ts2 = [ 1.25V1 / V1 ]2 × Ts1         [ As V2 = 1.25V1 ]
Ts2 = [ 1.5625 ] Ts1   
[ Ts2 / Ts1 ]  = [ 1.5625 / 1 ]  
[ Ts2 – Ts1 / Ts1 ] = [ 1.5625 – 1 / 1 ]  × 100 %
                            = 56.25 %
( c ) If the supply voltage is decreased by 10%, the starting torque decreases by 19%
Ts a V2  
[ Ts2 / Ts1 ]  = [ V2 / V1 ]2 
Ts2 = [ V2 / V1 ]2 × Ts1
Ts2 = [ 0.9V1 / V1 ]2 × Ts1         [ As V2 = 0.9V1 ]
Ts2 = [ 0.81 ] Ts1   
[ Ts2 / Ts1 ]  = [ 0.81 / 1 ]  
[ Ts2 – Ts1 / Ts1 ] = [ 0.81 – 1 / 1 ]  × 100 %
                             = – 19 %
( d ) If the supply voltage is decreased by 5%, the starting torque decreases by  9.75%
Ts a V2  
[ Ts2 / Ts1 ]  = [ V2 / V1 ]2 
Ts2 = [ V2 / V1 ]2 × Ts1
Ts2 = [ 0.95V1 / V1 ]2 × Ts1         [ As V2 = 0.9V1 ]
Ts2 = [ 0.9025 ] Ts1   
[ Ts2 / Ts1 ]  = [ 0.9025 / 1 ]  
[ Ts2 – Ts1 / Ts1 ] = [ 0.9025 – 1 / 1 ]  × 100 %
                             = – 9.75 %


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