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.

T_s 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

dT_s / dR₂ = 0

              = d { [ KF E₂ / [ R₂² + X₂² ] }  / dR₂  = 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

T_s = KFE₂ R₂ / [ R₂² + X₂² ]

Putting R₂ = X₂ in the torque equation

T_s = KFE₂ X₂ / [ X₂² + X₂² ]

T_{s ( 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.

        T_{s ( MAX )} a ( 1 / X₂ )

Ratio of starting torque to maximum torque

T_s = KF E₂ R₂ / [ R₂² + X₂² ] and

T_{s ( MAX )} = KFE₂  / 2X₂

T_s / T_{s ( MAX )} = ( KF E₂ R₂ ) ( 2X₂ ) / ( KFE₂ ) ( R₂² + X₂² )

                    = 2 R₂ X₂ / ( R₂² + X₂² )

Multiplying and dividing equation by X₂²

T_s / T_{s ( MAX )} = { 2 R₂ / X₂ } / ( R₂² / X₂²  + 1 )

Putting R₂ / X₂ = a

T_s / T_{s ( MAX )} = 2a / ( a² + 1 )

Effect of Supply voltage on the starting torque

The starting torque T_s = 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.

        T_s 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%

 T_s a V²  

[ T_s2 / T_s1 ]  = [ V₂ / V₁ ]² 

T_s2 = [ V₂ / V₁ ]² × T_s1

T_s2 = [ 1.1V₁ / V₁ ]² × T_s1         [ As V₂ = 1.1V₁ ]

T_s2 = [ 1.21] T_s1   

[ T_s2 / T_s1 ]  = [ 1.21 / 1 ]  

[ T_s2 – T_s1 / T_s1 ] = [ 1.21 – 1 / 1 ]  × 100 %

                           = 21 %

( b ) If the supply voltage is increased by 25%, the starting torque increases by  56.25%

T_s a V²  

[ T_s2 / T_s1 ]  = [ V₂ / V₁ ]² 

T_s2 = [ V₂ / V₁ ]² × T_s1

T_s2 = [ 1.25V₁ / V₁ ]² × T_s1         [ As V₂ = 1.25V₁ ]

T_s2 = [ 1.5625 ] T_s1   

[ T_s2 / T_s1 ]  = [ 1.5625 / 1 ]  

[ T_s2 – T_s1 / T_s1 ] = [ 1.5625 – 1 / 1 ]  × 100 %

                            = 56.25 %

( c ) If the supply voltage is decreased by 10%, the starting torque decreases by 19%

T_s a V²  

[ T_s2 / T_s1 ]  = [ V₂ / V₁ ]² 

T_s2 = [ V₂ / V₁ ]² × T_s1

T_s2 = [ 0.9V₁ / V₁ ]² × T_s1         [ As V₂ = 0.9V₁ ]

T_s2 = [ 0.81 ] T_s1   

[ T_s2 / T_s1 ]  = [ 0.81 / 1 ]  

[ T_s2 – T_s1 / T_s1 ] = [ 0.81 – 1 / 1 ]  × 100 %

                             = – 19 %

( d ) If the supply voltage is decreased by 5%, the starting torque decreases by  9.75%

T_s a V²  

[ T_s2 / T_s1 ]  = [ V₂ / V₁ ]² 

T_s2 = [ V₂ / V₁ ]² × T_s1

T_s2 = [ 0.95V₁ / V₁ ]² × T_s1         [ As V₂ = 0.9V₁ ]

T_s2 = [ 0.9025 ] T_s1   

[ T_s2 / T_s1 ]  = [ 0.9025 / 1 ]  

[ T_s2 – T_s1 / T_s1 ] = [ 0.9025 – 1 / 1 ]  × 100 %

                             = – 9.75 %

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