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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
I_{2} Cos F_{2}
= KF
I_{2} Cos F_{2
} ( As E_{2
}a
F_{2}
)
= KF
[ E_{2} / Z_{2} ] [ R_{2} / Z_{2} ]
= KF
E_{2} R_{2} / Z_{2}^{2}
= KF E_{2} R_{2}
/ [ R_{2}^{2 }+ X_{2}^{2} ]……………… ( 1 )
Where
K = Constant
F = Stator flux
in weber
E_{2} =
Rotor emf at standstill condition
R_{2} =
Rotor resistance in ohm
R_{2} =
Rotor resistance in ohm
X_{2} =
Rotor reactance in ohm
Z_{2} =
Rotor impedance in ohm
Condition
for maximum starting torque
dT_{s} /
dR_{2} = 0
= d { [ KF
E_{2} / [ R_{2}^{2 }+ X_{2}^{2} ]
} / dR_{2 }= 0
= [ R_{2}^{2 }+ X_{2}^{2}
] KFE_{2}
– KFE_{2}
[ 2R_{2} ] / [ R_{2}^{2 }+ X_{2}^{2} ]^{2}
= 0
= KFE_{2} {
R_{2}^{2 }+ X_{2}^{2} – 2R_{2}^{2}
} = 0
The rotor induced emf and flux should not be zero therefore
{ R_{2}^{2
}+ X_{2}^{2} – 2R_{2}^{2} } = 0
{ X_{2}^{2}
– R_{2}^{2} } = 0
( X_{2}
– R_{2 })( X_{2} + R_{2} ) = 0
If
( X_{2}
+ R_{2} ) = 0, X_{2} = –
R_{2 }is not possible
Therefore ( X_{2}
– R_{2 }) = 0 resulting X_{2} = R_{2}………..( 2 )
When the rotor
resistance is equal to rotor reactance, the starting torque becomes maximum.
Maximum starting
torque
T_{s }=
KFE_{2}
R_{2} / [ R_{2}^{2 }+ X_{2}^{2} ]
Putting R_{2}
= X_{2} in the torque equation
T_{s }=
KFE_{2}
X_{2} / [ X_{2}^{2 }+ X_{2}^{2} ]
T_{s ( MAX ) }=
KFE_{2} / 2X_{2}………….( 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 / X_{2} )
Ratio
of starting torque to maximum torque
T_{s} = KF
E_{2} R_{2} / [ R_{2}^{2 }+ X_{2}^{2}
] and
T_{s ( MAX ) }=
KFE_{2} / 2X_{2}
T_{s} / T_{s
( MAX )} = ( KF E_{2} R_{2} ) ( 2X_{2}
) / ( KFE_{2}
) ( R_{2}^{2 }+ X_{2}^{2} )
= 2 R_{2} X_{2}
/ ( R_{2}^{2 }+ X_{2}^{2} )
Multiplying and
dividing equation by X_{2}^{2}
T_{s} / T_{s
( MAX )} = { 2 R_{2} / X_{2} } / ( R_{2}^{2 }/
X_{2}^{2 } + 1 )
Putting R_{2}
/ X_{2} = a
T_{s} / T_{s
( MAX )} = 2a / ( a^{2} + 1 )
Effect of Supply
voltage on the starting torque
The starting
torque T_{s} = KFE_{2} R_{2} / [ R_{2}^{2
}+ X_{2}^{2} ]
- As the rotor
resistance R
_{2} and rotor reactance X_{2} and stator flux F
are constant, the starting torque is directly proportional to supply voltage.

T_{s} a
V^{2} ( As flux F
a
V and E_{2} a V_{ } )
( a ) If the
supply voltage is increased by 10%, the starting torque increases by 20%
T_{s} a V^{2}
[ T_{s2}
/ T_{s1} ]_{ } = [ V_{2}
/ V_{1} ]^{2}
T_{s2 }=
[ V_{2} / V_{1} ]^{2} × T_{s1}
T_{s2 }=
[ 1.1V_{1} / V_{1} ]^{2} × T_{s1 } [ As V_{2} = 1.1V_{1} ]
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
V^{2}
[ T_{s2}
/ T_{s1} ]_{ } = [ V_{2}
/ V_{1} ]^{2}
T_{s2 }=
[ V_{2} / V_{1} ]^{2} × T_{s1}
T_{s2 }=
[ 1.25V_{1} / V_{1} ]^{2} × T_{s1 } [ As V_{2} = 1.25V_{1} ]
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
V^{2}
[ T_{s2}
/ T_{s1} ]_{ } = [ V_{2}
/ V_{1} ]^{2}
T_{s2 }=
[ V_{2} / V_{1} ]^{2} × T_{s1}
T_{s2 }=
[ 0.9V_{1} / V_{1} ]^{2} × T_{s1 } [ As V_{2} = 0.9V_{1} ]
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
V^{2}
[ T_{s2}
/ T_{s1} ]_{ } = [ V_{2}
/ V_{1} ]^{2}
T_{s2 }=
[ V_{2} / V_{1} ]^{2} × T_{s1}
T_{s2 }=
[ 0.95V_{1} / V_{1} ]^{2} × T_{s1 } [ As V_{2} = 0.9V_{1} ]
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 %

_{s}a F I

_{2}Cos F

_{2}

_{2}Cos F

_{2 }( As E

_{2 }a F

_{2})

_{2}/ Z

_{2}] [ R

_{2}/ Z

_{2}]

_{2}R

_{2}/ Z

_{2}

^{2}

_{2}R

_{2}/ [ R

_{2}

^{2 }+ X

_{2}

^{2}]……………… ( 1 )

_{2}= Rotor emf at standstill condition

_{2}= Rotor resistance in ohm

_{2}= Rotor resistance in ohm

_{2}= Rotor reactance in ohm

_{2}= Rotor impedance in ohm

_{s}/ dR

_{2}= 0

_{2}/ [ R

_{2}

^{2 }+ X

_{2}

^{2}] } / dR

_{2 }= 0

_{2}

^{2 }+ X

_{2}

^{2}] KFE

_{2}– KFE

_{2}[ 2R

_{2}] / [ R

_{2}

^{2 }+ X

_{2}

^{2}]

^{2}= 0

_{2}{ R

_{2}

^{2 }+ X

_{2}

^{2}– 2R

_{2}

^{2}} = 0

_{2}

^{2 }+ X

_{2}

^{2}– 2R

_{2}

^{2}} = 0

_{2}

^{2}– R

_{2}

^{2}} = 0

_{2}– R

_{2 })( X

_{2}+ R

_{2}) = 0

_{2}+ R

_{2}) = 0, X

_{2}= – R

_{2 }is not possible

_{2}– R

_{2 }) = 0 resulting X

_{2}= R

_{2}………..( 2 )

_{s }= KFE

_{2}R

_{2}/ [ R

_{2}

^{2 }+ X

_{2}

^{2}]

_{2}= X

_{2}in the torque equation

_{s }= KFE

_{2}X

_{2}/ [ X

_{2}

^{2 }+ X

_{2}

^{2}]

_{s ( MAX ) }= KFE

_{2}/ 2X

_{2}………….( 3 )

_{s ( MAX )}a ( 1 / X

_{2})

_{s}= KF E

_{2}R

_{2}/ [ R

_{2}

^{2 }+ X

_{2}

^{2}] and

_{s ( MAX ) }= KFE

_{2}/ 2X

_{2}

_{s}/ T

_{s ( MAX )}= ( KF E

_{2}R

_{2}) ( 2X

_{2}) / ( KFE

_{2}) ( R

_{2}

^{2 }+ X

_{2}

^{2})

_{2}X

_{2}/ ( R

_{2}

^{2 }+ X

_{2}

^{2})

_{2}

^{2}

_{s}/ T

_{s ( MAX )}= { 2 R

_{2}/ X

_{2}} / ( R

_{2}

^{2 }/ X

_{2}

^{2 }+ 1 )

_{2}/ X

_{2}= a

_{s}/ T

_{s ( MAX )}= 2a / ( a

^{2}+ 1 )

_{s}= KFE

_{2}R

_{2}/ [ R

_{2}

^{2 }+ X

_{2}

^{2}]

_{2}and rotor reactance X_{2}and stator flux F are constant, the starting torque is directly proportional to supply voltage._{s}a V

^{2}( As flux F a V and E

_{2}a V

_{ })

_{s}a V

^{2}

_{s2}/ T

_{s1}]

_{ }= [ V

_{2}/ V

_{1}]

^{2}

_{s2 }= [ V

_{2}/ V

_{1}]

^{2}× T

_{s1}

_{s2 }= [ 1.1V

_{1}/ V

_{1}]

^{2}× T

_{s1 }[ As V

_{2}= 1.1V

_{1}]

_{s2 }= [ 1.21] T

_{s1 }

_{s2}/ T

_{s1}]

_{ }= [ 1.21 / 1 ]

_{ }

_{s2}– T

_{s1}/ T

_{s1}] = [ 1.21 – 1 / 1 ] × 100 %

_{s}a V

^{2}

_{s2}/ T

_{s1}]

_{ }= [ V

_{2}/ V

_{1}]

^{2}

_{s2 }= [ V

_{2}/ V

_{1}]

^{2}× T

_{s1}

_{s2 }= [ 1.25V

_{1}/ V

_{1}]

^{2}× T

_{s1 }[ As V

_{2}= 1.25V

_{1}]

_{s2 }= [ 1.5625 ] T

_{s1 }

_{s2}/ T

_{s1}]

_{ }= [ 1.5625 / 1 ]

_{ }

_{s2}– T

_{s1}/ T

_{s1}] = [ 1.5625 – 1 / 1 ] × 100 %

_{s}a V

^{2}

_{s2}/ T

_{s1}]

_{ }= [ V

_{2}/ V

_{1}]

^{2}

_{s2 }= [ V

_{2}/ V

_{1}]

^{2}× T

_{s1}

_{s2 }= [ 0.9V

_{1}/ V

_{1}]

^{2}× T

_{s1 }[ As V

_{2}= 0.9V

_{1}]

_{s2 }= [ 0.81 ] T

_{s1 }

_{s2}/ T

_{s1}]

_{ }= [ 0.81 / 1 ]

_{ }

_{s2}– T

_{s1}/ T

_{s1}] = [ 0.81 – 1 / 1 ] × 100 %

_{s}a V

^{2}

_{s2}/ T

_{s1}]

_{ }= [ V

_{2}/ V

_{1}]

^{2}

_{s2 }= [ V

_{2}/ V

_{1}]

^{2}× T

_{s1}

_{s2 }= [ 0.95V

_{1}/ V

_{1}]

^{2}× T

_{s1 }[ As V

_{2}= 0.9V

_{1}]

_{s2 }= [ 0.9025 ] T

_{s1 }

_{s2}/ T

_{s1}]

_{ }= [ 0.9025 / 1 ]

_{ }

_{s2}– T

_{s1}/ T

_{s1}] = [ 0.9025 – 1 / 1 ] × 100 %

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