- The design of DC series motor starter is same as that of DC shunt motor starter.
- The flux is almost constant in the DC shunt motor whereas the flux changes with load current in the DC series motor.

Let

I_{1} =
Maximum current

I_{2} =
Minimum current

Ф_{1} =
Flux per pole for I_{1}

Ф_{2} =
Flux per pole for I_{2}

( I_{1}
/ I_{2} ) = α and ( Ф_{1} / Ф_{2} ) = β

*Case ( 1 ) Flux
increases non – linearly *

( Ф_{1}
/ Ф_{2} ) ≠ ( I_{1} / I_{2} )

When the starter
arm on nth stud

I_{1} =
( V – E_{b} ) / R_{n}…….. ( 1 )

Where E_{b}
= Back emf produced when current I_{2} passes through nth stud

Now

When the starter
arm moves form nth to ( n + 1 )th stud, the value of back emf increases.

E_{b}’ =
( Ф_{1} / Ф_{2} ) E_{b}

= β E_{b}

Now

I_{1} = (
V – E_{b}’ ) / R_{n+1}

R_{n+1}
= ( V – E_{b}’ ) / I_{1}

= V – β E_{b}
/ I_{1}

= V – β ( V – I_{2}R_{n}
) / I_{1}

= V – β V + β I_{2}R_{n}
/ I_{1}

= V ( 1 – β ) +
β I_{2}R_{n} / I_{1}

= V ( 1 – β ) /
I_{1} + β R_{n} ( I_{2} / I_{1} )

R_{n+1} = R_{1} ( 1 – β ) + (
β / α ) R_{n} ( Where α = I_{2}
/ I_{1 })

Now substitute n
– 1 in place of n, we get

R_{n} =
R_{1} ( 1 – β ) + R_{n} _{– 1} …….. ( 2 )

Therefore the
resistance between n^{th} and ( n +1 )^{th} stud

r_{n} =
R_{n} – R_{n} _{+ 1}

= ( β / α )( R_{n}
_{+ 1} – R_{n} )

= ( β / α )( r_{n}
_{–} _{1} )

r_{n} /
r_{n} _{– 1} = ( β / α ) = γ

Obviously, the
section resistance from a geometric progression series

r_{2} =
γr_{1}

r_{3} =
γr_{2}

r_{4} =
γr_{3} and

so on…………

Now substitute N
= 2 in the equation ( 2 ) to find out r_{1}

R_{2 }=
R_{1} ( 1 – β ) + ( β / α ) R_{1}

r_{1} =
R_{1} – R_{2}

= R_{1 }–
{ R_{1} ( 1 – β ) + ( β / α ) R_{1} }

= R_{1 }(
β – γ )………. ( 3 )

Where R_{1 }=
V_{1} / I_{1}

*Case ( 2 ) Flux
increases linearly*

( Ф_{1 }/
Ф_{2} ) = ( I_{1} / I_{2} ) = α = β

The value of γ =
β / α = 1

All the stud
resistances have same value in this condition

r_{1 }=
R_{1} ( β – 1 ) { From equation
( 3 ) }….. ( 4 )

Also

r_{1 }=
( R_{1} – r_{a} ) / n….. ( 5 )

from equation (
4 ) and ( 5 )

β = 1 + 1/n ( 1
– [ r_{a} / R_{1} ] )

All the sections
have same value

r = ( R_{1}
– r_{a} ) / Number of sections

**Example****Determine the
number of studs and resistance of the section of a DC series motor starter form
following data :**

**6 kW, 415 V DC
series motor, starting current varies from 1.4 to 2 times full load current,
flux increases 15% for variation of current from minimum to maximum, motor
resistance 1.7 ohm, motor efficiency 80%.**

Solution

Full load
current I_{f} = 6000 / 415 × 0.80

Maximum current
I_{2} = 2 × 18 = 36 Amp

Minimum current
I_{1} = 1.4 × 18 = 25.2 Amp

Therefore I_{1}
/ I_{2} = 36 / 25.2 = 1.42 Say α

Also

( Ф_{1 }/
Ф_{2} ) = 1.15 = β ( Given )

R_{1} = V1 / I1 =
415 / 36 = 11.52 ohm

If we substitute
value of n = 2, 3 …. in the following equation.

Rn = R_{1}
( 1 – β ) + ( β / α ) R_{n} _{–} _{1}

We get

R_{2 }=
R_{1} ( 1 – β ) + ( β / α ) R_{1}

= 11.52 ( 1 –
1.15 ) + ( 1.15 / 1.42 ) ( 11.52 )

= 7.53 ohm

R3 = R_{1}
( 1 – β ) + ( β / α ) R_{2}

= 4.29 ohm

R4 = R_{1}
( 1 – β ) + ( β / α ) R_{3}

= 1.70 ohm

Here, the value
of R4 is equal to armature resistance therefore number of studs are 4 and
number of steps are 3.

r_{1} =
R_{1} – R_{2} = 4.0 ohm

r_{2} =
R_{2} – R_{3} = 3.24 ohm

r_{3} =
R_{3} – R_{4} = 2.59 ohm

**OR**

r_{2} =
γr_{1 } and r_{3} = γr_{2
}and so on..

Where γ = ( β /
α )

*You may also
like : *

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by voltage harmonics and current harmonics?

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