Design of DC Series Motor Starter

• 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

I₁ = Maximum current

I₂ = Minimum current

Ф₁ = Flux per pole for I₁

Ф₂ = Flux per pole for I₂

( I₁ / I₂ ) = α and ( Ф₁ / Ф₂ ) = β

Case ( 1 ) Flux increases non – linearly

( Ф₁ / Ф₂ ) ≠ ( I₁ / I₂ )

When the starter arm on nth stud

I₁ = ( V – E_b ) / R_n…….. ( 1 )

Where E_b = Back emf produced when current I₂ 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’ = ( Ф₁ / Ф₂ ) E_b

       = β E_b

Now

I₁ = ( V – E_b’ ) / R_n+1

R_n+1 = ( V – E_b’ ) / I₁

        = V – β E_b / I₁

        = V – β ( V – I₂R_n ) / I₁

        = V – β V + β I₂R_n / I₁

        = V ( 1 – β ) + β I₂R_n / I₁

        = V ( 1 – β ) / I₁ + β R_n ( I₂ / I₁ )

 R_n+1 = R₁ ( 1 – β ) + ( β / α ) R_n      ( Where α = I₂ / I₁ )

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

R_n = R₁ ( 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

r₂ = γr₁

r₃ = γr₂

r₄ = γr₃ and

so on…………

Now substitute N = 2 in the equation ( 2 ) to find out r₁

R₂ = R₁ ( 1 – β ) + ( β / α ) R₁

r₁ = R₁ – R₂

    = R₁ – { R₁ ( 1 – β ) + ( β / α ) R₁ }

    = R₁ ( β – γ )………. ( 3 )

Where R₁ = V₁ / I₁

Case ( 2 ) Flux increases linearly

( Ф₁ / Ф₂ ) = ( I₁ / I₂ ) = α = β

The value of γ = β / α = 1

All the stud resistances have same value in this condition

r₁ = R₁ ( β – 1 )  { From equation ( 3 ) }….. ( 4 )

Also

r₁ = ( R₁ – r_a ) / n….. ( 5 ) 

from equation ( 4 ) and ( 5 )

β = 1 + 1/n ( 1 – [ r_a / R₁ ] )

All the sections have same value

               r = ( R₁ – 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 I₂ = 2 × 18 = 36 Amp

Minimum current I₁ = 1.4 × 18 = 25.2 Amp

Therefore I₁ / I₂ = 36 / 25.2 = 1.42 Say α

Also

( Ф₁ / Ф₂ ) = 1.15 = β ( Given )

R₁ = V1 / I1 = 415 / 36 = 11.52 ohm

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

Rn = R₁ ( 1 – β ) + ( β / α ) R_{n – 1}

We get

R₂ = R₁ ( 1 – β ) +  ( β / α ) R₁

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

= 7.53 ohm

R3 = R₁ ( 1 – β ) +  ( β / α ) R₂

= 4.29 ohm

R4 = R₁ ( 1 – β ) +  ( β / α ) R₃

= 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.

r₁ = R₁ – R₂ = 4.0 ohm

r₂ = R₂ – R₃ = 3.24 ohm

r₃ = R₃ – R₄ = 2.59 ohm

             OR

r₂ = γr₁  and r₃ = γr₂ and so on..

Where γ = ( β / α )

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