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05/12/2020

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


design-0f-dc-series-motor-starter


Case ( 1 ) Flux increases non – linearly

( Ф1 / Ф2 ) ≠ ( I1 / I2 )

When the starter arm on nth stud

I1 = ( V – Eb ) / Rn…….. ( 1 )

Where Eb = 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.

Eb’ = ( Ф1 / Ф2 ) Eb

       = β Eb

Now

I1 = ( V – Eb’ ) / Rn+1

Rn+1 = ( V – Eb’ ) / I1

        = V – β Eb / I1

        = V – β ( V – I2Rn ) / I1

        = V – β V + β I2Rn / I1

        = V ( 1 – β ) + β I2Rn / I1

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

 Rn+1 = R1 ( 1 – β ) + ( β / α ) Rn      ( Where α = I2 / I)

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

Rn = R1 ( 1 – β ) + Rn – 1 …….. ( 2 )

Therefore the resistance between nth and ( n +1 )th stud

rn = Rn  Rn + 1

    = ( β / α )( Rn + 1 – Rn )

    = ( β / α )( rn 1 )

rn / rn – 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 – ra ) / n….. ( 5 ) 

from equation ( 4 ) and ( 5 )

β = 1 + 1/n ( 1 – [ ra / R1 ] )

All the sections have same value

               r = ( R1 – ra ) / Number of sections


flux-increases-linearly-in-the-dc-series-motor-starter-design
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 If = 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 ) Rn 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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