## 27/11/2020

### Hopkinson Test

Introduction

• This test is also known as back to back test. This test is carried out for two identical shunt machines.
• The two machines are mechanically coupled and so adjusted that one machine act as a motor and the other machine acts as a generator.
• The mechanical output of the motor drives the generator and the electrical output of the generator is given back to the motor therefore it is also called as Regenerative test
• As the generator output is not sufficient to drive the motor because of losses, the losses are supplied from the supply mains.

Procedure

• The switch S is kept open before the supply is given to machine M1.
• When supply is given machine M1, it acts as a motor and machine M2 acts as generator.
• The function of the field regulator F1 is to set the speed of the motor and that of F2 is to set the rated output voltage of the generator.
• The voltage of the generator is adjusted till the voltmeter V1 shows zero, after switch S is closed.
• Now both machines are run in parallel. The generator will “Float” neither taking nor giving current to the supply under this condition.
• Any load can be put on the generator by increasing the field current of the generator or decreasing field current of the motor.
• The machine with less excitation acts as a motor and that of more excitation acts as a generator because the back emf of the motor is less than generated emf of the generator.

Let

V = Supply voltage

I1 = Line current

I2 = Motor shunt field current

I3 = Generator output current

I4 = Generator shunt field current

Ra = Armature resistance

Case ( 1 ) : Both machines have same efficiency

Motor input = V ( I1 + I3 )

Motor output = V ( I1 + I3 ) × η

Now the Generator input = Motor input

= V ( I1 + I3 ) × η…… ( 1 )

Generator efficiency η = Generator output / Generator input

Generator output = η × Generator input

= η2 × V ( I1 + I3 )  { from equation ( 1 ) }…..( 2 )

But Generator output = VI3……..( 3 )

From equation ( 2 ) and ( 3 )

VI3 = η2 × V ( I1 + I3 )

η2 = ( I1 + I3 )  / I3

η = √ {( I1 + I3 )  / I3 }

Case ( 2 ) : Both machines have different efficiency

To obtain accurate results, the armature copper loss and field copper loss are determined separately and stray losses are assumed to be equal in both machines.

Armature copper losses = ( I1 + I3 – I2 )2 × Ra

Shunt copper loss = VI2

Generator

Armature copper losses = ( I3 + I4 )2 × Ra

Shunt copper loss = VI4

But total losses of both machines are supplied by the main supply

Power drawn by the supply = VI1

Input power = Total copper losses + Stray losses

Therefore

Stray losses of both machines Ws

= VI1 – [ VI2 + VI4 + ( I1 + I3 – I2 )2 × Ra + ( I3 + I4 )2 × Ra ]

Stray losses per machine = Ws / 2

( a ) Motor efficiency

Motor input = V × ( I3 + I4 )

Total losses Wm = ( I1 + I3 – I2 )2 × Ra  + VI2 + Ws / 2

Motor efficiency

= [ Output power / Input power ]  × 100%

= [ Input power – Losses / Input power ]  × 100%

= { [ V × ( I3 + I4 ) – Wm ] / V × ( I3 + I4 ) } × 100%

( b ) Generator efficiency

Generator output = VI3

Total losses Wg = ( I3 + I4 )2 × Ra + VI4 + Ws / 2

Generator efficiency

= [ Output power / Input power ]  × 100%

= [ Output power / Output power + losses ]  × 100%

= [ VI3 / ( VI3 + Wg ) ]  × 100%

• Power required is small as compared to full load power of both machines
• Commutation condition and temperature rise can be observed at any load condition
• Any change in stray losses is taken into account because of full load condition

• Two identical DC Shunt machines requires

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