Testing of DC Machine /Hopkinson's Method or Regenerative Method or Back to Back Test Method

 

Hopkinson's Method 

In this method, two identical DC machines are both mechanically and electrical coupled, and are tested simultaneously. One of the machines is run as a motor, whereas the other as a generator.

the connection diagram of Hopkinson's test is given below:-

For performing the test, machine I is started as a DC shunt motor and brought to rated speed with switch 'S' open. Both machines run at same speed as both are mechanically coupled.

The field current of the generator (machine II) is so adjusted that its output terminal voltage changes and becomes equal to that supplied to the motor (machine I) terminals. 

At this time voltmeter V2  reads zero voltage. the switch 'S' is closed at that instant.

Under this condition, the generator will neither taking nor giving current to the supply.

After this state is achieved, any desired load can then be put on the generator by controlling the induced EMF of the machines.

If2  > If1 , machine II  acts as generator and machine I  as motor  ( Ea2 > Ea1 ).

As the generator induced EMF is more than that of the motor, current is supplied to the motor from the generator.

Machine I  drives Machine II  and Machine II  supplies electrical power to Machine I , Hence Hopkinson's test is called a regenerative test.

I = Total current supplied from the mains as read by the armature A1.

I2 = Generator armature current as read by the ammeter A2.

If1 = Motor field current as read by the armature A3.

If2 = Generator field current as read by the ammeter A4.

Vt = Supply voltage as read by the voltmeter V1.

R1  and  R2 = Hot resistance of armature circuits of the two machines.

Motor armature current, I1 = I + I2

Power drawn from supply by the two machines excluding field losses of both machines = Vt I

Motor armature circuit copper loss = I1^2  R1

Generator armature circuit copper loss = I2^2  R2

→ Input to motor armature = Vt I1

If motor efficiency = 𝜂m

Motor output = 𝜂m . Vt I1 = Pom

→ Generator output = Vt I2

If generator efficiency =  𝜂g

Generator input = Vt I2 /𝜂g  = Pig

→If Rotational Losses are neglected between the two machines then mechanical output of motor will be equal to electrical input to the generator.

Pom = Pig  

𝜂m . Vt I1  =  Vt I2 /𝜂g 

𝜂m . 𝜂g  = I2 /I1  ,    if   𝜂m  =  𝜂g   =  𝜂

𝜂 =  (I2 /I1)^1/2  =  (generator armature current / motor armature current)^1/2

If Rotational losses are not neglected,

therefore, Total iron + Rotational losses for two machines = Vt I - (I1^2  R1  +  I2^2  R2 )

Total iron + Rotational losses per machines  Wo  = 1/2{Vt I  -(I1^2  R1  +  I2^2  R2 )}

For a motor,

Motor input power Pim = Vt I1  +  Vt If1

Motor losses Wm = Wo  +   I1^2  R1  +  Vt If1

Motor output power  Pom = Pim   -  Wm  

  

Hence, motor efficiency  𝜂m = Pom /Pim

   

                      𝜂m  ={Vt I1  -  Wo  -  I1^2  R1} / {Vt I1  +  Vt If1}

For a generator,

Generator output power  Pog = Vt I2

Generator input power  Pig = Pog  +  Wg

Generator losses Wg = Wo  +  I2^2  R2  +  Vt If2

Hence, generator efficiency  𝜂g = Pog /Pig

  

                        𝜂g =Vt I2  / {Vt I2  +  Wo  +  I2^2  R2  +  Vt If2}