Hopkinson's test | circuit diagram | Advantages and disadvantages - Engineeringforu

Hopkinson's test gives the advantages of both brake test and swinburne test. Hopkinson's test is a regenerative test and it is also called back to back test.
A back to back connected motor generator set is anologous to simple pendulum. In back to back connected between mechanical and electrical forms but in simple pendulum energy is converted between potential and kinetic forms.
By this method, full-load test can be carried out on two identical shunt machines. The connection diagram for hopkinson test on a pair of shunt machines is as shown in figure below.
It requires two identical machines which is  mechanically coupled and their fields are so adjusted that one of them acts as a motor and the other as a generator. The electric power produced by the generator is utilized by the motor is driving the generator. The power taken from the supply is that required to overcome the losses only. Two identical machines of any size can be tested under full-load conditions and for long duration to study their performance regarding commutation, temperature rise etc conveniently.

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Hopkinson test circuit diagram

Hopkinson's test procedure:

  • Machine M is started through a starter and its field rheostat is adjusted so that it runs at normal speed. The machine M will drive machine G. The switch S is initially kept open.
  • The excitation of machine G is gradually increased till the voltmeter V reads zero. Then switch S is closed. Machines G is now floating neither taking any current from the supply nor delivering any current. Any desired load can be put on the set by adjusting the shunt field regulators. The machine with lower excitation will act as a motor and the other machine will act as generator.
Input to motor armature = Vt I1
If ηm  is the motor efficiency then the motor output = ηm Vt I1
Generator output = Vt I2
If ηg  is the generator efficiency then the generator input = Vt I2 / ηg
Since the generator is driven by the motor, we can say that
Motor output = Generator input

ηm Vt I1  =  Vt I2 / ηg
ηm ηg = I2 / I1  
If the efficiencies of the two machines are assumed equal i.e, ηm = ηg = η , then
η = √I2 /I1 
But the efficiencies of the two machines are not equal, because of the following two reasons:

  • The motor armature current is greater than the generator armature current and thus the armature circuit loss in motor is more than the armature circuit loss in generator.
  • The generator field current is greater than the motor field current. Since both the machines are running at the same speed, the generator iron losses are more than the motor iron losses.
The above two reasons are taken into consideration in calculating the efficiency.
armature circuit loss in motor  = I12 ra 
armature circuit loss in generator = I22 ra 

Power drawn from the supply = Vt I
No load rotational loss in two machines Wo = Vt I - (I12 ra + I22 ra)   = Vt I - ra (I12  + I22 )
If resistance of generator armature circuit is not equal to resistance of motor armature circuit then

Wo  = Vt I - I12 ram -  I22 rag
Where ram is the resistance of motor armature circuit and
           rag is the  resistance of generator armature circuit.
Assume that no load rotational loss for each machine is same.
No load rotational loss for each machine = W / 2

Calculation of efficiency in back to back test:

Generator efficiency calculation in hopkinson's test:
Generator output = Vt I2
Generator loss = Wg = (W / 2 ) + Vt If2 + I22 ra
Generator efficiency = Output / input = Output / (output + losses)
                                  = Vt I / (Vt I + Wg )
 ηg  = Vt I / (Vt I + Vt If2  + I22 r+ (W / 2 ))

Motor efficiency calculation in hopkinson's test:
Motor input = Vt I1 + Vt If1
Generator loss = Wm = (W / 2 ) + Vt If1 + I12 ra
Motor efficiency = Output / input = (input - losses) / input
                            = Vt I / (Vt I + Wm )
                            = (Vt I1 + Vt If1 - Wm) / (Vt I1 + Vt If1)
                            = (Vt I1 + Vt If1- (W / 2 ) - Vt If1 -  I12 r/ (Vt I1 + Vt If1)

η= (Vt I1 -  I12 ra- (W / 2 ) ) / (Vt I1 + Vt If1)

Advantages of Hopkinson's test:

  1. Hopkinson test is very economical due to the power regeneration.
  2. The actual performance of the machines regarding commutation and temperature rise can be conveniently studied because the machines can be tested under full-load conditions for long duration.
  3. It is accurate when compared to swinburne test since the stray losses are calculated at each load current.

Disadvantages of Hopkinson's test:

  1. It requires two identical machines.
  2. The results are not accurate when compared to brake test this is because of during total stray losses equally among two machines whose excitations are different.
  3. In hopkinson test the armature emf's of two machines must be different to make the power generation possible.

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