## Hopkinson Test

Table of Contents

Hopkinson Test

Another effective technique for evaluating a DC machine’s efficiency is **Hopkinson’s Test** (**Hopkinson Test**). It is a full load test, and two identical machines that are coupled to one another are needed. The mechanical power for one of these two machines is provided by a generator, and the mechanical power for the other device is provided by a motor that drives the generator. Hopkinson test is also known as the back-to-back test or regenerative test for this method of back-to-back driving the motor and the generator.

No external power source would have been required if there are no machine losses. But in order to give the motor the correct input voltage, we need an additional voltage source because of the generator’s decreased output voltage. In order to reduce the internal losses of the motor-generator set, power is drawn from an external source. Hopkinson’s test is also known as the heat run test, back to back test, and regenerative test.

## Connection Diagram of Hopkinson Test

The Hopkinson test circuit connection is shown in the figure below. There is a coupling between an identical motor and generator. The machine starts as a motor when it is turned on. The machine’s shunt field resistance is altered to allow the motor to run at its rated speed.

By adjusting the shunt field resistance connected across the generator, it is now possible to bring the generator voltage to the same level as the supply voltage. The voltmeter indicates that the generator and supply voltages are equal because it reads zero when it is connected across the switch. By adjusting the field currents of the motor and generator, the machine can run at rated speed and under the desired load.

### Calculation of Efficiency by Hopkinson Test

Let, V = supply voltage of the machines.

Then,

I_{1} = The current from the generator

I_{2} = The current from the external source

And, Generator output = VI_{1}â€¦â€¦â€¦â€¦â€¦â€¦(1)

Let, both machines are operating at the same efficiency â€˜Î·â€™.

Then, Output of motor =

From equation 1 an 2 we get,

Now, in case of motor, armature copper loss in the motor =

.

R_{a} is the armature resistance of both motor and generator.

I_{4} is the shunt field current of the motor.

Shunt field copper loss in the motor will be = VI_{4}

Next, in case of generator armature copper loss in generator =

I_{3} is the shunt field current of the generator.

Shunt field copper loss in the generator = VI_{3}

Now, Power drawn from the external supply = VI_{2}

Therefore, the stray losses in both machines will be

Let us assume that the stray losses will be same for both the machines. Then,

Stray loss / machine = W/2

### Efficiency of Generator

Total losses in the generator,

Generator output = VI_{1}

Then, efficiency of the generator,

### Efficiency of Motor

Total losses in the motor,

Then, efficiency of the motor,

### Advantages of Hopkinson Test

The merits of this test areâ€¦

- This test requires very small power compared to full-load power of the motor-generator coupled system. That is why it is economical. Large machines can be tested at rated load without much power consumption.
- Temperature rise and commutation can be observed and maintained in the limit because this test is done under full load condition.
- Change in iron loss due toÂ fluxÂ distortion can be taken into account due to the advantage of its full load condition.
- Efficiency at different loads can be determined.

### Disadvantages of Hopkinson Test

The demerits of this test are

- It is difficult to find two identical machines needed forÂ
**Hopkinson test** - Both machines cannot be loaded equally all the time.
- It is not possible to get separate iron losses for the two machines though they are different because of their excitations.
- It is difficult to operate the machines at rated speed because field currents vary widely.