0
\$\begingroup\$

Is there any way to predict which machine will act as a motor and which acts as a generator before applying the supply voltage?

enter image description here

I cannot find a way to predict what will be the behaviour(motor or generator ) of both machines after applying the external voltage source.

To me both machines in this test seems very symmetrical and we already know that for Hopkinson's test both machines should be identical. That is why I cannot find a way to predict the behaviour after applying the voltage source.

Are there any factors prior (or after) to applying voltage source which makes one of them behave as a motor and other as a generator?

Or is there not a way predict their behaviour behaviour before applying the voltage source?

\$\endgroup\$
2
  • 1
    \$\begingroup\$ I don't know much about the Hopkinson's test, but at the beginning the switch S is open, hence the machine on the right is not powered by your voltage source and will act as a generator. \$\endgroup\$ – DavideM Dec 17 '20 at 8:32
  • \$\begingroup\$ As switch S is kept open, only the left engine will start and thus act as the motor. When you close the switch when the motor is running at full speed, the supply will not have to add too much current anymore as the generator started to supply some of the current. Note the orientation of the motor and generator. \$\endgroup\$ – Ananas_hoi Dec 17 '20 at 9:07
3
\$\begingroup\$

You don't predict. You choose.

The diagram shows a rheostat in the field coils of the motors.

You turn the field current of the designated generator to zero (maximum resistance on the rheostat) and set the field current of the designated motor high enough for it to run (rheostat resistance somewhere in the middle.)

With that setup, you apply power and your motor turns. The motor drives the generator, but the generator doesn't produce any power because its field windings aren't energized.

Next you adjust the motor field current to make the motor turn at its rated speed.

Now you can increase the field current on the designated generator and carry out the test.


Hopkinson's test is described in some detail on this site, which also seems to be the source of the diagram you used.

The description of the test on that site seems fairly clear and complete.

\$\endgroup\$
2
  • \$\begingroup\$ Thanks for answer , actually I didn't pay attention to that there is switch S which is open open but after comments and your answer I understand how it works ,just a doubt what will happen if both supply and switch S simultaneously closed ? \$\endgroup\$ – user215805 Dec 17 '20 at 16:00
  • \$\begingroup\$ It depends on the field current settings (as described above.) If the field current settings are about equal and S is closed at power up, then there's no guessing which of the two will play motor and which will play generator. \$\endgroup\$ – JRE Dec 17 '20 at 16:19
2
\$\begingroup\$

A DC motor simultaneously acts as a motor and a generator.

The generator part manifests itself as a voltage referred to as back EMF.

The value of back EMF is proportional to rotation speed times field flux.

Assuming both of the motors are identical and the rotational speed is the same. The amount of EMF is proportional to the field flux.

So the machine with the higher flux and hence EMF acts as the generator and the machine with the lower field flux acts as the motor.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.