Hopkinson’s Test Method For Testing DC machine

This as an elegant method of testing d.c machines. Here it will be shown that while power drawn
from the supply only corresponds to no load losses of the machines, the armature physically
carries any amount of current (which can be controlled with ease). Such a scenario can be
created using two similar mechanically coupled shunt machines. Electrically these two machines
are eventually connected in parallel and controlled in such a way that one machine acts as a
generator and the other as motor. In other words two similar machines are required to carry out
this testing which is not a bad proposition for manufacturer as large numbers of similar machines
are manufactured


Connect the two similar (same rating) coupled machines as shown in figure 40.6. With switch S
opened, the first machine is run as a shunt motor at rated speed. It may be noted that the second
machine is operating as a separately excited generator because its field winding is excited and it
is driven by the first machine. Now the question is what will be the reading of the voltmeter connected across the opened switch S?

The reading may be

(i) either close to twice supply voltage or
(ii) small voltage. In fact the voltmeter practically reads the difference of the induced

voltages in the armature of the machines. The upper armature terminal of the generator may have either + ve or negative polarity. If it happens to be +ve, then voltmeter reading will be small
otherwise it will be almost double the supply voltage.

   Since the goal is to connect the two machines in parallel, we must first ensure voltmeter reading is small. In case we find voltmeter reading is high, we should switch off the supply,reverse the armature connection of the generator and start afresh. Now voltmeter is found to read small although time is still not ripe enough to close S for paralleling the machines. Any attempt to close the switch may result into large circulating current as the armature resistances are small. Now by adjusting the field current Ifg of the generator the voltmeter reading may be adjusted to zero (Eg ≈ Eb) and S is now closed. Both the machines are now connected in parallel as shown in figure 40.7

Loading the machines

After the machines are successfully connected in parallel, we go for loading the machines i.e.,
increasing the armature currents. Just after paralleling the ammeter reading A will be close to zero as Eg ≈ Eb.

Now if Ifg is increased (by decreasing Rfg), then Eg becomes greater than Eb and
both Iag and Iam increase, Thus by increasing field current of generator (alternatively decreasing
field current of motor) one can make Eg > Eb so as to make the second machine act as generator
and first machine as motor. In practice, it is also required to control the field current of the motor Ifm to maintain speed constant at rated value. The interesting point to be noted here is that Iag and Iam do not reflect in the supply side line. Thus current drawn from supply remains small (corresponding to losses of both the machines).

The loading is sustained by the output power of
the generator running the motor and vice versa. The machines can be loaded to full load current
without the need of any loading arrangement.

Once Prot is estimated for each machine we can proceed to calculate the efficiency of the
machines as follows,

Efficiency of the motor

As pointed out earlier, for efficiency calculation of motor, first calculate the input power and
then subtract the losses to get the output mechanical power as shown below,

Efficiency of the generator

For generator start with output power of the generator and then add the losses to get the input
mechanical power and hence efficiency as shown below

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