How hopkinson test is done...in which parameter it depend...and abstract for hopkinson test

hopkinson test wikipedia

Hopkinson's test is another useful method of testing the efficiency of a dc machine. It is a full load test and it requires two identical machines which are coupled to each other. One of these two machines is operated as a generator to supply the mechanical power to the motor and the other is operated as a motor to drive the generator. For this process of back to back driving the motor and the generator, Hopkinson's test is also called back-to-back test or regenerative test.

If there are no losses in the machine, then no external power supply would have needed. But due to the drop in the generator output voltage we need an extra voltage source to supply the proper input voltage to the motor. Hence, the power drawn from the external supply is therefore used to overcome the internal losses of the motor-generator set.

Here is circuit connection for the Hopkinson's test shown in figure below. A motor and a generator, both identical, are coupled together. When the machine is started it is started as motor. The shunt field resistance of the machine is adjusted so that the motor can run at its rated speed. The generator voltage is now made equal to the supply voltage by adjusting the shunt field resistance connected across the generator. This equality of these two voltages of generator and supply is indicated by the voltmeter as it gives a zero reading at this point connected across the switch. The machine can run at rated speed and at desired load by varying the field currents of the motor and the generator.

Calculation of Efficiency by Hopkinson's Test

Let, V = supply voltage of the machines.
Then, Motor input = V(I1 + I2)
I1 = The current from the generator
I2 = The current from the external source
And, Generator output = VI1..................(1)
Let, both machines are operating at the same efficiency 'η'.
Then, Output of motor = η x input = η x V(I1 + I2)
Input to generator = Output of the motor = η X V(I1 + I2)
Output of generator = η x input = η x [η x V(I1 + I2)] = η2
V(I1 + I2)..................(2)
From equation 1 an 2 we get,
VI1 = η2 V(I1 + I2) or I1 = η2 (I1 + I2)


Now, in case of motor, armature copper loss in the motor = (I1 + I2 - I4)2 Ra.
Ra is the armature resistance of both motor and generator.
I4 is the shunt field current of the motor.
Shunt field copper loss in the motor will be = VI4
Next, in case of generator armature copper loss in generator = (I1 + I3)2Ra
I3 is the shunt field current of the generator.
Shunt field copper loss in the generator = VI3
Now, Power drawn from the external supply = VI2
Therefore, the stray losses in both machines will be
W = VI2 - (I1 + I2 - I4)2 Ra + VI4 + (I1 + I3)2 Ra + VI3
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, WG = (I1 + I3)2 Ra + VI3 + W/2
Generator output = VI1
Then, efficiency of the generator,

Efficiency of Motor

Total losses in the motor, WM = (I1 + I2 - I4)2 Ra + VI4 + W/2
Motor input = V(I1 + I2)
Then, efficiency of the motor,

Advantages of Hopkinson's Test

The merits of this test are…
1. This test requires very small power compared to full-load power of the motor-generator coupled system. That is why it is economical.
2. Temperature rise and commutation can be observed and maintained in the limit because this test is done under full load condition.
3. Change in iron loss due to flux distortion can be taken into account due to the advantage of its full load condition
Disadvantages of Hopkinson's Test

The demerits of this test are
1. It is difficult to find two identical machines needed for Hopkinson's test.
2. Both machines cannot be loaded equally all the time.
3. It is not possible to get separate iron losses for the two machines though they are different because of their excitations.
4. It is difficult to operate the machines at rated speed because field currents vary widely.

Procedure

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.

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http://studentbank.in/report-hopkinson-test-wikipedia

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