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Connection Diagram of Hopkinson's TestHere is a 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 TestLet, 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 GeneratorTotal losses in the generator, WG = (I1 + I3)2 Ra + VI3 + W/2 Generator output = VI1 Then, efficiency of the generator,
Efficiency of MotorTotal 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 TestThe merits of this test are…
- 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 to flux 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's TestThe demerits of this test are
- It is difficult to find two identical machines needed for Hopkinson's 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.
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