Hopkinson’s Test _ Electrical4u.pdf

11/29/2014 Hopkinsons Test | Electrical4u

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Hopkinsons Test

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Hopkinsons test is another useful method of testing the efficiency of a dc machine. Itis 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 powerto the motor and the other is operated as a motor to drive the generator. For thisprocess of back to back driving the motor and the generator, Hopkinsons test is alsocalled back-to-back test or regenerative test.

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

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Connection Diagram of HopkinsonsTest

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



Calculation of Efficiency by HopkinsonsTestLet, 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

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And, Generator output = VI1(1)

Let, both machines are operating at the same efficiency .

Then, Output of motor = x input = x V(I1 + I2)

Input to generator = Output of the motor = X V(I1 + I2)

Output of generator = x input = x [ x V(I1 + I2)] = 2

V(I1 + I2)(2)

From equation 1 an 2 we get,

VI1 = 2 V(I1 + I2) or I1 = 2 (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 Hopkinsons TestThe merits of this test are1. 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 limitbecause this test is done under full load condition.

3. Change in iron loss due to flux distortion can be taken into account due to theadvantage of its full load condition

Disadvantages of Hopkinsons TestThe demerits of this test are

1. It is difficult to find two identical machines needed for Hopkinsons 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 aredifferent because of their excitations.

4. It is difficult to operate the machines at rated speed because field currents varywidely.

Objective Questions on DC Generator (MCQs)



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Hopkinson’s Test _ Electrical4u.pdf

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