Monday, February 22, 2016
Hopkinson's Test or Regenerative Test On DC Motor
In the previous post we have seen how to determine the efficiency of DC machines using brake test. Hopkinson's test is also a test of finding the efficiency of a dc motor. Hopkinson's test or regenerative test is a full load test and it requires two identical machines which are coupled to each other.In this test two identical d.c. machines mechanically coupled to each other and simultaneously tested.One is operated as generator another one as motor,hence we can find efficiency of two dc machines simultaneously.So output power of dc machines are going to be wasted.The mechanical output of motor given to generator through shaft to shaft mechanical coupling.And generator's electrical power supplied to run the motor,where losses will be supplied by external power source.
If there are no losses in the motor-generator set,the electrical power from the generator and mechanical output from motor are enough to run motor,generator respectively.So no need of any external power supply to the motor.But due to losses, the generator output is not sufficient to drive the motor. Thus motor takes current from the supply to account for losses.
Observe circuit diagram of Hopikinson's test. The two shunt dc machines are connected in parallel. In that two machines,one is started as a motor another one operated as generator.Here the only rotor connections are mentioned,stator connections are not shown for simplicity.
Connection Diagram of Hopkinson's Test
First switch S is kept open. The other machine which is coupled to first will act as load on first which is acting as motor. Thus second machine will act as a generator.With the help of field rheostat speed of the motor is adjusted to normal value.Note down the observed voltmeter readings.With the help of generator field rheostat voltage of the generator is adjusted up to voltmeter reading is zero.This is to make sure generator voltage is having same magnitude and polarity of that of supply voltage.By making this we can prevent heavy circulating current flowing in the local loop of armatures on closing the switch.
Now close the switch S. The two machines can be put into any load by adjusting their field rheostats. The generator current I2 can be adjusted to any value by increasing the excitation of generator or by reducing the excitation of motor. The various reading shown by different ammeters are noted for further calculations.
The input to the motor is nothing but the output of the generator and small power taken from supply. The mechanical output given by motor after supplying losses will in turn drive the generator.
Calculation of Efficiency by Hopkinson's Test
Let V = Supply voltage
Motor power Input = V(I1 + I2)
Generator power Input = VI1
We can determine the efficiency of DC machines in two cases.
Case 1: Assuming that the efficiency of both the machines are same.
Case 2: Assuming both the machines has same iron loss, friction loss and windage loss.
Case 1:
Assuming that the efficiency of both the machines are same.
Motor output power = η x Motor Input power
= η V(I1+I2)
i.e., Motor Input power = Generator Input
Now the Generator output = η x generator Input
= η x ηV(I1 +I2)
= η2 V(I1+I2)
VI1 = η2V(I1+I2)
∵ Generator output η = √ {I1 / (I1+I2)}
Note: The above expression is used to determine the efficiency satisfactorily perfect for a rough test. If case need to find more accuracy then the efficiency of the two machines can be determined separately using the below expressions.
Case 2:
Assuming both the machines has same iron loss, friction loss and windage loss.
However the iron loss, friction loss and windage loss of both the machines will be same due to both the machines are identical. On this notion we can find the efficiency of each machine.
It is not necessary to assume that the efficiency of both the machines are same. It is due to that both the DC machines don’t have the same armature winding and the field winding.
Let,
Ra = Armature winding resistance of individual machines.
I3 = Shunt field current Generator G
I4 = Shunt fief current of Motor M
Generator armature copper loss = (I1+I32) Ra
Motor armature copper loss = (I1 + I2 – I42) Ra
Shunt field copper loss in G = VI3
Shunt field copper loss in M = VI4
Power drawn from the DC source is VI2 and is equal to the total losses of motor and generator.
VI2 = Motor and Generator total losses
To get the iron loss, friction and windage loss subtract the armature copper loss and shunt copper loss of both the machine from VI2.
Total losses of 2 machines (M & G)
= VI2 – [(I1+I3)2Ra + (I1+I2-I42Ra+VI3+VI4)] = W
To find the individual machine losses divide by 2
i.e. Total losses of each machine = W/2
To find the efficiency of Motor in hopkinson's test
Input motor power = V(I1 + I2)
Total Losses = (I1+I2-I42)Ra + VI4 + (W/2)
= Wm
Efficiency of Motor ηm = (Input – Losses) / Input
= [V(I1+I2) – Wm] / [V(I1+I2)]
To find the efficiency of Generator in hopkinson's test
Generator output power = VI1
Total Losses = (W/2) + (I1+I32)Ra + VI3
= Wg
Efficiency of Generator ηg = VI1 / (VI1+Wg)
Advantages of Hopkinson's Test or Regenerative Test On DC Motor
The various merits of Hopkinson's test are,
1. This test is very economical because it requires only very small power just to compensate the losses which is very small value when compared to full-load power of the motor-generator coupled system.
2. This is performed at full load condition so we can take flux distortion into account.
3. There is no need for arranging any actual load. Similarly by changing the field currents of two machines, the load can be easily changed and a load test over complete range of load can be taken.
4. This test is better suited in case of large machines.
Disadvantages of Hopkinson's Test or Regenerative Test On DC Motor
The various demerits of Hopkinson's test are,
1. It is a big task to check for two identical machines needed for Hopkinson's test.2. We can not load two machines 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 excitation.
4. It is difficult to operate the machines at rated speed because field currents vary widely.
5.The machines are not loaded equally in case of small machines which may lead to difficulty in analysis.