Regenerative or Hopkinson’s test is a specialized method used to assess the performance and characteristics of large electrical machines, such as DC generators and motors. This test is particularly valuable in understanding the behavior of devices under load conditions, offering insights into efficiency, losses, and other critical parameters. In the previous post, we have seen how to determine the efficiency of DC machines using Swinburn’s test. The Regenerative or Hopkinson’s Test method is another method to determine a DC machine’s efficiency. This method saves power and gives a much more accurate DC machine result. We need similar DC machines and supply voltage to carry out this test.
Working Principle
Two similar DC Shunt machines are coupled mechanically and parallel connected across the DC source. In both the machines, one is running as a motor, and another is running as a generator by varying the shunt field excitation. The electrical power given to the engine motor is converted to mechanical energy, the rest going to the other motor losses.
This converted mechanical power is given to the Generator. The electrical energy generated from the Generator is returned to the Motor except for the power wasted as generator loss. Thus, the electric power taken from the DC supply is the sum of motor and generator losses, which can be directly read through the ammeter and voltmeter. Since the input power from the DC supply is equivalent to the energy needed to provide the losses of two machines, this test can be done only with a small amount of power.
We can load the machine by varying the machine field strength. Therefore the total loss of the device we can determine at any load. The machine’s commutation performance and the temperature rise can be observed since the device tested at full load condition.
The figure above shows the typical circuit diagram of a Hopkinson test. Two similar DC shunt machines are coupled mechanically and parallel connected across the DC source. By varying each machine field strength 1st, machine M can run as a motor, and machine G can run as a generator. The Motor draws current I1 from the Generator and current I2 from the DC source. Thus the current input to the Motor is (I1+I2). Electrical power from the DC source is VI2, equivalent to the total losses (motor and Generator losses). The shunt field current of the Motor is I4, and the Generator is I3.
Calculation
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.
Assuming Equal Machine Efficiencies
Assuming Equal Iron, Friction, and Windage Losses
Assuming Equal Machine Efficiencies
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)
The above expression determines the efficiency satisfactorily perfect for a rough test. If the case needs to find more accuracy, then the efficiency of the two machines can be defined separately using the below expressions.
Assuming Equal Iron, Friction, and Windage Losses
It is not necessary to assume that the efficiency of both machines is the same. It is that both the DC machines don’t have the same armature winding and field winding. However, the iron loss, friction loss, and windage loss of both devices will be the same due to both machines are identical. On this notion, we can find the efficiency of each machine.
Power drawn from the DC source is VI2 and is equal to the total losses of the Motor and Generator.
VI2 = Motor and Generator total losses
To get the iron loss, friction, and windage loss, subtract both machines’ armature copper loss and shunt copper loss 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
Calculating the Efficiency of a Motor and Generator
Calculating the efficiency of a motor and a generator is crucial for assessing their performance in converting electrical and mechanical energy. Efficiency serves as a valuable indicator of how effectively these machines operate.
Let’s delve into the efficiency calculations for both the Motor and Generator:
The efficiency of the Motor
Input Motor Power: The input power to the Motor is given by the expression V(I1 + I2), where V is the voltage and I1, I2 are the currents.
Total Losses (Wm): The total losses in the Motor are calculated as (I1 + I2 – I4²)Ra + VI4 + (W/2), where Ra is the armature resistance and W represents mechanical losses.
Motor Efficiency (ηm): The efficiency of the Motor is calculated using the formula: ηm = [V(I1 + I2) – Wm] / [V(I1 + I2)].
Input motor power = V(I1 + I2)
Total Losses = (I1+I2-I42)Ra + VI4 + (W/2)
= Wm
The efficiency of Motor ηm = (Input – Losses) / Input
= [V(I1+I2) – Wm] / [V(I1+I2)]
The efficiency of the Generator
Generator Output Power: The Generator’s output power is given by VI1, where V is the voltage and I1 is the current.
Total Losses (Wg): The total losses in the Generator are computed as (W/2) + (I1 + I3²)Ra + VI3, where Ra is the armature resistance, and W represent losses.
Generator Efficiency (ηg): The generator efficiGeneratoralculated as ηg = VI1 / (VI1 + Wg).
Generator output power = VI1
Total Losses = (W/2) + (I1+I32)Ra + VI3
= Wg
Efficiency of Generator ηg = VI1 / (VI1+Wg)
Conclusion
In conclusion, the regenerative or Hopkinson’s test offers a powerful means to unveil the potential of large electrical machines by providing accurate insights into their efficiency, losses, and performance characteristics under varying load conditions. This specialized testing method empowers engineers and researchers to make informed decisions regarding machine design, operation, and application optimization.
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