DC machines, like motors and generators, are used in various electrical applications. The main function of the generator is to convert the power from mechanical to electrical, though the motor is used to convert the power from electrical to mechanical. The DC generator’s input power is electrical, while the output power is mechanical.

In this article, we will be going through Hopkinson’s, First, we will start our article with the introduction of Hopkinson’s test, then we will go through its connection diagram and then we will calculate the efficiency of the machine, motor and generator, At last, we will conclude our article with advantages, disadvantages, applications with some faqs.

What is Hopkinson’s Test?

A full-load test that is used to test the efficiency of a DC machine is known as Hopkinson’s test. An alternate name for this test is back-to-back, heat run, or regenerative test. This test makes use of two machines that are mechanically and electrically linked to each other. From these machines, one acts as a motor, while the other works as a generator. The generator gives mechanical power to the electric motor, though the motor is used to drive the generator.

Accordingly, the o/p of one machine is used as an input to another machine. At the point when these machines run on full load, the input supply can be equivalent to the entire losses of the machines. Assuming there are no losses inside any machine, there is no requirement for an external power supply. In any case, in the event that the o/p voltage of the generator is dropped, we want an extra voltage source to give the motor an appropriate o/p voltage. Hence, the power that is drawn from the external supply can be used to overcome the losses of the machines.

Connection Diagram of Hopkinson’s Test

The electrical supply is given to the first machine, and this machine acts as a motor. At first, switch S is kept open. Thus, the input is provided to the motor only. The speed of the motor is changed in accordance with the evaluated speed with the help of a field regulator.

Connection Diagram of Hopkinson’s test

The second machine functions like a generator. At the point when we give a contribution to the motor, it will begin turning. Furthermore, the two machines are connected on a similar shaft. In this way, the generator produces electrical power. The result of a generator is adjusted to its rated power with the help of a field regulator.

In any case, the switch S is open. A voltmeter is connected across the switch. At the point when the voltage created by the generator is equivalent to the supply voltage, this voltmeter demonstrates a zero reading. What’s more, at this stage, close the switch S.

Now the generator will supply the motor. What’s more, the electrical power provided by the input is used to meet the losses of the two machines

At the point when the generator is connected to a motor, the excitation of the generator is increased. It causes its EMF to rise to levels higher than the supply voltage. As the motor is located, the speed decreases. The speed of the motor and output voltage of the generator are changed by the field controllers.

Calculation of the Efficiency of the Machine by Hopkinson’s Test

Calculation of efficiency by using the Hopkinson Test.

• Input voltage = V
• Current drawn from the supply = I₁
• Generator armature current = I₂
• Motor armature current = I₁ + I₂
• Motor field current = I₃
• Generator field current = I₄
• Armature resistance for motor = R_m
• Armature resistance for the generator = R_g
• Power drawn from the supply = V I₁

The power drawn from the supply is equal to the losses of the two machines. The losses of DC machines are:

• Copper loss
• Iron loss
• Mechanical loss

Let us assume that the iron and mechanical losses for each machine are Wc.

Armature copper loss for the motor;

P_am = (I₁ + I₂)² R_m

Armature copper loss for generator;

P_ag = I₂₂ R_g

The sum of all losses is equal to the power drawn from the supply.

VI₁ = 2Wc + P_am + P_ag

VI₁ = 2W_C + (I1+ I2)²R_m +I₂₂R_g

Wc = ½ (VI₁ – (I₁ + I₂₎² R_m – I₂₂ R_g

Efficiency of Motor

There are three types of motor losses:

• Armature copper losses, (P_am)
• Shunt field copper losses (P_fm)
• Iron and mechanical losses (W_c)

The current that goes through the shunt field winding is I₃.

Consequently, shunt field copper loss;

P_fm = V I₃

Total loss of motor

P_tm = Wc + P_am + P_fm

P_tm = Wc + (I₁+ I₂)² R_m + V I₃

Total input power of the motor

P_i = V(I₁ + I₂) + V(I₃)

P_i = V( I₁ + I₂ + I₃)

Therefore, a motor’s efficiency is

η = (input power minus total loss minus input power) x 100

η = (Pi – P_tm / Pi) x 100

Efficiency of Generator

The current that goes through the field winding of the generator is I₄.

Hence, the shunt field copper loss

P_fg = V I₄

Total loss for the generator

P_tg = W_c + P_ag + P_fg

P_tg = W_c = I ₂₂R_g + V_I4

The generator’s output

Po = V I ₂

Hence, the efficiency of a generator

= (output/output + total losses) x 100

= (Po / Po + Ptg) x 100

Advantages of Hopkinson Test

• Accurate Efficiency Measurement: Contrasted with different techniques, Hopkinson’s Test gives a highly accurate measurement of efficiency under full-load conditions, including stray losses.
• Economical: It doesn’t need additional power sources as the tested machine generates the power for the other machine, making it economical.
• Realistic Testing: The machines work under their actual working circumstances, considering a more reasonable assessment of execution and losses.
• Account for Iron Losses: The test considers the change in iron losses because of flux distortion at full load, which gives a more accurate efficiency calculation.
• Efficiency at Different Loads: You can determine the efficiency across different load conditions by changing the excitation currents.

Disadvantages of Hopkinson Test

• Double Machine Requirement: The test requires two identical machines, which can be difficult and costly to obtain, particularly for large machines.
• Challenge of Sharing Loads: Keeping up with equal load dividing among the two machines can be testing and requires exact control of the field flows.
• Separate Iron Loss Measurement: Acquiring the individual iron losses of each machine during the test is troublesome.
• Control Complexity: Exact control of the machines’ velocities and field current is necessary, making the test setup and execution complex.
• Temperature Rise: Delayed testing can lead to a huge temperature rise in the machines, possibly influencing the accuracy of the results.

Applications of the Hopkinson Test

• Testing the efficiency of new or repaired electrical machines.
• Looking at the performance of different machine designs.
• Exploring the effect of working conditions on machine efficiency.

Conclusion

Despite requiring two identical machines and complex control, the Hopkinson Test is a highly accurate and effective method for evaluating electrical machines’ full-load performance, particularly DC motors and generators. Its assets lie in practical testing conditions, representing iron losses, and variable load analysis.

However, careful consideration is required regarding difficulties with load sharing, iron loss separation, and temperature rise. While options like separation of losses and dynamometer tests exist, the Hopkinson Test’s accuracy and efficiency stay unparalleled for requesting applications. The Hopkinson Test is still a useful tool for precise efficiency measurement and analysis, particularly in high-performance machines.

Hopkinson Test – FAQs

Which type of machines is the Hopkinson Test suitable for?

Primarily it is used for DC motors and generators, however can be adjusted for different types of electrical machines with changes.

Could the Hopkinson Test be used for AC machines?

Yes, with extra instrumentation and alterations to represent AC conduct. However, because of its complexity and the availability of other methods, it is less common.

Are there any options in alternatives to the Hopkinson Test?

Yes, methods like separation of losses test and regenerative dynamometer test exist. Each has its own advantages and disadvantages depending upon the particular application and accessible assets.

What are the safety precaution during the Hopkinson Test?

Proper electrical isolation and grounding are crucial, alongside cautious checking of machine temperatures and currents to prevent accidents.