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 = Rm
- Armature resistance for the generator = Rg
- 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;
Pam = (I1 + I2)2 Rm
Armature copper loss for generator;
Pag = I22 Rg
The sum of all losses is equal to the power drawn from the supply.
VI1 = 2Wc + Pam + Pag
VI1 = 2WC + (I1+ I2)2Rm +I22Rg
Wc = ½ (VI1 – (I1 + I2)2 Rm – I22 Rg
Hopkinson Test
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.
Table of Content
- Hopkinson’s Test
- Connection Diagram
- Calculation of the Efficiency of the Machine
- Efficiency of Motor
- Efficiency of Generator
- Advantages of Hopkinson Test
- Disadvantages of Hopkinson Test
- Applications of the Hopkinson Test