EXPERIMENT 11

11.1Object: To Perform a Hopkinson Test on Two Identical D.C. Shunt Machines

11.2Experimental Setup

This test is called a back-to-back test, in which two identical D.C. shunt machines are required. These two machines are coupled mechanically and connected electrically in parallel. One of the machines runs as a motor and the other as the generator. Mechanical energy of the motor drives the generator, and electrical energy of the generator is used in supplying most of the input to the motor.

If the machines were lossless, they would run with the help of external power input. The losses are supplied from the external supply source.

Figure 11.1

The circuit diagram is connected as shown in Figure 11.1 with instruments. The range of the voltmeters and ammeters will depend on the ratings of the machines. Read the nameplates and find the ranges of the instruments.

The motor is run from the supply mains, and rated current is fed to the motor. The voltage of the generator is adjusted so that the voltmeter connected across the switch K reads zero volts. This will ensure the same magnitude and polarity of the generator output voltage as that of the supply. The readings of the instruments are noted.

EXPERIMENTAL RESULTS

The symbols used in the previous table are:

Im	:	motor current
Imf	:	motor field current
VL	:	line voltage
IL	:	line current
Eag	:	generator voltage
Igf	:	generator field current
Ig	:	generator current

11.3Calculations

The copper loss in the two machines can be calculated as follows:

(i) Copper loss in motor field

= V_LI_mf watts

(ii) Copper loss in motor armature

= R_a(I_m – I_mf)²

= R_aI_am² watts

(iii) Copper less in generator field

= V_-LI_gf watts

(iv) Copper loss in generator armature

= R_a(I_g – I_gf)²

= R_aI_ag² watts

Total copper loss will be

Pc = RaIam2 + RaIag2 + VLIgf + VLImf

Pc = Ra(Iam2 + Iag2) + VL(Igf + Imf)

(11.1)

Input power to the two machines will be

P = VLIL watts

(11.2)

Constant loss = Iron+Friction+Windage

Pi = VLIL – Pc

(11.3)

The loss for one machine will be:

Pc /2 and Pi/2

(11.4)

This experiment is made to find the losses in the machines when the machines have large capacity and when large power will be consumed in testing.

The efficiency will be:

11.4Questions and Answers on the Experiment

Q1. What are other names for the Hopkinson test?

Ans. Back-to-back test and regenerative test.

Q2. What are the advantages of the Hopkinson test?

Ans. Large capacity generators can be tested in a laboratory without wasting a large amount of energy in the loads. Large loads are avoided and actual load is applied on the machine.

Q3. How much power is taken from the supply?

Ans. To meet the losses in windings and core, the copper loss and iron loss are fed from the supply.