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1. UNIT-III
TESTING OF D.C. MACHINES
by
Kumar Saliganti
Assistant Professor (C)
skjntum@gmail.com
Department of Electrical and Electronics Engineering
JNTUH College of Engineering Manthani

2. TESTING OF DC MACHINES

3. CONTENTS :
 Objectives Of Testing
 Methods of Testing – direct, indirect and
regenerative testing
 Brake test
 Swinburne’s test
 Hopkinson’s test
 Field’s test
 Separation of stray losses in a d.c. motor

4. OBJECTIVES OF TESTING :
 A DC machine has to be tested for
proper fabrication and trouble free
operation.
 From the tests one can determine
the external characteristics needed
for application of these machines.
 Also, one can find the efficiency,
rating and temperature rise of the
machine.

5. METHODS FOR TESTING
 There are different methods for testing
of DC Machines. They are :
1. Direct Method
2. Indirect Method
3. Regenerative method

6. (1) DIRECT METHOD :
 In this method, the DC machine is
loaded directly by means of a brake
applied to a water cooled pulley coupled
to the shaft of the machine.
 It is not practically possible to arrange
loads for machines of large capacity. So
this method is used only for testing of
small dc machines .
 Brake Test is an example of direct test.

7. (2) INDIRECT METHOD :
• In this method of testing, the losses are
determined without actual loading of the
machine.
• If the total losses in the machine are
known the efficiency can be calculated.
• Swinburne’s Test is an example of
Indirect Test.

8. (3) REGENERATIVE METHOD :
• This method requires two identical dc
machines. One of the machines is operated
as a motor and drives the other machine as a
generator.
• The electrical output of the generator is
feedback into the supply. Thus the power
drawn from the supply is small only to
overcome the losses of two machines.
• Hence the large machines can be tested at
rated load without consuming much power
from the supply.
• Hopkinson’s Test is an example of
Regenerative Method .

9. BRAKE TEST OR LOAD
TEST:
 To assess the rating of a machine a load test has to
be conducted.
 When the machine is loaded, certain fraction of the
input is lost inside the machine and appears as
heat, increasing the temperature of the machine.
 If the temperature rise is excessive then it affects
the insulations, ultimately leading to the breakdown
of the insulation and the machine.
 The load test gives the information about the
efficiency of a given machine at any load condition.
Also, it gives the temperature rise of the machine.
 The load test alone can give us the proper
information of the rating and also can help in the
direct measurement of the efficiency.

10. Brake
Test
Precaution: While performing this test with series machines care should be
taken that brake applied is tight failing Which the motor will attain
dangerously high speed and get damaged

11. Let , V= supply voltage measured by
voltmeter V
I = Input current measured by ammeter A
W1 and W2 = Spring balance reading in Kg
N = Speed of armature in rpm
r = Radius of pulley
The net force acting on pulley is OR
(W1W2) Kg
9.81(W1  W 2)Newton

12. N-m
Output of the Motor =
Therefore, the torque developed by the motor
T  (W1W 2)* r kg-m OR
T 9.81*(W1W2)*r
T * w
 9.81*(W1W 2)* r *
2 N
watt …..............(1)
…
Input to motor …………................................ (2)
Efficiency of Motor= output/input,
60
VIL
 
2N * 9.81* (W1  W 2) * r
60 *VIL

13. LIMITATIONS :
1. Large amount of power is required to test a large
machine and the entire output power is wasted at
the mechanical brake.
2. Non- availability of large capacity load for testing
large motors in the laboratories.

14. Indirect method of
Testing
● In this method, the m/c under test is not directly
loaded for determining its efficiency but its
performance characteristics is determined by using the
data obtained in no load test performed on the m/c.

15. Swinburne's test
● In this test, the machine under test is run as a motor
although it may be generator
● At no load, we apply the rated voltage across its
terminals and adjust its field current to run the motor
at its rated speed
● Under this condition its
1.No load line current Io
2.Field current Ish
3.Rated voltage VL are recorded
From which either constant losses or stray losses are
computed.

16. Swinburne's test

17. No load arm current of the motor is Iao  Io  Ish
Constant losses (Pc)=No load input – No load arm
copper loss
2
Extra:
Therefore,
stray losses = input on no load – shunt field copper
loss – armature copper loss
V*IoIsh2
RshIao2
Ra
V*Io Iao Ra

18. From the detail plate of m/c, its full load current is known.
So ,let full load line current is IL
Then, input to motor on full load = watts
Its armature current on full load,
VI L
I a  I L  Ish
Full load armature copper loss  Ia 2
Ra
 (I  Ish ) 2
Ra
L
Efficiency of the motor on full load = (input-losses)/input
V I  Pc (I  Ish)2
Ra
L L
VIL

19. If machine under the test is generator having IL its full load
output current and V is the load voltage or terminal voltage
Then, output of generator on full load = watts
Its armature current on full load,
VIL
I a  I L  Ish
Full load armature copper loss  Ia 2
Ra
 (I  Ish ) 2
Ra
L
Efficiency of the generator on full load = output/(output+losses)
V I
sh
L
L
L
V I  Pc (I  I )2
Ra


20. This is a regenerative test which requires two identical dc shunt
machines coupled mechanically and connected electrical in parallel.
One of the machines is operated as a motor and other as a
generator.
In this test, the mechanical power output of the motor is given to
the generator and the electrical power output of the generator is
given to the motor. Hence, this test is also called regenerative test or
back-to-back test.
The power input from the supply is only to meet the losses.
Therefore, this test is economical.
In this test stray losses are measured.
Hopkinson's test

21. Hopkinson's test
Fig. Circuit diagram for performing Hopkinson's test on two dc shunt
machines

22. Observation Table
Total losses in both machines = VILo
VILo = Iag
2Rag + Iam
2Ram + stray losses in both machines
Stray losses (Ws) = VILo - Iag
2Rag - Iam
2Ram
Stray losses for each machine = Ws/2
Input
voltage
v volts
Current
drawn
from the
supply
I1A
Gen.
Arm.
Current
I2 A
Motor
Arm.
Current
(I1+I2) A
Motor
field
Current
I3 A
Gen.
Field
Current
I4 A

23. Efficiency of the generator = output / (output+losses)
V Iag
V I  Iag
2Rag + Ishg
2Rshg  Ws/2

ag
Efficiency of the motor = (input - losses) / input
V (Iam +Ishm)  Iam
2Ram  Ishm
2Rshm  Ws/2

V (Iam +Ishm )

24.  Advantages of Hopkinson’s test
 The temperature rise can be estimated during the test.
 The efficiency of the machine can be accurately determined
at various loads.
 The commutation conditions can be checked under rated load
conditions.
 Disadvantages of Hopkinson’s test
 Availability of two identical dc machines.
 It is impossible to separate out iron losses of the two
machines
which are different because of different excitations.

25. By above fig. small series machines can be tested by direct load test, but the
large series
machines cannot be tested by Swinburne’s test because series motor can not
run at
No- load due to dangerously high speed .
Field’s test is applicable to two similar series machines. These two machines are
mechanically coupled and electrically isolated. One of machines are run as a
motor
and drives the other machine as a generator.
A variable load is connected across the generator terminal. The output power of
Field's test (for series motors)
Fig. Circuit diagram for performing Field’s test on dc series
machines

26. Observation Table
Supply
Voltage
V
Motor
Armature
Current
I1
Generator
Armature
Current
I2
Generator
Output
Voltage
V2
Motor
Voltage
V1
Let V1 = Supply voltage
I1 = Current taken by motor
I2 = Load current
V2 = Terminal p.d. of generator
Ra, Rse = Armature and series field resistance of each machine
Power taken from supply = V1I1
Output obtained from generator = V2 I2
Total losses in both the machines, WT = V1I1 - V2 I2
V1I1 - V2 I2 = Iam
2 (Ram + Rsem + Rseg) + Iag
2Rag + stray losses in both

27. Efficiency of the generator = output / (output+losses)
Efficiency of the motor = (input - losses) / input
V2 I2  Iag
2Rag + Iam
2Rseg  Ws/2

V2 I2
 Iam
2 (Ram + Rsem)  Ws/2

V1 I1
V1 I1
Motor Efficiency
Generator Efficiency
Stray losses (Ws) = V1I1 - V2 I2 - Iam
2 (Ram + Rsem + Rseg) - Iag
2Rag
Stray losses for each machine = Ws/2
The stray losses are equally divided between the machines because of
their equal excitation and speed.

28.  Advantages of Field’s test
 The actual performance of the machine is verified, i.e.
temperature rise and commutation.
 Stray losses are considered and they are equally divided, which
is justified.
 Disadvantages of Field’s test
 Two identical dc series machines are required. hence, cost is
very high.
 The entire power drawn from the supply is wasted across load
resistance of generator.

29. Separation of stray losses in a d.c. motor test
16A
220V DC
SUPPLY
16A
(0-300V)
V
(0-300V)
A
A
FUSE
300 Ω /2A
300 Ω /2A
Z
L A
V
Min
(0-2A)
Z
ZZ
1
2
AA
3- POINT STARTER
M
DPST
A
(0-20A)

30. S.No
Field current (If)
(Amps)
Armature Voltage
(Va)
(volts)
Armature
Current (Ia)
Speed (N)
(r.p.m)
Input
VIa
Constant
losses
(VIa-Ia2Ra)
S.No
Field current (If)
(Amps)
Armature Voltage
(Va)
(volts)
Armature
Current (Ia)
Speed (N)
(in r.p.m)
Input
VIa
Output
losses
(VIa-Ia2Ra)
Field control method
Armature control method
Observation Table

31. At a given excitation, friction losses and hysteresis are proportional to
speed. Windage losses and eddy current losses on the other hand are
both proportional to square of speed. Hence, for a given excitation (field
current) we have,
Friction losses = AN Watts
Windage losses = BN2 Watts Hysteresis losses = CN Watts
Eddy current losses = DN2 Watts Where, N = speed.
For a motor on no load, power input to the armature is the sum of the
armature copper losses and the above losses. In the circuit diagram,
Power input to the armature = V*Io watts.
Armature copper losses = Iao2*Ra watts V*Io – Ia2*Ra = (A + C)N +
(B + D)N2
Ws/N = (A+C) + (B+D)N.

32. The graph between Ws/N & N is a straight line, from which (A+C) and
(B+D) can be found.
In order to find A, B, C and D separately, let the field current be changed
to a reduced value If, and kept constant at that value. If, voltage is applied
to the armature as before,
we have ,
Ws = (A+C1) N + (B+D1) N2
(at the reduced excitation, friction and windage losses are still are AN and
BN2, but hysteresis losses become C1N and eddy current losses become
D1N2. We can now obtain (A+C) and (B+D) as before.
Now,
C/C1 = (flux at normal excitation/flux at reduced excitation) D/D1 = (flux at
normal excitation/flux at reduced excitation)
So, if we determine the ratio (flux at normal excitation/flux at reduced excitatio
we can find
A, B, C, D, C1, & D1