UNIT-III-Testing of DC Machines.ppt

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1. UNIT-III
TESTING OF D.C. MACHINES
by
Kumar Saliganti
Assistant Professor (C)
[email protected]
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
test

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:Whileperformingthistestwithseriesmachinescareshouldbe
takenthatbrakeappliedistightfailingWhichthemotorwillattain
dangerouslyhighspeedandgetdamaged

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

12. N-m
Output of the Motor=
Therefore, the torque developed by themotor
T (W
1W
2)*rkg-mOR
T 9.81*(W
1W
2)*r
T *w
9.81*(W
1W
2)*r*
2N
watt…..............(1)
…Inputtomotor …………................................(2)
Efficiency of Motor=output/input,
60
VI
L

2N*9.81*(W
1W
2)*r
60 *VI
L

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
●Inthismethod,them/cundertestisnotdirectly
loadedfordeterminingitsefficiencybutits
performancecharacteristicsisdeterminedbyusingthe
dataobtainedinnoloadtestperformedonthem/c.

15. Swinburne'stest
●Inthistest,themachineundertestisrunasamotor
althoughitmaybegenerator
●Atnoload,weapplytheratedvoltageacrossits
terminalsandadjustitsfieldcurrenttorunthemotor
atitsratedspeed
●Underthisconditionits
1.NoloadlinecurrentIo
2.FieldcurrentIsh
3.RatedvoltageV
Larerecorded
Fromwhicheitherconstantlossesorstraylossesare
computed.

16. Swinburne'stest

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

18. From the detail plate of m/c, its full load current isknown.
So ,let full load line current isI
L
Then, input to motor on fullload= watts
Its armature current on fullload,
VI
L
I
aI
LIsh
Full load armature copperlossIa
2
Ra
(IIsh )
2
Ra
L
Efficiency of the motor on full load =(input-losses)/input
VIPc(IIsh)
2
Ra

LL
VI
L

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

20. Thisisaregenerativetestwhichrequirestwoidenticaldcshunt
machinescoupledmechanicallyandconnectedelectricalinparallel.
Oneofthemachinesisoperatedasamotorandotherasa
generator.
Inthistest,themechanicalpoweroutputofthemotorisgivento
thegeneratorandtheelectricalpoweroutputofthegeneratoris
giventothemotor.Hence,thistestisalsocalledregenerativetestor
back-to-backtest.
Thepowerinputfromthesupplyisonlytomeetthelosses.
Therefore,thistestiseconomical.
Inthisteststraylossesaremeasured.
Hopkinson'stest

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

22. Observation Table
Total losses in both machines = VI
Lo
VI
Lo= I
ag
2
R
ag
+ I
am
2
R
am+ stray losses in both machines
Stray losses (W
s) = VI
Lo-I
ag
2
R
ag
-I
am
2
R
am
Stray losses for each machine = W
s/2
Input
voltage
vvolts
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)
VIag
V II
ag
2
R
ag
+ I
shg
2
R
shgW
s/2

ag
Efficiency of the motor =(input-losses)/ input
V (I
am+I
shm)I
am
2
R
am
I
shm
2
R
shmW
s/2

V (I
am+I
shm)

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
the
Field'stest (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
LetV
1 = Supply voltage
I
1= Current taken by motor
I
2= Load current
V
2= Terminal p.d. of generator
Ra, Rse = Armature and series field resistance of each machine
Power taken from supply = V
1I
1
Output obtained from generator = V
2I
2
Total losses in both the machines, W
T = V
1I
1-V
2I
2
V
1I
1-V
2I
2 = I
am
2
(R
am
+ R
sem
+ R
seg) + I
ag
2
R
ag+ stray losses in both
machines

27. Efficiency of the generator =output/(output+losses)
Efficiency of the motor =(input -losses) / input
V
2I
2
I
ag
2
R
ag
+ I
am
2
R
segW
s/2

V
2I
2
I
am
2
(R
am
+R
sem)W
s/2

V
1I
1
V
1I
1
Motor Efficiency
Generator Efficiency
Stray losses (W
s) = V
1I
1-V
2I
2 -I
am
2
(R
am+ R
sem+ R
seg) -I
ag
2
R
ag
Stray losses for each machine = W
s/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
300Ω/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 (I
f)
(Amps)
Armature Voltage
(Va)
(volts)
Armature
Current (Ia)
Speed (N)
(r.p.m)
Input
VIa
Constant
losses
(VIa-Ia
2
Ra)
S.No
Field current (I
f)
(Amps)
Armature Voltage
(Va)
(volts)
Armature
Current (Ia)
Speed (N)
(in r.p.m)
Input
VIa
Output
losses
(VIa-Ia
2
Ra)
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 = BN
2 Watts Hysteresis losses = CN Watts
Eddy current losses = DN
2 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 = Iao
2
*Ra watts V*Io –Ia
2
*Ra = (A + C)N +
(B + D)N
2
W
s/N = (A+C) + (B+D)N.

32. ThegraphbetweenW
s/N&Nisastraightline,fromwhich(A+C)and
(B+D)canbefound.
InordertofindA,B,CandDseparately,letthefieldcurrentbechanged
toareducedvalueI
f,andkeptconstantatthatvalue.If,voltageisapplied
tothearmatureasbefore,
we have ,
W
s= (A+C
1) N + (B+D
1) N
2
(at the reduced excitation, friction and windage losses are still are AN and
BN
2
, but hysteresis losses become C
1N and eddy current losses become
D
1N
2. We can now obtain (A+C) and (B+D) as before.
Now,
C/C
1 = (flux at normal excitation/flux at reduced excitation) D/D
1 = (flux at
normal excitation/flux at reduced excitation)
So, if we determine the ratio (flux at normal excitation/flux at reduced excitation)
we can find
A, B, C, D, C
1, & D
1