The document describes different testing methods for DC machines. It discusses the simple/direct test method, Swinburne's indirect test method, and Hopkinson's regenerative test method. The simple/direct test method calculates efficiency by directly loading the DC machine, but it is only suitable for small machines. Swinburne's method measures no-load losses to determine efficiency indirectly. Hopkinson's method couples two identical DC machines together to test them simultaneously, with one acting as a motor and the other as a generator.

1. TRUBA INSTITUTE OF ENGG. & INFO. TECHNOLOGY
BHOPAL
A
TECHNICAL SEMINAR
ON
“TESTING OF DC MACHINE”
SUBMITTED TO:- SUBMITTED BY:-
MR. PRIYANK NEMA MRITYUNJAI KUMAR ANAND
ROLL NO. 0114EX101035
DEPARTMENT OF ELECTRICAL & ELECTRONICS ENGG.
SESSION-2012-13

2. CONTENTS
1. Testing method of dc m/c.
2. simple/direct test .
(a) Explanation,(b) Numerical (C)
Disadvantage .
3 .Swinburne’s test.
4.Hopkinsion’s test

3. TESTING OF DC MACHINE
Q:- Why testing are required ?
Ans-Machines are tested for finding out losses,
efficiency and temperature rise. For
small m/c we used DIRECT LOADING
test and for large shunt m/c, INDIRECT MET-
HOD are used.

4. TESTING METHOD ARE
 Simple / Direct test.
 Swinburne’s / Indirect Method test.
 Hopkinson’s / Regenerative / Back - to -
Back / Heat –run test.

5. Simple / Direct test
 This method is suitable
only for SMALL DC
m/c. In direct, method
the DC m/c is
subjected to rated load
and the entire o/p
power is wasted. The
ratio of o/p power to i/p
power gives the
efficiency of DC m/c.

6. EFFICIENCY MOTOR
 Motor o/p={ω(s1-s2)r× 9. 81}
 Motor i/p= VtI.
 ή(motor)={o/p}/{i/p}.
= [{ω(s1-s2)r× 9. 81}×100]/VtI
where; s1&s2 are the tension on the belt.
ω=2πn(motor speed in rad/sec
r=radius of pulley in meters(=1/2 out
side pulley dia+1/2 belt thickness)
Vt=terminal voltage & I=line current.

7. NUMERICAL
 A full load break test on a small d.c m/c shunt
motor ,give the following data
Spring balance readings 25 kg & 9 kg
Out side pulley diameter 19.5 cm
Belt thickness 0.50 cm
Motor speed 1500 rpm
Applied voltage 230 v
Line current 12.5 A
Calculate the shaft torque, shaft power & motor
efficiency at rated load.

8. SOLUTION
 Shaft torque=(S1-S2)×Effective radius
=(25.9)[(19.5/2)+(0.5/2)]×(1/100)kg-m
=16×0.1×9.81 Nm=15.7 Nm
Shaft power=ω×Shaft torque
=[(2π×1500)/60]×15.7=2470w
Motor i/p=Vti=230×12.5=2875w
ή(motor efficiency at rated
load)=[2470/2875]×100=85.6%

9. DISADVANTAGES
 The spring balance readings are not steady.
 Friction torque does not const. At a particular
setting of handwheels H1&H2

10. INDIRECT METHOD
 In this method ,no load m/c losses are first
measured by a suitable test and then
additional losses on load are determined
from the m/c data ,in order to calculate the
m/c efficiency .
 The simplest method of measuring the no
load m/c losses is by SWINBURNE’S
METHOD.

11. SWINBURNE’S METHOD
 As this is no load
test ,it can’t be
performed on a dc
series motor
 In this method, the
m/c whether it is a
MOTOR or
GENERATOR, is run
as a no load shunt
motor at rated
speed

12.  If Ia0 and If are the no load armature and
field current respectively .Then power
absorbed by the armature (=Vt×Iao) is equal
to the no load rotational loss W0 plus a small
amount armature circuit loss Ia0^2 ×ra.
 No load rotational loss W0=Vt×Iao- Ia0^2 ×ra.
 Here armature circuit resistance ra includes
the brush contact resistance also.
 Shunt field loss=Vt×If.

13.  Let IL be the load current at which m/c
efficiency is required.
 Generator efficiency
Generator o/p=Vt×IL
Armature current Ia=IL+If
Armature circuit loss= Ia^2×ra
ra=armature circuit resistance when hot.
Total loss=W0+ Ia^2×ra+Vt×If
ή(Generator)=[1-{(W0+
Ia^2×ra+Vt×If)}/{(Vt×IL+ W0+ Ia^2×ra+Vt×If)}]

14.  Motor efficiency
When m/c is working as a motor then
Ia=IL-If
Motor i/p=Vt×IL
ή(motor efficiency)=[1-
{(W0+Ia^2×ra+If×Vt)}/{(Vt×IL)}]

15. NUMERICAL
 A 400 v,20kw dc shunt motor takes 2.5 A
when running light. For an armature
resistance of 800Ω and brush drop 2V,find full
load efficiency.

16. SOLUTION
 Field current If=400/800=.5A
No load armature current Ia0=2.5-0.5=2.0A
Constant no load rotational loss
W0=Vt×Ia0-Ia0^2×ra-Vb×Ia.
=400×2.0-2^2×0.5-2×2=794w.
Total constant loss=W0+Vt×If=794+400×0.5
=994w
At full load, let IL=Line current=Ia+If

17. Power i/p=Power o/p +total loosses
Vt×IL =20,000+994+Ia^2×ra+Vb×Ia
400(Ia+0.5)=20,000+994+Ia^2+2×Ia
Ia=56.22A
Armature ohmic losses=56.22^2×0.5=1580.34w
Brush drop losses=56.22×2=112.44w.
Total losses=994+1580.34+112.44=2686.78w.
Power i/p=20,000+2686.78=22686..78w.
ή(motor at full load)
=(20.000/22686.78)×100=88.157%

18. ADVANTAGES
 Low power is required for testing even large
m/c, since only no load losses are to be
supplied from the main.
 It is convenient and economical.
 The efficiency can be calculated at any load
because constant losses are known.

19. DISADVANTAGES
 As the test is on no load, it doesn’t indicate whether
the commutation on full load is satisfactory and
whether the temperature rise would be within
specified limit.
 THIS TEST CAN’T BE APPLIED FOR A SERIES
MOTOR because speed of series motor is very high at
no load ,it is not possible to run series motor on no
load.
 Note:-In comparison to other the armature cu
losses is so small that it may be neglected &
constant loss may be take equal to no load i/p

20. Regenerative/Hopkinson’s
method.
 In this method ,two identical d.c m/c are
coupled ,both mechanically & electrically and
are tested simultaneously. One of the m/c
made to run as a MOTOR and it derives the
other m/c as GENERATOR .

21.  For this test m/c 1 is as
a dc shunt motor by a
starter & brought upto
rated speed with
switch S open .Bothe
the m/c run at same
speed ,because these
are MECHANICALLY
coupled .

22. TO PERFORM THE TEST FOLLOWING
PROCEDURE IS ADOPTED
 M/c 1(motor) is started through starter & its
field rheostat is adjusted so that it runs at
normal speed. The m/c 1(Motor) will drive m/c
2(Generator).The switch S is initially kept
open.
 The excitation of m/c 2 is gradually increased
( by decreasing the field circuit resistance).till
the volt metre 1 reads ZERO.Then switch S is
closed.

23.  M/c 2 is now floating neither taking any
current from the supply nor delivering any
current .Any desired load can be put on the
set by adjusting the shunt field regulators.The
m/c with lower excitation will act as a MOTOR
and other m/c will act as a GENERATOR.
Let V=supply voltage
I2=Armature current of m/c 2(Generator)
I1= Armature current of m/c 1(motor) .
If2=Field current of m/c 2(Generator).
If1=Field current of m/c 1(Motor).
Ra=Armature resistance of each m/c

24.  Motor i/p power =V(I+I2)=V×I1
 Generator o/p power=V×I2. ............(1)
If both the m/c have same efficiency ή,then
o/p of motor =ή×i/p
=ήV(I+I2)= V×I1 =Generator i/p .
o/p of Generator =ή×i/p .
=ή×ήV×I1=ή^2V(I1)............(2)
From equ (1) & (2).
ή^2V(I1)=V×I2.
sqrt{(I2/(I1)}

25. EFFICIENCY
Armature circuit loss in Generator =I2^2×ra .
Armature circuit loss in Motor=I1^2×ra .
Power drawn from supply= V×I.
No load rotational loss in two m/c=W0=
V×I-ra(I1^2+I2^2) .
No load rotational loss for each m/c=W0/2 .
Generator o/p= I2^2×ra.
Generator loss=Wg=(W0/2)+V×If2+ I2^2×ra.
ή(g)=[1-(Wg)/(V×I2+Wg)]

26.  Motor i/p=V(I1+If1)
Total motor losses Wm=(W0/2)+ V×If1
+I1^2×ra .
ή( Motor)=[1-(Wm)/V(I1+If2)]

27. ADVANTAGE
 Total power taken from the supply is very
low. Therefore this method is very
economical.
 The temperature rise and the commutation
condition can be checked under rated load
condition.
 Large m/c can be tested at rated load
without consuming much power from the
supply.
 Efficiency at different load can be determine.

28. DISADVANTAGES
 The main disadvantage of this method is the
necessity two practically identical m/c to be
available

29. THANK
YOU....