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.
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
