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Separation of losses in a dc shunt motor
1. Separation Of Losses in a DC Shunt Motor
Dept Of E&E Engg Page 1
Separation Of
Losses In A Dc
Shunt Motor
2. Separation Of Losses in a DC Shunt Motor
Dept Of E&E Engg Page 2
Contents
SL. NO TITLE PAGE NO.
01 INTRODUCTION 03
02 CIRCUIT DIAGRAM 04
03 APPARATUS REQUIRED 04
04 PRECAUTIONS 05
05 PROCEDURE 05
06 TABULAR COLUMN 05
07 CALCULATIONS 06
08 GRAPH 09
09 CONCLUSION 09
3. Separation Of Losses in a DC Shunt Motor
Dept Of E&E Engg Page 3
INTRODUCTION-
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*Ia watts.
Armature copper losses = Ia2*Ra watts
V*Ia – Ia2*Ra = (A + C)N + (B + D)N
W/N = (A+C) + (B+D)N.
The graph between W/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 seperately, let the field current be changed to a reduced value IIf,
and kept constant at that value. If, voltage is applied to the armature as before,
we now have ,
W/N = (A+C1) N + (B+D1) N2
(at the reduced excitation, function 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 excitation) we can find
A, B, C, D, C1, & D1
4. Separation Of Losses in a DC Shunt Motor
Dept Of E&E Engg Page 4
CIRCUIT DIAGRAM-
APPARATUS-
SL.NO NAME TYPE RANGE QUANTITY
01 DC
Voltmeter
M.C 0-300V 1
02 DC
Ammeter
M.C 0-2.5/5A
0-2.5/5A
1
04 Variable
Rheostat
--
--
138Ω/8.5A
700Ω/1.5A
1
1
03 Tachometer -- -- 1
5. Separation Of Losses in a DC Shunt Motor
Dept Of E&E Engg Page 5
PRECAUTIONS-
1. Keep armature rheostat (38Ω, 8.5A) at its maximum resistance position (cut in position)
2. Keep field rheostat (700 Ω, 1.5A) at its minimum resistance position (cutout position)
PROCEDURE-
1. Connections are made as per circuit diagram.
2. Keep armature rheostat (38Ω, 8.5A) at its maximum resistance position and keep field
rheostat (700 Ω, 1.5A) at its minimum resistance position
3. Switch on the main supply.
4. Start the motor with the help of 3 point starter
5. Keep the field current to certain constant value with the help of shunt field rheostat.
(say 1.0 Amps)
6. Gradually cutout the (38Ω, 8.5A) rheostat connected in series with the armature of the
motor in suitable steps and at each step of increase, note down the voltmeter reading
and speed of the motor.
7. Bring back the armature rheostat to its original position.
8. Switch off the supply.
TABULAR COLUMN-
Armature resistance=3.33 Ω(Ra)
Full excitation: Field current= 1.2Amps
Armature voltage, volts (v) 160 170 180 190
Armature current, Amps(Ia) 0.24 0.2 0.2 0.2
Speed, N(rpm) 1280 1350 1440 1528
W = V*Ia – I2a*Ra, watts 33.20 33.86 35.86 37.6
W/N (watts/rpm) 29.85*10-3 25.08*10-3 24.4*10-3 24.78*10-3
6. Separation Of Losses in a DC Shunt Motor
Dept Of E&E Engg Page 6
Reduced excitation: Field current= 0.8Amps
Armature voltage, volts (v) 167 170 180 190
Armature current, Amps(Ia) 0.12 0.118 0.12 0.12
Speed, N(rpm) 1454 1477 1566 1618
W = V*Ia – I2a*Ra, watts 19.99 20.01 21.55 22.75
W/N (watts/rpm) 13.74*10-3 13.55*10-3 13.76*10-3 14.06*10-3
CALCULATION-
For motor on no load, power input to the armature is the sum of the armature copper losses
and the above losses.
Power input to the armature=V.Ia watts.
Armature copper losses=Ia2.Ra watts.
Stray losses, W = V.I−Ia2.Ra = (A+C)N+(B+D)N2
W/N = (A+C)+(B+D)N
Friction losses + Hysteresis losses = AN+CN watts
= (A+C)N watts
From graph, OR = A+C = 9.8*10-3 watts/rpm
OP = A+C1 = 10.4*10-3 watts/rpm
Windage losses + Eddy current losses = BN2 + DN2 watts
= (B+N)N2 watts
From graph,
line TR Tanϴ =
∆W/N
∆N
= B+D =
3∗10 −3
1100
= 2.7272*10-6 watts/rpm2`
7. Separation Of Losses in a DC Shunt Motor
Dept Of E&E Engg Page 7
line SP Tanϴ1 =
∆W/N
∆N
= B+D1 =
5∗10−3
500
= 10*10-6 watts/rpm2
Eb = V-IaRa and Eb = ØZN P = K ØN
K ØN = Eb,
K Ø =
𝐸𝑏
𝑁
=
𝑉−𝐼𝑎𝑅𝑎
𝑁
At 4th reading
At full excitation, value of KØ =
𝐸𝑏
𝑁
=
𝑉−𝐼𝑎𝑅𝑎
𝑁
=
190−(0.2∗3.33)
1528
= 0.129145
At reduced excitation, value of KØ1 =
𝐸𝑏
𝑁
=
𝑉−𝐼𝑎𝑅𝑎
𝑁
=
190−(0.12∗3.33)
1618
= 0.1171819
𝐾Ø
𝐾Ø1 =
Ø
Ø1 =1.1020899
𝐶
𝐶1 = [
Ø
Ø1]1.6
C1 = C[
Ø1
Ø
]1.6 = 0.8555958C
OR = A+C = watts/rpm−
OP = A+C1 = A+C[
Ø1
Ø
]1.6
OP−OR = A+C [
Ø1
Ø
]1.6− A+C = C([
Ø1
Ø
]1.6- 1)
8. Separation Of Losses in a DC Shunt Motor
Dept Of E&E Engg Page 8
C =
𝑂𝑃−𝑂𝑃
([
Ø
1
Ø
]1.6− 1)
=
0.6∗10−3
144 .4042∗10−3 = 0.0041555
A = OR – C =9.8 ∗ 10−3
− 4.155 ∗ 10−3
= 0.005645
𝐷
𝐷1 = [
Ø1
Ø
]2 = 0.823314715
D1 = D[
Ø1
Ø
]2
Tanϴ = B+D watts/rpm
Tanϴ1 = B+D1 = [
Ø1
Ø
]2 watts/rpm
Tanϴ1 − Tanϴ = B+D [
Ø1
Ø
]2 – (B+D) = D ([
Ø1
Ø
]2- 1)
D =
Tanϴ1 − Tanϴ
([
Ø
1
Ø
]2− 1)
=
7.2728 ∗10−6
176 .6852∗10−3 = 0.000041162
B = Tanϴ − D = 2.7272 ∗ 10−6
− 0.000041162 = 0.000038434
At rated speed (1500rpm)
Friction losses = A* 1500 = 8.4675 watts
Windage (air friction) losses = B* 15002 = 86.4765 watts
Hysteresis losses = C* 1500 = 6.23325 watts
Eddy current losses =D* 15002 = 92.6145 watts
Total stray losses = total mechanical losses + iron losses
= friction losses + Windage losses + Hysterisis
losses + Eddy current losses
=193.79175 watts
9. Separation Of Losses in a DC Shunt Motor
Dept Of E&E Engg Page 9
GRAPH-
CONCLUSION-
From this experiment, hence we dertermined the different losses.
