Theory of Time 2024 (Universal Theory for Everything)
SWINBURNE’S TEST
1. SWINBURNE’S TEST
DEPT. EEE KLS’s VDRIT HALIYAL Page 1
ABSTRACT:
An electric motor is a devise using electrical energy to produce mechanical energy to produce
mechanical power nearly by interaction of magnetic fields and current carrying conductors.In
this method we are going to test the d.c shunt motor. The reverse process that of using
mechanical power output and including its losses we get its efficiency.
3. SWINBURNE’S TEST
DEPT. EEE KLS’s VDRIT HALIYAL Page 3
INTRODUCTION:
Swinburne’s test is an indirect method of testing of dc machines. In this
method losses are measured separately and the efficiency at any desired load is predetermined.
Machines are tested for finding out losses, efficiency and temperature rise. For small machines
direct loading test is performed. For large shunt machines, indirect methods are used like
Swinburne’s test. Very large capacity machines which cannot be tested with actual loading can
be tested by this method and its performance characteristics is determined by using the data
obtained from Swinburne’stest.
CHARACTERISTICS OF D.C SHUNT MOTOR:
This characteristic is also known as electrical characteristic. We know that
torque is directly proportional to the product of armature current and field flux, Ta ∝ ɸ.Ia. In DC
series motors, field winding is connected in series with the armature, i.e. Ia = If. Therefore, before
magnetic saturation of the field, flux ɸ is directly proportional to Ia. Hence, before magnetic
saturation Ta α Ia2. Therefore, the Ta-Ia curve is parabola for smaller values of Ia.
After magnetic saturation of the field poles, flux ɸ is independent of armature
current Ia. Therefore, the torque varies proportionally to Ia only, T ∝ Ia.Therefore, after
magnetic saturation, Ta-Ia curve becomes a straight line. The shaft torque (Tsh) is less than
armature torque (Ta) due to stray losses. Hence, the curve Tsh vs Ia lies slightly lower. In DC
series motors, (prior to magnetic saturation) torque increases as the square of armature current,
these motors are used where high starting torque is required.
We know the relation, N ∝ Eb/ɸ For small load current (and hence for small
armature current) change in back emf Eb is small and it may be neglected. Hence, for small currents
speed is inversely proportional to ɸ. As we know, flux is directly proportional to Ia, speed is
inversely proportional to Ia. Therefore, when armature current is very small the speed becomes
dangerously high. That is why a series motor should never be started without some mechanical load.
But, at heavy loads, armature current Ia is large. And hence, speed is low which results in decreased
back emf Eb. Due to decreased Eb, more armature current is allowed.
Aim: to perform the Swinburnes test on D.C machine to find the losses.
APPARATUS REQUIRED:
Sl.no Item Type Range Quantity
1 Voltmeter M.c 0-300v 1
2 Ammeter M.c 0-2.5/5a 2
3 Rheostat ---- 700/1.5a 1
4 Tachometer ---- ----- 1
5. SWINBURNE’S TEST
DEPT. EEE KLS’s VDRIT HALIYAL Page 5
WORKING PRINCIPLE:
The machine is run on no load at its rated voltage.
At starting some resistance is connected in series with the armatrure which is cut when
motor attains sufficient speed.
Now the speed of the machine is adjusted to the rated speed with the help of shunt field
rheostsat.
PRECAUTIONS:
1. keep field rheostat (700ohm,1.5A) at its minimum resistance position (cutout position).
PROCEDURE:
1. connection are made as per circuit diagram.
2. keep the field rheostat (700ohm,1.5A) to the minimum resistance position.
3. switch on the main supply.
4. start the motor with the help of 3point starter.
5. bring the motor to its rated speed by variying the rheostat connected in series with field of the
motor
6. then note down the readings of the voltmeter and ammeters.
7. bring back all the rheostat to its original positions.
8. switch off the supply.
TABULAR COLUMN:
Sl.no
Voltage v
(volts)
Field current
If=Ish(amps)
No load arm.
Current
Iao(amps)
No load current
Io=Ish+Iao
(amps)
1 200 1 1.2 2.2
6. SWINBURNE’S TEST
DEPT. EEE KLS’s VDRIT HALIYAL Page 6
CALCULATION:
Armature resistance of the motor(Ra)=3.6ohms
Armature current (Iao) =Io-Ish =1.2Amps
Input to the motor on no-load =V*Io=440 watts
Armature copper loss =Iao2*Ra =5.184 watts
Shunt field copper loss=V*If =200 watts
Constant losses (Wc) =N.L input to motor-N.L armature copper loss
=V*Io-(Iao2*Ra)=434.816 watts
EFFICIENCY AS MOTOR AT FULL LOAD
Full load rated current (Ifl) =10 amps
Armature current (Ia)=Ifl-Ish =9 amps
Input to the motor at full load =Vm*Ifl=2200 watts
Armature copper loss =Ia2*Ra =291.6 watts
Total losses (Wt) ={constant losses(Wc) + armature copper loss}
= 434.816 + 291.6 watts
= 726.416 watts
Output=input-total losses
= Vm*Ifl-total losses(Wt)
= 1473.584 watts
Efficiency = output *100
Input
= 1473.584 * 100
2200
= 66.98%
EFFICIENCY AS GENERATOR AT FULL LOAD
Full load rated current (Ifl) =10 amps
Armature current (Ia) =Ifl+Ish=11 amps
Output of generator at full load=Vm*Ifl=2200 watts
7. SWINBURNE’S TEST
DEPT. EEE KLS’s VDRIT HALIYAL Page 7
Armature copper loss =Ia2*Ra=435.6 watts
Total losses(Wt) ={constant losses + armature copper loss}
= 434.816 + 435.6
= 870.416 watts
Input = output + total losses
= V*Ifl + total losses(Wt)
= 3070.416 watts
Efficiency = output *100
input
= 2200 *100
3070.416
= 71.65%
Machine Full load Half load 3/4th load 1/4th load
As motor 66.98 55.23 64.42 19.46
As generator 71.65 66.08 70.36 53.45
ADVANTAGES:
The power required to test a large machine is small. Thus, this method is an economical
and convenient method of testing of dc machines.
As the constant loss is known the efficiency can be predetermined at any load.
The machine is not required to be loaded i.e., only test to be carried out is the no load
test.
LIMITATIONS:
Machines having constant flux are only eligible for swinburne’s test.
Series machine cannot run on light loads, and the values of speed and flux varies greatly.
Thus, the swinburne’s test are not applicable for series machines.
8. SWINBURNE’S TEST
DEPT. EEE KLS’s VDRIT HALIYAL Page 8
MATLAB PROGRAM
clc;
clear all;
% Initialising module
Ra=3.6; %Armature resistance of motor
V=input('enter the value of supply voltage V= ');
Ish=input('enter the value of field current Ish= ');
Ia0=input('enter the value of noload arm current Ia0= ');
Vm=input('enter the value of voltage Vm= ');
%calculation
I0=Ish+Ia0;
Min=V*I0;%input to the motor on noload
Acu=Ia0^2*Ra;%armature cu loss
Scu=V*Ish;%shunt filed cu loss
Wc=Min-Acu;%constant losses
%calculation of motor effiency
Ifl=[10 7.5 5 2.5]
Ish=1.2;
for ii=1:length(Ifl)
Ia=Ifl(ii)-Ish; %motor current
Mi=Vm*Ifl(ii); %input to the motor
acu=Ia^2*Ra; %armature copper loss in motor
Wt=Wc+acu; %total loss in motor
O=Mi-Wt; %output of the motor
eff=(O*100)/Mi %efficiency of the motor
end
%calculation of generator effiency
for ii=1:length(Ifl)
Ia=Ifl(ii)+Ish; %armature current
Go=Vm*Ifl; %generator output
Acug=Ia^2*Ra; %armature copper loss in generator
Wtg=Wc+Acug; %total losses in generator
Gi=Go+Wtg; %generator input
effg=(Go*100)/Gi %efficiency of generator
end
9. SWINBURNE’S TEST
DEPT. EEE KLS’s VDRIT HALIYAL Page 9
OUTPUT:
enter the value of supply voltage V=200
enter the value of field current Ish=1
enter the value of noload arm current Ia0=1.2
enter the value of voltage Vm=220
eff =
66.89
eff =
55.15
eff =
64.50
eff =
14.54
effg =
71.72
effg =
66.15
effg =
70.34
effg =
53.48
CONCLUSION:
By conducting this experiment it can be concluded that the theoretically obtained
results and results obtained from MATLAB programming are almost equal fro motor and
generator. Hence results are verified.
