This test is performed in order to
find out
 Constant losses
 Efficiency at any load
 The Swinburne test is no load test in order to find out constant losses of the DC shunt motor or DC compound motor.
 As it is no load test, it is not applicable for DC series motor.
 The rated supply voltage is given to the DC machine. The rated speed of the DC machine is set by field rheostat.
Let
DC supply voltage = V
No load input current = I_{o}
Field current = I_{Sh}
No load armature current I_{ao} = I_{o}  I_{sh}
The no load input power = V I_{o}
The no load input power supply the
following losses
Iron losses in the core
Friction and windage losses and
No load armature copper losses = I_{o}^{2}R_{a}
Where R_{a} is hot armature
resistance ( hot armature resistance at any temperature is find out by R_{t}
= R_{o} ( 1 + α_{o}t )
 Resistance at 55^{o }C R_{55} = R_{o}( 1 + 55α_{o} )
 Resistance at 30^{o }C R_{30} = R_{o}( 1 + 30α_{o} )
 Resistance at 55^{o }C = R_{55 }/R_{30} = ( 1 + 55α_{o} ) / ( 1 + 30α_{o} )
Where α_{0} is temperature co – efficient of resistance = 1 / 234.5 at 0^{o}C
Constant Losses
 The constant losses can find out by subtracting no load armature copper losses from no load input power.
 The constant losses is constant from no load to full load therefore the DC machine efficiency is find out at any load condition.
Constant losses W_{c}
= No load input power – No load
armature copper losses
=VI_{o} – I_{o}^{2}R_{a}
Efficiency : DC Motor
Input power = VI ( where I is full
load current )
Output power = Input power – ( Armature
copper losses + Field copper losses + Constant losses )
= VI – ( I_{a}^{2}R_{a}
+W_{c} )
Efficiency of DC motor = ( Output
power / Input power ) × 100 %
Efficiency : Generator
Output power = VI
Input power = Output power + Losses
= VI +I_{a}^{2}R_{a} + W_{c}
Efficiency of DC Generator = ( Output
power / Input power ) × 100 %
Observation table
Supply voltage 
No load input current 
Field current 
Speed 
Hot armature resistance 
V 
I_{o} 
I_{SH} 
N 
R_{a} 





Calculation
Motor efficiency at full load current
Input power = VI
= __________
Output power = Input power + armature
copper losses + constant losses
W_{0} = W_{i} + I_{a}^{2}R_{a} + W_{c}
= __________
DC Motor efficiency = ( W_{0}
/ W_{i} ) × 100%
DC Generator at full load efficiency
Output power = VI
= __________
Input power = VI + I_{a}^{2}R_{a}
+ W_{c}
= __________
DC Generator efficiency = ( W_{0}
/ W_{i} ) × 100%
= __________
Observation
Whether the efficiency of DC machine
as DC motor and DC Generator is same?
Conclusion
The efficiency of the DC Machine can
find out at any load condition.
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Transformer
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and hysteresis loss in the DC Machines?
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