There are mainly two types of losses in the DC
Machines
( 1 ) Variable losses : Armature copper loss and Field copper loss
( 2 ) Constant losses : Iron losses and Friction and winding losses
 We can find out constant losses and variable losses separately with the help of both Field test and Hopkinson test. We can separate out following test.
Iron losses and friction losses
 Figure shows circuit diagram for this test.
 The variable DC voltage at armature terminal is applied from the Ward – Leonard set or say DC chopper.
 It is necessary to feed the armature and field winding from separate DC source.
Procedure
 The DC motor is run at no load.
 The variable DC voltage is applied to armature and field current is set such that speed of motor remains constant.
 The friction losses are constant at constant speed.
 The total armature copper losses are very small therefore input is dissipated in constant losses at no load.
Supply voltage V_{A} 
Armature current I_{A} 
Field current I_{F} 
Total input power V_{A}I_{A} 




Armature copper losses I_{A}^{2}R_{A} 
Constant losses WC = V_{A}I_{A}
– I_{A}^{2}R_{A} 
Friction losses W_{f} 
Iron losses W_{i} 




V_{A} = Armature voltage
I_{A} = Armature current
V_{A}I_{A} = Total input
Then constant losses W_{C} = V_{A}I_{A}
– I_{A}^{2}R_{A}
 Figure shows total constant losses for various field current.
 If we extend curve A up to y – axis, the DE represent friction and windage losses.
For any field current I_{f}
DE = FG represent friction and windage losses and
GH = Total iron losses
Hysteresis and Eddy current losses
 It is very easy to draw curve for total iron losses for various field current.
 The hysteresis loss is directly proportional to supply frequency whereas the eddy current loss is directly proportional to square of supply frequency for a given voltage.
 Let the total iron losses W_{i1} at normal speed is
W_{i1} = W_{h} + W_{e}………. ( 1
)
 Now , repeat the same test at half speed ( or say twice the normal speed ) therefore the total iron losses at half speed is
W_{i2} = 0.5W_{h} + 0.25W_{e}………..
( 2 )
Comparing equation ( 1 ) and ( 2 )
W_{e} = 2 ( W_{i1} – 2W_{i2} )
 The curve A represent total iron losses at normal speed, curve B represent twice iron losses at half speed.
PQ = half of total eddy current loss at normal speed
 Draw the curve C such that point on each point on it represent PQ = QR.
 The hysteresis loss at normal speed is represented by RS. Therefore
PS = Total iron losses at normal speed
PR = Total eddy current loss at normal speed
RS = Total hysteresis loss at normal speed
Example
The hysteresis and eddy current losses in the DC
machine are 580 watt and 240 watt respectively at 1440 RPM. At what speed, the
total iron losses are halved if the flux density remains constant?
Solution
Total iron loss
W_{i }=
W_{h} + W_{e}……. ( 1 )
W_{h} = 500 watt and W_{e }= 240 watt
N = 1440 RPM = 1440 / 60 = 24 rps
Now W_{i} = AN + BN^{2}……..( 2 )
From equation ( 1 ) and ( 2 )
W_{h} = AN and W_{e }= BN^{2}
W_{h} = AN
500 = A ( 24 )
Therefore A = 500 / 24 = 24.16
W_{e }= BN^{2}
240 = B ( 24 )^{2}
B = 240 / ( 24 )^{2}
= 10 / 24
Let us consider that the N rps will be speed at total
iron losses becomes halved
Total iron losses = ( 580 + 240 ) / 2 = 410 Watt
From equation ( 2 )
W_{i} = AN + BN^{2}
410 = 24.16 ( N ) + 10 / 24 ( N^{2 })
10 N^{2} + 580 N – 9840 = 0
Solving this equation
N = 13.72 rps
= 13.72 * 60
= 823 RPM
The total iron losses will be halved at 823 RPM.
You may also like :
Construction & Working of Earthing Transformer
Compare BLDC Motor & Conventional DC Motor
No comments:
Post a comment