27/11/2020

How to separate out the Eddy current loss and Hysteresis loss in the DC Machines?

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

VA

Armature current

IA

Field current

IF

Total input power

VAIA

 

 

 

 

 

Armature copper losses

IA2RA

Constant losses WC = VAIA – IA2RA

Friction losses Wf

Iron losses

Wi

 

 

 

 

 If

VA = Armature voltage

IA = Armature current

VAIA = Total input

Then constant losses WC = VAIA – IA2RA

  • 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 If

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 Wi1 at normal speed is

      Wi1 = Wh + We………. ( 1 )

  • Now , repeat the same test at half speed ( or say twice the normal speed ) therefore the total iron losses at half speed is

      Wi2 = 0.5Wh + 0.25We……….. ( 2 )

Comparing equation ( 1 ) and ( 2 )

      We = 2 ( Wi1 – 2Wi2 )

  • 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

Wi  = Wh + We……. ( 1 )

Wh = 500 watt and We = 240 watt

N = 1440 RPM = 1440 / 60 = 24 rps

Now Wi = AN + BN2……..( 2 )

From equation ( 1 ) and ( 2 )

Wh = AN and We = BN2

Wh = AN

500 = A ( 24 )

Therefore A = 500 / 24 = 24.16

We = BN2

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 )

Wi = AN + BN2

410 = 24.16 ( N ) + 10 / 24 ( N2 )

10 N2 + 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.

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