Saturday, June 11, 2011

Practicle Transformer on No Load

       Actually in practical transformer iron core causes hysteresis and eddy current losses as it is subjected to alternating flux. While designing the transformer the efforts are made to keep these losses minimum by,
  1. Using high grade material as silicon steel to reduce hysteresis loss.
  2. Manufacturing core in the form of laminations or stacks of thin lamination to reduce eddy current loss.
       Apart from this there are iron losses in the practical transformer. Practically primary winding has certain resistance hence there are small primary copper loss present.
       Thus the primary current under no load condition has to supply the iron losses i.e. hysteresis loss and eddy current loss and a small amount of primary copper loss. This current is denoted as Io.
Now the no load input current Io has two components :
  1. A purely reactive component Im called magnetising component of no load current required to produce the flux. This is also called wattless component.
  2. An active component Ic which supplies total losses under no load condition called power component of no load current. This also called wattful component or core loss component of Io.
       Th total no load current Io is the vector addition of Im and Ic.

        In practical transformer, due to winding resistance, no load current Io is no longer at 90with respect to V1. But it lags V1 by angle Φo which is less than 90o . Thus cos Φo is called no load power factor of practical transformer.
Fig 1.  Practical transformer on no load

        The phasor diagram is shown in the Fig. 1. It can be seen that the two components Io are,

       This is magnetising component lagging V1 exactly by 90.

       This is core loss component which is in phase withV1.
       The magnitude of the no load current is given by,

       While Φo = no load primary power factor angle
       The total power input on no load is denoted as Wo and is given by,

       It may be denoted that the current is very small, about 3 to 5% of the full load rated current. Hence the primary copper loss is negligibly small hence Ic is called core loss or iron loss component. Hence power input Wo on no load always represent the iron losses, as copper loss is negligibly small. The iron losses are denoted as Pi and are constant for all load conditions.

Example 1 : The no load current of a transformer is 10 A at a power factor 0f 0.25 lagging, when connected to 400 V, 50 Hz supply. Calculate,
a) Magnetising component of the no load current
b) Iron loss and c) Maximum value of flux in the core.
    Assume primary winding turns as 500.
Solution : The given value are, = 10 A, cos = 0.25, = 400 V  and f = 50 Hz
a)                         Im = Io sin Φo =  magnetising component
                            Φo = cos-1(0.25) = 75.522
...                          Im = 10 x sin (75.522o ) = 9.6824 A
b)                         Pi = iron loss = power input on no load
                             = Wo = V1 Io cos Φo = 400 x 10 x 0.25
                             = 1000 W
c) On no load,       E1 = V1 = 400 V  and    N = 500
Now                     E1 = 4.44 f Φm N
...                          400 = 4.44 x 50 x Φm x 500
...                          Φm = 3.6036 mWb

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