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 I_o.

Now the no load input current I_o has two components :

1. A purely reactive component I_m called magnetising component of no load current required to produce the flux. This is also called wattless component.
2. An active component I_c 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 I_o.

       Th total no load current I_o is the vector addition of I_m and I_c.

        In practical transformer, due to winding resistance, no load current I_o is no longer at 90^o  with respect to V₁. But it lags V₁ by angle Φ_o which is less than 90^o . Thus cos Φ_o 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 I_o are,

       This is magnetising component lagging V₁ exactly by 90^o  .

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

       While Φ_o = no load primary power factor angle
       The total power input on no load is denoted as W_o 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 I_c is called core loss or iron loss component. Hence power input W_o on no load always represent the iron losses, as copper loss is negligibly small. The iron losses are denoted as P_i 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)                         I_m = I_o sin Φ_o =  magnetising component
                            Φ_o = cos^-1(0.25) = 75.522^o 
.^..                          I_m = 10 x sin (75.522^o ) = 9.6824 A
b)                         P_i = iron loss = power input on no load
                             = W_o = V₁ I_o cos Φ_o = 400 x 10 x 0.25
                             = 1000 W
c) On no load,       E₁ = V₁ = 400 V  and    N₁  = 500
Now                     E₁ = 4.44 f Φ_m N₁ 
.^..                          400 = 4.44 x 50 x Φ_m x 500
.^..                          Φ_m = 3.6036 mWb

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