Separation of Core Losses Test


Separation of Core Losses Test
       It is seen that the core losses of transformer includes,
1. Hysteresis loss
2. Eddy current loss
       For a given volume and thickness of laminations, these losses depend on the operating frequency, maximum flux density in the core and the voltage.
       The hysteresis loss is given by Steinmet'z relation,
              Ph = Kh Bm1.67 f v    watts
i.e.          Ph = A Bm1.67 f    watts                                                            ..............(1)
where     A = constant assuming constant voltage
The eddy current loss is given by,
              Pe = Ke Bm2 f2 t2 watts
i.e.          Pe = B Bm2 f2   watts                                                                       .........(2)
where     B = constant for given thickness t of core
       Thus the total core loss becomes,
             Pi = Ph + Pe = A Bm1.67 f +B Bm2 f2                                                   .........(3)
       Practically conduct two tests on transformers at two different frequencies f1 and f2, keeping maximum flux density in the core same. The results are to be used in the equations (1), (2) and (3) to obtain the constants A and B. Thus the core losses i.e. iron losses can be separated into hysteresis and eddy current losses.
Example 1 : A single phase transformer shows 63 W core losses at 40 Hz while 110 W  at 60 Hz. Both the tests are performed at same value of maximum flux density in the core. Find hysteresis and eddy current losses at 50 Hz frequency.
Solution : Pi1 = 63 W, f1 = 40 Hz,  Pi2 = 110 W,   f2 = 60 Hz,    Bm   is same.
       As Bm is same, it can be absorbed in the constants A and B. Thus we can write,
                   Ph = Af    while Pe = B f2  
...                Pi = Ph + Pe = Af + B f2  
...                 63 = A x 40 + B x (40)2                                                      ...........(1)
and              110 = A x 60 + B x (60)2                                                    ............(2)
       Solving (1) and (2) we get,
                     A = 1.0584,     B = 0.0129
       Thus two losses at 50 Hz are,
                      Ph = Af = 1.0584 x 50 = 52.92 W
                       Pe = B f2 = 0.0129 x 502 = 32.25 W

3 comments:

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  3. little or no or no or no voltage and power transformers ar at intervals the foremost utilized in low voltage electrical and electronic circuit styles tend to use copper conductors and it's have succeeding mechanical strength and smaller conductor size than same Al sections.toroidal transformer winding machine

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