Frequency Variation in 50 Hz Transformer

tasneemahmd

Read further in the book and let us know! Which book is it anyway?

It be magic Jim, plain and simple.....

http://www.edaboard.com/thread261091.html

http://info.ee.surrey.ac.uk/Workshop/advice/coils/power_loss.html

Iron losses DECREASE with frequency increase. This is why standby generators on airplanes are 400 HZ to cut down on core weight of the generator. I designed transformer for years and the 50Hz core had to be larger than 60 Hz cores. The book is correct about the 25 Hz.

I think the answer is that for a given core size, you can get more power through a transformer at a higher frequency. 400 hertz transformers are smaller for a given power rating, which is why 400 hertz is used in applications such as aircraft where weight is important. Audio transformers are smaller and RF transformers much smaller.

The amount of hysteresis power loss increases with frequency:

Ref: http://en.wikipedia.org/wiki/Transformer#cite_note-Heathcote.2C_pp._41-42-47

"Hysteresis lossesEach time the magnetic field is reversed, a small amount of energy is lost due to hysteresis within the core. According to Steinmetz's formula, the heat energy due to hysteresis is given by, and,hysteresis loss is thus given bywhere, f is the frequency, η is the hysteresis coefficient and β_max is the maximum flux density, the empirical exponent of which varies from about 1.4 to 1 .8 but is often given as 1.6 for iron."

Bottom line: Increase in power loss does not necessarily mean lower efficiency. If the increase in power transferred is even greater, the efficiency can be higher.

According to [for instance]:

NASA TECHNICAL NOTE [NASA TN D-3114]:

HYSTERESIS AND EDDY-CURRENT LOSSES OF A TRANSFORMER LAMINATION VIEWED AS AN APPLICATION OF THE POYNTING THEOREM by John Barranger Lewis Research Center Cleveland, Ohio:

".....according to Maxwell's equations the total loss or core loss is

Pc = Ph+ Pe formula (9) where Ph is the hysteresis loss and Pe is the eddy-current loss."

Ph=kh*Bm^2*f formula (58)

Pe=ke*Bm^2*f^2 formula( 40-42).

If supply voltage remains the same, for the same column cross section area:

B1*f1=B2*f2

PW1 =kh*B1^2*f1+ke*(B1*f1)^2

Pw2=kh*B2^2*f2+ke*(B2*f2)^2 and if we shall use B2=B1*f1/f2 we shall get:

Pw2=kh*(B1^2*f1)*f1/f2+ke(B1*f1)^2

That means eddy current loss remains constant but hysteresis losses decrease by f1/f2.

If f1=25 Hz and f2=60 Hz the hysteresis losses will be 25/60=0.417 [only 41.7%].

1. could not quite get how eddy current remains constant.

Regs,

Thats what I meant by funny. Broaden your view, will ya?

http://www.carbonlighthouse.com/2011/03/03/three-phase-power/

Btw, I am not the OP for this one.