The answer is on the web:
https://www.bayt.com/en/specialties/q/48539/what-would-happen-if-a-power-transformer-designed-for-operation-on-50-hz-frequency-were-connected-to-a-5-hz-frequency-source-of-the-same-voltage/
For designed, constant working flux density(v/f= const.), for lower frequency, applied voltage must reduced, for same voltage the magnetizing current would be very high(core may saturate), and would damage the transformer(Wi=(Wh+Wi)=Af+B(f)(f), valid if v/f is constant)........
If the current is the same, the secondary voltage will be also divided by 10 since the secondary voltage is= 4.44. n. phi. f.
f is the frequency,
phi: the magnetic flux,
n: number of rolls
N1 is responsible for producing flux in core, hence applied source voltage would be V1/10 to to obtained same flux density, at f = 5Hz transformer woud not damage......
Dear Ahmad Kalaie! If the transformer has a magnetic core. When reducing the transformer frequency, magnetic induction in the magnetic core of the transformer is increased (Bm = sqrt (2)* U1/(2*pi*f*w1*S); U1 - voltage on the primary winding, pi = 3.14, f - frequency, w1 - the number of turns of the primary winding, S - the rod cross-sectional area of the magnetic circuit). Reducing the frequency of 10 times lead to magnetic saturation of the transformer. Magnetic induction will rise to the level of saturation of the magnetic material. Critical will increase losses in the magnetic circuit. Increase the no-load current of the transformer. This emergency operation for the transformer.
The frequency is inversely proportional to the magnetic flux, hence the eddy current will increase and thus the transformer will get an extra heating which can damage it.
Probably, you are nearing dc, and the transformer may saturate, take care
If the transformer has a magnetic core. When reducing the transformer frequency, magnetic induction in the magnetic core of the transformer is increased (Bm = sqrt (2)* U1/(2*pi*f*w1*S); U1 - voltage on the primary winding, pi = 3.14, f - frequency, w1 - the number of turns of the primary winding, S - the rod cross-sectional area of the magnetic circuit). Reducing the frequency of 10 times lead to magnetic saturation of the transformer. Magnetic induction will rise to the level of saturation of the magnetic material. Critical will increase losses in the magnetic circuit. Increase the no-load current of the transformer. This emergency operation for the transfor
Transformers have high inductance. The inductive impedance is directly proportional to frequency. ZL=jwL where w=2*3,14*f. Therefore, reducing the frequency by a factor of 10 then ZL is reduced by a factor of 10. Now his is not a problem so far. BUT reducing the impedance then current is increased. Power Losses = I*I*ZL and as a result reactive power losses are also increased by a factor of 10.
So, looking at the current and power ratings of this transformer, can it handle a current and power increase of up to 10 times higher??
It will draw very high magnetizing current as magnetizing reactance will reduce by order of 10.
New magnetizing impedance by will be equal to Rated magnetizing impedance/10.
Hence magnetizing current will increase 9-10 times of rated magnetizing current. This will produce high flux and increase the magnetic flux density (more than rated Bmax), which will increase the losses.