High Frequency Operation of Power Transformer

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Forget the secondary voltage. This transformer will glow bright red, then burn to the ground. Increasing the inductive reactance by 1200% (X_L = 2∏fL) does not produce optimal results.

Don't waste your valuable time, this guy is having fun here. Better focus on serious posters. Not this joker.

. . . . . . . . . choke . . . choke . . . choke . . . choke . . . . . . . . . . . .

looking at your previous questions.. you are in some sort of training centre, Yes?

A clear case of homework, methinks!

I fully agree with the answers of Anonymous Post#1 and pwr2thepeople, in addition to that I want to add that it is TIME PASS BOTANA question( Enjoying himself by seeing other's time being wasted).

The question could have been somewhat meaning full, if he had said 400Hz instead of 600Hz. Because 400 Hz is used for AC machines in Airplanes (aero plane) to reduce the weight of all AC equipment more than 10 times that of 50Hz and that is a requirement of plane to have lowest possible self weight.

However, let us assume that the question is asked sincerely & positive sense after doing some home work.

The answer is that transformer (T/F) will burn like a candle immediately after 600Hz Supply is connected (with some condition).

The following information provides the burning reason:

1. Eddy current losses in watts in 600Hz T/F increases by 144 times that of 50Hz.
2. Hysteresis losses in watts in 600Hz T/F increases by 12 times that of 50Hz.
3. Copper losses in watts in 600Hz T/F increases by 6 times that of 50Hz.

Example: A 5KVA, 200/1000V, 50Hz single phase transformer with the following losses:

Eddy current losses = 80watts; Hysteresis losses = 40watts; Copper losses =10watts (No load & due to magnetization current in primary winding)

The above losses are at 50Hz, If 600Hz and 12*200V=2400volt applied to primary of same transformer. Then the new losses will be as follows:

1. Eddy current losses We =80*144 = 11520watts

2. Hysteresis losses Wh = 40*12= 480watts

3. Copper losses Wc = 10*6 = 60watts

The total losses = We+Wh+Wc =11520+480+60=12060watts

The total losses will be more than twice the transformer power rating. These losses will create tremendous heat in T/F and burns the T/F immediately after 600Hz power supply is applied.

Mushtaq Hussain-INDIA

The analysis is given below:

The equations related to Transformer are as follows:

Transformer Voltage equation Vt= 4.44 f Ø T----------------------------------------------1

Eddy current loss in watts We = K1 B² f² …………………………….…….………………..……..2

Hysteresis losses in watts Wh = K2 B^1.6 f ………………………………….……………………3

Copper losses in watts Wc = I² R ………………………………….……………..…….…………4

Where: K1 & K2 are constants

Vt = design transformer voltage for 5oHz

f = 50 Hz is the design frequency of the transformer (T/F)

Ø = Flux in the T/F in Wb

T = Turns of the T/F for design voltage

A = T/F core area

B = Ø/A = Flux density in the core & is limited to 1.6 for linearity of B in hysteresis loop.

Having defined the above parameters, 600Hz with same T/F voltage supply can't apply directly to T/F.

600Hz can only be apply in two conditions that

Either 1. T/F iron core shall be decreased OR 2. Supply voltage shall be increased

1. T/F designed iron core area shall be reduced by 12 time that is A*(50Hz/600Hz) as

Vt = 4.44 f Ø T = 4.44 f BA T ( B flux density is required to limit to 1.6 for linear operation)

Vt = 4.44 f BA T= 4.44 * 50Hz *(600Hz/600Hz) B A T = 4.44* 600Hz (50Hz/600Hz) B A T or

Vt = 4.44 * 600Hz (50Hz/600Hz) B A T = 4.44 *600Hz B A T (1/12) = 4.44 * 600Hz B T( A/12)

Hence T/F iron core area shall be reduced from A to A/12. This can't be done as T/F is built.

Now we can investigate second condition to apply to 600Hz supply to T/F

2. The supply voltage shall be increased by 12; Vt(600Hz/50Hz) times the T/F design Vt as

Vt = 4.44 * 600Hz (50Hz/600Hz) B A T = 4.44 600Hz B A T (1/12) OR 12* Vt = 4.44 600Hz B A T

Vs = 12*Vt = 4.44 600Hz B A T

As Vs is external supply voltage which can be assumed to be available with 600Hz supply Voltage for T/F.

Once having established the supply voltage rating of 600Hz to be applied to T/F.

The affect of the 600Hz on the T/F loss as

A. Eddy current loss in watts from equation-2 as We = K1 B² f²

Increase in ratio of We loss as We(600Hz)/ We(50Hz) = K1 B² (600)²/k1 B²(50)²

As K1 and B² are constant

We(600Hz)/ We(50Hz) = (600)²/²(50)²=(600/50)²= 12²=144

We(600Hz)/ We(50Hz) =144,

Hence eddy current losses increases by 144 time ……………………………………………..6

B. Hysteresis losses in watts Wh = K2 B^1.6 f

Wh(600Hz)/Wh(50Hz)=600Hz/50Hz = 12

Hence hysteresis loss increases by 12 time …………………………………………..………..7

C. Copper losses in watts Wc = I² R

50Hz winding conductor will have uniform current density over the round cross section of winding conductor around 4Amps/mm². In 600Hz there will be predominate skin effect in which the induced current in the conductor due to the main current of conductor repulse main current away from center of the circular cross section of the conductor and limited to edge circle of the conductor . Main current passing area reduces & will be near the edge of the conductor or in imaginary the circular cross section of the conductor becomes like donut shape area with small width of donut. So skin effect will reduce the conductor cross section by 5 to 6 times. Result is T/F winding resistance increases by 5 to 6 times that of 50Hz resistance.

Wc(600Hz)/Wc(50Hz) = I²R(600)/I²R(50Hz) if I=Current constant by assumption

Wc(600Hz)/Wc(50Hz)= R(600Hz)/R(50Hz) = 5 to 6 ; say 6times

Wc(600Hz)/Wc(50Hz = 6 ……………………………………………………………………..8