Important Formula of Design of Transformer - 3

  

Relation of height of window to width of window is 2 to 4 Hw / Ww = 2 to 4

 

Area of yoke = ( 1.15 to 1.25 ) Agi for hot rolled steel = Agi for cold rolled steel

 

The diameter of circumscribing circle is directly proportional to √ Ai Where Ai = Core Area

 

The length of mean turns of conductors wound on cold rolled steel is 0.817 times that of hot rolled steel Volume of conductor in cold rolled steel is 18.3% lesser than that of hot rolled steel = ( 1 – 0.817 ) × 100

 

MMF of primary winding = NPIP MMF of secondary winding = NSIS Voltage regulation of transformer = Er Cos Φ + Ex Sin Φ + ½ ( Er Sin Φ – Ex Cos Φ )2 Where Er = Per unit resistance     = IPRP / VP Ex = Per unit reactance      = IPXP / VP

 

The average value of flux density in the duct = Ba / 2

 

Short circuit fault current = I / Ex Where I = RMS fault current Ex = Per unit leakage reactance

 

Ratio of maximum radial force under worst fault condition to maximum radial force under normal condition Fr ( max ) f / Fr ( max ) n = 2 / Ex2 Where Fr ( max ) f = Radial force under fault condition Fr ( max ) n = Radial force under normal condition If one winding is only 5% shorter than other winding, it produces large axial force.

 

No load current I0 = √ ( Iw2 + Iµ2 ) Where Iw = Loss component of no load current Iµ = Magnetizing current

 

RMS value of magnetizing current Iµ = AT0 ( N0 I0 ) / √ 2 NP Where AT0 = Total primary turns NP = Primary turns Magnetizing current Iµ = Magnetizing VA per kg × Weight of core / Number of phases × Voltage per phase

 

Eddy current loss We = kf2Bmax2 If the supply voltage is kept constant , ( f Bmax ) remain constant therefore the eddy current loss remains constant in spite of frequency is changed

 

Hysteresis loss Wh =  k fBmax2 If the supply voltage is constant, product of f Bmax remains constant Wh = k ( f Bmax ) Bmax Let f Bmax = k1 constant Bmax = k1 / f Wh = kk12 / f If the supply voltage is kept constant, hysteresis loss decreases with increases in the frequency.

 

The leakage reactance increases with increase in frequency. The effective resistance increases with increases in frequency.

 

The copper losses don’t depend on supply frequency.

 

Effect of frequency on flux density if the supply voltage is kept constant V2 = 4.44 f2Φmax2N2 V1 = 4.44f1Φmax1N1 Let us consider that frequency f1( 50 Hz ) changes to f2 ( 40 Hz ) Φmax2f2 = Φmax1f1 Φmax2 = ( f1 / f2 ) Φmax1 = ( 50 / 60 ) Φmax1 Φmax2 = ( 0.82 ) Φmax1

 

The eddy current losses do not affect with change in frequency.

 

Copper loss varies as the square of the output WC2 / WC1 = ( output in 2nd case )2 / ( output in 1st case )2 The power rating of transformer increases with increase in frequency.

 

Specific heat dissipation due to convection of oil λconv = 40.3 ( θ / H )1.4 Where θ = Temperature difference of surface relative to oil H = Height of dissipating surface

 

Effect of Wc / Wi on overload capacity of transformer Case 1 : Wc / Wi = 1 Case 2 : Wc / Wi = 2 At 20% overlaod Case 1 Total loss at overload / Total load at full load [ ( 2 )2 × 1 ] + 1 / ( 1 + 1 ) = 2.5 Case 2 Total loss at overload / Total load at full load [ ( 2 )2 × 1 ] + 1 / ( 2 + 1 ) = 3 Therefore the overload losses in the two cases are 2.5 : 3 therefore the transformer with higher Wc / Wi is less as compared to sustain overload

 

Heating time constant Th = Gh / Sλ Where G = Weight h = Specific heat λ = Specific heat dissipation Th = Gh θ / Q     ( ⸫ Sλ = Q / θ ) The heating time constant of oil is about 30 times to that of winding.

 

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