80 likes | 267 Views
Output of Transformer. Let, Ø m = main flux, Wb ; B m = maximum flux density, Wb/m 2 ; δ= current density, A/m 2 ; A gi = gross core area, m 2 ; A i = net core area, m 2 = stacking factor * gross core area A c = area of copper in window A w = window area, m 2 ;
E N D
Output of Transformer • Let, • Øm= main flux, Wb ; • Bm= maximum flux density, Wb/m2 ; • δ= current density, A/m2 ; • Agi= gross core area, m2; • Ai= net core area, m2= stacking factor * gross core area • Ac= area of copper in window • Aw= window area, m2; • D= distance between core centers, m; • d= diameter of circumscribing circle, m; • Kw= window space factor; • F=frequency, hz;
Et= emf per turn, v; • Tp,Ts= no. of turns in primary & secondary winding respectively; • Vp,Vs= terminal volt. In primary & secondary wdg. respectively, V; • Ip,Is= current in primary & secondary wdg. respectively, A; • ap,as= area of conductor of primary & secondary wdg. respectively, m2; • li= mean length of flux path in iron, m; • Lmt= length of mean turn of x’mer windings, m; • Gt=weight of active iron, kg; • Ge=weight of copper, kg; • gt=weight per m3 of iron, kg; • ge=weight per m3 of copper, kg; • pi=loss in iron per kg, W; • pe=loss in copper per kg, W;
Single phase Transformers • The volt. induced in a x’mer wdg. Wth T turns & excited by a source having frequency f Hz is given by: Voltage per turn Et = E/T=4.44fØm. • The window in a single phase x’mer contains one primary & secondary wdg. • Therefore, Total copper area in window Ac= copper area of primary wdg. + Copper area of secondary wdg. = primary turns × area of primary conductor + area of secondary conductor = Tpap+Tsas.
Taking the current density δ to be the same in both primary & secondary wdg. ap=Ip/δ & as =Isδ • Therefore, • Total conductor area in window Ae=Te. Ip/δ+Ts. Is=(TeIp+TsIs)/δ • = 2AT/ δ • As TeIp= TsIs=AT if we neglect magnetizing mmf.