2 Analysis of heat transfer in a single-phase transformer

Assignment 2 4-6u() t U ( t) = e()t() t dφ( t)dψ=mcos ω = = N( 2.5 )dt dtIdeally it is assumed that all the magnetic flux links with the winding and flux can be expresseddirectly from the voltage that is applied to the winding.φU=mm m( 2.6 )ω Nm() t Φ sin ( ω t) = sin( ω t) = B A sin( ω t)The maximum value of magnetic flux Φ m is related to the cross-section area of the pure magneticconductor A m and the maximum flux density BmB that is defined by the material ability to conductSimilar to the magnetic circuit, the flow in themagnetic flux and the magnetic saturation B satB .electric circuit is defined by current I m (maximum value) in the single turn which is related to thetotal Ampere turns NI m i.e. magnetomotive force (mmf), the cross-section area of the pure electricconductor A e and the maximum current density J m that is defined by device’s ability to conduct heatflow and the thermal limit ϑ coil .iNIJ Acos ( 2.7 )NNmm e() t = I ( ω t + ϕ) = cos( ω t + ϕ) = cos( ω t + ϕ)mBy substituting the maximum values of voltage and current in the equation of the apparent powerthe size of the transformer can be expressed through the nominal power and the magnetizationfrequency the allowed current and flux density.1 1S = UmIm= ω BmJmAeAm= P2 2( 2.8 )The power of transformer depends on the electric loading J m , the magnetic loading BmB and themagnetization frequency ω as well as the geometry such as cross section areas of electric A e andmagnetic circuit A m .By assuming equal magnetomotive forces in the primary and the secondary winding, thus theinfinitely permeable core, the active power can be taken equal to the apparent power, which is themaximum power that the core can operate at a certain temperature limit. The transferred power isless due to losses. The losses can be estimated according to the loss density and the geometry of theelectric as well as the magnetic circuit.P = A l p + A l p = P + P( 2.9 )losse eem mmcufeThe conductor loss for the direct current P cu is expressed through the power loss density, whichdepends on resistivity ρ and the current density square J 2 , and the volume of the conductor V e .22 ⎛ JAe⎞ leN2 1 2cu= I R = ⎜ ρ = ρ J Aele = ρ JmVe( 2.10 )N Ae2P⎝⎟⎠NThe remagnetization loss in the magnetic conductor for the symmetric sinusoidal excitation can befound from the specific loss data k fe at certain magnetization frequency and magnetic induction, andcore volume V mfefe( Bm) VmpfeVmP = k , ω =( 2.11 )The specific core loss is calculated according to the polynomial curve fitting of the losscharacteristics at 50 Hz.AAV0090 Electrical Drives - project, TTU, 2008