2 Analysis of heat transfer in a single-phase transformer

Assignment 2 2-6Initialization― Cooling conditions― Magnetic: B -> p fe― Electric: J k,ρ 0 -> p cuFind temperature― Coil hot-spot ϑ max― Coil average ϑ aveTarget― |ϑ target - ϑ max|≤0.05― iter ≥ max_iterResult visualization― Temperature plot (bmp)― Electric loading (txt)Obtain new values― ρ= ρ 0(1+α(ϑ ave-ϑ 0))― p cu=0.5 ρ (J k+1) 2 Kf― iter=iter+1Obtain current density― if ϑ target - ϑ max < - 40then J k+1=J k x 0.5― if ϑ target - ϑ max > 40then J k+1=J k x 2― elseJ k+1=J k W(ϑ target - ϑ max)Figure 2.2 Iterative hot-spot computation loop. Flowchart of the iterative current density computation in wherethe target hot spot temperature of the coil is estimated. The hot-spot temperature is obtained from aline that is defined through the cross-section of a winding. The weight factor W is chosen so that theconverging process is as well as fast as stabile. This flowchart does not indicate the solution when corelosses only give the target temperature or moreThe thermal models: EC model and FE model― Are two dimensional (2D), which means that heat transport along z axis and the heatgenerated in/dissipated from the end turns are neglected― Has the height of the transformer lamination stack equal to one meter (per unit of meter)― Assume the same loss density in the primary and the secondary windings― Assume naturally cooled sides h=20 W/Km 2 and ambient temperature of 40 0 C2.2 Geometry parameterizationThe geometric model was formulated so that transformer length and insulation thickness togetherwith proportion factors will define the whole geometry. Additional parameters, such as proportionfactors specify the geometric proportions between the different parts of the transformer.― Ks (variable) – proportion between the slot length and core limb length― Kw (=0.5) – proportions between primary and secondary windingThe length of the leg in the magnetic core is expressed by using the relative slot length k s and thelength needed to form a one electromagnetic pole. For the single-phase core and shell type oftransformers the number of poles N p =2.ll( − k )trc= 1s( 2.1 )NpSimilarly the length of the slot is formulated according to the length of a single electromagnetic poleand the relative slot length.AAV0090 Electrical Drives - project, TTU, 2008

Assignment 2 3-6ltrls= ks( 2.2 )NpThe width of the slot is given by the total width of the transformer w tr minus the width of themagnetic core yokes. The width of the magnetic back-core (yoke) equals to the length of themagnetic leg-core. The slot width for the shell type of transformer isws= w − l( 2.3 )trcFigure 2.3 The geometry parameterization of the shell type of transformer. The upper figure shows the magneticflux flow plane (xy-plane) and the lower figure shows the electric current flow plane (xz-plane).2.3 Size equationsThe size of transformer is related to the power of the electromagnetic device and the allowed ‘flow’densities such as current density, flux density and the loss density. In an ideal transformer themagnetic coupling between the windings is perfect. There is the same core flux φ(t) that links eachturn of each winding. Apparent power of an ideal lossless transformer is expressed asS 1=2 U I m m( 2.4 )here is assumed sinusoidal variation of voltage and current and instead of rms values the peakvalues has been used instead. Considering the lossless electromagnetic circuit the voltage of theelectric circuit can be directly linked to the induced back electromotive force (emf) and themagnetic flux in the magnetic circuit.AAV0090 Electrical Drives - project, TTU, 2008

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

Assignment 2 5-62( 2.5517B−1.2936B+ 0.8143) ⋅ 7700p ( 2.12 )fe=mmFinally, the efficiency can be found from the input power and the loss power.P − Plossη =( 2.13 )PLua-script as well as m-script EMK_task_1.lua and EMK_task_1.m, calculate hot spottemperature of the winding ϑmax, core temperature ϑcc, the specific copper loss pcu, maximumcurrent density Jcm and so on as a function of proportion. Some of the variables are not estimatedwith the thermal equivalent circuit.for kh = 1,9,1 doendx[m]ϑmax[°C]ϑave[°C]pcu[W/m 3 ]Jcm[A/m 2 ]Aw[m 2 ]ϑcc[°C]The result of the computations is written into file tmp_heat_fem.txt and tmp_heat_emc.txt.2.4 EC modelEquivalent circuit model has been given in EMK_task_1.m.The thermal conductivity network consists of 11 elements that represent the heat dissipation in thesymmetric part of the transformer. The cooling condition through the natural convection is takeninto account in the elements 4 and 11. Nodal network represents the temperature over 9 nodepoints. The copper losses are applied into node 2 and 3, the core losses to node 1 and the ambient(reference) temperature is described in nodes 5 and 9.9iter[-]1165876981012141110131211512167233445Figure 2.4 The thermal equivalent circuit of the single-phase shell-type transformer. Thermal elements in theend-turns are excluded in order to make EC model and FE models comparable.AAV0090 Electrical Drives - project, TTU, 2008

Assignment 2 6-6Code that shows the formulation of thermal equivalent circuit and the elements in the circuit% thermal equivalent circuit - topology matrix% [element(1) node(n) node(m) thermal conductivity]Tec = [ 1 1 2 0.5*h_c * 0.5*w_s / (0.5*l_c/tc_fe+ins/tc_ins+0.5*l_w1/tc_win);2 2 3 0.5*h_c * 0.5*w_s / (0.5*l_w1/tc_win+ins/tc_ins+0.5*l_w2/tc_win);3 3 4 0.5*h_c * 0.5*w_s / (0.5*l_w2/tc_win+ins/tc_ins+0.25*l_c/tc_fe);4 4 5 0.5*h_c * 0.5*w_s / (0.25*l_c/tc_fe+1/Aconv);5 1 6 0.5*h_c * 0.5*l_c / ((0.5*w_s+0.25*l_c)/tc_fe);6 2 7 0.5*h_c * l_w1 / ((0.5*w_s-ins)/tc_win+ins/tc_ins+0.25*l_c/tc_fe);7 3 8 0.5*h_c * l_w2 / ((0.5*w_s-ins)/tc_win+ins/tc_ins+0.25*l_c/tc_fe);8 6 7 0.5*h_c * 0.25*l_c / ((0.5*l_c+ins+0.5*l_w1)/tc_fe);9 7 8 0.5*h_c * 0.25*l_c / ((0.5*l_w1+ins+0.5*l_w2)/tc_fe);10 8 4 0.5*h_c * 0.5*l_c / ((0.5*w_s+0.5*l_c)/tc_fe);11 6 9 0.5*h_c * 0.5*Ltr / ((0.25*l_c)/tc_fe+1/Aconv);];2.5 FE modelFinite element model has been given in EMK_task_1.lua.The xy-plane cross-section of a shell type of transformer is the base geometry for the heat transferanalysis. Notice that the loop calculations are commented out and initially the relative slot openingis selected 50%.2.6 AssignmentBased on the outcome from FE and EC model (that takes at least 5 minutes)― Show Jcm=f(ks) for these deferent models― Calculate a transferred power P=f(ks) for these deferent models― Calculate copper and core losses and the sum of these losses Ploss=f(ks)― Estimate the efficiency for these deferent modelsAnalyze the outcome of the FE analysis and if necessary introduce additional calculations in orderto evaluate the thermal conductance shown below:a)ϑ cuQ cuG th1ϑ ambc)ϑ cu1Q cu1G th1ϑ ambb)G th3ϑ cuQ cuG th1ϑ ambG th6ϑ cu2Q cu2G th4G th2G th5Q feϑ feG th2ϑ feQ feG th3Carry out this study for an optimal transformer and as far you are able i.e. start from circuit a,continue with b and if you are able then also circuit c.AAV0090 Electrical Drives - project, TTU, 2008