energies
Article Leakage Inductances of Transformers at Arbitrarily Located Windings
Marcin Jaraczewski * and Tadeusz Sobczyk
Faculty of Electrical and Computer Engineering, Cracow University of Technology, 24 Warszawska Street, PL 31-155 Cracow, Poland; [email protected] * Correspondence: [email protected]; Tel.: +48-12-628-2658
Received: 14 November 2020; Accepted: 2 December 2020; Published: 7 December 2020
Abstract: The article presents the calculation of the leakage inductance in power transformers. As a rule, the leakage flux in the transformer window is represented by the short-circuit inductance, which affects the short-circuit voltage, and this is a very important factor for power transformers. This inductance reflects the typical windings of power transformers well, but is insufficient for special transformers or in any case of the internal asymmetry of windings. This paper presents a methodology for calculations of the self- and mutual-leakage inductances for windings arbitrarily located in the air window. It is based on the 2D approach for analyzing the stray field in the air zone only, using discrete partial differential operators. That methodology is verified with the finite element method tested on real transformer data.
Keywords: power transformers; winding asymmetry; 2D stray field analysis; 2D discrete partial differential operators; energy-based approach; leakage inductances
1. Introduction In the long history of research on power transformers, countless books have been published, for example [1–7]. Theoretically, basic construction and operational problems were solved there. In addition to single-phase and three-phase transformers, new types of special power transformers for power system control, power electronics, and electric and traction have been designed. Two approaches are used to analyze the properties of transformers. The first is based on Maxwell’s field equations. The second is based on Kirchhoff’s equations with mutual-inductances. The field approach is mainly used for transformer design problems, the second for operational problems when the transformer is part of the system. The field approach uses the design data of magnetic circuits and windings, the circuital approach requires relations between the linked fluxes of lumped inductances of the transformer windings and the winding currents. Due to the magnetic non-linearity of the transformer ferromagnetic core, these relations are non-linear, which can only be determined by solving Maxwell’s equations [8–15]. Nowadays, software packages allow us to combine the field and the circuit approaches and calculate output values like currents or voltages at transient or steady-states that are interesting for engineers [16]. However, this is costly and time-consuming and seems inappropriate for solving operational problems in systems with power transformers. Therefore, simplified methods for calculating the lumped parameters of a transformer are often used [17–22]. To overcome the difficulties of 2D or 3D non-linear magnetic field computation in the whole transformer magnetic circuit, the relations between linked fluxes ψn and currents in of elementary
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Energies 2020, 13, 6464 2 of 20 ψ μμ μ σσ σ 1 L1,1 (ii ) L 1,2 ( ) L 1,N ( i ) L 1,1 L' 1,2 L' 1,N i1 ψ μμ σσ 2 L()ii L () L' L' i2 windings, which=+ are necessary2,2 for the Kirchho 2,Nff’s equations 2,2 of transformers, 2,N cani ={ii be , ,..., presented i } in 11 N (1) the form: ψ (sym)) Lμσ (i ) (sym L' i N µµ µ N,N σ σ σ N,N N ψ L (i) L (i) L (i) L L L i 1 1,1 1,2 ··· 1,N 1,1 1,2 1,N 1 µ µ 0σ ··· 0σ Theψ2 above two matricesL (i) can beL calculated(i) separatelyL using a simplifiedL imethod.2 The first matrix 2,2 ··· 2,N 02,2 ··· 02,N . = . . + . . . i = i1, i1, ... , iN (1) has inductances,. due to fluxes in.. the core,. and the second one.. results. from .the leakage{ field} in the . . . . µ μ σ ψN (sym ) L L()(i)i (sym ) L iN transformer windows. Inductances N,Nnm, are non-linear, taking into0N,N account the saturation of the ferromagnetic core, and may depend on all winding currents. This matrix can be determined using The above two matrices can be calculated separately using a simplified method. The first matrix circuit representation of ferromagnetic core with non-linear lumped parameters, omitting fluxes in has inductances, due to fluxes in the core, and the second one results from the leakage field in the the air. The second matrix, containingµ the leakage inductances, can be found taking into account transformer windows. Inductances L (i) are non-linear, taking into account the saturation of the n,m →∞ σ ferromagneticonly the window core, zone and mayand dependassuming on that all windingμFe currents.. The leakage This matrix inductances can be determinedLnm, are constant using circuitbecause representation the window zone of ferromagnetic is magnetically core linear. with non-linear lumped parameters, omitting fluxes in the air. The second matrix, containing the leakage inductances, can be found taking into account only the windowFor symmetrically zone and assuming designed that powerµ transformers,. The leakage with inductances magnetic circuitsLσ are and constant winding because locations the Fe → ∞ n,m windowsketched zone in Figure is magnetically 1, the matrices linear. in (1) take extremely simple forms, which are well known from booksFor and symmetrically textbooks. Commonly, designed power those transformers, matrices are with determined magnetic by circuits three and lumped winding parameters: locations σ sketchedNon-linear in Figuremagnetizing1, the matricesinductance in (1)L( takeμμi ) extremely and two simpleleakage forms, inductances which are of wellprimary known Lp from and books and textbooks.σ Commonly, those matrices are determined by three lumped parameters: secondary winding Ls . The inductance L(μμi ) represents relation between the linkedσ flux in the Non-linear magnetizing inductance Lµ(iµ) and two leakage inductances of primary Lp and secondary limb and theσ magnetizing current. The sum of leakages inductances determines the short-circuit winding Ls . The inductance Lµ(iµ) represents relation between the linked flux in the limb and theinductance magnetizing Lsc , current.and consequently, The sum of leakagesthe short-circuit inductances volt determinesage. It is a the very short-circuit important inductance technical
Lparametersc, and consequently, of power transformers. the short-circuit The short-circuit voltage. It is in aductance very important is calculated technical from parameter the magnetic of power field transformers.in the transformer The short-circuitwindow with inductance two windings is calculated on the same from limb, the magnetic assuming field that in the the ampere-turn transformer windowcompensates with and two the windings magnetic on thepermeability same limb, of assuming the ferromagnetic that the ampere-turncore is sufficiently compensates high. The and field the magneticanalysis in permeability the 1D transformer of the ferromagnetic window model core provid is suesffi ratherciently simple high. formulae The field for analysis the short in thecircuit 1D transformerinductance. windowThis can modelbe found provides in books rather on transformers, simple formulae and for is still the shortrecommended circuit inductance. for designers. This The can be found in books on transformers,σ σ and is still recommended for designers. The leakage inductances leakage inductances Lp and Ls are not calculated, but estimated based on the short circuit Lσ and Lσ are not calculated, but estimated based on the short circuit inductance [19,20], often just inductancep s [19,20], often just by dividing the short-circuit inductance by 2. This approach is by dividing the short-circuit inductance by 2. This approach is sufficient for symmetrically located sufficient for symmetrically located windings, when they cover, more or less, the limbs, and the windings, when they cover, more or less, the limbs, and the Rogowski factor can correct the length Rogowski factor can correct the length of the magnetic field path in the air. Figure 2a illustrates such of the magnetic field path in the air. Figure2a illustrates such a case. Figure2b shows the position a case. Figure 2b shows the position of windings when HV windings are modified, for voltage of windings when HV windings are modified, for voltage regulation purposes, by eliminating some regulation purposes, by eliminating some numbers of turns in the middle part. Figure 3a presents a numbers of turns in the middle part. Figure3a presents a case when a short circuit in the HV winding in case when a short circuit in the HV winding in phase ‘A’ arise. For the cases ‘b’ and ‘c’ leakage phase ‘A’ arise. For the cases ‘b’ and ‘c’ leakage inductances cannot be estimated from 1D field analysis. inductances cannot be estimated from 1D field analysis.
1 4 2 5 3 6
A B C
Figure 1. Locations of three phase transformer windings. Figure 1. Locations of three phase transformer windings.
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phase Aphase B phase Aphase B phase A phase B
LV LV HV HV
w pip w i p p wssi
w scsci w scsci
(a) (b) (c)
FigureFigure 2. 2.Examples Examples of the location location of of windings windings in inthe the air airwindow window of three of three phase phase transformers: transformers: (a) (a)Symmetrical; Symmetrical; (b ()b at) at voltage voltage regulation; regulation; (c) (atc) short at short circuit circuit of a of part a part of a ofwinding. a winding.
w w p1 wp1 w i w i p1 2 ip ws1i s1 ws1i s1 2 ip w i w i 2 p 2 p ws1i s1 ws1i s1
wi wi wi wi wp2pi wp2pi wpip wpip
w i w i ws2i s2 ws2i s2 w p1 w i w i w p1 p1 i ws2i s2 ws2i s2 p1 i 2 ip 2 ip
(a) (b)
FigureFigure 3. 3.Windings Windings locationlocation inin rectifierrectifier transformers:transformers: (a (a) )3/6 3/6 phase phase transformers; transformers; (b) ( b3/6) 3 /phase6 phase transformerstransformers for for 24 24 pulses pulses rectifier. rectifier.
RectifierRectifier transformers transformers for for multi-phase multi-phase voltages voltages and and currents currents [7] [7] can can have have windings windings that that are muchare shortermuch than shorter the than limb’s the height, limb’s andheight, the and Rogowski the Rogows factorki cannot factor becannot applied. be applied. Figure 2Figurea shows 2a anshows example an ofexample the location of the of thelocation windings of the in windings the transformer in the transformer window for window Yyd 3/6-phase for Yyd transformers 3/6-phase transformers for 12 pulses for 12 pulses rectifiers and Figure 2b for Yzyd 3/12-phase transformers for 24 pulses rectifiers. The rectifiersfor 12 pulses and Figure rectifiers2b for and Y zFigureyd 3/12-phase 2b for Yz transformersyd 3/12-phase fortransformers 24 pulses rectifiers.for 24 pulses The rectifiers. position The of the windingsposition loses of the characteristicwindings loses symmetry the characte for classicristic three-phasesymmetry for power classic transformers, three-phase see power Figure 1. Attransformers, least the 2D fieldsee Figure is necessary 1. At least tofind the 2D leakage field is inductances. necessary to find leakage inductances. AnAn analytical analytical solution solution of theof magneticthe magnetic vector vect potentialor potential in transformer in transf airormer windows air windows with arbitrarily with arbitrarily located balanced pairs of windings is presented in [21]. The zero Neumann boundary locatedarbitrarily balanced located pairs balanced of windings pairs isof presentedwindings inis [presented21]. The zeroin [21]. Neumann The zero boundary Neumann conditions boundary can conditions can be stated for such cases on the ferromagnetic boundaries, which means that the beconditions stated for suchcan be cases stated on thefor ferromagneticsuch cases on boundaries,the ferromagnetic which boundaries, means that which the magnetizing means that current the magnetizing current is omitted. We can then calculate the short circuit inductance for any pair of is omitted. We can then calculate the short circuit inductance for any pair of windings. However, windings. However, this approach cannot be used to determine the leakage inductances of this approach cannot be used to determine the leakage inductances of individual windings and the individual windings and the mutual-inductances caused by the stray field. To find these mutual-inductances caused by the stray field. To find these inductances, 2D or 3D field calculations inductances, 2D or 3D field calculations are typically needed, including the entire magnetic circuit are typically needed, including the entire magnetic circuit with the ferromagnetic core and the air with the ferromagnetic core and the air zone. [11–15,22]. zone. [11–15,22,23]. Lσ This paper presents a 2D numerical approach for determining all leakage inductances Lnm, , bothσ This paper presents a 2D numerical approach for determining all leakage inductances Ln,m, self, as well as mutual, for arbitrarily located windings considering the magnetic field in the air bothself, self, as well as well as mutual, as mutual, for forarbitrarily arbitrarily located located windings windings considering considering the magnetic the magnetic field in field the inair the window only, when surrounded by ferromagnetic walls with high permeability. Difficulties in airwindow window only, only, when when surrounded surrounded by by ferromagnetic ferromagnetic walls walls with with high high permeability. permeability. Difficulties Difficulties in in
Energies 2020, 13, 6464 4 of 20 Energies 2020, 13, x FOR PEER REVIEW 4 of 20 formulating the boundary conditions for the case of one winding in the air window has been omitted based on the resultsresults ofof testtest calculationscalculations usingusing thethe 2D2D FiniteFinite ElementElement MethodMethod (FEM).(FEM). FerromagneticFerromagnetic plates with with sufficiently sufficiently high high magnetic magnetic permeability permeability around around the the air airwindow window have have been been included included for that for thatcalculations calculations at zero at zero Dirichlet Dirichlet condition condition on onextern externalal boundaries. boundaries. To To fulfill fulfill the the conditions conditions that that were followed for those tests, additional ampere-turns have been added at thethe boundaries.boundaries. It allowed us to obtain the magnetic vector potential distribution distribution in in the the air air window window area area the the same as form the FEM method, with susufficientlyfficiently good accuracy. Self- and mutual-leakagemutual-leakage inductances can be calculated using an energy approach. This time, however, each pair of windings is represented by two self-inductances and one mutual-inductancemutual-inductance instead of one short-circuit inductance. In In this this way, way, it it is possible to determine all elements of thethe leakageleakage inductioninduction matrixmatrix inin equationequation ((1)).((1)). When the elementaryelementary windings areare connected,connected, constitutingconstituting the the final final transformer’s transformer’s windings, windings, the the matrix matrix of constraintsof constraintsC can C becan defined be defined and and the the final final inductance inductance matrix matrix takes takes the the form, form,
=⋅TT + ⋅ LLC= C (()L Lµ + Lσ) CC (2)(2) · μσ· The paper is is organized organized as as follows: follows: In In the the second second se section,ction, numerical numerical tests tests are are presented, presented, leading leading to toa formulation a formulation of ofboundary boundary conditions conditions on on ferroma ferromagneticgnetic walls walls on on the the air air window for unbalancedunbalanced ampere-turns in in the the air air window. window. For For this this purpose, purpose, the the 2D 2D FEM FEM has has been been applied. applied. In Section In Section 3, the3, thefinite-difference finite-difference equations equations have have been been formulated formulated for for magnetic magnetic potential potential inin 2D 2D air air zone zone using the Discrete Partial Partial Differential Differential Operat Operatorsors (DPDO), (DPDO), which which are are presented presented in Appendix in Appendix A. ResultsA. Results have have been beencompared compared to those to those from from FEM, FEM, confirming confirming satisfac satisfactorytory accuracy. accuracy. Section Section 44 presentspresents energy-basedenergy-based calculations of self- and mutual-inductances, due to fluxes fluxes in the air window. window. The The results results for for two two cases: cases: Windings typicaltypical location location of of a threea three phase phase power power transformers transformers and for and short-circuited for short-circuited part of apart winding of a arewinding presented are presented in Section in5 .Section Finally, 5. conclusions Finally, conclusions are given are in given Section in6 .Section 6.
2. Calculations of Magnetic Field in the Transformer’s AirAir WindowWindow withwith FEMFEM
To determinedetermine thethe leakageleakage inductanceinductance matrixmatrixL σL,σ the, the self- self- and and mutual-inductances mutual-inductances for for each each pair pair of windingsof windings should should be be calculated. calculated. As As an an example, example, a pair a pair of of arbitrarily arbitrarily located located windings windings is illustratedis illustrated in Figurein Figure4. 4.
d → ⋅ μFe w11i
μ0
h ⋅ w 22i
Figure 4. 2D problem for a transformer window with two arbitrarily located windings. Figure 4. 2D problem for a transformer window with two arbitrarily located windings.
Calculation of self-inductances is not a trivial problem because the magnetic field field in the air window zone should be generated by one winding only,only, omittingomitting a magneticmagnetic core. It is not possible to formulate the boundary conditions for the equation of the magnetic potential for such a case, as it has been explained in [24]. In 2D, the magnetic potential in the air window is described by the partial differential equation of the form:
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∂∂22Axy(, ) Axy (, ) +=−+μ(,)(,)()jxy jxy (3) ∂∂xy2201 2 where Axy(, ) is the magnetic potential, j1 (,xy ) and j2 (,xy ) are ampere-turns densities of respective windings.Energies 2020 The, 13, 6464 windings with currents i1 and i2 have numbers of5 of turns 20 w1 and w2 respectively. Taking into account the ferromagnetic, the zero Dirichlet conditions can be determined on the external hascircumference been explained in [24of]. Inthe 2D, ferromagneti the magnetic potentialc core. in the air However, window is described by omitting by the partial the ferromagnetic differential equation=∞ of the form: core, i.e., assuming μFe , it is not possible to formulate the boundary conditions only for the air ∂2A(x, y) ∂2A(x, y) + = µ (j1(x, y) + j2(x, y)) (3) window. The zero Neumann conditions∂x2 are∂y 2applicable− 0 only in the case of balanced ampere-turns ww⋅+ii ⋅ = 0 [23]. To determine the boundary conditions for non-balanced ampere-turns in the 11 2 2where A(x, y) is the magnetic potential, j1(x, y) and j2(x, y) are ampere-turns densities of respective transformer window,windings. some The windings numeric with test currents wasi1 and carriedi2 have out. numbers The of turnsmagneticw1 and wpotential2 respectively. was calculated for Taking into account the ferromagnetic, the zero Dirichlet conditions can be determined⋅= on the external the case when onlycircumference one winding of the ferromagnetic exists core. in However, the air by window, omitting the ferromagnetic i.e., w022 core,i i.e., assuming, but the ferromagnetic µ = , it is not possible to formulate the boundary conditions only for the air window. The zero Fe ∞ walls are includedNeumann too. conditionsThe 2D are finite applicable elements only in the soft case ofware balanced has ampere-turns been usedw i here,+ w iassuming= 0 [23]. zero Dirichlet 1 · 1 2 · 2 To determine the boundary conditions for non-balanced ampere-turns in the transformer window, conditions at the outer boundaries of the ferromagnetic core. The permeability of ferromagnetic μFe some numeric test was carried out. The magnetic potential was calculated for the case when only have been risingone up winding to 10 exists 000 in⋅ the μ air in window, successive i.e., w i calculations.= 0, but the ferromagnetic Figure walls 5 shows are included the too. magnetic potential 0 2 · 2 The 2D finite elements software has been used here, assuming zero Dirichlet conditions at the outer distribution Axy(, ) in the air zone only when one winding generates the field at the highest value boundaries of the ferromagnetic core. The permeability of ferromagnetic µFe have been rising up to 10 000 µ0 in successive calculations. Figure5 shows the magnetic potential distribution A(x, y) in the of magnetic permeability· μFe . air zone only when one winding generates the field at the highest value of magnetic permeability µFe.
Figure 5. Magnetic potential()A(x, y)/µ0 in the air window was obtained from the finite element method. Figure 5. Magnetic potential Axy, /μ0 in the air window was obtained from the finite element Figure6 presents the components of the magnetic field Hx(x, y) and Hy(x, y) along boundaries of method. the air window, where
1 d 1 d Hx(x, y) = A(x, y) Hy(x, y) = A(x, y) µ0 dy −µ0 dx () () Figure 6 presents the components of the magnetic field Hxyx , and Hxyy , along boundaries of the air window, where
()= 1d () ()=− 1d () Hxyx , Axy, Hxyy , Axy, μd0 y μ0 dx
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() () Figure 6. Components of the magnetic field Hxyx , and Hxyy , along boundaries. Figure 6. Components of the magnetic field Hxy(), and Hxy(), along boundaries. Figure 6. Components of the magnetic field Hx x(x, y) and Hyy(x, y) along boundaries. Results of FEM calculations show that the tangential components H t of a magnetic field on the
air windowResultsResults ofboundaries of FEM FEM calculations calculations keep consta shownt thatvalues. the tangentialIts value can componentscomponents be determinedH Ht oft ofafrommagnetic a magnetic the integral field field on form on the the airof airwindowthe window Maxwell boundaries boundaries equation, keep keep constant consta values.nt values. Its Its value value can can be determinedbe determined from from the the integral integral form form of theof Maxwell equation, the Maxwell equation, I Hld(,)dd= jxyxy t (4) H dl = j(x, y)dxdy (4) lSHlt d(,)dd= x jxyxy t S (4) l lS The curve l''l is the circumference of the air window and S is''S the surface limited by this curve. The curve 0 0 is the circumference of the air window and0 0 is the surface limited by this The value of that component equals Ht = w1 i1=⋅/(2 (h + d ⋅)) in considered case. curve.The The curve value ''l of isthat the component circumference equals of theHw· t11 air window· i /(2(h+d)) and ''S in isconsidered the surface case. limited by this In this paper it is proposed to modify the ampere-turns=⋅ function ⋅ in the air window instead of solving curve.In The this value paper of it that is proposed component to equalsmodify Hwthet11 ampere-turnsi /(2(h+d)) function in considered in the air windowcase. instead of the Equation (3) with zero Neumann boundary conditions when omitting j2(x, y). This modification solvingIn this the paperEquation it is (3)proposed with zero to modify Neumann the ampere-turnsboundary conditions function when in the omitting air window jxy2 (, instead ). This of results in ampere-turns that are now balanced. To eliminate the ferromagnetic core from the computation solvingmodification the Equation results in(3) ampere-turns with zero Neumann that are boundarynow balanced. conditions To eliminate when omitting theair ferromagnetic Fejxy2 (, ). Thiscore a general formula for tangential components on both sides of the boundary Ht Ht = 0 has been modificationfrom the computation resultsair in aampere-turns general formula that for are tangenti now balanced.al components To eliminate on both the sides ferromagnetic− of the boundary core changed to H 0 = Kz, where Kz is a surface current in z direction, which equals to the constant air−= Fe t air −= fromHHtt the computation0 has− been a generalchanged formula to HKt for0 tangentiz , whereal components 0 K0 z is a surface on both current sides ofin the''z boundarydirection, component Ht found using FEM air−= Fe air −= HHwhichtt equals0 tohas the been constant changed component to HKt H0t foundz , where using FEMK z is a surface current in ''z direction, K = H = w i /(2(h + d)) (5) which equals to the constant componentz ==⋅Htt found1 1using FEM Kizt11Hw· /(2(h+d)) (5) Ki==⋅Hw /(2(h+d)) That additional surface current is zt11 located on thethe air window boundary, as it is presented(5) in Figure7. FigureThat 7. additional surface current is located on the air window boundary, as it is presented in Figure 7. d j (,xyd ) h 1 j (,xy ) h 1
μ0
μ0
Kz
Kz
Figure 7. Additional surface current on the air boundaries. Figure 7. Additional surface current on the air boundaries. The exciting function is now balanced and satisfies the relation, The exciting function is now balanced and satisfies the relation,
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Energies 2020, 13, x FOR PEER REVIEW 7 of 20 The exciting function is now balanced and satisfies the relation, jxy(, ) d xy d=∇⋅ Kxy (, ) d xy d (1 ) = →( ) x 0 Figure 8. j (x y) j (x y) y = = Figure 8. Ampere-turnAmpere-turn function function jxy1(,, ) and and thethe modifiedmodified excitationexcitationmod jxymod (,, for ) for d /d/dx2dxy 2. . The solution of that equation can be predicted in the form of the double Fourier series of cosine The solution of that equation can be predicted in the form of the double Fourier series of cosine functions [24], functions [24], X∞ X∞ x x A(x, y) = ∞∞Ar,s cos(r π ) cos(s π ) at A 0 (8) ·· · dxx· · · h 0,0 ≡ =⋅⋅⋅⋅⋅⋅r=0 s=0 ≡ Axy(, ) Ars, cos(π r )cos(π s ) at A00,0 (8) r=00 s= dh The function jmod(x, y) is rather inconvenient for the Fourier analysis, so to solve Equation (6) a special numerical approach has been applied. To find the magnetic potential the DPDO [25,26] have been developedThe function for thejmod series(,xy ) (8). is Fullrather consideration inconvenient leading for the to Fourier the DPDO analysis, are presented so to solve in Appendix EquationA (6). Thea special DPDO numerical have been approach successfully has applied been applied. for solving To find 1D boundary the magnetic problems potential [24–26 the]. DPDO [25,26] haveThe been Finite-Di developedfference for Equations the series (FDE) (8). Full can consideration be rather easily leading written to using the DPDO operators,are presented and itin takes the form, Appendix A. The DPDO have been successfully applied for solving 1D boundary problems [24-26]. (2) (2) The Finite-Differenceµ EquationsD(2) A (FDE)= J canwhere be ratherD(2) = easilyD written+ D )using DPDO operators, and(9) 0 · · − xx yy it takes the form, The vectors A and J are composed like vectors z and g in AppendixA. The matrix D(2) is singular μ ⋅⋅=−DA(2) J DDD(2)=+() (2) (2) and has only one eigenvalue equal0 to zero, so its where rank is (( R + 1) (xxS + 1 yy) 1). As it has been mention(9) · − before the series should satisfy the condition A = 0, i.e., (2) The vectors A and J are composed like0,0 vectors z and g in Appendix A. The matrix D ⋅− is singular and has only one eigenvalue equalRX+1 SX +to1 zero, so its rank is ((R+1) (S+1) 1) . As it has been A = 0= (10) mention before the series should satisfy the conditionn,m A00,0 , i.e., n=1 m=1 R+1 S+1 = A0nm, (10) nm==11 EnergiesEnergiesEnergies 20202020 2020,, 1313, 13,, x 6464, FORx FOR PEER PEER REVIEW REVIEW 8 8of8 of of20 20 20 It Itis, is, in in fact, fact, the the constraint constraint equati equationon for for the the elements elements of of the the vector vector A A, which, which can can be be expressed expressed in in thethe form: Itform: is, in fact, the constraint equation for the elements of the vector A, which can be expressed in the form: A −−−− A1,11,1A1,1 11111 1 − · · · − AAA1,2 A 1 1,2 1,2 1,2 A1,2A 1 1 = ====1,2= . =⋅ =⋅ = A ACAACA. C CC AC (11)(11)(11) . .. . · . . AR+1,S+1AA + + A 1 R+1,S+1R 1,S 1 AR+1,S+R+1,S+1A1 R+1,S+1 11 where C is the constraint matrix. Finally, the FDEs can be written in the form, wherewhere C C is isthe the constraint constraint matrix. matrix. Finally, Finally, the the FDEs FDEs can can be be written written in in the the form, form, TTTTT((2)2)(2) T µμ0 μ⋅⋅⋅⋅=−⋅C()CD⋅⋅⋅⋅=−⋅()CDD CAC CAAC = CJC CJ J (12) 0C· 0C· · · − · (4)(4) CDTT ⋅⋅=T ⋅⋅=(2)(2)(2) CD (2) (2)(2) TheTheThe matrix matrix matrix C CDD C CD= CDC is isisnon-singular, non-singular, non-singular, and and the the new new FDEs FDEsFDEs (12) (12)(12) have havehave a auniquea uniqueunique solution. solution.solution. · · C TheTheThe algorithm algorithm forfor for solving solving solving the the FDEthe FDE FDE (12) (12) has(12) beenhas has been prepared been prepared prepared in MATLAB, in in MATLAB, MATLAB, and the and resulting and the the resulting magnetic resulting magneticpotentialmagnetic distributionpotential potential distribution distribution in the air windowin in the the air areaair window window is presented area area is in ispresented Figurepresented9, denotedin in Figure Figure as 9, DPDO. 9,denoted denoted These as as DPDO. results DPDO. TheseareThese compared results results are to are thecompared compared results to of to the FEM the results withresults aof reducedof FEM FEM with with mean a reduceda valuereduced to mean zero.mean value value to to zero. zero. FigureFigureFigure 9. 9.MagneticMagnetic Magnetic potentialpotential potentialA (Axyx ,(,Axyy(,)/ )/µ )/inμ μ thein in airthe the windowair air window window obtained obtained obtained using theusing using Finite the the ElementFinite Finite Element MethodElement (FEM) and Discrete Partial Differential Operators (DPDO). MethodMethod (FEM) (FEM) and and Discrete Discrete Partial Partial Differential Differential Operators Operators (DPDO). (DPDO). The magnetic field distribution, TheThe magnetic magnetic field field distribution, distribution, q =+=+222 222 HHxy(xHxy(,, y(,) )= ) (( HH ( Hxyx (,(xy x (,, xyy )))) ))+ ( H ( (H H (,y xy( (,x xy, ))y)) )) (5)(13)(5) hashashas been been determined determined basedbased based onon onA (Axy x(,Axy, y(,). ) The. )The. The first first first order order order DPDO DPDO DPDO has has beenhas been been used used forused that.Thefor for that.The that.The results, results, results, both bothofboth FEM of of FEM and FEM DPDO,and and DPDO, DPDO, are shown are are shown shown in Figure in in Figure Figure10. 10. 10. FigureFigureFigure 10. 10.10. Magnetic MagneticMagnetic field fieldfield Hxy H(,Hxy(x(,, y )) )in in in thethe the airair air windowwindow window obtainedobtained obtained withwith with FEMFEM FEM andand and DPDO.DPDO. DPDO. Energies 2020, 13, 6464 9 of 20 Energies 2020, 13, x FOR PEER REVIEW 9 of 20 TheThe di differencesfferences between between solutions solutions of of magnetic magnetic field field distributions distributions are are presented presented in in Figure Figure 11 11 and and seemseem to to be be su sufficientlyfficiently small. small. FigureFigure 11.11. DiDifferencefference betweenbetween magneticmagnetic fieldfieldH Hxy(x(,, y) resulting ) resulting from from DPDO DPDO and and FEM. FEM. andand the the solution solution can can be be predicted predicted as as the the Fourier Fourier series series (7). (7). Application Application of of the the DPDO DPDO leads leads to to the the requiredrequired solution. solution. 4. Co-Energy Based Calculations of Self- and Mutual-Leakage Inductances 4. Co-Energy Based Calculations of Self- and Mutual-Leakage Inductances σ σ σ The inductances Ln n, σLm m, and Ln m in theσ Formula (1) can be calculated based on field distribution The inductances , L , Lσ , and, L in the Formula (1) can be calculated based on field in the air window, obtainednn, withmm, applicationnm, of the second order DPDOs presented in AppendixA. Twodistribution formulae in of the magnetic air window, co-energy obtained stored with in a thepplication air can beof the used second here. order The first DPDOs one determinespresented in co-energyAppendix from A. Two the distribution formulae of of magnetic the magnetic co-energy field: stored in the air can be used here. The first one determines co-energy from the distribution of the magnetic field: 1 →→→ → Eco = (B H) dV (14) E=2 1y (BH ⋅· ) d V co 2 V (6) V andand the the second second is is based based on on lumped lumped inductances inductances of of coils, coils, 11σ2σ2σ⋅+ ⋅ +⋅⋅ E=L1coσ 22nn , ()2iiii n1 σ L mm , () m2 L nmσ , n m (15) Eco = L (in) + L (im) + L in im (15) 2 n,n · 2 m,m · n,m · · Inductances can be calculated in three steps: − Inductances can be calculatedσ in three steps: = = Calculate inductance Lnn, from the co-energy Eco,n at innI and im 0 , σ i = i = - Calculate inductance Ln,n from the co-energy1 Eco,σ2n⋅at n In and m 0, E=L(I)co,nnnn2 , (16) 1 σ 2 Eco,n = L (In) (16) 2 n,n · − σ = = Calculate inductance Lmm, from the co-energy Eco,m at in 0 and immI , σ - Calculate inductance L from the co-energy Eco,m at in = 0 and im = Im, m,m 1 σ2⋅ E=L(I)co,mmmm2 , (17) σ 1 σ 2 − Calculate inductance L from theE co-energyco,m = L mE,m (I mat) i = I and i = I , (17) mm, 2 co,·nm n n mm −− =⋅⋅σ σ EEco co,n E co,mmmnm LII , (18) - Calculate inductance Lm,m from the co-energy Eco,nm at in = In and im = Im, In 2D the formula of the co-energy takes the form, σ Eco Eco,n Eco,m = Lm,m In Im (18) −hd − · · 2 EHxyHxyx=μ1 ⋅⋅ l() (,)2 +() (,)) ddy co2 0eq ( x y (19) 00 Energies 2020, 13, 6464 10 of 20 In 2D the formula of the co-energy takes the form, h d Z Z 1 2 2 Eco = µ leq (Hx(x, y)) + Hy(x, y) dxdy (19) 2 0 · · 0 0 where leq is an equivalent length of the air window in z-direction, usually it is the circumference taken in the middle of the air window. 5. Results of Teste Calculations 5.1. Test 1 As the first test, the leakage inductance matrix Lσ of a typical power transformer has been determined. The transformer is shown schematically in Figure1. The inductance matrix can be predicted in the form Lσ Lσ 0 Lσ Lσ 0 1,1 1,2 1,4 1,5 σ σ σ σ σ L2,2 L2,3 L2,4 L2,5 L2,6 Lσ 0 Lσ Lσ 3,3 3,5 3,6 Lσ = σ σ L L 0 4,4 4,5 ( ) σ σ sym L5,5 L5,6 σ L6,6 Design data, more or less related to a power transformer of about 60MVA, 110/60kV, Yd11 have been used for calculations. Its air window and the winding locations are shown in Figure 12, keeping the proportions. The following data are used for computations: Air window dimensions h = 1750 mm, d = 480 mm, diameter of limbs dlimb = 580mm. The equivalent length of widows in z-direction equals to circumference of a circle of a diameter leq = p (580 + 480)mm. The winding’s dimensions are: · HV h = 1550 mm, d = 70 mm, LV-h = 1650 mm,d = 50 mm and the distance between HV and − 1 1 2 2 δ = LVEnergies windings 2020, 13, is x FOR PEER30mm. REVIEW Numbers of turns are: HV–720 and LV–240. 11 of 20 4 1 2 5 A B Figure 12. Air window with windings. Figure 13. Modified ampere-turn function for winding ‘1′. Figure 14 presents distributions of magnetic potential A(,xy ) and magnetic field Hxy(, ) in the air window obtained from FEM and the DPDO. Energies 2020, 13, x FOR PEER REVIEW 11 of 20 4 1 2 5 A B Energies 2020, 13, 6464 11 of 20 The magnetic vector potential in the air window has been determined by solving equation (12) using the second order DPDO, presented in AppendixA. The modified ampere-turn functions have been determined for individual cases. An exemplary winding ‘10 is presented in Figure 13. The magnetic field distribution in the air window has been obtained using the first order DPDO. The discretization mesh, with ∆x = 10 mm and ∆y = 50 mm has been chosen, i.e., the number of points in x’ direction is 49 and in y direction is 36, which leads to 1764 points in the whole area. The leakage inductances have beenFigure obtained 12. Air from window equalities with (14)–(16)windings. for an arbitrary value of currents. FigureFigure 13. 13.Modified Modified ampere-turn ampere-turn function function for for winding winding ‘1 ‘10.′. ( ) ( ) FigureFigure 14 14 presents presents distributions distributions of magneticof magnetic potential potentialA x , yA(,xyand ) magneticand magnetic field H fieldx, y Hxyin(, the ) air in windowEnergies 2020 obtained, 13, x FOR from PEER FEM REVIEW and the DPDO. 12 of 20 the air window obtained from FEM and the DPDO. FigureFigure 14. 14. DistributionsDistributions of of magnetic magnetic potential potential AAxy((,x, y ) and and magneticmagnetic fieldfieldH Hxy(x(,, y) in ) inthe the air air zone. zone. The components of magnetic field Hx(x, y) and Hy(x, y) are shown in Figure 15. The value of Hx(x) The components of magnetic field Hxyx (, ) and Hxy(, ) are shown in Figure 15. The value of x =y H (y) on the boundaries in direction is constant and equals Hx 50A/m and=±y on the boundaries in Hxx () on the boundaries in direction x is constant and ±equals H50A/mx and Hyy () on the direction y is constant also and equals Hy = 50A/m. ± =± boundaries in direction y is constant also and equals H50A/my . . Figure 15. Components Hxyx (, ) and Hxyy (, ). The values of leakage inductances have been calculated using relations (14)–(16) for the number of turns of HV winding ‘1′. The inductance values are presented in the matrix below. The symmetry of windings has been taken into account, which is reflected in matrix structure. 203.06 93.21 0 153.41 30,07 0 203.06 93.21 31.10 153.41 30,07 203.06 0 31.10 153.41 L =⋅10−3 H σ − 195.95 26.74 0 (sym ) 195.95 − 26.74 195.9 5 The short circuit inductance, when accounting for the couplings between the windings on the = same limb only, equals Lsc 82.19 mH . The same matrix calculated from expressions presented in [24], obtained in 1D approach, has the form, Energies 2020, 13, x FOR PEER REVIEW 12 of 20 Figure 14. Distributions of magnetic potential Axy(, ) and magnetic field Hxy(, ) in the air zone. The components of magnetic field Hxyx (, ) and Hxyy (, ) are shown in Figure 15. The value of =± Hxx () on the boundaries in direction x is constant and equals H50A/mx and Hyy () on the Energies 2020, 13, 6464 =± 12 of 20 boundaries in direction y is constant also and equals H50A/my . . Figure 15. Components Hx(x, y) and Hy(x, y). Figure 15. Components Hxyx (, ) and Hxyy (, ). The values of leakage inductances have been calculated using relations (14)–(16) for the number of turnsThe of values HV winding of leakage ‘10. The inductances inductance have values been are calculated presented using in the relations matrix below. (14)–(16) The for symmetry the number of windingsof turns of has HV been winding taken into‘1′. The account, inductance which values is reflected are presented in matrix in structure. the matrix below. The symmetry of windings has been taken into account, which is reflected in matrix structure. 203.06 93.21 0 153.41 30.07 0 203.06 93.21 0 153.41 30,07 0 203.06 93.21 31.10 153.41 30.07 203.06 93.21 31.10 153.41 30,07 203.06 0 31.10 153.41 3 Lσ = 10 H 195.95 26.74 0 − 203.06 0 31.10 153.41 · −3 L =⋅− 10 H σ (sym) 195.95− 26.74 195.95 26.74− 0 195.95 (sym ) 195.95 − 26.74 The short circuit inductance, when accounting for the couplings between195.9 5 the windings on the same limb only, equals Lsc = 82.19 mH. The same matrix calculated from expressions presented in [24], obtainedThe in short 1D approach, circuit inductance, has the form, when accounting for the couplings between the windings on the = same limb only, equals Lsc 82.19 mH . The same matrix calculated from expressions presented in 134.20 30.97 0 92.91 24.78 0 [24], obtained in 1D approach, has the form, − 134.20 30.97 24.78 92.91 24.78 − − 134.20 0 24.78 92.91 = 3 Lσ − 10− H 138.33 80.52 0 · − (sym) 138.33 80.52 − 138.33 The corresponding short circuit impedance equals Lsc = 86.72 mH. The differences between those matrices seem to be rather high, but the short circuit inductances are almost equal. 5.2. Test 2 For the second test, the leakage inductance matrix has been determined when the short circuit of a part of HV winding occurs, as it is presented in Figure 16. The same data as for Test 1 have been assumed, but 20% of HV winding is short circuited. For that test, the other windings have been omitted. For that case, the leakage inductance matrix takes the form, σ σ σ L L L 1,1 1,2 1,sc L = σ σ σ L2,2 Lsc,2 σ (sym) Lsc Energies 2020, 13, x FOR PEER REVIEW 13 of 20 134.20 30.97 0 92.91− 24.78 0 −− 134.20 30.97 24.78 92.91 24.78 134.20 0− 24.78 92.91 L =⋅10−3 H σ − 138.33 80.52 0 (sym ) 138.33− 80.52 138.33 = The corresponding short circuit impedance equals Lsc 86.72 mH . The differences between those matrices seem to be rather high, but the short circuit inductances are almost equal. 5.2. Test 2 For the second test, the leakage inductance matrix has been determined when the short circuit of a part of HV winding occurs, as it is presented in Figure 16. The same data as for Test 1 have been assumed, but 20% of HV winding is short circuited. For that test, the other windings have been omitted. For that case, the leakage inductance matrix takes the form, σσσ LLL1,1 1,2 1,sc = σσ Energies 2020, 13, 6464 Lσ2,2sc,2LL 13 of 20 ()sym Lσ sc and values of respective inductancesinductances are,are, 229.84 47.31− 44.22 229.84 47.31 44.22 = − Lσ =Lσ 21.7821.78 5.38 5.38 (sym) 66.80 (sym) 66.80 These inductances cannot be found from a 1D approach. To confirmconfirm it, the magneticmagnetic potential distribution in the short circuit coilcoil casecase isis shownshown inin FigureFigure 1717.. 2 1' A B 1sc Figure 16. Windings with short circuit. Energies 2020, 13, x FOR PEER REVIEW 14 of 20 FigureFigure 17. 17.The The current current density density and and magnetic magnetic potential potential distribution distribution for for short short circuited circuited winding winding ‘1 ‘10.′. 6. Conclusions 6. Conclusions This paper presented a method for calculations of self- and mutual-leakage inductances of power This paper presented a method for calculations of self- and mutual-leakage inductances of transformer windings with arbitrary winding locations in the air windows. It was possible to obtain power transformer windings with arbitrary winding locations in the air windows. It was possible to inductance values based on 2D calculations of the magnetic field only in the air zone, using discrete obtain inductance values based on 2D calculations of the magnetic field only in the air zone, using partial differential operators for 2D periodic functions. The boundary conditions which allowed for discrete partial differential operators for 2D periodic functions. The boundary conditions which the elimination of the transformer ferromagnetic core were determined based on the test calculations allowed for the elimination of the transformer ferromagnetic core were determined based on the test of the magnetic field distribution using FEM that included a transformer core of sufficiently high calculations of the magnetic field distribution using FEM that included a transformer core of permeability. It followed that the surface current should be located on the boundaries. sufficiently high permeability. It followed that the surface current should be located on the boundaries. To find the solution predicted in the air zone, the discrete partial differential operators of the first- and the second orders have been developed and presented in Appendix A. They allowed us to easily create the finite-difference equations determining the 2D magnetic vector potential distribution in the air window, and consequently, the distribution of the magnetic field strength. The results are sufficiently accurate comparing to FEM. The energy-based method has been successfully used to calculate the leakage inductances, both the self- and mutual-inductances. The matrix of such inductances has been calculated for the design data of a real three-phase power transformer. For a case when a part of one winding on one limb is short circuited, the inductances have also been calculated. Author Contributions: Conceptualization, T.S. and M.J.; methodology, T.S. and M.J..; software, M.J.; validation, T.S. and M.J.; formal analysis, T.S. and M.J..; investigation, T.S. and M.J..; resources, T.S..; data curation, T.S. and M.J..; writing—original draft preparation, T.S..; writing—review and editing, T.S. and M.J..; visualization, M.J.; supervision, T.S.; project administration, T.S.; funding acquisition, T.S. All authors have read and agreed to the published version of the manuscript. Funding: This research, which was carried out under the theme Institute E-2, was funded by the subsidies on science granted by the Polish Ministry of Science and Higher Education. Conflicts of Interest: The authors declare no conflict of interest. Appendix A Discrete Differential Operators for Double Cosine Fourier Series Discrete Partial Differential Operators (DPDOs) are developed for the two-variable periodic function zxy(, ), which is defined in the area 02π< Energies 2020, 13, 6464 14 of 20 To find the solution predicted in the air zone, the discrete partial differential operators of the first- and the second orders have been developed and presented in AppendixA. They allowed us to easily create the finite-difference equations determining the 2D magnetic vector potential distribution in the air window, and consequently, the distribution of the magnetic field strength. The results are sufficiently accurate comparing to FEM. The energy-based method has been successfully used to calculate the leakage inductances, both the self- and mutual-inductances. The matrix of such inductances has been calculated for the design data of a real three-phase power transformer. For a case when a part of one winding on one limb is short circuited, the inductances have also been calculated. Author Contributions: Conceptualization, T.S. and M.J.; methodology, T.S. and M.J.; software, M.J.; validation, T.S. and M.J.; formal analysis, T.S. and M.J.; investigation, T.S. and M.J.; resources, T.S.; data curation, T.S. and M.J.; writing—original draft preparation, T.S.; writing—review and editing, T.S. and M.J.; visualization, M.J.; supervision, T.S.; project administration, T.S.; funding acquisition, T.S. All authors have read and agreed to the published version of the manuscript. Funding: This research, which was carried out under the theme Institute E-2, was funded by the subsidies on science granted by the Polish Ministry of Science and Higher Education. Conflicts of Interest: The authors declare no conflict of interest. Appendix A Discrete Differential Operators for Double Cosine Fourier Series Discrete Partial Differential Operators (DPDOs) are developed for the two-variable periodic function z(x, y), which is defined in the area 0 < x < 2π, 0 < y < 2π, and can be expanded in the Fourier series with a limited number of terms, XR XS z(x, y) = Zr,s (cos(r x/2) cos(s y/2)) (A1) · · · · r=0 s=0 Firstly, the unique relations between the function values and the Fourier coefficients can be found. To determine that relation the set of ((R + 1) (S + 1)) points uniformly distributed in the area has × been chosen, as shown in Figure A1. x1 = ∆x/2 y1 = ∆y/2 x2 = ∆x/2 + ∆x y2 = ∆y/2 + ∆y x3 = ∆x/2 + 2∆x y3 = ∆y/2 + 2∆y (A2) . . . . x + = π ∆x/2 y = π ∆y/2 R 1 − S+1 − where ∆x = π/(R + 1),∆y = π/(S + 1). Proper ordering of the point set x , y allows writing relations between the set of function { n m} values z and the set of Fourier coefficients Z in the form { n,m} { r,s} z = T Z (A3) · where z is a vector of the function values z , ordered, as follows, { n,m} h iT h i z = z z z wherez = z z z 1 2 ··· R+1 n n,1 n,2 ··· n,S+1 Energies 2020, 13, x FOR PEER REVIEW 15 of 20 RS =⋅⋅⋅⋅() zxy( , ) Zrs, cos( r x / 2) cos( s y / 2) (A1) r=00 s= Firstly, the unique relations between the function values and the Fourier coefficients can be found. To determine that relation the set of ()(R+1)× (S+1) points uniformly distributed in the area has been chosen, as shown in Figure A1. =Δ =Δ xx/21 yy/21 =Δ +Δ =Δ +Δ xx/2x2 yy/2y2 =Δ + Δ =Δ + Δ xx/22x3 yy/22y3 (A2) =−Δ =−Δ xπx/2R+1 yπy/2S+1 Energies 2020, 13, 6464 15 of 20 where Δx=π / (R+1) , Δy=π / (S+1) . y π yS+1 ...... ym ...... y2 ...... π y1 ...... x x xx x 1 2 n R+1 Figure A1. Discretization of the function’s area Figure A1. Discretization of the function’s area TheProper vector orderingZ contains of the the point Fourier set coe{xnmffi ,ycients } allows of the writing series (A1),relations between the set of function values {znm, } and the seth of Fourier coefficientsiT {Zrs, } inh the form i Z = Z Z Z whereZr = Z Z Z 0 1 ··· R r,0 r,1 ··· r,S zTZ= ⋅ (A3) The T hyper-matrix has the form, Where z is a vector of the function values {znm, } , ordered, as follows, cos(0 x1) B cos(1T x1) B cos(R x1) B zzz=· · z · · z =z··· z· z· cos(0 12x ) B cos R+1(1 xwhere) B nn,1cos n ,2(R x ) nB ,S+1 2 2 2 T = ·. · ·. · ···. ·. · (A4) . . . . . . . . The vector Z containscos( 0thex Fourier+ ) B coefficientscos(1 x + of) theB seriescos (A1),(R x + ) B · R 1 · · R 1 · ··· · R 1 · T The B matrix is squareZZZ and= [] has dimension Z((S where+ 1) Z(S =+1ZZ)). It takes the Z form, 01 R × rrrr,0 ,1 ,S The T hyper-matrix has costhe( form,0 y ) cos(1 y ) cos(S y ) 1 1 1 · · ··· · cos(⋅⋅0 y2) cos( ⋅⋅1 y2) cos(S ⋅⋅y2) cos(0 x11 )BB cos(1 x ) cos(R x 1 ) B B = . · . · ···. . · (A5) ⋅⋅. ⋅⋅. . . ⋅⋅ cos(0 x. 22 )BB cos(1. x ). cos(R. x 2 ) B T= (A4) cos(0y ) cos(1 y ) cos(S y ) · S+1 · S+1 ··· · S+1 ⋅⋅ ⋅⋅ ⋅⋅ cos(0 xR1++ )BB cos(1 x R1 ) cos(R x R1 + ) B The T matrix has dimension ((R + 1) (S + 1)). Its inverse matrix can be calculated from × the formula,The B matrix is square and has dimension ()(S+1)× (S+1) . It takes the form, 1 T 1 T T− = T T − T · · The (TT T) matrix is a diagonal of the form, · h i h i TT T = ((R + 1) (S + 1)) diag C (1/2) C (1/2) C = CC = diag 1 1/2 1/2 · · · S · S ··· · S S ··· 1 The inverse matrix T− can be determined from the relation, 1 T 1 1 T 1 C− = (T T − = T− (T )− · · Obtaining, h i 1 = 1 T = 1 1 1 1 T T− C− T (R+1) (S+1) diag CS− 2 CS− 2 CS− T · · · h · i ··· · · C 1 = diag 1 2 2 S− ··· Energies 2020, 13, 6464 16 of 20 Therefore, the inverse relation to (A3) can be written as, 1 h i Z = diag C 1 2 C 1 2 C 1 TT z (A6) (R + 1) (S + 1) · S− · S− ··· · S− · · · If the first partial derivatives exist, they have the Fourier series forms, XR XS XR XS z0x(x, y) = ( r Zr,s) (sin(r x) cos(s y)) = (Z0x,r,s) (sin(r x) cos(s y)) (A7) − · · · · · · · · · r=0 s=0 r=0 s=0 XR XS XR XS z0y(x, y) = ( s Zr,s) (cos(r x) sin(s y)) = (Z0y,r,s) (cos(r x) sin(s y)) (A8) − · · · · · · · · · r=0 s=0 r=0 s=0 It can be used to find a relation between the Fourier coefficients of the series (A1) and (A7), (A8). They can be written in the forms, Z0 = R Z and Z0 = R Z (A9) x x · y y · Z0x and Z0y are vectors of the Fourier coefficients of the respective series of partial derivatives (A7), (A8), ordered analogously, h iT h i Z0x = Z Z Z where Z0x,r = Z r Z Z 0x,0 0x,1 ··· 0x,R 0x, ,0 0x,r,1 ··· 0x,r,S and h iT h i Z0y = Z Z Z where Z0y,r = Z r Z Z 0y,0 0y,1 ··· 0y,R 0y, ,0 0y,r,1 ··· 0y,r,S The matrices Rx and Ry stand for the differential operators of a 2D periodic function in the frequency domain with respect to ‘x’ and ‘y’. They are diagonal. The matrix Rx is constituted by (R + 1) diagonal Rx,r matrices, h i R = diag R R R x x,0 x,1 ··· x,R Each of them is diagonal too and has dimension ((S + 1) (S + 1)) and takes the form, × h i R = r diag 1 1 1 x,r − · ··· The matrix Ry is also constituted by (R + 1) matrices Ry,r, h i R = diag R R R y y,0 y,1 ··· y,R Each matrix Ry,r is diagonal too and have (S + 1) elements, h i R = diag 0 1 S y,r − ··· Analogous to (A3), using the same notations, the relations, Z0x = Tx Z0x and z0y = Ty Z0y (A10) · · Energies 2020, 13, 6464 17 of 20 can be created. The matrices Tx and Ty have the forms, sin(0 x1) B sin(1 x1) B sin(R x1) B · · · · ··· · · sin(0 x ) B sin(1 x ) B sin(R x ) B · 2 · · 2 · ··· · 2 · Tx = . . . . (A11) . . . . . . . . sin(0 x + ) B sin(1 x + ) B sin(R x + ) B · R 1 · · R 1 · ··· · R 1 · cos(0 x1) By cos(1 x1) By cos(R x1) By · · · · ··· · · cos(0 x ) By cos(1 x ) By cos(R x ) By · 2 · · 2 · ··· · 2 · Ty = . . . . (A12) . . . . . . . . cos(0 x + ) By cos(1 x + ) By cos(R x + ) By · R 1 · · R 1 · ··· · R 1 · The matrix By has the form, sin(0 y ) sin(1 y ) sin(S y ) · 1 · 1 ··· · 1 ( ) ( ) ( ) sin 0 y2 sin 1 y2 sin S y2 = · · ··· · By . . . . (A13) . . . . sin(0 y ) sin(1 y ) sin(S y ) · S+1 · S+1 ··· · S+1 Having combined the relations (A3) and (A10), the following equations can be written, 1 (Tx Z0x) = Tx Rx T− (T Z) · · · 1 · · Ty Z = Ty R T (T Z) · 0y · y · − · · These lead to relations, (1) (1) Z0 = D z Z0 = D z (A14) x x · y y · (1) (1) The matrices Dx and Dy are the sought first order discrete partial-differential operators, (1) (1) T (1) (1) T D = Tx R T D = Ty R T (A15) x · x · y · y · where (1) 1 1 h (1) (1) i (1) h i R = R C− = diag 0 R R R = r diag 2 4 4 x x · (R+1) (S+1) x,1 ··· x,R x,r − · ··· · (1) = 1 = 1 (1) (1) (1) Ry Ry C− (R+1) (S+1) diag Ry,S 2 Ry,S 2 Ry,S · · · ··· · ( ) h i R 1 = diag 0 2 1 2 S y,S − · ··· · (1) (1) The matrices Rx and Ry are singular. The second partial derivatives have the Fourier series forms, XR XS XR XS 2 z00 (x, y) = ( r Zr,s) (cos(r x) cos(s y)) = (Z00xx,r,s) (cos(r x) cos(s y)) (A16) xx − · · · · · · · · · r=0 s=0 r=0 s=0 XR XS XR XS 2 z00 (x, y) = ( s Zr,s) (cos(r x) cos(s y)) = (Z00yy,r,s) (cos(r x) cos(s y)) (A17) yy − · · · · · · · · · r=0 s=0 r=0 s=0 XR XS XR XS z00 (x, y) = (r s Zr,s) (sin(r x) sin(s y)) = (Z00xy,r,s) (sin(r x) sin(s y)) (A18) xy · · · · · · · · · · r=0 s=0 r=0 s=0 Energies 2020, 13, 6464 18 of 20 The relation between the Fourier coefficients of the series (A1) and (A16)–(A18) can be written in the forms, Z00 = R ZZ0 = R ZZ00 = R Z (A19) xx xx · yy yy · xy xy · where Z00xx, Z00yy and Z00xy are vectors of the Fourier coefficients of the respective series of the second partial derivatives (A16)–(A18), ordered analogously as it has been used in first derivatives. The matrices Rxx, Ryy and Rxy have the following forms. The matrix Rxx is constituted by (R + 1) diagonal R matrices, xx,n h i R = diag R R R xx xx,0 xx,1 ··· xx,R Each of them is diagonal too and has dimension ((S + 1) (S + 1)) and takes the form, × h i R = r2 diag 1 1 1 x,r − · ··· The matrix Ryy is also constituted by (R + 1) matrices Ryy,r, h i R = diag R R R yy yy,0 yy,1 ··· yy,R Each matrix Ryy,r is diagonal too and has (S + 1) elements, h i R = diag 0 1 22 S2 yy,r − ··· The matrix Rxy is constituted by (R + 1) matrices Rxy,r, h i R = diag R R R xy xy,0 xy,1 ··· xy,R The matrices Rxy,r are diagonal too and have (S + 1) elements, h i R = r diag 0 1 2 S xy,r · ··· Analogous to (A10), using the same notations, the relations, z00 = Txx Z00 z00 = Tyy Z00 and z00 = Txy Z00 (A20) xx · xx yy · yy xy · xy can be created. The matrices Txx and Tyy fulfill the relation Txx = Tyy = T and the matrix Txy can be easy predicted as, sin(0 x1) By sin(1 x1) By sin(R x1) By · · · · ··· · · sin(0 x ) By sin(1 x ) By sin(R x ) By · 2 · · 2 · ··· · 2 · Txy = . . . . (A21) . . . . . . . . sin(0 x + ) By sin(1 x + ) By sin(R x + ) By · R 1 · · R 1 · ··· · R 1 · The mathematical operations described below, 1 (T Z00xx) = T Rxx T− (T Z) · · · 1 · · T Z00yy = T Ryy T− (T Z) · · · 1· · Txy Z = Txy R T (T Z) · 00xy · y · − · · lead to relations defining the second order partial differential operators, (2) (2) (2) z00 = D z z00 = D z z00 = D z (A22) xx xx · yy yy · xy xy · Energies 2020, 13, 6464 19 of 20 where ( ) ( ) ( ) ( ) ( ) ( ) D 2 = T R 2 TT ; D 2 = T R 2 TT ; D 2 = T R 2 TT (A23) xx · xx · yy · yy · xy xy · xy · where ( ) h ( ) ( ) i R 2 = R C 1 = 1 diag 2 2 xx xx − (R+1) (S+1) 0 2 Rxx,1 2 Rxx,R · · · ··· · ( ) h i R 2 = r2 diag 1 2 2 xx,r − · ··· (2) = 1 = 1 (2) (2) (2) Ryy Ryy C− (R+1) (S+1) diag Ryy,S 2 Ryy,S 2 Ryy,S · · · ··· · ( ) h i R 2 = 2 diag 0 12 S2 yy,S − · ··· h i (2) = 1 = 1 (2) (2) (2) = Rxy Ryy C− (R+1) (S+1) diag 0 2 Rxy,1 2 Rxy,R Rxy,r 2 r diag 0 1 S · · · ··· · · · ··· (2) (2) (2) The matrices Rxx , Ryy and Ryx are singular. 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