Energy of the magnetic field, permanent magnets, forces, losses

Transcription

1 Energy of the magnetic field Let use model of a single coil as a concentrated resistance and an inductance in series Switching the coil onto a constant voltage source by the voltage equation () U = u () t u () t i() t R d t R + L = + ψ dt R i(t) L U Coil with concentrated parameters The consumed energy dw within the time dt is a sum of that in resistance and inductance: dw=dw R +dw m =Ui(t)dt=i (t)rdt+i(t)dψ(t) Portion of total energy i (t)rdt dissipates in the resistance of coil, while the rest i(t)dψ(t) is storing in the magnetic field Latter will be retrieved when the current decrease reduces the magnetic field During the power-on process the flux linkage ψ(t) increases from zero to Ψ (the current i(t) increases from to I ), therefore the full magnetic energy W m stored becomes: W m Ψ () = i t dψ In case of linear function of ψ(i) (eg iron-less coil) the inductance L=const, dψ=ldi and Ψ =LI, whereby I W L i() = m t di = LI I = = Ψ Ψ L the energy stored in the magnetic field of a coil calculated from the flux linkage, further at constant current it is proportional to the inductance In a circuit containing ferromagnetic material (eg coil with iron core) the function ψ(i) is nonlinear, L const, hence the integral giving the magnetic energy may not be simplified ψ ψ Ψ idψ i I i Ψ idψ i The energy stored in a coil if the core is non-ferromagnetic ferromagnetic I i

2 VIVEM Alternating current systems 4 After the coil under consideration disconnected from the voltage source the stored energy will retrieved, since the induced voltage produced due to the decreasing the flux linkage acts for conservation the current ie for delay of current decrease, according to Lenz's law This statement is true for interruption any inductive circuit, therefore such operations require attention and carefulness If the magnetic circuit is linear and homogeneous then the magnetic energy often determined with the space variables Substituting the equations Ψ=NΦ=NA and Θ=NI=l the magnetic energy W = I = NA l Ψ = V, N where V=Al the volume of space analysed The energy stored in the unit volume (ie the energy density) w W = = = = V If the magnetic circuit is non-linear, but homogeneous (if const eg solenoid, toroid with iron core) then ψ ψ Φ l W i() t d N d l = ψ = ψ = NdΦ = A d = V d N l, and the energy stored in the unit volume w = d Latter equation is true for individual points of inhomogenious space thus in general case for the magnetic energy in a given volume V: W = ddv V Energy of coupled coils Consider two coupled coils without ferromagnetic cores, when the current in the first coil I =const while no current flows in the second one The magnetic energy stored in the first coil is: W = LI Now, increasing the current of the second coil i (t) from zero to I smoothly an induced voltage u i appears in the first coil because of changing the flux linkage ψ : dψ u M di i = = dt dt I =áll I = i I u t Initial state Vary of the current in the second coil

3 If the fluxes of the coils tend towards the same direction (ie ψ =ψ +dψ ), then the voltage u i according to Lenz's low tries to decrease current I (to keep the full flux linkage of coil unchanged) Therefore to keep current I constant an energy input needed depending on vary of current i (t): dw=u i I dt=m I di The necessary energy input during the change of i (t): cs I W = M I di = M I I The magnetic energy stored in the second coil is: W = LI The total stored energy of the two coupled coils is: W = L I + M I I + L I This stored energy usually does not depend on the sequence of power-on, switching onto a source the first coil after the second one is on the total stored energy becomes: W = L I + M I I + L I The sign of the middle part, representing the coupling, depends on the direction of magnetisation of currents I and I, thus MII < > Determination of the leakage coefficient from magnetic energy In coupled coils a portion of energy is stored in the common field Ψ m and the rest is stored in the leakage field Ψ l When creating a desired value of flux an extra energy required which is storing in the leakage field I Ψ l Ψ l I Ψ Ψ m =Ψ +Ψ The coupled coils considered Say current I in coil is not enough to produce the required flux linkage Ψ (Ψ >Ψ =I L ), thus contribution of coil is needed, adding flux linkage Ψ =I M created by current I : Ψ =I L + I M In this way while producing Ψ the flux leakage Ψ s developed and magnetic energy is storing in the leakage of coil The total stored magnetic energy of the coils as written previously W = L I + M I I + L I Calculate the energy W * (and current I * ) which able to produce the required flux linkage Ψ only using coil alone In this case flux leakage Ψ s is not developing, not stores energy because coil remains current-less 3

4 VIVEM Alternating current systems 4 Ψ M I I L L I = = + The energy accumulated by developing the flux linkage Ψ becomes: M W L I L I I L I M L I L I M I I M = = + + = + + L I The energy W s used to produce flux leakage of coil in the previous case equal to the difference W-W * : M Ws = W W = LI LL The term in bracket is the leakage coefficient of coil : σ = M, that is LL Ws = σ LI ecause M L L, therefore <σ < Explanation of leakage coefficient: the current of coil I produces leakage flux in inductance σ L while the same current produces a mutual flux linking coil in inductance (-σ )L : Ψ s =I σ L and Ψ =I (-σ )L, since Ψ =Ψ s +Ψ =I L Otherwise the leakage coefficient is the ratio of leakage flux to total flux produced by a coil: ψ s σ = ψ The energy in the leakage field is σ -times the energy stored in the magnetic field W produced by coil In a case when the flux Ψ desired with contribution of coil the energy to produce the leak- age field of coil can be calculated The leakage coefficient of coil is: σ = M LL Permanent magnets After a magnetising process the magnetic field of permanent magnet materials remains for long time without any excitation and can be demagnetised only with high opposite coercitive field The permanent magnet materials, also designated as a magnetically hard materials, have coercive force generally greater than, A/m, whereas the magnetically soft materials may be easily demagnetised When magnetising a permanent magnet () material of closed ring form up to the saturation value max then after ceasing the excitation a remanent magnetisation (flux density) r remains Since the excitation Θ becomes zero according to the excitation law the field intensity in the permanent magnet becomes zero thereby the magnetic energy W m becomes also zero Opening an air gap in the ring l + =, according to the excitation law since no excitation enclosed The new value of field intensity inside the permanent magnet: = = l l, here l the center line along the permanent magnet Therefore a negative field intensity developed and the flux density decreased form the remanent value to ' 4

5 l operating line r r * ' c Ring form permanent magnet with air gap - demagnetisation curve of a permanent magnet If the leakage flux is negligible Φ s = then the flux in the permanent magnet equal to that in the air gap Φ =Φ or A = A, that is A = A Substituting into the formula of the excitation law above a linear function ( ) called A operating line obtained: = = a, A l If the leakage flux is not negligible then the flux in the air gap is less than that in the permanent magnet Applying the definition σ = Φ s : Φ Φ =Φ -Φ s =Φ -σφ =(-σ)φ ( σ) A A ereby = and = σ = ( σ) a A A lv The operating zone of permanent magnet is the demagnetisation curve ( ) in the second quadrant The operating point of the magnet is represented by the intersection of the operating line with the - curve (magnetisation curve + law of excitation) The measure of air gap depends on the application Important property of a permanent magnet is the magnitude of flux density * r obtained after decreasing the field intensity to zero when the air gap is closed again, its rate to the initial r The operating zone of permanent magnet is usually along the saturated zone of - curve thus the permeability is taken as or close to Permanent magnet alloys Alnico magnets Alnico alloys basically consist of aluminium (Al), nickel (Ni), copper (Cu), iron (Fe) in some product of cobalt (Co) and titanium (Ti) Ceramic (ferrite) magnets The most known chemical compositions contain barium and strontium eg ao 6 Fe O 3 and SrO 6 Fe O 3 5

6 VIVEM Alternating current systems 4 Iron-Chrome-Cobalt magnets These alloys are primarily the compositions of iron (Fe), chromium (Cr) and cobalt (Co) Some grades may also contain additions of vanadium (V), silicon (Si), titanium (Ti), zirconium (Zr), manganese (Mn), molybdenum (Mo) or aluminium (Al) Rare earth magnets The rare earth elements used: samarium (Sm), neodymium (Nd), praseodymium (Pr), dysprosium (Dy) The transition metals used: iron (Fe), copper (Cu), cobalt (Co), zirconium (Zr), hafnium (f), The rare earth magnets currently fall into three families of materials: - rare earth + cobalt 5, eg samarium-cobalt SmCo 5, - rare earth + transition metal 7, eg samarium-cobalt Sm Co 7, - rare earth + iron alloys, eg neodymium-iron-boron Nd Fe 4 The aim of different combinations is to improve properties like temperature dependence, remanence and coercitivity A special material is the silver-magnese-aluminium Ag-Mn-Al alloy, a ferromagnetic composition of non-ferromagnetic elements Optimal exploitation of hard magnets The magnetic circuit with permanent magnets usually contain soft magnet segments and air gap as well ecause of the high charge of high quality magnets general requirement the use of the minimum volume Neglecting the flux leakage and the excitation of soft magnet segments =- l and Φ =Φ = A, where index refers to the permanent magnet Φ Substituting = = the volume of permanent magnet V : A Φ V = l A = = A Φ r optimal operating pont c Graphical determination of optimal operation point If the volume of the air gap and the air gap flux are given the value of permanent magnet material is minimal when the energy product (product ) is the maximum: 6

7 V min = c ( ) The approximate maximum of the energy product ( ) max often determinated graphically Lifting force of permanent magnet The energy of a closed (air gap-less) magnet is zero since = max F m x dx F ex Calculation the lifting force of permanent magnet When an air gap is opening due to external force F ex the field intensity will be not zero and the invested mechanical energy W mech will transformed to magnetic energy W magn and loss W loss For changes of energy: dw mech =dw magn +dw loss, where dw mech the vary of invested mechanical energy, dw magn the vary of magnetic energy, dw loss the vary of loss energy Suppose the loss and the leakage are negligible, then dw loss =, φ =φ =φ, here φ the flux in air gap, φ the flux in permanent magnet The vary of invested mechanical energy: dw mech =F ex dx=-f m dx, here F ex the external force, F m the lifting force of permanent magnet The negative sign means that according to the reference of co-ordinate x in the figure F m acts to decrease dx The magnitude of F m can be calculated by the method of virtual work The principle of virtual work A particle or rigid body is in equilibrium if the resultant of forces in action is zero (In this case forces act thus F ex +F m =) To solve this equilibrium of forces may be used the method of virtual work Virtual work: work done by the acting forces (F ex, F m ) for a kinematically admissible virtual displacement dx The acting forces are in equilibrium if the total work done by the forces is zero A particle or rigid body is in equilibrium under the action of a number of forces if the total work done by the forces is zero for any virtual displacement of the particle or rigid body A physical system under the action of forces is in equilibrium only if the total virtual is zero: F ex dx+f m dx= 7

8 VIVEM Alternating current systems 4 The unknown force in this case F m determined using this principle because the virtual work done by the other force (F ex dx) is in equilibrium with the unknown one may be calculated from the change of energy The vary of the magnetic energy dw magn is stored both in the permanent magnet (dw ) and in the air gap (dw ): dw magn = dw + dw The total energy stored in the permanent magnet is W = V d thus its vary dw =V d =l A d =l dφ since V =A l The total energy stored in the air gap is W V V = = Due to the displacement dx of the closing plate both the volume of the air gap and the field density is changing thus W dw V dx W dx = +, thereby The volume of the air gap and its change: V =A, dv =A dx, thus dv dw dx dx V d dx dx Adx Vd = + = + = Adx + d Φ The equation of energy balance becomes: Fexdx = ldφ + Adx + d Φ = ( l + ) d Φ + Adx Since according to the law of excitation l + = in steady state the lifting force of permanent magnet F = m A Lifting force of electromagnet In this case the magnetic field is created by an excited coil According to the energy conservation principle the sum of the supplied electric energy and the work of the external force equal to the sum of the stored magnetic energy and the energy of losses This balance is true for changes as well: dw electric +dw mech =dw magn +dw loss When supplying from a dc source the current is determined by the dc resistance of the excitation coil thus the excitation considered to be constant Θ = i li = const As a consequence i the increase of the air gap decreases the flux and the field intensity in the air gap whereas the decrease of the air gap increases the flux and the field intensity The induced voltage through the vary of flux linkage dψ means an electric energy u i idt during a time dt, which energy acts against the flux change Thus to perform the change this energy have to be counteracted with an energy from external source dw u Idt N d φ electric = i = Idt = NIdφ dt 8

9 I F m x dx F ex Calculation the lifting force of electromagnet The mechanical work performed by the external force F ex is: dw mech = F ex dx The stored energy in the magnetic circuit (in iron and air gap) changes through the virtual displacement of closing plate Neglecting the flux leakage the magnetic energy in the iron W = V d vary iron iron iron iron iron when the flux density changes dw iron =V iron iron d iron =l iron iron dφ, while the magnetic energy in the air gap W V V = = vary when both the flux density and the volume of air gap change: W dw = V dx W dx + The volume of the air gap and its change: V =A, dv =A dx, thus dv dw dx dx V d dx dx Adx Vd = + = + = Adx + d Φ The vary of the losses may be neglected since the excitation current is constant (I coilr=const), thus the equation of energy balance dw electric +dw mech =dw iron +dw Substituting the components: NIdΦ + Fk dx = liron irondφ + Adx + d Φ = ( liron iron + ) d Φ + Adx According to the excitation law Θ=Ni= iron l iron +, hence F dx force of electromagnet in steady state: F = m A, this formula is the same as that of permanent magnet Adx ex = so the lifting 9

10 VIVEM Alternating current systems 4 Losses du to the flux change Keeping a magnetic field constant do not consumes energy (see permanent magnets) owever, the changing magnetic field (flux) causes losses in the iron core of magnetic circuit (termed as iron losses) producing increase its temperature The two components of iron losses P Fe which differ by nature are: - hysteresis losses, - eddy current losses, P Fe = P hyst + P eddy For non-sinusoidal changes the losses due to higher harmonics are calculated separately Iron losses in sinusoidal magnetic field a) ysteresis losses A simple interpretation of hysteresis losses is that the molecular magnets inside the ferromagnetic material are continuously rearranging due to the alternating change of external field intensity and this rearranging occurs internal friction The stored magnetic energy in the unity volume differs according to integral w = d differs by zones of hysteresis loop - m m r - r 4 - m m m r 3 - r The consumed and retrieved magnetic energy along the hysteresis loop rising side falling side Inside zone - r m ( m ) the field intensity is positive, the flux density increases thus d >, thus the energy density also positive w >, consumed Inside zone m r ( m ) the field intensity is positive, the flux density decreases thus d <, thus the energy density negative w <, retrieved 3 Inside zone r - m ( - m ) the field intensity is negative, the flux density increases thus d <, thus the energy density also positive w >, consumed 4 Inside zone- m - r (- m ) the field intensity is negative, the flux density increases thus d >, thus the energy density negative w <, retrieved During a whole period the difference between the consumed and retrieved energy density is proportional to the area formed by the hysteresis loop which is lost to heat in the ferromagnetic material - m - m m

11 According to formula by Steinmetz the area formed by the hysteresis loop: w m =γ x max, where γ specific for material, x specific for material and depends on max, usual value x=,7- Thus the power of hysteresis losses is proportional to frequency and also to the volume of the ferromagnetic material if the field is considered to be uniform: P hyst =γ max x fv k hyst Ψ f - m m r w m - r - m m The difference between the consumed and retrieved energy density is the area of the hysteresis loop The value of coefficient k hyst defined for a given magnetic circuit taking into account that the flux linkage Ψ substituted as rms or maximum value b) Eddy current losses The currents inside conductive bodies (eg iron core) that are a result of the induced voltage are called eddy currents Say the resistance of the eddy current path is R eddy, then the losses from eddy currents I eddy, heating the iron core may be approximated as P eddy =I eddy R eddy I eddy ψ(t) P eddy R eddy Formation of eddy currents Charles Proteus Steinmetz (865-93) American engineer, researcher (German origin Karl August Rudolf Steinmetz)

12 VIVEM Alternating current systems 4 To reduce the eddy current losses the iron cores are laminated and made from material of increased resistance eg from alloy with few percent of silicon If the flux density vector is parallel to a thin ferromagnetic sheet, the possible current loops have small areas, the enclosed magnetic flux is small, and the induced voltage in them is also small If the flux density vector is normal to the sheet, large current loops (as large as the sheet itself) are possible, enclosing large flux The ferromagnetic sheets are insulated each from other using either thin layers of suitable material (eg stain) or insulating surfaces creating in the rolling process When the flux is sinusoidal the induced voltage U eddy is proportional to the frequency d U ψ örv f dt Ψ and the eddy current is proportional to the voltage Ieddy U eddy, thus P eddy =k eddy Ψ f The value of coefficient k eddy defined for a given magnetic circuit taking into account that the flux linkage Ψ substituted as rms or maximum value Separating the hysteresis and eddy current losses For develoent and diagnostics may be necessary to separate the components of iron losses by means of measurement in the range f f f Using supply of variable voltage and frequency, providing Ψ=const P Fe = P eddy + P hyst =k eddy Ψ f +k hyst Ψ f=fψ (k eddy f+k hyst ), or rearranging ( k f khyst ) PFe fψ = + eddy P f Fe Ψ Ψ=const k eddy f k hyst f f f Separating the hysteresis and eddy current losses PFe The quotient has a constant component and one with linear dependence of frequency fψ Coefficients k eddy f and k hyst are according to the figure Composed by: István Kádár March 4

13 Questions for self-test Determine the magnetic energy of an iron-less coil Determine the magnetic energy of a coil with iron core 3 Determine the magnetic energy in a given volume with space variables 4 Determine the magnetic energy density with space variables 5 Determine the magnetic energy in two coupled coils 6 Determine the magnetic flux leakage of two coupled coils from magnetic energy 7 Illustrate and explain the () curve of a permanent magnet 8 ow do you mean the optimal exploitation of permanent magnet? 9 What energy product means? Determine the lifting force of permanent magnet ow do you use method of virtual work? Determine the lifting force of electromagnet 3 Explain the components of iron losses 4 Explain the hysteresis losses and the frequency dependence 5 Explain the eddy current losses and the frequency dependence 6 ow to separate hysteresis and eddy current losses from measurement data? 3