Characteristic Analysis Of The Peak Braking Force And The Critical .

Energies 2019, 12, 2622 4 of 15 In order to get the concrete expressions of A, the general expressions of the magnetic vector potential A in z direction can be obtained by the variable separation method [17]. P Azn (x, y) Az (x, y) n 1 Azn (x, y) G( y) e jn πτ x n n 1, 3, 5 . . . (6) where G(y)n is the function of y and j is the imaginary number unit. In Figure 1, the exciting current in region I can be equivalent to the linear current density at the interface between region I and region II, according to the principle of equal magnetic potential [18,19]. This equivalence will have influence on the results of magnetic field calculation in region I, but does not affect the accuracy of magnetic field calculation in region II, III, and IV. Additionally, the braking force calculation is carried out in region III, so it can ensure the correctness of the result of the braking force calculation. According to this equivalence, it can be considered that there is no exciting source in region I. According to the principle of eddy current braking, eddy current can be induced in the induction plate of region III. In order to solve Equation (3), the expression of eddy current density is given by the equation below. Jw σvB(III) y (7) where σ is the induction plate conductivity, v is the running speed of eddy current braking device, and the direction of v is parallel to the x-axis, according to Figure 1. Combined with Equation (6) and Equation (7), the expression of Poisson equation in Equation (3) in each region can be given by the following formulas. 2 A(i)zn 2 n πτ A(i)zn 0 i I, II, IV 2 A(i)zn n πτ A(i)zn µ0 σv x i III y2 2 A(i)zn y2 (8) where i is the region number. Combining Equation (6) and Equation (8), the expressions of the nth harmonic of magnetic vector potential in each region can be calculated by the following formulas. nπ y n π y jn π x A(i)zn C (i)n e τ D(i)n e τ e τ π A(i)zn C(i)n eαy D(i)n e αy e jn τ x i I, II, IV (9) i III where C(i)n and D(i)n are the constants to be determined, and the expression of α is given by Equation (10). π α n τ s 1 jµ0 σv n πτ (10) Then, the expressions of flux density B can be calculated by Equations (9) and (10). 2.3. Magnetic Field Boundary Condition In order to calculate the coefficients C(i)n and D(i)n in Equation (9), connection conditions on different interfaces are given by the following formulas. ( B(i) y B(i I) y H(i)x H(i I)x K i I, II, III (11) (i)(i I)z where K(i)(i I)z is the z-component of the linear current density at the interface of region i and i I.

Energies 2019, x FOR PEER REVIEW Energies 2019, 12,12, 2622 5 5ofof1515 where K(i)(i I)z is the z‐component of the linear current density at the interface of region i and i I. Since 0,then thenthe the distribution of K(II)(III)z 0, 0, KK(III)(IV)z (III)(IV)z 0, of linear linearcurrent currentdensity densityK(I)(II)z K(I)(II)zat atthe theinterface interface SinceK(II)(III)z ofofregion I and region II is analyzed below. region I and region II is analyzed below. The Thedistribution distributioncurve curveofoflinear linearcurrent currentdensity densityatatthe theinterface interfaceofofregion regionI and I andregion regionIIIIisisshown shown ininFigure inin Table 1. 1. Figure2,2,and andthe theabscissa abscissaexpressions expressionsofofthe thecurves curvesare areshown shown Table y/(A/m) b o x1 x5 x2 x3 x4 x/(m) ‐b Figure Figure2.2.Distribution Distributionofofequivalent equivalentlinear linearcurrent currentdensity densityatatthe theinterface interfaceofofregion regionI and I andregion regionII.II. Table 1. Abscissa expression in Figure 2. Table 1. Abscissa expression in Figure 2. Parameter Value Parameter Value x1 c/2 x1 c/2 x2 c/2 bc x3 x 2 τ c/2 (c/2bc bc ) x 3 ‐ ( c bc) x4 τ/2 c/2 x5 x 4 τ ‐c/2 x5 Fourierdecomposition decompositionofofthe thecurve curveininFigure Figure2 2shows showsthat thatthere thereare areonly onlyodd oddcomponents componentsininthe the Fourier Fourier expression, which is shown below. Fourier expression, which is shown below. K(I)(II)z X an cos n π x n 1, 3 , 5. (12) K(I)(II)z n 1 an cos n x n 1, 3, 5 . . . (12) τ n 1 The expression of an is shown below. The expression of an is shown below. 2bk an (13) 2bk an (13) π The relationship between the linear current density constant b and the electric density J, the The relationship linear currentcoil, density constant b and J, the widththe bc of the exciting and the full factor kf ofthe the electric coil slot density is given by cross‐sectional area S,between the area S, the width bc of the exciting coil, and the full factor kf of the coil slot thecross-sectional equation below. is given by the equation below. JSk f JSk b f (14) b (14) bcbc k kisisthe the formula below. thelinear linearcurrent currentdensity densitydistribution distributioncoefficient, coefficient,which whichisisgiven givenbyby the formula below. Z c Z τ c c 2 π 2 2c b cbc π π 2 cos nn xxdx k k c cos dx dx ( cc b ) coscos n n τx x dx τ c 2 τ τ (2 c bc ) 2 (15) (15) 2 nthharmonic harmonicexpression expressionofof AccordingtotoEquation Equation(12), (12),and andfor forconvenience convenienceofofcalculation, calculation,the thenth According lineardensity densityatatthe theinterface interfaceofofregion regionI and I andregion regionII IIisisgiven givenbybythe theformula formulabelow. below. linear jnπ x K(I)(KII)zn aan eejn τ x (I)(II)zn n 2.4. Eddy Current Braking Force n . n 1,13, , 35, 5. (16) (16)

Energies 2019, 12, 2622 6 of 15 2.4. Eddy Current Braking Force In order to calculate the expression of braking force, the expressions of eddy current density and flux density in region III need to be calculated. The nth harmonic expression of eddy current density Jwn in the induction plate can be obtained by combining Equation (7) along with Equations (9)–(16). Jwn h π i π 0 0 2jn πτ µ0 σvan e(n τ α)δ eαy e(n τ α)δ 2αbg e αy n πτ 2(n πτ δ0 αb g ) α)[1 e ] (n πτ α)(e 2αb g e 2n πτ δ0 π e jn τ x (17) where δ0 is the modified air gap length and bg is the induction plate thickness. The magnetic density in region III is the sum of the magnetic density produced by exciting current and eddy current. Because the magnetic field generated by the eddy current in the induction plate does not work on the induction plate itself, there is only the magnetic field generated by the exciting current working on the induction plate in region III when calculating the braking force. The magnetic density in region III can be simplified to that in the static case (v 0). Combined with Equations (9)–(16), in region III, the nth harmonic expression of magnetic density in the y direction is given by the equation below. π π π A(III)zn en τ y e2n τ d e n τ y jn π x B(III) yn e τ (18) jµ0 an π x 1 e2n τ d v 0 where d δ bg . Then, according to the ampere force formula [20], the nth harmonic calculation formula of braking force in the induction plate is given by the equation below. FWn Z lp w · Re d Jw B(III) yn δ dy (19) where lp 2pτ, which is the total length of eddy current braking device, p is the number of pole-pairs, w is the vertical paper depth of the device, and * represents the complex conjugate of a variable. The nth harmonic expression of braking force is calculated using the equation below. FWn h i n πτ lp wµ0 an 2 Msin(2 Nb g ) Nsin h(2 Mb g 2 2 π π 2 MY1 NY2 n τ Y3 MY2 NY1 n τ Y4 s (n πτ ) M 2 n πτ (n πτ ) 2 µ0 2 σ2 v2 1 2 2 s 2 N (20) (n πτ ) n πτ (n πτ ) 2 µ0 2 σ2 v2 1 2 2 Y1 cos h(n πτ δ0 ) sin h(Mb g ) cos(Nb g Y2 cos h(n πτ δ0 ) cos h(Mb g ) sin(Nb g (21) Y3 sin h(n πτ δ0 ) cos h(Mb g ) cos(Nb g Y4 sin h(n πτ δ0 ) sin h(Mb g ) sin(Nb g Therefore, the expression of total braking force is given by the equation below. FW X n 1 FWn n 1, 3, 5 . . . (22)

Energies 2019, 12, 2622 7 of 15 In this paper, the expression of eddy current braking force (22) has been deduced, which is the key to analyze the peak braking force and the critical speed. 3. Finite Element Verification According to the design requirement of a maglev project, a set of rated parameters is selected to design an FEA model of eddy current braking to verify the accuracy of the analytical formula of the braking force. The design parameters of the analytical and FEA model are shown in Table 2. Table 2. Basic parameters of the eddy current braking model. Parameter number of pole pairs p maximum operating speed vmax exciting current density J static air gap magnetic density B0 the fill factor kf height of exciting coil hc width of exciting coil bc pole pitch τ air gap length δ height of yoke hs height of pole hp width pole bp thickness of induction plate bg thickness of back iron bs model depth w Energies 2019, 12, x FOR PEER REVIEW induction plate conductivity σ setting simulation time Value 6 272 m/s 12 A/mm2 1.5 T 0.687 115 mm 15.3 mm 100 mm 10 mm 120 mm 120 mm 69.2 mm 5 mm 30 mm 200 mm 106 S·m 1 6 ms 8 of 15 Since the linear brake is analyzed in this paper, there will be entry and exit effects when the Note: The model is simulated 160 times in the setting simulation time of 6 ms, which means the model is simulated brakeevery moves. However, in Reference [6], it is mentioned that, if the brake exceeds four or more poles, 0.0375 ms. the effects are insignificant. In this paper, the brake has 12 poles, so the entry and exit effects of the device are neglected the analytical calculation. Since the linearinbrake is analyzed in this paper, there will be entry and exit effects when the According to the model parameters in Table 2,that, the FEA model ofor eddy current brake moves. However, in Reference [6], given it is mentioned if thesimulation brake exceeds four more poles, braking is set in Ansoft Maxwell, as shown in Figure the effects areup insignificant. In this paper, the brake has3. 12 poles, so the entry and exit effects of the In are Figure 3, the in upper part is thecalculation. eddy current braking device, which is composed of a yoke, device neglected the analytical poles,According and exciting coils. The lower part is the in rail, which is composed of an model induction platecurrent and a to the model parameters given Table 2, the FEA simulation of eddy back iron. braking is set up in Ansoft Maxwell, as shown in Figure 3. Magnetic yoke hs Pole hp Coil bc bp hc Induction plate bs bg Back iron Figure 3. FEA model of eddy current braking. Figure 3. FEA model of eddy current braking. In Figure 3, the upper part is the eddy current braking device, which is composed of a yoke, poles, and exciting The lower And it iscoils. assumed that: part is the rail, which is composed of an induction plate and a back iron. And it is assumed that: is simulated in 2D, and the Cartesian coordinate system and the 1) The magnetic field International Unit System are used; (1) The magnetic field is simulated in 2D, and the Cartesian coordinate system and the International 2) The lateral end effect has been neglected. The magnetic field distributes uniformly along the Unit System are used; z‐axis. The current density vector J and the magnetic potential vector A have only z‐axis components. (2) The lateral end effect has been neglected. The magnetic field distributes uniformly along the z-axis. In the FEA software, the eddy current braking model is simulated by the transient solver to The current density vector J and the magnetic potential vector A have only z-axis components. verify the accuracy of the braking force analytical formula given in Equations (20), (21), and (22). Furthermore, Figure 4 and Figure 5 are the static magnetic line distribution and the static magnetic density distribution of the FEA simulation, respectively.

Figure 3. FEA model of eddy current braking. Figure 3. FEA model of eddy current braking. And it is assumed that: And it ismagnetic assumed that: 1) The field is simulated in 2D, and the Cartesian coordinate system and the 1) The magnetic field simulated in 2D, and the Cartesian coordinate system and the International Unit System areisused; Energies 2019, 12, 2622 8 of 15 International Unit System arehas used; 2) The lateral end effect been neglected. The magnetic field distributes uniformly along the 2)The Thecurrent lateral end effect has been Thepotential magneticvector field A distributes along the z‐axis. density vector J andneglected. the magnetic have onlyuniformly z‐axis components. z‐axis. The current density vector J and the magnetic potential A have ‐axis components. In FEAsoftware, software, theeddy eddy current braking model isvector simulated byonly the ztransient to In the FEA the current braking model is simulated by the transient solversolver to verify In the FEA software, the eddy current braking model is simulated by the transient solver to verify the accuracy of the force braking force analytical formula given in(20), Equations (20), and (22). the accuracy of the braking analytical formula given in Equations (21), and (22).(21), Furthermore, verify the accuracy of the braking force analytical formula given in Equations (20), (21), and (22). Furthermore, 4 and 5 are thedistribution static magnetic linestatic distribution the static magnetic Figures 4 and Figure 5 are the staticFigure magnetic line and the magneticand density distribution of Furthermore, Figure 4 and Figure 5 are the static magnetic line distribution and the static magnetic density the FEA simulation, respectively. the FEAdistribution simulation, of respectively. density distribution of the FEA simulation, respectively. Figure Figure 4. 4. Distribution Distribution field field diagram diagram of of the the static static magnetic magnetic line. line. Figure 4. Distribution field diagram of the static magnetic line. Figure 5. 5. Distribution Distributionfield field diagram diagram of of static static magnetic magnetic density. density. Figure Figure 5. Distribution field diagram of static magnetic density. Figures 5PEER can REVIEW verify accuracy of the model its simulation. Additionally, it can9 of also Energies 2019, 12, xand FORFigure 15 Figure 44and 5 can the verify the accuracy of the and model and its simulation. Additionally, it Figure 4 Figure andfrom Figure 5the can verify the of the model and its simulation. Additionally, it be seen from 5 that gap magnetic density ofdensity the model is slightly less than 1.5 T, which is can also be seen Figure 5air that the air accuracy gap magnetic of the model is slightly less than 1.5 can also be seen from Figure 5 that the air gap magnetic density of the model is slightly less than 1.5 T, which due to theofsaturation magnetic in the corebe that cannot be neglected in the FEA due to theissaturation magnetic of circuit in thecircuit core that cannot neglected in the FEA simulation. simulation. Figure 6 shows the analytical and simulation curves of the variation of eddy current braking force Figure with speed. 6 shows the analytical and simulation curves of the variation of eddy current braking force with speed. 4 x 10 4 Analytical solution FEA solution FW (N) 3 2 1 0 0 50 100 150 200 250 272 v (m/s) Figure 6. 6. Eddy Eddycurrent current braking braking force force varying varying with with speed. speed. Figure It can can be be seen seen from from Figure Figure 66 that that the the analytical analytical solution solution is is in in good good agreement agreement with with the the FEA FEA It solution, but butthe theanalytical analyticalresult resultisisslightly slightly higher than FEA solution at high speeds. is solution, higher than thethe FEA solution at high speeds. This This is due due to the assumption that the relative permeability of the ferromagnetic region is infinite in the to the assumption that the relative permeability of the ferromagnetic region is infinite in the analytical analytical calculation, while of the ferromagnetic regionatincreases at high speed the FEA. calculation, while the loss of the the loss ferromagnetic region increases high speed in the FEA.inThen, the braking force decreases in the FEA at high speed. From Figure 6, the correctness of the braking force analytical formula is verified by this simulation. Additionally, it takes about 3 hours to run one working point (corresponding to one simulation point in Figure 6, and there are 20 FEA simulation points in Figure 6) whose setting simulation time is 6 ms by FEA. In contrast, the result of one working point can be obtained immediately by the

Energies 2019, 12, 2622 9 of 15 Then, the braking force decreases in the FEA at high speed. From Figure 6, the correctness of the braking force analytical formula is verified by this simulation. Additionally, it takes about 3 h to run one working point (corresponding to one simulation point in Figure 6, and there are 20 FEA simulation points in Figure 6) whose setting simulation time is 6 ms by FEA. In contrast, the result of one working point can be obtained immediately by the subdomain method calculation. In addition, the running time of simulation is also related to the accuracy of the model partition. The finer the partition, the longer the running time of simulation will be. 4. Parameter Analysis Equation (22) can be regarded as a function FW (v) of eddy current braking force and speed. Since the skin effect of the induction plate is considered in this paper, the analytical formula of braking force becomes very complicated, which makes it difficult to obtain the expressions of the peak braking force and critical speed point by differentiating the function FW (v) with respect to v. Therefore, the dichotomy method is used to calculate the root derivative of function FW (v), which is the critical speed vm . Then the peak braking force FWm can be obtained. Let the first derivative of function FW (v) be F0 W (v). The root of F0 W (v) is vm , which is the critical speed and is also the stationary point of FW (v). We take the velocity interval as [v1 , v2 ] in the velocity range from 0 m/s to 272 m/s to make the following formula. F0W (v1 ) · F0W (v2 ) 0

of eddy current density and magnetic density in the induction plate are obtained. Then, the analytical formula of the eddy current braking force is obtained by the Ampere force formula. The results of finite-element analysis confirm the validity of the analytical calculation. The methods of improving