In the specific case of FLIP, the conveying medium is incompressible fluid with constant conductivity, the fluid velocity is much less than the speed of light, and the electromagnetic frequency is below 1000 Hz (the displacement current can be completely ignored). Therefore, Maxwell’s equations and its constitutive equations can be modified into the following form: ∇ × H = J ∇ × E = − ∂ B ∂ t ∇ ⋅ B = 0 B = μ H J = σ E + V × B (1) Where H is the magnetic field intensity (A/m), J is the electric current density (A/m²), E is the electric field intensity (V/m), B is the magnetic induction intensity (T), u is the permeability of the medium, σ is the conductivity of the medium (S/m), V is the fluid velocity (m/s).

Based on Equation 1, the magnetic vector potential A is introduced. And add the following two equations: B = ∇ × A E = − ∂ A ∂ t (2)

The longitudinal end effect is directly related to the performance and characteristics of FLIP, and theoretical analysis is carried out below. In order to simplify the mode, some basic assumptions are introduced as follows.

1) The permeability of stator core is infinite, and the conductivity is zero.

The FLIP can be regarded as the unfolded ALIP, so the one-dimensional field model of FLIP considering longitudinal end effect is similar to ALIP (Zhang et al., 2022). Figure 2 shows a bilateral FLIP model considering longitudinal end effects alone. The whole model is divided into three calculation regions which includes entrance region I, coupling region (effective region) II, and exit region III. Based on the principle of equality of magnetomotive force (Ku et al., 1985), the primary magnetic potential is replaced by a surface current layer j _s (x, t). The upstream region of the inlet and the downstream region of the outlet are also represented by the virtual current layers j _k1 and j _k3 in the plural form. When the surface current layer j _s (x, t), j _k1 and j _k3 on one side are considered as zero, all the formulas are applicable to FLIPs in the form of single-sided winding arrangement.

The traveling-wave current layer of the three regions is defined as j k 1 = j σ 13 E z 1   x < 0 j s = J smax e j ω t − k x 0 < x < L s j k 3 = j σ 13 E z 3   x > L s (3) Where j is the imaginary symbol. k = π/τ is the wave number, τ is the pole pitch (m). J _smax is the amplitude of the traveling wave current layer (A/m). J _smax = 2 m 1 N 1 k w 1 I 1 , m ₁ is the number of phases, N ₁ is the winding per pole per phase series parameter, k _w1 is the fundamental winding factor, I ₁ is the phase current of each branch of the winding (A), p is the logarithm of the poles. ω is the angular velocity (rad/s), L _s is the effective region length (m), E _z1 and E _z3 represent the electric field intensities in the corresponding regions. σ ₁₃ is the virtual conductivity in regions I and III. σ 13 = 0.73 2 / g e ω μ 0 = k f 2 g e / ω μ 0 Where, g _e is the half-thickness of the electromagnetic air gap (m), u ₀ is the vacuum permeability,k _f = 0.73/g _e.

As shown in Figure 2, the reference coordinates are fixed at the primary. According to assumptions (1) and (5), along rectangular loop paths 1, 2, and 3 in regions I, II, and III, it is obtained by applying Abe’s loop theorem: g e u 0 ∂ ∂ x B y 1 = j f 1 z + j k 1 + j w 1 z   x < 0 (4) g e u 0 ∂ ∂ x B y 2 = j f 2 z + j s + j w 2 z   0 < x < L s (5) g e u 0 ∂ ∂ x B y 3 = j f 3 z + j k 3 + j w 3 z   x > L s (6) Where B _y1, B _y2, and B _y3 are the magnetic field intensities in regions I, II, III, j _f1z, j _f2z, j _f3z are the z components of the line induction current density of liquid metal in the corresponding regions, respectively. j _w1z, j _w2z, j _w3z are the z components of the line induction current density of the wall in the corresponding regions, respectively.

The magnetic vector potential A of each region can be defined as A z 1 = A 11 x e j w t   x < 0 A z 2 = A 22 x e j w t   0 < x < L s A z 3 = A 33 x e j w t   x > L s (7)

From Equations 1, 2 and 7, Formulas 4–6 can be written separately as g e μ 0 d 2 A 11 d x 2 − σ f d f ν f d A 11 d x − j ω σ f d f 1 + k w − ω σ 13 A 11 = 0 (8) g e μ 0 d 2 A 33 d x 2 − σ f d f ν f d A 33 d x − j ω σ f d f 1 + k w − ω σ 13 A 33 = 0 (9) g e μ 0 d 2 A 22 d x 2 − σ f d f ν f d A 22 d x − j ω σ f d f 1 + k w A 22 = − J s ⁡ max e j k x (10) Where, k _w = σ _f d _f/σ _w d _w is the wall eddy current coefficient, which represents the eddy current effect of the conductive pump ditch wall. d _w is the pump channel wall thickness (m). σ _f is the fluid conductivity (S/m), 2d _f is the pump channel thickness (m), ν _f is the fluid velocity (m/s).

The solutions of Equations 8–10 can be written as A 11 x = c 11 e r k 1 x   x < 0 A 22 x = c t e − j k x + c 21 e r c 21 x + c 22 e r c 22 x   0 < x < L s A 33 x = c 33 e r k 3 x   x > L s (11) Where c ₁₁, c ₂₁, c ₂₂, c ₃₃ are the integration constants depending on the boundary conditions. G is the quality factor of the pump, which reflects the ability of pump to convert one type of energy into another. R _m is the magnetic Reynolds number, which characterizes the dynamics of the MHD. s is the slip, s = (v _s-v _f)/v _s. v is the traveling wave velocity, v _s = 2pτ.

From Equations 1, 7 and 11 the distribution of magnetic field B _y1, B _y2, B _y3 and electric field E _z1, E _z2, E _z3 in the three regions can be written as B y 1 = − r k 1 c 11 e r k 1   x < 0 B y 2 = B y 2 c t + B y 2 c 1 + B y 2 c 2   0 < x < L s B y 3 = − r k 1 c 11 e r k 1   x > L s (12) Where, B _y2ct = jkc _te^j(ωt-kx), B _y2c1 = -c ₂₁e ^rc21x+jωt , B _y2c2 = -c ₂₂e ^rc22x+jωt represent the no-end effect component, the inlet end effect component and the outlet end effect component, respectively. E z 1 = − j ω c 11 e r k 1 x + j ω t   x < 0 E z 2 = − j ω c t e j ω t − k x − j ω c 21 e r c 21 x + j ω t − j ω c 22 e r c 22 x + j ω t   0 < x < L s E z 3 = − j ω c 12 e r k 3 x + j ω t   x > L s (13)

On the boundary line between regions I, II, and III (x = 0, x = L _s), tangential magnetic field continuity and tangential electric field continuity should be satisfied. B y 1 | x = 0 = B y 2 | x = 0   B y 2 | x = L s = B y 3 | x = L s E z 1 | x = 0 = E z 2 | x = 0   E z 2 | x = L s = E z 3 | x = L s (14)

From Equations 12–14, the specific expressions for c ₁₁, c ₂₁, c ₂₂, c ₃₃ can be solved. The detailed expressions are listed in the Supplementary Appendix section.

The inner core of ALIP is a cylinder, and the outer core is a ring, so the iron core of ALIP is broken only in the length direction. However, the core of FLIP is limited in the length direction and width direction, so there are both longitudinal end effect and transverse end effect, and the theoretical analysis and characteristic calculation of FLIP are more complex. Figure 3 shows the analytical model of FLIP considering only the transverse end effect.

The whole model is divided into liquid metal region I, pump trench wall region II and short circuit strip region III. The definition of the equivalent current layer js is consistent with the longitudinal end-effect coupling region. Ignoring the influence of the longitudinal end effect, the basic assumption of the model that considers the transverse end effect is basically consistent with the longitudinal end effect.

According to Assumption (4) and (5), the form of the magnetic field B in coupling region is defined as B y 1 = B 1 z e j   ω t − k x (15) Where B _y1 is the y components of air gap flux density in the region I.

Along rectangular loop paths 1 in regions I, it is obtained by applying Abe’s loop theorem g e μ 0 ∂ ∂ z B y 1 = − j f x − j w 1 x (16)

Similarly, it can be obtained in the longitudinal section: g e μ 0 ∂ ∂ z B y 1 = j fz + j w 1 z − j s (17) Where j _fx, j _fz are distributed to represent the x-direction and z-direction components of the current density of the liquid metal wire in the upper half of region I, j _w1x and j _w1z are distributed to represent the x-direction and z-direction components of the half-line current density of the pump trench pipeline in region I.

Simultaneous Equations 1, 16, 17, which are obtained after collation: ∂ 2 ∂ x 2 B y 1 + ∂ 2 ∂ z 2 B y 1 − μ 0 σ fs V x δ ∂ ∂ x B y 1 − μ 0 σ fs δ ∂ ∂ t B y 1 − μ 0 σ ws δ ∂ ∂ t B y 1 = − μ 0 g e ∂ ∂ x j s (18)

Where, σ _fs = σ _f d _f is Surface conductivity.

From Equations 3, 15, Formula 18 can be written as ∂ 2 ∂ z 2 B 1 z − k 2 + j ω μ 0 σ fs δ s + k w І B 1 z = j k μ 0 g e J smax (19) Where k _wⅠ = σ _w d _w/σ _f d _f is the eddy current effect coefficient in region I. Similarly, in regions II and III can be obtained as k w π = σ w d w + d f / σ f d f , k w ш = σ s d s / σ f d f

In the fluid region, B ₁(z) is an even function with respect to z. Thus, the full solution form of Equation 19 is given as Equation 20: B y 1 x , z , t = B 1 ⁡ cosh α 1 z + c t 1 e j ω t − kx (20) Where B ₁ is the integration constant to be determined. α 1 2 = k 2 + j ω μ 0 σ fs g e s + k w І c t 1 = − j μ 0 J smax k g e 1 + j G s + k w І

From Equations 17, 18, 20, the z and x components of the line current density of liquid metal (j _fz, j _fx) and wall (j _w1z, j _w1x) in region I can be written as, respectively j fz = 1 1 + k w Ι ε 1 B 1 ⁡ cosh α 1 z + c j 1 e j ω t − k x j fx = − g e α 1 μ 0 1 + k w Ι B 1 ⁡ sinh α 1 z e j ω t − k x j w 1 z = 1 1 + k w Ι ε w 1 B 1 ⁡ cosh α 1 z + c w 1 e j ω t − k x j w 1 x = − k w 1 g e α 1 μ 0 1 + k w Ι B 1 ⁡ sinh α 1 z e j ω t − k x

Combining Equation 20 and the fifth equation in maxwell’s Equation 1, the z components and x components of the electric field intensity in region I are obtained in the following forms: E z 1 = 1 σ fs k w Ι 1 + k w Ι ε w 1 B 1 ⁡ cosh α 1 z + c w 1 e j ω t − k x E x 1 = − g e α 1 μ 0 σ fs 1 + k w Ι B 1 ⁡ sinh α 1 z e j ω t − k x

Since the walls of the two sides of the pump communication channel and short circuit strip are located outside the core, the induced magnetic field generated by the current can be ignored. The magnetic field in region II is considered to be zero, then: ∇ × E = 0 (21)

Curl the fifth equation in maxwell’s Equation 1 and then combine it with Equation 21: ∂ ∂ x j w 2 z − ∂ ∂ z j w 2 x = 0 (22) Where, j _w2z, j _w2x are the z components and x components of the current density in the pipe line of the pump trench wall in the upper half of region II, respectively.

According to the current continuity theorem: ∂ ∂ x j w 2 x + ∂ ∂ z j w 2 z = 0 (23)

Take the derivative of Equations 22, 23 with respect to x and y, respectively, and add them together: ∂ 2 ∂ x 2 j w 2 z + ∂ 2 ∂ z 2 j w 2 z = 0 (24)

The general solution of Equation 24 is given as Equation 25: j w 2 z = A 1 e k z + A 2 e − k z e j ω t − k x (25) Where A ₁ and A ₂ are the integration constants to be determined.

From Equations 23, 25, j _w2x can be written as j w 2 x = j A 2 e − k z − A 1 e k z e j ω t − k x (26)

Combining Equations 25, 26, and the fifth equation in maxwell’s Equation 1, the z components and x components of the electric field intensity in region II are obtained in the following forms: E z 2 = 1 σ fs k w π A 1 e k z + A 2 e − k z e j ω t − k x E x 2 = j σ fs k w π A 2 e − k z − A 1 e k z e j ω t − k x

There is also no magnetic field in region III, so the resolution process is similar to that in region II. Thus, the expressions for each component of the induced eddy current and electric field in region III can be written as j s 3 z = A 3 e k z + A 4 e − k z e j ω t − k x j s 3 x = j A 4 e − k z − A 3 e k z e j ω t − k x E z 3 = 1 σ fs k w ш A 3 e k z + A 4 e − k z e j ω t − k x E x 3 = j σ fs k w ш A 4 e − k z − A 3 e k z e j ω t − k x Where j _s3z, j _s3x are the z components and x components of the current density of the short-circuiting bar in the upper half of region III, respectively. E _3z, E _3x are the z components and x components of the electric field in region III, respectively.

The complex integration constants B ₁, A ₁, A ₂, A ₃, and A ₄ in the solution process need to be determined through boundary conditions. Tangential electric field continuity and normal current continuity should be satisfied on the boundary lines of the pump groove wall in contact with liquid metal and short circuit strip (z = c and z = c + t _w), respectively. In addition, it should be satisfied that the normal current at the outer boundary of the short circuit strip (z = c + t _w + t _s) is zero. The form of the boundary conditions can be written as E x 1 | z = c = E x 2 | z = c   j fz + j w 1 z | z = c = j w 2 z | z = c E x 2 | z = c + t w = E x 3 | z = c + t w     j w 2 z | z = c + t w = j s 3 z | z = c + t w j s 3 z | z = c + t w + t s = 0

The detailed expressions for B ₁, A ₁, A ₂, A ₃, and A ₄ are listed in the Supplementary Appendix section.

The mathematical analysis of longitudinal dynamic end effect and second transverse end effect of FLIP are carried out in Section 3.1, 3.2, respectively. In this section, the influence of longitudinal and transverse end effect on the air gap magnetic field, head and energy conversion efficiency will be further analyzed based on a practical FLIP (Bluhm et al., 1964).

In the literature, the FLIP adopts the arrangement of double-sided core and single-sided winding. This arrangement of pumps is often used in some high temperature application environments because of the low requirement for heat insulation. The operating temperature is 673.15 K and the voltage source is used for power supply. The specific parameters of the FLIP are shown in Table 1.

Based on the analytical model in Section 3.1, the total complex power S _i1 transferred from the primary to the secondary and the air gap can be divided into three parts. The complex power S _i2 generated by the equivalent current layer j _s in the coupling area II, the complex power S _ik1 and S _ik3 generated by the equivalent virtual current layer in the upstream area I of the inlet and the downstream area III of the outlet. The three-part complex power can be written as S i 2 = 2 c ∫ 0 L s − j s * E z 2 d x = P i 2 + j Q i 2 S ik 1 = 2 c ∫ − L 1 0 − j k 1 * E z 1 d x = P ik 1 + j Q ik 1 S ik 3 = 2 c ∫ L s L s + L 3 − j k 3 * E z 3 d x

Therefore, the total complex power S _i1 can be written as S i 1 = S i 2 + S ik 1 + S ik 3 = P i 1 + j Q i 1 (28)

The secondary induced electromotive force (EMF) E ₁ can be calculated using Equation 28. E 1 · = − S i 1 / m 1 I 1 ·

Therefore, the excitation reactance X́_m considering the longitudinal end effect can be calculated using the total reactive power of the primary input secondary: X m ′ = m 1 E ˙ 1 2 Q i 1 (29)

From X _m in Equation 27 and X ^́ _m in Equation 29, the correction factor K _x for the excitation reactance considering the longitudinal end effect can be written as K x = X m ′ / X m

The Joule heat P _w of the upper half of the pump ditch pipe can be calculated using the wall eddy current j _w: P w = c σ s k w Ι ∫ − L 1 0 j w 1 * × j w 1 d x + ∫ 0 L s j w 2 * × j w 2 d x + ∫ L s L s + L 3 j w 3 * × j w 3 d x (30)

The wall resistance R w ′ and correction factor K _w(s) considering the longitudinal end effect can be written as R w ′ = m 1 E 1 2 / P w , K w = R w ′ / R w

The power of liquid metal is mainly divided into mechanical power P _mec and thermal power P _rf. The calculation of P _rf is similar to the Joule heat of the wall in Equation 30. P rf = c σ s k w Ι ∫ − L 1 0 j f 1 * × j f 1 d x + ∫ 0 L s j f 2 * × j f 2 d x + ∫ L s L s + L 3 j f 3 * × j f 3 d x (31)

In the pump communication channel, the Lorentz force generated by the interaction of the magnetic field and the induced eddy current is the only driving force for the liquid metal. The time-averaged electromagnetic force density f _em(x) can be written as f em x = − Re j * × B 2 d f = f em 1 x + f em 2 x + f em 3 x Where, f _em1(x), f _em2(x), f _em3(x) are the time-averaged electromagnetic force densities in regions I, II, III, respectively. f em 1 x = − R e j f 1 ∗ × B y 1 2 d f   x < 0 f em 2 x = − R e j f 2 ∗ × B y 1 2 d f   0 < x < L s f em 3 x = − R e j f 3 ∗ × B y 3 2 d f   x > L s (32)

The electromagnetic pressure difference between the inlet and outlet ΔP _em can be obtained by integrating the electromagnetic force along the flow direction x: Δ P em = ∫ l f em x d x (33)

The mechanical power P _mec of the planar electromagnetic pump considering the longitudinal end effect can be written as P mec = Δ P em · Q (34)

The total impedance of the liquid metal series branch can be written as R f / s ′ = m 1 E 1 2 P mec / 2 + P rf (35)

The current I _f of the liquid metal branch can be obtained according to Ohm’s law: I f = E 1 · / R f / s ′ (36)

From Equations 31, 34–36, the equivalent resistance Ŕ _fmec, Ŕ _fΔ and correction factors K _fmec(s), K _fΔ(s) for the thermal and mechanical power of the liquid metal considering the longitudinal end effect can be written as, respectively R fmec ′ = P mec / 2 m 1 I f 2 , R f Δ ′ = P rf m 1 I f 2 K fmec s = R fmec ′ / R fmec , K f Δ s = R f Δ ′ / R f Δ

Let the magnetic flux at each pole be ϕ(t), which can be written in the following form: ϕ t = ∫ − c c ∫ 0 τ B y 1 x , z , t d x d z

The instantaneous value e 1 t of the single-side primary per-phase potential can be written as e 1 t = − W 1 k w 1 d d t ϕ t = − 2 E 1 e j ω t (37)

The total induced EMF E ₁ in the non-magnetically conductive air gap in Equation 37 can be written as E 1 = 2 2 j ω W 1 k w 1 ϕ t

Therefore, the complex power S _i1 of the secondary considering the transverse end effect can be obtained as S i 1 = m I 1 − E 1 2 3 = P i 2 + j Q i 2

Similarly, the excitation reactance X˝_m and its correction factor C _x considering the transverse end effect can be written as X m ″ = m 1 E 1 2 / Q i 2 , C x = X m ″ / X m

The Joule heat generated by the eddy current effect of the short circuit strip is included in the wall Joule heat, and the total Joule heat P _w of the upper half of the pipe can be written as P w = P w 1 + P w 2 + P s 3 Where, P w 1 = 1 2 σ fs k w 1 ∫ 0 L s ∫ − c c j w 1 z * × j w 1 z + j w 1 x * × j w 1 x d z d x P w 2 = 1 σ fs k w 2 ∫ 0 L s ∫ c c + t w j w 2 z * × j w 2 z + j w 2 x * × j w 2 x d z d x P s 3 = 1 σ fs k w 3 ∫ 0 L s ∫ c + t w c + t w + t s j s 3 z * × j s 3 z + j s 3 x * × j s 3 x d z d x

The wall resistance R˝_w considering transverse end effects and the corresponding correction factor C _w(s) can be written as R w ″ = m 1 E 1 2 / P w , C w s = R w ″ / R w

The liquid metal induced eddy Joule heating P _rf in the upper half of the inner region I can be expressed as P rf = 1 2 σ fs ∫ 0 L s ∫ − c c j fx * × j fx + j fz * × j fz ) d z d x

The electromagnetic differential pressure ΔP _em and mechanical power P _mec of the planar electromagnetic pump can be calculated as Δ P em = − L s 2 A s ∫ − c c j fz × B y 1 * d z P mec = Δ P em · Q

The liquid metal branch resistance (R _f/s)˝ and branch current RMS |I _f| considering the lateral end effect can be expressed as R f / s ″ = m 1 E 1 2 P mec + P rf I f = E 1 / R f / s ″

Similar to the longitudinal end effect, the resistance R˝_fmec, R˝_fΔ corresponding to the mechanical and thermal power of the liquid metal considering the transverse end effect, and its corresponding correction factor C _fmec(s), C _fΔ(s) can be expressed as R fmec ″ = P fmec / 2 m 1 I f 2 , R f Δ ″ = P rf m 1 I f 2 C fmec s = R fmec ″ / R fmec , C f Δ s = R f Δ ″ / R f Δ

Based on the longitudinal end effect correction factor K(s) and transverse end effect correction factor C(s) for each impedance obtained in Sections 4.1, 4.2, each impedance value considering the two end effects can be expressed as r w = K w s C w s R w x m = K x s C x s R w r fmec = K fmec s C fmec s R fmec r f Δ = K f Δ s C f Δ s R f Δ

The excitation branch and the secondary total impedance can be expressed as Z 2 = 1 1 j x m + 1 r w + 1 r mec + r f Δ

The total voltage segment input impedance Z _i can be written as Z i = R 1 + jX 1 + Z 2

If the voltage is known, the stator phase current I ˙ 1 and the secondary induced electromotive force E ˙ 1   can be obtained as I 1 · = U 1 · / Z i − E 1 · = I 1 · Z 2 (38)

The branch currents I ˙ m , I ˙ w , I ˙ f can be written as, respectively I m · = − E 1 · / j x m I w · = − E 1 · / r w I f · = − E 1 · / r fmec + r f Δ (39)

The total complex power S _i of the single-side winding input can be written as S i = m 1 I 1 · 2 Z i = P i + jQ i Where, P _i and Q _i are the total active and reactive power, respectively. P i = m 1 I 1 · 2 · Re Z i = P r 1 + P rw + P rf + P mec (40) Q i = m 1 I 1 · 2 · Im Z i Where, P _r1 is the Joule heating of the one-sided winding, P _rw is the Joule heat of the upper half of the wall and the short-circuit strip, P _rf is the liquid metal Joule heat of the upper half, P _mec is half of the total mechanical power of the liquid metal.

The current of each branch has been derived in Equations 38, 39, so the specific expression of the power of each part can be easily derived according to the equivalent circuit model. P r 1 = m 1 I · 1 2 r 1 , P rw = m 1 I · w 2 r w P rf = m 1 I · f 2 r f Δ , P mec = m 1 I · f 2 r fmec (41)

From the mechanical power P _mec in Equation 41, the electromagnetic pressure difference ΔP _em of the electromagnetic pump can be obtained as Δ P em = 2 P mec / Q (42)

The Darcy-Weisbach formula is adopted for calculating the frictional loss ΔP _m of liquid metal in rectangular pump groove. When the flow regime of the liquid metal is turbulent (Re > 2,320), the explicit form of the Colebrook-White equation (Fang et al., 2011; Hafsi, 2021) is used to solve the resistance coefficient λ. Δ P m = λ l d ρ f v f 2 2 (43) λ = 64 / Re  Re ≤ 2320 λ = 1.613 ln 0.234 Δ D h 1.1007 − 60.525 R e 1.1105 + 56.291 R e 1.0712 − 2   3000 < Re < 10 8 Where, l is the length of the pipe (m), d is the feature size (m), ρ _f is the fluid density (kg/m³), v _f is the average velocity in the fluid tube (m/s), Re = ν _f D _h/ν is the Reynolds number of the liquid metal, D _h is the equivalent diameter of the rectangular flow channel (m), ν is the kinematic viscosity (m²/s), Δ is the absolute roughness (m).

From Equations 42, 43, the head of the planar electromagnetic pump ΔP _d can be obtained as Δ P d = Δ P em − Δ P m (44)

The power factor cosφ can be written as cos ⁡ φ = Re Z i / Z i

From Equations 40, 44, the energy conversion efficiency η of the FLIP can be calculated as η = Δ P d · Q 2 P i

In this section, the accuracy and applicability of all equations of the established equivalent circuit model is checked by means of two sets of experimental data: a) Variable voltage condition and b) Variable temperature condition.