Question 5.8.3: Set up the equations of motion for the actuator shown in Fig...

Set up a coordinate system with f_{fld} positive in the direction of increasing x as shown. As the motion is linear, we use the first-order system
of equation (5.86)

\frac{di(t)}{dt} = \frac{1}{L(x)} [V_{s}(t) – i(t) (R+R_{coil}) – \frac{∂Λ(i,x)}{∂x} v(t)]
\frac{dv(t)}{dt} =\frac{1}{m} F(i,x)
\frac{dx(t)}{dt} = v(t).                       (5.173)

To solve Eq. (5.173) we need expressions for L(x), ∂Λ(i, x)/∂x and F(i, x). As the independent variables are i and x, we work with the coenergy W_{fld}^{c}(i, x). From Eq. (5.139) we have

Eq. (5.139): W_{fld}^{c} = \left\{\begin{matrix} \frac{∂Λ(i,x)i}{2}=\frac{L(x)i^{2}}{2} \text{(linear motion)} \\  \frac{∂Λ(i,θ)i}{2}=\frac{L(θ)i^{2}}{2} \text{(rotational motion)}\end{matrix} \right.

W_{fld}^{c} ( i,x) = \frac{Λ(i,x)i}{2}.

= \frac{L(x)i^{2}}{2}.            (5.174)

Inductance: We determine the inductance using Eq. (3.75),

Eq. (3.75): L = \left\{\begin{matrix} \frac{1}{I^{2}} \int_{V} A⋅J dv \\ \frac{1}{I^{2}} \int_{V} B⋅H dv. \end{matrix} \right.  (linear system)

L = \frac{1}{i^{2}} \int_{V} B ⋅ H dv.             (5.175)

As the core and plunger have a high permeability, H is negligible in these elements and Eq. (5.175) reduces to an integration over the gap region and the nonmagnetic sleeve regions,

L = \frac{1}{i^{2}} \int_{V_{g}} B ⋅ H dv + \frac{2}{i^{2}}\int_{V_{s}} B⋅H dv,             (5.176)

where V_{g} and V_{s} are the volumes of the gap and sleeve regions, respectively. The factor of 2 in the second integral takes into account the identical integrations over each sleeve region. Notice that V_{g} = A_{g}x and V_{s} = A_{s}s, where A_{g} and A_{s} are the cross-sectional areas of the gap and sleeve, respectively. We need to determine H_{g} and H_{s}. To this end, apply Eq. (3.141) to the dotted path in Fig. 5.15. This gives

Eq. (3.141): \oint_{C} H⋅ dl = \sum\limits_{i=1}^{m} H_{i} l_{i} = I_{tot}.

H_{g} x + H_{s} s = ni.              (5.177)

Further, as there is no flux leakage Φ_{g} = 2Φ_{s},or

B_{g}A_{g} = 2B_{s} A_{s} .              (5.178)

Combining Eqs. (5.177) and (5.178) with B_{g} = μ_{0}H_{g} and B_{s} = μ_{0}H_{s} gives

H_{g} = \frac{ni}{(x+(A_{g}/2A_{s})s)},                            (5.179)

and

H_{s} = \frac{A_{g}}{2A_{s}} \frac{ni}{(x+(A_{g}/2A_{s})s)}.                            (5.180)

Next, substitute Eqs. (5.179) and (5.180) into Eq. (5.176), which gives the inductance

L(x) = \frac{μ_{0}n^{2}A_{g}}{(x+(A_{g}/2A_{s})s)}             (5.181)

Coenergy: Once we know the inductance, the coenergy is obtained easily from Eq. (5.159)

Eq. (5.159): W_{fld}^{c} ( i,x) = \frac{L(x)i^{2}}{2}.

W_{fld}^{c} (i, x) = \frac{μ_{0}i^{2}n^{2}A_{g}}{2[x+(A_{g}/2A_{s})s]}.            (5.182)

Use Eqs. (5.137) and (5.138) to obtain Λ and f_{fld}. We find that

Eq.(5.138):
f_{fld} = \frac{∂W_{fld}^{c} (i, x)}{∂i} (linear motion)
T_{fld} = \frac{∂W_{fld}^{c} (i, θ)}{∂θ} (rotational motion)

Λ = \frac{∂W_{fld}^{c} (i, x)}{∂i}

= \frac{μ_{0}in^{2}A_{g}}{[x+(A_{g}/2A_{s})s]}.            (5.183)

and

f_{fld} = \frac{∂W_{fld}^{c} (i, x)}{∂x}

= \frac{μ_{0}i^{2}n^{2}A_{g}}{2[x+(A_{g}/2A_{s})s]^{2}}.            (5.184)

Notice that f_{fld} is negative, which implies that the direction of f_{fld} is opposite to the direction of increasing x (i.e., opposite to the direction of increasing air gap). The total force F(i, x) is a sum of the forces due to the field and the spring,

F(i, x) = f_{fld}(i,x) + f_{s}(x),            (5.185)

where

  f_{s}(x) = -k(x -x_{0}).            (5.186)

Here, x_{0} is the initial position of the plunger, which is assumed to be the equilibrium position for the spring. When x > x_{0} (x  <  x_{0}), f_{s}(x) tends to move the mass into (out of) the circuit. Finally, substitute Eqs. (5.181), (5.184), (5.185), and (5.186) into Eq. (5.173) to obtain

\frac{di(t)}{dt} = \frac{(x+(A_{g}/2A_{s})s)}{μ_{0} n^{2}A_{g}}  \{ V_{s}(t) – i(t) (R+R_{coil}) + \frac{μ_{0}i n^{2}A_{g}}{[x+(A_{g}/2A_{s})s]^{2}} v(t) \}
\frac{dv(t)}{dt} = -\frac{1}{m} \{ \frac{μ_{0}πi^{2} n^{2}d}{2g} \frac{h^{2}}{(2h-x)^{2}}+ k(x-x_{0}) \}
\frac{dx(t)}{dt} = v(t).                        (5.187)

This nonlinear first-order system is solved subject to the initial conditions x(0) = x_{0}, v(0) = v_{0} and i(0) = i_{0}.