Question 5.9.1: Determine the equations of motion for the axial-field actuat...

This device is a moving coil actuator and is governed by Eqs. (5.102),

\frac{di(t)}{dt} = \frac{1}{L} [V_{s}(t) – i(t)(R+ R_{coil}) + \int_{coil} [(ω ×r) × B_{ext}] ]⋅dl

\frac{dω(t)}{dt} = \frac{1}{j_{m}} [i(t) \int_{coil} [r ×(dl × B_{ext})] ⋅ \hat{z} + T_{mech}(θ)]                   (5.188)

\frac{dθ(t)}{dt} = ω(t).

Here, L and R_{coil} are the inductance and resistance of the coil, and J_{m} is its moment of inertia above the z-axis. We need to evaluate the induced voltage and torque integrals in Eq. (5.188). We use a cylindrical coordinate system at rest with respect to magnets as shown in Fig. 5.16. Before we begin, we state some simplifying assumptions.

Assumptions: The first assumption is that the magnetic field due to the magnet is essentially uniform and constant across the coil, that is,

B_{mag} = \left\{\begin{matrix} -B_{mag} \hat{z} \qquad 0 ≤ θ≤ θ_{m}\\ B_{mag} \hat{z} \qquad -θ_{m} ≤ θ≤ 0, \end{matrix} \right.                    (5.189)

where –θ_{m} and +θ_{m} define the angular positions of the radial edges of the magnet. In reality, the field distribution above the magnet varies with height, and from point to point at a fixed height. However, an average value of B_{mag} that is adequate for this  analysis can be obtained using three-dimensional FEA, closed-form analysis [11], or measured data if the fabricated magnet is available. The second assumption is that each of the coil’s two radial edges rotates over a separate magnetic pole throughout the entire range of motion.

Induced voltage: We determine the voltage induced at the terminals of the coil by summing the voltage induced across its various segments. For reference purposes, Fig. 5.18 shows the electrical connection of a single turn. The coil has four segments: an upper and a lower segment, and two side segments (Fig. 5.19). We consider a single turn of the coil and evaluate the side (radial) segments first. We need to determine (ω × r) × B_{ext} along these segments. First, we know that

ω(t) = ω(t) \hat{z}.

Then from Fig. 5.19 we have

r = r \hat{r}.

Thus,

ω × r = ω(t) r \hat{Φ}    (side segments).

Note that B_{ext} is in the z-direction,

B_{ext} = \left\{\begin{matrix} -B_{ext} \hat{z} \qquad (left  side) \\ B_{ext} \hat{z} \qquad (right  side)  \end{matrix} \right.

Therefore,

(ω × r) × B_{ext} = \left\{\begin{matrix} -ω(t) r B_{ext} \hat{r} \qquad (left  side) \\ ω(t) r B_{ext} \hat{r} \qquad (right  side)  \end{matrix} \right.            (5.190)

We evaluate

\int_{coil} [(ω ×r) × B_{ext}] ⋅dl

along the left and right sides where

dl = dr \hat{r}            (side segments).

We obtain

\underbrace{\int [(ω ×r) × B_{ext}] ⋅dl}_{ left  side} = – \int_{R_{1}}^{R_{2}} ω(t) B_{ext} r dr

= – ω(t) B_{ext} \frac{R_{2}^{2} – R_{1}^{2}}{2},                      (5.191)

and

\underbrace{\int [(ω ×r) × B_{ext}] ⋅dl}_{ right  side} =  \int_{R_{2}}^{R_{1}} ω(t) B_{ext} r dr

= – ω(t) B_{ext} \frac{R_{2}^{2} – R_{1}^{2}}{2},                      (5.192)

Now we consider the top and bottom segments. For these we have

(ω ×r) × B_{ext}  α  \hat{r} .

Because dl α \hat{Φ} , it follows that

 [(ω ×r) × B_{ext}] ⋅ dl α  \hat{r}⋅ \hat{Φ}  =0            (top/bottom segments).

Therefore,

\underbrace{\int [(ω ×r) × B_{ext}] ⋅dl}_{ top/bottom   segments} =  0,

which shows that these segments do not contribute to the induced voltage. Finally, add Eqs. (5.191) and (5.192), and multiply by n (the number of turns) to obtain an expression for the induced voltage

\int_{coil} [(ω ×r) × B_{ext}] ⋅dl = – ω(t) n B_{ext} (R_{2}^{2} – R_{1}^{2}).            (5.193)

Torque: Only the side segments of the coil contribute to the torque. We make a brief argument to show that the upper and lower segments make no contribution. Consider the torque on an element of the upper or lower segment,

dT_{z} = [r × (dl ×B_{ext})] ⋅ \hat{z},

where r = r \hat{r} and dl α \hat{Φ} . Notice that

dl × B_{ext} ∝ \hat{Φ}  ×B_{ext} ∝ \hat{r}.

This gives

dT_{z} ∝ [\hat{r} × \hat{r}] ⋅ \hat{z} = 0,

which shows that the upper and lower segments generate no torque. Now consider the left- and right side segments. For these, we have dl = dr \hat{r}, and

dl × B_{ext} = \left\{\begin{matrix}  B_{ext} dr\hat{Φ} \qquad (left  side) \\ – B_{ext} dr \hat{Φ} \qquad (right  side)  \end{matrix} \right.

Therefore,

r × (dl ×B_{ext}) = \left\{\begin{matrix}  B_{ext}r dr\hat{z} \qquad (left  side) \\ – B_{ext} r dr \hat{z} \qquad (right  side)  \end{matrix} \right.                   (5.194)

We evaluate

\int [r × (dl ×B_{ext})] ⋅ \hat{z}

along the left- and right side segments with the integration in the direction of the current flow. This gives

\underbrace{\int [r × (dl ×B_{ext})] ⋅ \hat{z}}_{ left  side} =  \int_{R_{1}}^{R_{2}}  B_{ext} r dr

=  B_{ext} \frac{R_{2}^{2} – R_{1}^{2}}{2},                      (5.195)

and

\underbrace{\int [r × (dl ×B_{ext})] ⋅ \hat{z}}_{ right  side} = – \int_{R_{2}}^{R_{1}}  B_{ext} r dr

=  B_{ext} \frac{R_{2}^{2} – R_{1}^{2}}{2},                      (5.196)

The torque on the coil is ni times the sum of Eqs. (5.195) and (5.196),

\int_{coil} [r × (dl ×B_{ext})] ⋅ \hat{z} = ni B_{ext} (R_{2}^{2} – R_{1}^{2}).                 (5.197)

Equations of motion: We are finally ready to write the equations of motion. Substitute the integrals (5.193) and (5.197) into Eq. (5.188) and obtain

\frac{di(t)}{dt} = \frac{1}{L} [V_{s}(t) – i(t)(R+ R_{coil}) – n B_{ext} (R_{2}^{2} – R_{1}^{2}) ω(t)]

\frac{dω(t)}{dt} = \frac{1}{j_{m}} [i(t) n B_{ext} (R_{2}^{2} – R_{1}^{2})  + T_{mech}(θ)]

\frac{dθ(t)}{dt} = ω(t).                 (5.198)

These equations are solved subject to the initial conditions of Eq. (5.89). Finally, from Eq. (5.198) we see that the electrical constant K_{e} equals the torque constant K_{t} (see Eq. (5.100)). Specifically,

Eq. (5.100): Rotational Motion
V_{s}(t) = i(t)(R+ R_{coil}) + L \frac{di(t)}{dt}+ K_{e} ω(t)
j_{m}\frac{d^{2}θ(t)}{dt^{2}} = i(t) K_{t} + T_{mech}(θ)

K_{e} =K_{t}= n B_{ext} (R_{2}^{2} – R_{1}^{2}).

Calculations: We apply Eq. (5.198) to an actuator with the mechanical mechanism shown in Fig. 5.17. Here, the mechanical torque is supplied by the spring. To perform the analysis we need expressions for the coil inductance L and the mechanical restoring torque T_{mech} (θ).

Inductance: The inductance of the coil can be estimated by considering a short circular coil with a height h_{c} equal to that of the actuator coil, and with a radius r_{c},

r_{c} = \frac{R_{2} – R_{1}}{2}.          (5.199)

To first order, the flux through the center of the coil (through each turn) is

Φ(i) ≈ \frac{μ_{0}inπr_{c}^{2}}{(h_{c}^{2}+ (2r_{c})^{2})^{1/2}} .                (5.200)

Therefore, the flux linkage is Λ(i) = nΦ(i). Recall that the inductance is

L = \frac{dΛ(i)}{di}                         (5.201)

which gives

L ≈ \frac{μ_{0}n^{2}πr_{c}^{2}}{(h_{c}^{2}+ (2r_{c})^{2})^{1/2}} .                (5.202)

This approximation is appropriate if the mean circumferential arc subtended by the coil is approximately equal to R_{2} – R_{1}, which is the case for many practical designs. However, if this is not the case the inductance can be determined empirically, or calculated using three-dimensional FEA.

Mechanical Torque: The mechanical torque is provided by the spring mechanism shown in Fig. 5.17. The restoring force is

F(Y) = k_{s} (Y- Y_{0})+ F_{0},

where k_{s} is the spring constant, Y_{o} is the initial spring length, F_{o} is the force when Y = Y_{o}, and Y is the stretched length. The length Y_{o} is defined by the angle β. From geometry we have

γ = \sqrt{X^{2}+D^{2} – 2XD cos(β+θ)}

The restoring torque is given by

T_{mech}(θ) = DF(Y(θ)) cos(β+θ).               (5.203)

Analysis: We substitute Eqs. (5.202) and (5.203) into Eq. (5.198) and perform the analysis. The applied voltage V_{s}(t) is shown in Fig. 5.20 and the remaining parameters are as follows:

B_{mag} = 0.5 T
R + R_{coil} = 3 Ω
n = 100 turns
R_{1} =10 mm
R_{2} =15 mm
D = 10 mm
β = 45°
h_{c} =1.5 mm
k_{s} =1.5 N/mm
j_{m} = 6× 10^{-7}  kg⋅m^{2}.

The moment of inertia is computed using Eq. (5.77), which gives

Eq. (5.77): j_{m} = \int_{V} (x^{2} + y^{2}) ρ_{mass} dv

j_{m} = \frac{2}{3}(R_{2}^{3} – R_{1}^{3})ρ_{mass} A_{wire}+ ΔΦ_{coil}(R_{2}^{3} – R_{1}^{3})ρ_{mass} A_{wire},

where ρ_{mass} and A_{wire} are the density and area of the wire and ΔΦ_{coil} is the angular span of the coil (left- to right side) in radians. The response of the actuator is computed for the initial conditions i(0) = 0A, θ(0) = 0°, and ω(0) = 0 rad/sec. The current profile i(t) is shown in Fig. 5.21. The rotation angle θ(t) and angular velocity ω(t) are shown in Fig. 5.22. Notice that the angular velocity (t) passes through zero and turns negative as the angular position θ(t) peaks at approximately 6.5 ms.