Question 5.8.2: Consider the cylindrical actuator shown in Fig. 5.14. Assume...
Consider the cylindrical actuator shown in Fig. 5.14. Assume that the magnetic circuit is linear with μ >> μ_{0} , and that there is no flux leakage. Set up the equations of motion for the actuator [2].
First, choose a convention in which a positive f_{fld} is in the direction of increasing x as shown. As the motion is linear, we use the first-order system of equations (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) (5.158)
\frac{dx(t)}{dt} = v(t).
To solve Eq. (5.158) 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}. (5.159)
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. (5.160)
As the core and plunger have a high permeability, the H-field in these elements is negligible, and Eq. (5.160) reduces to an integration over the nonmagnetic sleeve regions,
L = \frac{1}{i^{2}} \int_{V_{top}} B ⋅ H dv +\frac{1}{i^{2}} \int_{V_{bot}} B ⋅ H dv, (5.161)
where the subscripts top and bot refer to the top and bottom sleeve regions, respectively. Notice that V_{top} ≈ πdgh and V_{bot} ≈ πdg(h – x). We need to determine H_{top} and H_{bot}. To this end, apply Eq. (3.141) to either of the dotted paths in Fig. 5.14. This gives
Eq. (3.141): \oint_{C} H⋅ dl = \sum\limits_{i=1}^{m} H_{i} l_{i} = I_{tot}.
H_{top} g + H_{bot} g = ni. (5.162)
In addition, as there is no flux leakage Φ_{top} = Φ_{bot}, or
B_{top}A_{top} = B_{bot} A_{bot} , (5.163)
where A_{top} = πdh and A_{bot} = πd(h – x). From Eq. (5.163) and the fact that B_{top} = μ_{0} H_{top} and B_{bot} = μ_{0} H_{bot}, we find that
H_{bot} = H_{top} \frac{h}{(h – x)} (5.164)
Substitute Eq. (5.164) into Eq. (5.162) and obtain
H_{top}= \frac{ni}{g} \frac{(h – x)}{(2h – x)}. (5.165)
Finally, substitute Eqs. (5.164) and (5.165) into Eq. (5.161), which gives the inductance
L(x) = \frac{μ_{0} n^{2}πdh}{g} \frac{(h – x)}{(2h – x)} (5.166)
Coenergy: Once we know the inductance, the coenergy is easily obtained from Eq. (5.159),
W_{fld}^{c} ( i,x) = \frac{μ_{0}πi^{2} n^{2}dh}{2g} \frac{(h – x)}{(2h – x)}. (5.167)
We use Eqs. (5.137) and (5.138) to obtain ∧ and f_{fld}, and find that
Eq (5.138):
f_{fld}=\frac{∂W_{fld}^{c} ( i,x)}{∂i} (linear motion)
f_{fld}=\frac{∂W_{fld}^{c} ( i,θ)}{∂θ} (rotational motion)
Λ = \frac{∂W_{fld}^{c} ( i,x)}{∂i}
= \frac{μ_{0}πi n^{2}dh}{g} \frac{(h – x)}{(2h – x)}, (5.168)
and
f_{fld}= \frac{∂W_{fld}^{c} ( i,x)}{∂x}
= – \frac{μ_{0}πi^{2} n^{2}d}{2g} \frac{h^{2} }{(2h – x)^{2}}, (5.169)
As x < h, Eq. (5.169) 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 spring,
F(i, x) = f_{fld}(i,x) + f_{s}(x). (5.170)
where
f_{s}(x) = -k(x -x_{0}). (5.171)
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 solenoid. Finally, substitute Eqs. (5.168), (5.169), (5.170), and (5.171) into Eq. (5.158) and obtain
\frac{di(t)}{dt} = \frac{g}{μ_{0}π n^{2}dh} \frac{(2h-x)}{(h-x)} \{ V_{s}(t) – i(t) (R+R_{coil}) +[ \frac{μ_{0}πi n^{2}d}{g} \frac{h^{2}}{(2h-x)^{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). (5.172)
This nonlinear first-order system is valid for 0 ≤ x < h. It is solved subject to the initial conditions x(0) = x_{0}, v(0) = v_{0} and i(0) = i_{0}.