Electrical Info: Articles | Reluctance Motors

The salient pole synchronous machine without any field or extiting winding runs at synchronous speed due to the reluctance torque and is termed Reluctance Motor. These motors find application in automatic control systems, telemetering devices and systems, signalisation circuits, sound film systems etc. requiring synchronised or simultaneous movements. Simplicity of construction is the main attraction of such motors. They are generally required in fractional kW sizes having ratings from a few watts to several hundreds of watts.

Reluctance motors are started automatically by the induction motor torque developed by the currents induced in the solid rotor of the motor. Pull into synchronisation takes place due to the reluctance torque which is developed due to the tendency of the rotating magnetic field to retain the running rotor in minimum reluctance position to field flux. The condition is met when the rotor runs in step with the field and when the pole-axis coincides with the magnetic flux axis. As soon as the rotor is pulled into synchronisation, the relative speeed between the rotating field and the rotor core vanishes and the 'Induction Motor' operation ceases. When the motor is loaded, the axis of the poles shift away from the stator poles (a rotationg field axis) towards lagging as shown in figure b.

In a machine with cylindrical rotor, as shown in figure c, the reluctance torque is not created because of the lack of any difference in position of the rotor relative to the stator field.

The magnetic field of a reluctance motor is produced due to the armature reaction magnetic flux and therefore the motor draws the reactive (lagging) current required for producing the magnetic field from the power circuit and consequently operates with low power factor.

The analysis of the motor can be done following the method of analysis of salient pole synchronous motor with zero excitation. Phasor diagram of both the machines are given below:

From the above phasor diagram, we see that—
I_a = I_d + I_q
U_t = U_d + U_q
U_t = U_f + jI_dX_d + jI_qX_q + I_aR_a (for synchronous motor)
U_t = jI_dX_d + jI_qX_q + I_aR_a (for reluctance motor)

and with the positions of phasors as in figure above, we have—
I_d = I_d + j0
I_q = 0 + jI_q
U_d = −U_d + j0
U_q = 0 + jU_q

So we get—
U_t cos δ = U_q = U_f + I_dX_d + I_qR_a (for synchronous motor)
U_t cos δ = U_q = I_dX_d + I_qR_a (for reluctance motor)

U_t sin δ = U_d = I_qX_q − I_dR_a

Now to get power equations, we have—
Complex power S = P + jQ = U_tI_a^* = (U_d + U_q) (I_d + I_q)^*
= (– U_d + jU_q) (I_d + jI_q)^* = (– U_d + jU_q) (I_d − jI_q)
= (– U_dI_d + U_qI_q) + j(U_qI_d + U_dI_q)

So, P = – U_dI_d + U_qI_q and Q = U_qI_d + U_dI_q

Using above equations and negecting armarure resistance (which is usually small, we get for synchronous motor—

$P = \frac{U_{t} U_{f}}{X_{d}} \sin δ + \frac{U_{t}^{2}}{2} (\frac{1}{X_{q}} - \frac{1}{X_{d}}) \sin 2 δ$

Q = - \frac{U_{t} U_{f}}{X_{d}} \cos δ + \frac{U_{t}^{2}}{2} (\frac{1}{X_{d}} - \frac{1}{X_{q}}) \cos 2 δ + \frac{U_{t}^{2}}{2} (\frac{1}{X_{d}} + \frac{1}{X_{q}})

and for reluctance motor—

P = \frac{U_{t}^{2}}{2} (\frac{1}{X_{q}} - \frac{1}{X_{d}}) \sin 2 δ

Q = \frac{U_{t}^{2}}{2} (\frac{1}{X_{d}} - \frac{1}{X_{q}}) \cos 2 δ + \frac{U_{t}^{2}}{2} (\frac{1}{X_{d}} + \frac{1}{X_{q}})

So the maximum power of reluctance motor occurs at δ = 45° and maximum power is given by—

P_{max} = \frac{U_{t}^{2}}{2} (\frac{1}{X_{q}} - \frac{1}{X_{d}})

Power angle curves of both the motors are given below—

The maximum power of a reluctance motor increases with the increase in X_d/X_q. Measures are taken to increase the ratio as large as possible and for this purpose, rotor may be assembled at steel plants with non-magnetic strips (like Aluminium) inserted between them as given below—

Single phase reluctance motors are also used in practice and these are like capacitor motors.

Problems: For a single phase reluctance motor find an expression for the instantaneous current and average torque in terms of reluctance and maximum flux. Neglect winding resistance.

Solution:

Reluctance of iron path is neglected. Reluctance of air gap varies with rotor position as shown in figure below.

An inspection of curve shows that the value of reluctance R at any space angle θ_r is given by—

R = \frac{R_{q} + R_{d}}{2} - \frac{R_{q} - R_{d}}{2} \cos 2 θ_{r}

If the applied voltage is sinusoidal—

ϕ = ϕ_{max} \cos ω t

The axis of rotor is angle δ behind the stator field axis due to mechanical load. If the position of rotor and stator at time 0 is as shown in figure (a) above, then at any time t, rotor position θ_r will be—

θ_{r} = (ω \cdot t - δ)

Now, flux $ϕ = \frac{mmf}{reluctance} = \frac{N \cdot i}{R}$ , so

i = \frac{ϕ \cdot R}{N} = \frac{ϕ_{max} cos ω t}{N} (\frac{R_{q} + R_{d}}{2} - \frac{R_{q} - R_{d}}{2} cos 2 θ_{r}) = \frac{ϕ_{max}}{2 N} [(R_{q} + R_{d}) \cos ω t - (R_{q} - R_{d}) cos (2 ω t - 2 δ) cos ω t] = \frac{ϕ_{max}}{2 N} [(R_{q} + R_{d}) cos ω t - \frac{R_{q} - R_{d}}{2} [cos (3 ω t - 2 δ) + cos (ω t - 2 δ)]] (by cos A cos B = \frac{1}{2} [cos (A + B) + cos (A - B)])

Now, $v = i \cdot r + \frac{ⅆ ψ}{ⅆ t}$ , neglecting r, we get—

v = \frac{ⅆ ψ}{ⅆ t} = \frac{ⅆ}{ⅆ t} (N \cdot ϕ) = \frac{ⅆ}{ⅆ t} (N \cdot ϕ_{max} \cos ω t) = - N ω ϕ_{max} \sin ω t

p_{inst} = v \cdot i = \frac{- ω ϕ_{max}^{2}}{2} [\frac{R_{q} + R_{d}}{2} \sin 2 ω t - \frac{R_{q} - R_{d}}{4} [sin (4 ω t - 2 δ) + \sin 2 δ]]

Now, first two terms within the bracket vary at two and four times the supply frequency, so their average is zero; thus average power is—

P = \frac{ω ϕ_{max}^{2}}{8} (R_{q} - R_{d}) \sin 2 δ

Average torque T = \frac{P}{ω} = \frac{ϕ_{max}^{2}}{8} (R_{q} - R_{d}) \sin 2 δ

Now, inductance $L = \frac{ψ}{I} = \frac{N ϕ}{I} and reluctance R = \frac{N I}{ϕ} from these, we get R = \frac{N}{ϕ / I} = \frac{N^{2}}{N ϕ / I} = \frac{N^{2}}{L}$ , so expression of torque in terms of inductances will be—

T = \frac{ϕ_{max}^{2} N^{2}}{8} (\frac{1}{L_{q}} - \frac{1}{L_{d}}) \sin 2 δ

or in terms of reactances,

T = \frac{ϕ_{max}^{2} N^{2} ω}{8} (\frac{1}{X_{q}} - \frac{1}{X_{d}}) \sin 2 δ

Now, $V = \sqrt{2} π f N ϕ_{max} = \frac{ω}{\sqrt{2}} N ϕ_{max}, so N^{2} ϕ_{max}^{2} = 2 \frac{V^{2}}{ω^{2}}$ , so expression of torque—

T = \frac{V^{2}}{4 ω} (\frac{1}{X_{q}} - \frac{1}{X_{d}}) \sin 2 δ

If I_d and I_q are the currents drawn when the rotor is held in minimum and maximum reluctance positions respectively, then—

T = \frac{V}{4 ω} (\frac{V}{X_{q}} - \frac{V}{X_{d}}) \sin 2 δ = \frac{V}{4 ω} (I_{q} - I_{d}) \sin 2 δ

The single phase and three phase reluctance motors can operate as generator as well, if mechanical power is supplied to the prime mover. As the mechanical power is increased, the rotor position advances in phase and leads the corresponding stator poles. The machine then starts generating and feeding power to the supply. The current circle diagram of a reluctance motor (for both motoring and generating operations) can be drawn in a similiar manner to that of a salient pole synchronous motor working under loss of excitation

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Reluctance Motors

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