## Derive the expression of torque in closely excited system., Mechanical Engineering

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Q. Derive the expression of torque developed in closely excited magnetic system. Clearly explain then assumption made.

Sol. Double - Excited System

A doubly - excited magnetic system has two independent sources of excitations. Examples of such systems are separately excited dc machines synchronous machine, loudspeakers, tachometers etc.Let us consider that both the stator and rotor have silency. Assumptions are as for a singly - excited system.

The flux linkage eq. for the two windings are

Ψ1 = L1I1 + Mi2

Ψ2 = L2I2 + Mi1

The instantaneous voltage eq. for the two coils are

V1 = R1i1 + d Ψ1/dt

V2 = R2i2 + d Ψ2/dt

Substituting the values Ψ1 and  Ψ2

V1 = R1i1 + d/dt + (L1i1) + d/dt (Mi2)

V2 = R2i2 + d/dt + (L2i2) + d/dt (Mi1)

Now the inductances are independent of currents and depend on the position of the root angle θm which is a function of time. Similarly, current are time dependent  and are not function of inductances. Therfore,

V1 = R1i1 + L1di1/dt + i1dL1/dt + Mdi2/dt + i2dM/dt

V2 = R2i2 + L2di12/dt + i2dL2/dt + Mdi1/dt + i1dM/dt

By multiplying we get,

V1i1 = R1i1+ L1i1di1/dt + i12dL/dt + i1Mdi2/dt + i1i2dM/dt

V2i2 = R2i2+ L2i2di2/dt + i22dL/dt + i2Mdi1/dt + i1i2dM/dt

Now we get,

(( v1i1 + v2i2 ) dt =(( R1i12 + R2i22 ) dt + (( L1i1di1 + L2i2di2 + i1Mdi2 + 2i1i2dM + i12dL1 + i22dL2 + i2Mdi1)

Also,   [Useful electrical energy input] =  (( v1i1 + v2i2 ) dt - (( R1i12 + R2i22 ) dt

[Energy to field storage in the electrical systems] + [Electrical to mechanical energy] = (( L1i1di1 + L2i2di2 + i1Mdi2 +2i1i2dM + i12dL1 + i22dL2 + i2Mdi1)

Stored energy in the Magnetic field

The instantaneous value of energy stored in the magnetic field depends on the inductance and current values at the instant considered. This energy may be found by considering the transductor to be stationary and the coils to be energized from zero current to the required instantaneous values of current. There is no mechanical output and Wem is zero. The inductance values are constant. Therefore terms dL1, dL2 and dM become zero

(dWfe = oi1(L1i1di1 + oi2(L2i2di2 + oi2,i2 ( (i2Mdi1 + i1 Mdi2 )

[Total Wfe] = 1/2L1i12 + 1/2L2i22 + Mi1i2

Electromagnetic Torque

If the transductor rotates, the rate of change of field energy with respect to time is given by differentiating.

dWfe/dt = 1/2L1 d/dt i12 + 1/2i12 dL1/dt + 1/2L2 di2/dt2 + 1/2i22 dL2/dt + i2i2 dM/dt + i1M di2/dt + i2M di/dt

dWfe/dt = L1i1 di1/dt  + 1/2i12 dL1/dt + L2i2 di2/dt + 1/2i22 dL2/dt + i2i2 dM/dt + i1M di2/dt + i2M di/dt

Integrated with respect to time

(dWfe = Wfe = ((L1i1di1 + 1/2i12dL1 + L2i2di2 + 1/2i22dL2) + i1i2dM + i1Mdi1

This is general eq. for a moving transducer in which L1, L2 and M, i1 and i2 are all varying with position and time. On comparing we get,

Wem = [Electrical to mechanical energy] =  ((1/2 i12dL1 + 1/2i22dL2 + i2i2dM)

Differentiating with respect to θm

dWem/d θm = ½ i12 dL1/d θm = ½ i22 dM/d θm

as only L1, L2 and M are dependent on θm

It includes the case of singly - excited system when one of the two current is equal to zero so that the expression for the torque becomes

Τe = i2/2 dl/d θm

The first two terms of the torque are reluctance torques or saliency torques. The last term i1i2 dM/dθ is called the co - alignment torque, that is two superimposed fields, that try to align.

For machines having uniform air gaps reluctance torque is not produced.

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