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

                                    V_{1 =} R₁i₁ + d Ψ₁/dt

                                    V_{2 =} R₂i₂ + d Ψ₂/dt

  Substituting the values Ψ₁ and  Ψ₂

                                    V_{1 =} R₁i₁ + d/dt + (L₁i₁) + d/dt (Mi₂)

                                     V_{2 =} R₂i₂ + d/dt + (L₂i₂) + d/dt (Mi₁)

Now the inductances are independent of currents and depend on the position of the root angle θ_m 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,

                           (( v₁i_{1 +} v₂i₂ ) dt =(( R₁i₁² + R₂i₂² ) dt + (( L₁i₁di₁ + L₂i₂di₂ + i₁Mdi₂ + 2i₁i₂dM + i₁²dL₁ + i₂²dL₂ + i₂Mdi₁)

         Also,   [Useful electrical energy input] =  (( v₁i_{1 +} 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 W_em is zero. The inductance values are constant. Therefore terms dL₁, dL₂ and dM become zero

        (dW_fe = _oⁱ₁(L₁i₁di₁ + _oⁱ₂(L₂i₂di₂ + _oⁱ₂^,i₂ ( (i₂Mdi₁ + i₁ Mdi₂ )

         [Total W_fe] = 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.

             dW_fe/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

             dW_fe/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

                       (dW_fe = W_fe = ((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,

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

               Differentiating with respect to θ_m

              dW_em/d θ_m = ½ i₁² dL₁/d θ_m = ½ i₂² dM/d θ_m

           as only L₁, L₂ and M are dependent on θ_m

       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

                       Τ_{e =} i²/2 dl/d θ_m

         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.