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

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.

Posted Date: 7/23/2012 1:12:01 AM | Location : United States







Related Discussions:- Derive the expression of torque in closely excited system., Assignment Help, Ask Question on Derive the expression of torque in closely excited system., Get Answer, Expert's Help, Derive the expression of torque in closely excited system. Discussions

Write discussion on Derive the expression of torque in closely excited system.
Your posts are moderated
Related Questions
Plate : These are also known as electrodes. There are two types of group of plates - Positive and Negative. Each group contains more than one plate. A group of positive plates is

An 8 kN.m moment is applied at the end of a block fixed into the ground as illustrated in thefigure below. The moment vector (double arrow head representation) acts in line with th

A pump station has been designed to lift water out of a 6 metre deep pit (vented to atmosphere) via a centrifugal pump mounted at ground level. Liquid conditions

It is a 3-D drawing used by draftsmen, architects etc

A pull up door is stuck in the closed position. A strap, fastened on the inside bottom and located at the center of the door is pulled outward at an angle of 45 deg above horizonta

Calculate the length of plastic deformation: Calculate the length of plastic deformation along 45o in case the applied stress is 63.25% of yield stress in the plate of abov

Resulting structure of steel: Heat treatments as mentioned below are given to thin steel strips for which TTT diagram is as shown in Figure 19. What will be the resulting stru

define elastomer.explain in details?

temparature stresses in composite bars

In designing piping systems, it is sometimes desirable to estimate the appropriate pipe length for a given diameter, pump power and flow rate. In such cases, if minor pipe losses a