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where k is a machine constant, IS is the rms value of the stator current, and If is the DC rotor current Now, the rotor angle can be expressed as a function of time by = 0 + m t (1762)
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where 0 is the angular position of the rotor at t = 0; the torque expression then becomes T = k 2IS If sin( e t) sin( m t + 0 ) (1763) 2 IS If cos[( m e )t 0 ] cos[( m + e )t + 0 ] =k 2 It is a straightforward matter to show that the average value of this torque, T , is different from zero only if m = e , that is, only if the motor is turning at the synchronous speed The resulting average torque is then given by T = k 2IS If cos( 0 ) (1764) Note that equation 1763 corresponds to the sum of an average torque plus a uctuating component at twice the original electrical (or mechanical) frequency The uctuating component results because, in the foregoing derivation, a singlephase current was assumed The use of multiphase currents reduces the torque uctuation to zero and permits the generation of a constant torque A per-phase circuit model describing the synchronous motor is shown in Figure 1733, where the rotor circuit is represented by a eld winding equivalent resistance and inductance, Rf and Lf , respectively, and the stator circuit is represented by equivalent stator winding inductance and resistance, LS and RS , respectively, and by the induced emf, Eb From the exact equivalent circuit as given in Figure 1733, we have VS = Eb + IS (RS + j XS ) (1765)
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Rf + RS + Eb
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where XS is known as the synchronous reactance and includes magnetizing reactance The motor power is Pout = S T = |VS ||IS | cos( ) (1766)
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Is + VS 0
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for each phase, where T is the developed torque and is the angle between VS and IS When the phase winding resistance RS is neglected, the circuit model of a synchronous machine can be redrawn as shown in Figure 1734 The input power (per phase) is equal to the output power in this circuit, since no power is dissipated in the circuit; that is: P = Pin = Pout = |VS ||IS | cos( ) Also by inspection of Figure 1734, we have d = |Eb | sin( ) = |IS |XS cos( ) Then |Eb ||VS | sin( ) = |VS ||IS |XS cos( ) = XS P (1769) (1768) (1767)
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Figure 1733 Per-phase circuit model
IS + Eb
XS + VS 0
IS VS jISXS Eb d
17
Introduction to Electric Machines
P Pmax
Generator 180 90 90 Motor <0 180 3 Vs Eb XS
Pmax = Pmax
Figure 1735 Power versus power angle for a synchronous machine
The total power of a three-phase synchronous machine is then given by |VS ||Eb | sin( ) (1770) P = (3) XS Because of the dependence of the power upon the angle , this angle has come to be called the power angle If is zero, the synchronous machine cannot develop useful power The developed power has its maximum value at equal to 90 If we assume that |Eb | and |VS | are constant, we can draw the curve shown in Figure 1735, relating the power and power angle in a synchronous machine A synchronous generator is usually operated at a power angle varying from 15 to 25 For synchronous motors and small loads, is close to 0 , and the motor torque is just suf cient to overcome its own windage and friction losses; as the load increases, the rotor eld falls further out of phase with the stator eld (although the two are still rotating at the same speed), until reaches a maximum at 90 If the load torque exceeds the maximum torque, which is produced for = 90 , the motor is forced to slow down below synchronous speed This condition is undesirable, and provisions are usually made to shut the motor down automatically whenever synchronism is lost The maximum torque is called the pull-out torque and is an important measure of the performance of the synchronous motor Accounting for each of the phases, the total torque is given by m T = |VS ||IS | cos( ) (1771) s where m is the number of phases From Figure 1734, we have Eb sin( ) = XS IS cos( ) Therefore, for a three-phase machine, the developed torque is 3 |VS ||Eb | P = sin( ) N-m (1772) T = s s XS Typically, analysis of multiphase motors is performed on a per-phase basis, as illustrated in the examples that follow