148 THE MAGNETIC CIRCUIT [ART. 47 



Squaring eqs. (71) and (72) and adding them together gives 



E= Vei 2 + e 2 2 . 1 ...... (75) 



Dividing eq. (72) by (71) results in 



tan <= e 2 /ei ....... (76) 



Consequently, the angle between E and i becomes known; namely, 



....... (76a) 



where <j>' is called the internal phase angle. Knowing E, the cor. 

 responding excitation M n is taken from the no-load saturation 

 curve of the machine ; from the triangle OFG we have then : 



Mf = M n * + M a 2 + 2M n M a sin <', . . . (77) 



where <>' is known from eq. (76a) . In numerical applications it is 

 convenient to express all the M's in kiloampere-turns. 



The diagram shown in Fig. 38 and the equations developed 

 above can be used for determining not only the phase characteris- 

 tics of a synchronous motor, but its overload capacity at a given 

 field current as well. This latter problem is of extreme importance 

 in the design of synchronous motors. The input into the machine, 

 per phase, is ei cos <j> ; the part ir of the line voltage is lost in the 

 armature, the part ix corresponds to the magnetic energy which is 

 periodically stored in the machine and returned to the line, without 

 performing any work. The remainder, E, corresponds to the use- 

 ful work done by the machine, plus the iron loss and friction. If 

 the armature possessed no resistance and no leakage reactance the 

 terminal voltage would be equal to E in magnitude and in phase 

 position. Thus, the expression Ei cos <$>', corrected for the core 

 loss in the armature iron, represents the input into the revolving 

 structure, per phase. The overload capacity of the machine is 

 determined by the possible maximum of this expression. 



The problem is complicated by the fact that the relation 

 between E and M n is expressed by the no-load saturation curve, 

 which is difficult to represent by an equation. The problem is 



1 In numerical applications it is more convenient to use the approximate 

 formula 



E=e l + ^/e l .......... (75o) 



obtained by the binomial expansion of expression (75) ; since all other terms 

 can be neglected when e 2 is small as compared to e,. 



