VII.  Synchronous  Motor 

16.  As  seen  in  the  preceding,  in  an  alternating-current  gen- 
erator the  field  excitation  required  for  a  given  terminal  voltage 
and  current  depends  upon  the  phase  relation  of  the  external 
circuit  or  the  load.  Inversely,  in  a  synchronous  motor  the 
phase  relation  of  the  current  into  the  armature  at  a  given  ter- 
minal voltage  depends  upon  the  field  excitation  and  the  load. 

Thus,  if  E  =  terminal  voltage  or  impressed  e.m.f.,  I  =  current, 
6  =  lag  of  current  behind  impressed  e.m.f.  in  a  synchronous 
motor  of  resistance  r  and  synchronous  reactance  XQ,  the  polar 
diagram  is  as  follows,  Fig.  62. 

OE  =  E  is  the  terminal  voltage  assumed  as  zero  vector.  The 
current  01  =  I  lags  by  the  angle  EOI  =  6. 

The  e.m.f.  consumed  by  resistance  isj9#'i  =  Ir.  The  e.m.f. 
consumed  by  synchronous  reactance,  OE'o  =  IxQ.  Thus,  com- 


142     ELEMENTS  OF  ELECTRICAL  ENGINEERING 


bining  OE'i  and  OE'o  gives  OE',  the  e.m.f.  consumed  by  the 
synchronous  impedance.  The  e.m.f.  consumed  by  the  synchro- 
nous impedance  OE'  and  the  e.m.f.  consumed  by  the  nominal 
generated  or  counter  e.m.f.  of  the  synchronous  motor  OEo, 
combined,  give  the  impressed  e.m.f.  OE.  Hence  OEo  is  one 
side  of  a  parallelogram,  with  OE'  as  the  other  side,  and  OE  as 
diagonal.  OEoo  .(not  shown),  equal  and  opposite  OE0,  would 
thus  be  the  nominal  counter-generated  e.m.f.  of  the  synchronous 
motor. 

In  Figs.  63  to  65  are  shown  the  polar  diagrams  of  the  syn- 
chronous motor  for  6  =  0  deg.,  6  =  60  deg.,  6  =  —  60  deg.  It 
is  seen  that  the  field  excitation  has  to  be  higher  with  lead- 


d    E' 


FIG.    62. — Vector    diagram    of 
synchronous  motor. 


FIG.    63. — Vector    diagram    of 
synchronous  motor.     0=0 


ing  and  lower  with  lagging  current  in  a  synchronous  motor, 
while  the  opposite  is  the  case  in  an  alternating-current  generator. 

In  symbolic  representation,  by  resolving  all  e.m.fs.  into  power 
components  in  phase  with  the  current  and  wattless  components 
in  quadrature  with  the  current  i,  we  have: 

the  terminal  voltage,  E  =  E  cos  6  +  jE  sin  6  =  Ep  +  jEq; 
the  e.m.f.  consumed  by  resistance,  E/i  =  ir, 
and  the  e.m.f.  consumed  by  synchronous  reactance,  E'Q  =  +  jix0. 

Thus  the  e.m.f.  consumed  by  the  nominal  counter-generated 
e.m.f.  is 

Eo  =  E  -  E'i  -  E'Q  =  (E  cos  0  -  ir)  +  j  (E  sin  6  -  ixQ) 
=  (Ep  -  ir)  +  j(Eq  -  ixQ); 


SYNCHRONOUS  MACHINES  143 

or,  in  absolute  values, 


V(Ecos  e  -  ir)2  +  (Esin  6  -  ix0)* 


=  V(Ep-  ij 
hence, 

E  =  i  (rp  +  xQq)  ±  \/EQ2  —  i2  (x0p  —  rq)z. 

The  power  consumed  by  the  synchronous  motor  is 

P  =  iEp; 

that  is,  the  current  times  the  power  component  of  the  impressed 
e.m.f. 

/" 

Eo 


FIG.  64. — Vector  diagram  of  syn-  FIG.  65. — Vector"  diagram  of  synchronous 
chronous  motor.  6  =  60  deg.  motor.  0  =  —  60  degrees. 

The  mechanical  power  delivered  by  the  synchronous  motor 
armature  is 

Po  =  i(Ep-ir); 

that  is,  the  current  times  the  power  component  of  the  nominal 
counter-generated  e.m.f.  Obviously  to  get  the  available  mechan- 
ical power,  the  power  consumed  by  mechanical  friction  and  by 
molecular  magnetic  friction  or  hysteresis,  and  the  power  of  field 
excitation,  have  to  be  subtracted  from  this  value  P0.