Synchronous Motors
            …Steady-state Operation
 • for field current less than value related to IA,min 
   armature current is lagging, consuming reactive 
   power, while when field current is greater than 
   value related to IA,min armature current is 
   leading, and supplying Q to power system
 • Therefore, by  adjusting the field current, 
   reactive power supplied to (or consumed by) 
   power system  can be controlled   
                              
            …Steady-state Operation
     Under-excitation & Over-excitation
 • If EA cosδ; projection of EA onto Vφ; < Vφ
    syn. motor has a lagging current & consumes Q 
   & since If is small, is said to be: underexcited
 • If EA cosδ; projection of EA onto Vφ; > Vφ it has a 
   leading current & supply Q (since If is large, 
   motor named overexcited
 •  
                                
    Syn Motor…Steady-state Operation
                       nd
                     2  Example
 • The 208, 45 kVA, 0.8 PF leading, Δ connected, 
   60 Hz syn. motor of last example is supplying a 
   15 hp load with an initial PF of 0.85 PF lagging. 
   If at these conditions is 4.0 A
  (a) sketch initial phasor diagram of this motor, & 
   find IA & EA 
  (b) motor’s flux increased by 25%, sketch new 
   phasor diagram of motor. what are EA, IA & PF ?
  (c) assume flux in motor varies linearly with If, 
                              
   make a plot of IA versus If for a 15 hp load
    Syn Motor…Steady-state Operation
                      ….Example
 •   Solution: 
 (a) Pin=13.69 kW,   
     IA=Pin/[3Vφ cosθ]=13.69/[3x208x0.85]=25.8 A
     θ=arc cos 0.85=31.8◦ A
     IA   = 25.8 /_-31.8◦ A
     EA=Vφ-j XS IA =208 – (j2.5)(25.8/_-31.8◦)=
         = 208 – 64.5/_58.2◦ = 182 /_-17.5 V
     Related phasor diagram shown next  
                                
     Syn Motor…Steady-state Operation
                        ….Example
 • (b) if flux φ increased by 25%, EA=Kφω will increase  
          by 25% too:
          EA2=1.25 EA1 =1.25(182)=227.5 V
     since EA sinδ1 is proportional to Power, it remains 
    constant when varying φ to a new level, so:
                                   
     EA sinδ1 =EA2 sinδ2