32 COMMERCIAL ELECTRICAL TESTING 



Calculations of Deceleration Core Loss on a 3000 Kw., 2300 

 Volt, 10-Pole, 60 Cycle, 3-Phase Generator 



Moment of inertia is equal to 705,000. 

 The normal speed of the turbine being 720, Si is 

 taken equal to 750 and S2 equal to 710. 



First curve (top), Fig. 8, taken with no field on the 

 machine. 



TI or time corresponding to Si = 6 1.6 seconds. 

 T z or time corresponding to S2 = 82.4 seconds. 

 7 2 = 82.4 Si 2 = 532900 



r,=61.6 S 2 2 = 504000 



DifT.=20.8 28900 



Substituting these values in formula 



2308 ... ,(Si 2 - S 2 2 ) 2308 ^ 705000 X 28900 



Kw.Loss= Wr^-^X- -^- 



= 226 



Second curve, taken with 77.4 amperes field current. 

 72 = 70.6-7^ = 52.1 = 18.5 



2308 705000X28900 

 Kw.Loss= X- -jgj- =2o4 



Third curve, taken with 103 amperes field current. 

 7 2 = 66. 1-7*1 = 48.2 = 17.9 



v 2308 w 705000X28900 



Kw.Loss= I ^X- - T =rg- 



Fourth curve, taken with 129 amperes field current. 

 T 2 = 60.5 -Ti = 44.4 = 16.1 



v T 2308^705000X28900 



Kw.Loss= I ^ rX - -jjj- 



Fifth curve, taken with 142 amperes field current. 

 T 2 = 58.2 -Ti = 42.8 =15.4 



TT T 2308 v 705000X28200 



Kw.Loss= X- - IST - 



From the saturation curve the volts armature corres- 

 ponding to the various field currents used can be obtained 

 and a core loss curve plotted between volts armature 

 as abscissas and core loss as ordinates. 



