22 
ALTERNATING GENERATORS AND SYN CHRONOUS MOTORS. 
193. 
194. 
195. 
196. 
19%. 
198. 
199. 
200. 
201. 
202. 
203. 
for the normal armature current of 75 amperes and for 
a power factor of cos P = .95. 
For the full cos P = .95 load, state the required number of 
field ampere turns per pole. (To be taken from the figure 
given in the preceding question. ) 
Assuming a winding depth of 1.5 inches and a thickness of 
about .25 inch for the metal bobbin and its insulation on 
each side of the pole core, calculate as a first approxi- 
mation the length of a mean turn of the field coil. 
Taking into consideration a margin. of 16% excitation, 
ealeulate for 100 volts excitation pressure the necessary 
conductor section of the field coils and choose the most 
suitable size of wire in S.W.G. ee 
Allowing for the maximum excitation a current density. 
of 1,600 amperes per square inch, calculate the necessary 
maximum excitation current and the resulting number of 
turns per pole. 
State the most suitable arrangement of turns, and make a 
sketch half full size, of two adjacent pole pieces with 
their respective windings. For this purpose assume that 
in the radial direction the total room taken up by the 
‘pobbins and end plates is .825 inch. 
Caleulate the excitation loss corresponding to the normal 
excitation of 4,060 ampere turns per pole. 
Calculate the hot armature resistance and determine the 
armature copper loss at full load. 
Calculate the weight of the teeth, and the iron losses in 
same. 
Calculate the weight of the armature core, and the iron 
losses in same. 
By means of Equation 8 of the Instruction Book, calculate 
the full load efficiency of this alternator. 
By means of Equation 41 of the Instruction Book caleulate 
_ the temperature rise on the field coils. For this purpose 
