ALTERNATING GENERATORS AND SYNCHRONOUS MOTORS. ag 
238. Plot the no-load and short-cireuit characteristics of this 
alternator, applying tte following scales: 
1,000 ampere turns excitation = 1 inch. 
1,000 volts terminal pressure = 1 inch. 
40 amperes armature current = 1 inch. 
239. In the same drawing made in answer to Question 238, de- 
termine graphically the voltage drop for the normal 
armature current ef 44 amperes, the power factor being 
cos P = .80. 
240. Assuming a total winding depth of 1.6 inch 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. 
241. Taking into account a margin of about 10 per. cent; exeita- 
- tion, caleulate the necessary conductor section of the field 
winding for 120 volts excitation pressure, choosing the 
most suitable size of S.W.G. wire. 
242. Allowing for the maximum excitation a current density of 
1,600 amperes per square inch, ealeulate the maximum 
necessary excitation current and the number of turns 
per pole. 
243. State the most suitable arrangement cf the exc'tation turns 
3 and make a sketch, half full size, of two adjacent pole 
pieces, showing their respective windings, assuming that 
in the radial direction the total room taken up by the 
bobbins and end plates is about .80 inch. 
244. Calculate the excitation loss corresponding to the full cos 
py = .80 load of the alternator. 
245. Caleulate the armature resistance (hot) and determine the 
armature copper loss at full load. 
246. Calculate the weight of the teeth and the iron losses in same. 
247. Calculate the weight of the armature core (without the 
teeth) and the iron losses in same. 
248. By means of Fquation 8 and Fig. 25 of the Instruetion 
Book, caleulate the full lead efficiency of this alternator. 
