CHAPTER XI 

 MAGNETIZATION CURVES 



1. A 4-kw., 110-volt, 2-pole generator (ISOOr.p.m.) with a cast- 

 iron field, has lengths and cross-sectional areas of its magnetic path 

 as follows: In the field, 110 cm., 470 sq. cm.; 2 gaps, 7 mm., 540 

 sq. cm.; armature, 20 cm., 165 sq. cm. The leakage coefficient 

 is 1.3. Plot magnetization curves for the gap, the armature, the 

 field and hence for the whole machine. Plot in terms of total 

 armature flux and field ampere-turns obtaining five points by 

 steps of 500,000 each. Give also on the same curves, scales of 

 e.m.f. and field current; the field turns being 2400 and the num- 

 ber of armature conductors 216. From the curve determine the 

 armature flux at no load. Find also the field currents to give 80, 

 100 and 120 volts at no load. (40 min.) 



2. A 150-kw., 6-pole, 250-volt generator, 450 r.p.m. (same as 

 in problem 6, Chap. XIV) has the following data for its magnetic 

 circuit: Yoke, cross-sectional area, 284 sq. cm., length, 66 cm.; 

 cores, area, 506 cm., length, 35.5 cm. each; armature, area, 684 

 sq. cm., length, 38 cm.; teeth, total area under pole face, 413 

 sq. cm., length, 4.1 each; gaps, area increased on account of 

 spreading to 658 cm., length, 0.8 cm. each; leakage coefficient, 

 1.17. Plot magnetization curves for the field, gaps and armature, 

 and from these for the whole machine, obtaining four points for 

 armature fluxes equal 1, 3, 5 and 7 X 10 6 lines. If there are 1800 

 field turns per pole and a total of 496 armature conductors, wound 

 6-path, lay off scales for field current and generated e.m.f. Find 

 the field current for 200, 250 and 275 volts at no load. The field 

 is all of cast steel. (50 min.) 



3. It is desired to use the machine described in problem 1 over- 

 compounded so that at full load it will generate 10% more voltage. 

 What per cent must the ampere-turns be increased to do this? 

 It is also desired to use this same machine to fill a special order for 

 a 90- volt machine to overcompound 10% at full load. What 

 ampere-turns will be needed at no load, and by what per cent must 

 they be increased at full load? (5 min.) 



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