ril. IV sir MOTORS 7 



Example 28. -A lour-pole lighting generator is 

 to be tested under the following condition*: The speed 



: I- in t he current 800 amperes, with a tenM 

 It* at the brushes. The remittance of the armature 

 is 0-00585 ohm. The heat-drop at foil load ia 17 volts. 

 Hence the induced tension of the generator must l - 

 volts, and the induction factor is therefore The 



generator will then ! giving out 800 amperes, and th.-n- 

 will consequently be a tonju- 1(I uu-li-pounds on 



the coupling letween the two dynamos. We have now to 

 find the induction factor >f the motor; assuming there are no 

 torque losses, it has to ti ^OOOineh-pOndsoftOftpM 



to the generator at 7*> r.j> m.. the tension <>f the line being 

 oils. Inserting these values in K.ju . and 



assuming the resistance to be the same as for the gene- 

 rator, we find the induction factor of the motor to be 1'."' 

 Togure the required torque the motor will then draw .- '' I : 

 amperes, 800 of whieh will lx supplied from tin- generator 

 and ;M 1 from the line. 



IK- rheek.-d thus; 



\Vatu ttupplictl from th< In.. . i 1 _''.< . 7,775 

 WatU l<mt in hmtini; the generator, 800* -00885 . :t 

 NYntt- !.>! in h. <itin k ' the motor, H81-P - -005H5 . 4.040 



Total watu lost in heat 



The .liM.-n-ii. , I* (In.- to the limit.-* of the acvunw 

 the -liile rule 



Sup|x^.-. n. .w.l hat ill.- ruii.-iit In.in the line a*actuall\ 



measured in 01 it. 1 1 H Mm-t il.-.ln. t : Hi amperes 



for the heat lusses, as the current in th.- motor i nw 



amperes. lea\inu r " ampere* for the t..r.|ii.- losnes 



in III.- tw.. iiuH-liin.-. or if divided equal I \ l-twe.-n them, a 



