38 SYNCHRONOUS MOTORS 



The power of the motor and that of the generator will be easily 

 obtained by noting that the phase-angle, which is with respect to E\, 

 becomes <j>+0. We will therefore have 



7> 2 = 2 7cos<5&, (8) 



P l =E l Icos<f>=E l Icos(<l>+0); (9) 



or, replacing 7 cos and 7 sin r/> by their values, I w and I d , we have 



(10) 



P 2 = -j[E l cos (r-0)-E 2 cos f] 



cos$[7ii cos (7- 0) E 2 cos 7-] 

 [isin(7 #)- 2 sinr] , 



which can also be written 



PI=~- [Ei cos f E 2 cos 



Numerical Example. Let us consider the case of a power-trans- 

 mission to a distance of 10 kilometers by two Mordey alternators of 37 

 kilovolt-amperes (2000 volts and 18.5 amperes), having a frequency 

 of loo cycles, the resistance being 3 ohms and the reactance 43 ohms. 1 

 Let us assume that the loss-allowance in the line is 200 volts at 

 20 amperes. The line will therefore have a resistance r' = io ohms. 

 If the conductors have a resistivity (p) of 2 microhms per centimeter- 

 cube, the sectional area of each conductor is 



2ft 2XlO- 6 X20Xl0 5 



s=-V= =0.40 sq.cm. 



r' 10 



=0.062 sq.in., 



and the corresponding diameter will be 7.15 mm. (0.28 inch). Let us 

 assume that the distance between the wires is 60 cm. (23.6 inches). 

 The tables for line-inductances 2 give, for the linear inductance, in 

 millihenrys per kilometer, the following values: 



2(0.8188 +0.1173) =1.872. 



1 This reactance was determined directly from the characteristic curves given 

 for these machines by Mr. Mordey. The value usually given for the inductance 

 of this machine is too low. 



1 See an article by the author in V Eclairagt Electrique, Oct.-Dec., 1894. 



