TRANSMISSION LINE 401 



the current is therefore 



P 5,100,000 



* 60,000X0.85 

 The vector diagram is drawn with the current 01 = I as horizontal. 



EG E 



4 $- 



FIG. 375. Single-phase transmission h'ne. 



The receiver e.m.f. OE = E leads the current by an angle <f> 

 and has two components 



OEi = EI = E cos < in phase with 7 and 

 OEz = E 2 = E sin <f> in quadrature ahead of 7. 



The voltage consumed in the resistance of the line is Ir in 

 phase with 7; the voltage consumed in the reactance of the line 

 is Ix in quadrature ahead of 7. 



The component of the generator e.m.f. in phase with 7 is 



EI + Ir = E cos + Ir 

 and the component in quadrature ahead of 7 is 



# 2 + Ix = E sin + 7z, 

 and therefore the generator e.m.f. is 



EG = V(E cos + 7r) 2 + (E sin + Ix) 2 , 

 or substituting the numerical values 



E G = V(60,000 X 0.85 + 100 X 20) 2 + (60,000 X 0.52 + 100 X 50) 2 

 = 64,000 volts. 



The e.m.f. consumed in the line is 



7 Vr 2 + x 2 = 100 V20 2 + 50 2 = 5400 volts. 

 The loss of power in the line is 



7 2 r = 100 2 X 20 2 = 200,000 volts 

 = 200 kilowatts. 



