FORMULAS FOR SHORT LINES 23 



F = 6600 - 72.73 X 8.47 - 54.55 X 7-14 



= 5594-5- 

 G = 54-55 X 8.47 - 72.73 X 7-14 



= - 57- 

 Voltage at receiver end 



= 5594.5 + 0.3 

 = 5595 approximately. 

 Drop in volts = 6600 5595 



= 1005 volts. 



Per cent drop = I0 X I0 * 



5595 

 = 17.96% of receiver voltage. 



PROBLEMS, CHAP. IV. 



(FORMULAS FOR SHORT LINES.) 



1. Find the voltage drop in the following case: 

 Length of line ......................... 3 miles. 



Spacing ............... ; ............... 2 feet. 



Conductor ............................ No. o copper cable. 



Load (at receiver end), 200 K.V.A., 2200 volts, 80% P.F., three 



phase, 60 cycles. (Prob. 2, Chap. III.) 



[Ans. 221 volts.] 



2. Find (a) the P.F. at the supply end, 



(6) the per cent efficiency of the line, for the case in Prob. i. 

 [Ans. \d) 78.7% P.F. (b) 92.3% efficiency.] 



3. Find the supply voltage in the following case: 



Length of line '. ............................. 5000 feet. 



Spacing ............................. . ...... 18 inches. 



Conductor, No. 4 copper wire of 1.312 ohms per mile. 

 Load (at receiver end), 75 Kw., 2000 volts, 95% P.F., single phase, 

 60 cycles. (Prob. 4, Chap. III.) 



[Ans. 2109 volts.] 



