'K FORMULAS 39 



4. Find, by the K formulas, the per cent voltage drop, the per 

 cent loss, and the power factor at the supply end of the following 

 line: 



Length of line 100 miles. 



Spacing 6 feet. 



Conductor No. oooo copper wire. 



Take r = 0.267, x = -7 2 7> & = 6 -3 X io~ 6 . 

 Load (at receiver end), 100 amperes per wire, 60,000 volts, 

 95% P.F., three phase, 60 cycles. [Problem of Fender and 

 Thomson, Proc. A.I. E. E., July, 1911.] 



Ans. 13.09% drop, 7.61% loss, 96.58% P.F. 

 [Calc. by series, 13.03% drop, 7.60% loss, 96.66% P.F. (Prob. 4, 

 Chap. VI).] 



5. Find, by the K formulas, the K.V.A. and voltage at the supply 

 end, and the efficiency of the following line: 



Length of line 250 Km. = 155.34 miles. 



Spacing 6 feet. 



Conductor No. ooo copper wire. 



Total resistance of one conductor 51.5 ohms. 



Total reactance of one conductor .... 48 . o ohms. 



Load (at receiver end), 15,000 K.V.A., 86,600 volts, 8o%fP.F., 



three phase, 25 cycles. 



[See page 91, "Application. of Hyperbolic Functions," by A. E. 

 Kennelly, University of London Press, 1912.] 



Ans. 15,130 K.V.A., 97,920 volts, 89.68%. 

 [By series, 15,153 K.V.A., 97,934 volts, 89.71%.] 



6. Find (a) star voltage at supply end at full load, 



(b) star voltage at supply end at no load, 



(c) regulation volts (star) at the supply end, 



(d) amperes per wire at supply end at full load, 



(e) power factor at supply end at full load, 



(f) loss in line at full load, 



(g) efficiency of the transmission line, 



(h) amperes per wire at supply end at no load (i.e., the 



"charging current"), 



(i) power factor at supply end at no load, 

 (j) loss in line at no load, 



