REGULATION CHART 13 



Laying a straightedge across the chart from the lo-foot spacing 

 point for 25 cycles and aluminum conductor, to the resistance point 

 for No. oo aluminum, the value of V for 95% P.F. is found to be 

 0.750. Then the line drop, in volts, is equal to 



rooox 15,000X80X0.750 _ t x x Q og x Q og 



100,000 



= 9000 240 

 = 8760 volts. 



The calculated value is 8810 volts (Chap. VI, Prob. i). The 

 error from the chart is 0.05% of line voltage. 



PROBLEM D. 



Find the regulation of the following single-phase line: 



Length of line 15 miles. 



Spacing 3 feet. 



Conductor No. o copper wire. 



Load (at receiver end), 300 K.V.A., 50% P.F., 



11,000 volts, single phase, 60 cycles. 

 From the chart, V = 0.851. 



on. e T> i *. IOO.OOO X 3OO X 15 X 2 X 0.8=51 , ~ 



Therefore Regulation = z *- = 6.33%. 



11,000 X 11,000 



[Calculated value, 6.40% (Chap. IV, Prob. A). Error from chart, 

 0.07% of line voltage.] 



PROBLEM E. 



Find the K.V.A. which can be delivered at the end of the following 

 line, with 8% regulation: 



Length of line 75 miles. 



Spacing 8 feet, regular flat spacing. 



Conductor No. oo aluminum cable. 



Character of load (at receiver end), 



88,000 volts, 85% P.P., three phase, 25 cycles. 

 Equivalent spacing s = 8 X 1.26 = 10.08 feet. 



V = 0.755- 



KVA = 8 X 88,000 X 88,000 

 100,000 X 0.755 X 75 

 = 10,000. 



