12 TRANSMISSION LINE FORMULAS 



Lay a straightedge from the 8-foot spacing point (60 cycles, copper 

 conductor) to the point on the resistance scale for No. 3 copper 

 cable. It is found to cross the 90% P.P. line at the reading 1.344. 

 Then, by the formula on the chart, 



, -TV , J .. 100,000 x 3000 x 100 x 1.344 



Per cent Regulation = *& 



66,000 X 66,000 



= 9.26%, or approximately 9.3%. 



The calculated value of the regulation of this line is 9.40% (Chap. 

 VI, Prob. 2), so that the error involved in using the chart is less than 

 I of i% of line voltage. 



The per cent line drop, according to the chart, is 

 9.26 100 x 2.16 x Ttf<y = 7-io%. 



As the calculated value is 7.08% (Chap. VI, Prob. 2), the error 

 from the chart is less than TQ of i% of the line voltage. 



PROBLEM B. 



To find the size of copper required to give approximately 10% 

 voltage drop in the following case: 



Length of line 3 miles - 



Flat spacing as in Fig. 7. Wires 2 feet apart. 

 Load (measured at receiver end), 250 K.V.A., 

 2200 volts, 85% P.F., three phase, 60 cycles. 

 First, find V from the formula on the chart. 



T/ 10 X 2200 X 2200 



V = = 0.64. 



100,000 X 250 X 3 



The equivalent spacing is 1.26 X 2, or 2.52 feet. The proper 

 spacing point will therefore be just below the spacing point for i\ 

 feet, copper conductor, 60 cycles. Lay a straightedge from this 

 point to the reading 0.64 on the line for 85% P.F. and it cuts the 

 resistance scale at 0.36 ohm per mile. The nearest size of copper is 



seen to be No. ooo. 



PROBLEM C. 



Find the voltage drop of the following two-phase line: 



Length of line 80 miles. 



Spacing 10 feet. 



Conductor No. oo aluminum cable. 



Load (measured at receiver end), 15,000 K.V.A., 

 100,000 volts, 95% P.F., two phase, 25 cycles. 



