LIGHTXIXG PROTECT lOX OF Bl'lUED TOLL CABLE 267 



the voltage between core and sheath at the stroke point due to a smusoidal 

 current Ji becomes, 



f'i(O) = Ti? = J^{£) J^(P + R/^^' (29) 



The corresponding voltage for a unit step current /i is: 



ri(0, /) = JiR(^ erf (Rt/i:)^ (30) 



where erf is the error function. 



For large values of time, when Rt/L > 1, (30) becomes 



The latter expression is valid when / exceeds about 2 milliseconds and 

 thus applies for the long duration current of a lightning stroke, since the 

 latter usually lasts for about 100 milliseconds. For a current of 1000 

 amperes, the core-sheath voltage for a cable of 1.4" diameter is about 700 

 volts. In many strokes the long duration current may be several hundred 

 amperes, and a substantial voltage may then exist between core and sheath 

 for .1 second or so. Thus, while this current component does not increase 

 the crest voltage, it substantially increases the likelihood of permanent 

 failure when the insulation is punctured by prolonging the current through 

 the puncture. 



1.7 Strokes to Ground Not Arcing to Cable 



Let it be assumed that the current enters the ground at the distance y 

 from a buried cable and that conditions are such that it does not arc to the 

 latter. The flow of current in the ground gives rise to an electric force in 

 the ground along the cable, and thus to currents in the sheath and to voltages 

 between core and sheath. When the earth is assumed to have uniform con- 

 ductivity, the earth potential at the distance r from the point where current 

 enters the ground is given by: 



Ve = JQoir) = Jp/Iwr (32) 



where 



p = Earth resistivity in meter-ohms 

 r = (.V + y )' = Distance in meters 



The sheath current and the voltage between sheath and core may in this 

 case be obtained from published formulas, provided propagation along 



