Question

lecture notes on voltage gradients of conductors

Answer 1

Abstract:The method of successive images is used to obtain accurate values of maximum surface voltage gradients for practical HVdc transmission lines. The results are presented in the form of graphs of gradient factors as a function of the principal line parameters. The gradient factor when multiplied by the line to ground voltage yields the maximum voltage gradient at the conductor surface. The range of values of parameters chosen covers transmission lines of voltages up to about ± 1200 kV. These graphs are also applicable for determining the center phase gradients on ac lines and therefore should be useful in making comparative studies between ac and dc systems. The results include an analysis of the sensitivity of the conductor surface voltage gradient to variations: in spacing between positive and negative conductors; in height; and in subconductor spacing. An optimum subconductor spacing is found for different numbers of conductors in the bundle as well as for different conductor sizes. However, it is shown that the gradient values are insensitive to fairly large variations in subconductor spacing about an optimum value.

Calculation of Conductor Surface Voltage Gradient:

Method obtained IEE Power Series 17, “High Voltage Engineering and Testing”. Appendix 3.4

 

E=(V/Ö3)*(b/(r*ln((a/Re)*2h/Ö(4h2+a2)))

 

Where,

b=(1+(n-1)r/R)/n

 

Re=Rn*Ö(nr/R)

 

R=S/(2*Sin(p/n))

 

E= Conductor Surface Voltage Gradient ( kV/cm)

V= Rated Voltage ( kV )

b= Factor for Multiple Conductors

r= Radius of Conductor ( cm )

R= Outside radies of bundle ( cm )

Re=Equivalent Radiues of bundle conductor ( cm )

S=  Distance between Component conductor centers ( cm )

a=  Phase Spacing ( cm )

h= Height of conductor above ground ( cm )

( This value is taken as the distance between phase connection/bus bar to the metalwork at earth potential)

n= Number of component conductors in bundle

 

Line Calculations:

V=1200 kV r= 16 cm S= 45.7 cm a= 24 m= 2400 cm h= 8 m = 800 cm n = 8   Therefore, R= 59.71 cm   b= 0.1485   Re= 699.38 cm   E= 9.99 kV/cm   R=S/(2*Sin(p/n))=45.7/(2*Sin(p/8))    =59.71 cm   b=(1+(n-1)r/R)/n    =(1+(7*16)/59.71)/8    =0.1485   Re=Rn*Ö(nr/R)      =(59.71*8)* Ö((8*16)/59.71)      =699.38 cm   Therefore, E=(V/Ö3)*(b/(r*ln((a/Re)*2h/Ö(4h2+a2))) =(1200/Ö3)*(0.1485/(16*ln((45.7/699.38)*1600 /Ö(4(800)2+(2400)2))) =9.99 kV/cm   hope that helps you 
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