- The maximum stress in the cable occurs at the surface of the conductor.
- The dielectric strength of the cable should be more than the maximum stress in order to safe working of the cable.

Maximum stress on the cable

g_{max} =
2V / { d Log_{e} ( D / d ) }…..( 1 )

- The internal sheath diameter D and working voltage have to be kept constant for design consideration therefore only variable parameter d is under consideration.
- The most economical diameter of conductor is achieved when stress on the conductor becomes minimum.
- The value of g
_{max}becomes minimum when { d Log_{e}( D / d ) } becomes maximum.

d /dd { d Log_{e}
( D / d ) } = 0

d {( d / D ) (
– D / d^{2} ) } + { Log_{e} ( D / d ) } × 1 = 0

{ Log_{e} ( D / d ) } – 1 = 0

{ Log_{e} ( D / d ) } = 1…. ( 2 )

( D / d ) = e^{1}

( D / d ) =
2.718

D = 2.178d

OR

d = D / 2.718….. ( 3 )

Maximum stress under this condition, using equation (
1 ) and ( 2 )

*g _{max} =
2V / d*

If we consider maximum stress, the most economical
conductor diameter

* d = 2V / g_{max}*….. ( 4 )

This is only theory point of view

*Practically Cable Diameter*

- If we calculate most economical conductor diameter considering maximum stress on the cable, it should be too small from the point of view of current density. Therefore the cable diameter is decided by considering current density.
- The diameter of cable is too much high for high voltage cable from point of view of current density considering above formula ( 4 ). The diameter of cable can be increased by
**Use of aluminum conductor instead of copper conductor because the diameter of aluminum conductor is larger than that of copper conductor for same current capacity****Use of hollow conductor**

*You may also like :*

Standard Alternating Voltage and applicable voltage
variations

Minimum Clearance between Transmission line conductor
and ground

What is ROW in the transmission line?

Compare : Distributor and Feeder

## No comments:

## Post a comment