1. Description of High Temperature Low Sag Conductor and its Technical Requirements
1.1 The offered HTLS Conductor shall be capable of providing the maximum ampacity of 600A at a continuous operating conductor temperature not exceeding the maximum permissible operating temperature for continuous operation of the offered HTLS Conductor and without exceeding the level of maximum permissible sag indicated below at Clause 1.2.1.
The physical and operating performance requirements of the transmission line with HTLS conductor are mentioned below. The bidder shall offer HTLS conductor complying with the specified requirements. The Bidder shall indicate particulars of the proposed conductor in the relevant GTP schedule of BPS along with calculations to establish compliance with the specified requirements.
1.2 Current Carrying Capacity/Ampacity Requirements
1.2.1 Each conductor shall be suitable to carry minimum 600 Amperes of 50 Hz alternating current under the ambient conditions & maximum conductor sag specified below while satisfying other specified technical requirements/ parameters: -
Elevation above sea level = 0 m Ambient température : 45 deg C Solar Absorption coefficient =0.8 Solar Radiation = 1045 watt/sq.m Emissivity Constant= 0.45
Wind velocity considering angle between wind & axis of conductor as 90 degrees = 0.56m/sec
Effective angle of incidence of sun’s rays= 90 deg
Maximum permissible Conductor sag for 300 m span (max. span) at steady state conductor temperature and nil wind corresponding to 50 Hz alternating current of 600 Amperes per conductor under ambient conditions specified above = Not exceeding the sag for ACSR Dog Conductor or existing sag of line, whichever is lower.
The calculations for Ampacity shall be based on IEEE Standard 738-2006 in SI units. Ratio of AC resistance & DC resistance for HTLS conductor shall be calculated on the basis of the formulae indicated as follows:-
Rac = Rdc X ( 1+ 0.00519 X (mr)n X k1 + k2) where, mr = 0.3544938/(Rdc)½
if mr< 2.8, then n = 4- 0.0616 + 0.0896 X mr – 0.0513 X(mr)2
if mr> 2.8 < 5.0, then n = 4+ 0.5363 -0.2949X mr +0.0097 X(mr)2
k1= {cos (90 (d/D)P)}2.35 where,