Home Page News Search Contact Us Language Bar فارسی English
 
 
 
Grounding Riddle No.22 - Main Earthing Conductor sizing calculation
Hi,
I urgently need the main earthing conductor sizing calculation of 132kV substation

Formula : I x {(TCAP x 10⁻⁴/tc ar rr) x ln [(Ko+Tm)/(Ko+Ta)}>⁻½

Fault current is 40kA in 3 sec.

I need this calculation for lead sheathed copper conductor. I donot know the values of lead sheathed copper conductor.

If anybody can help me, kindly send me an email to asifhammad@gmail.com

Thanks & Best Regards
Asif
Author : Asif - From: Abu dhabi
 
#1
Sat, February 6th, 2010 - 15:50
That mentioned Equation (which was not written correctly) reflect two basic assumptions
a) That all heat will be retained in the conductor (adiabatic process).
b) That the product of specific heat (SH) and specific weight (SW), TCAP, is approximately constant because SH increases and SW decreases at about the same rate. For most metals, these premises are applicable over a reasonably wide temperature range, as long as the fault duration is within a few seconds.
Therefore the thermodynamic behavior of conductors in fault duration ( 1 to 3 sec.) can not affected by conductor thin covering, and you can use submitted constant material values which illustrated in technical tables for required calculations in base of main uncoated conductors.



Also about mentioned fault duration (3 sec.), Relay malfunctions can result in fault duration in excess of primary clearing times. The backup clearing time is usually adequate for sizing the conductor. For smaller substations, this may approach 3 s or longer. However, because large substations usually have complex or redundant protection schemes, the fault will generally be cleared in 1 s or less.
Meanwhile for correct formula and table you can refer to IEEE 80 or following relations.




 
Author : Hamid - From: Iran- Firouzabad in Fars
 
Submit Your Answer

 
 
Change Language :