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Conductor Tension

Conductor Tension

Conductor Tension

(OP)
Hi Gents,
Could you advise me please what appropriate %CBL should be applied for existing 3/4/2.50 ACSR conductor with a span of 180m. For new conductors,the recommended %CBL is 20%.

regards,
Danilo

RE: Conductor Tension

Sorry. This Manufacturer gives the CBL in kN [as force -not effort!]. As rated conductor cross section is 34.36 mm^2 then CBL[kg/mm^2]=24.4kN*1000/9.8/34.36=72 kgf/mm^2 [extra strong indeed!].
If the sag will be 5.5 m [taken into consideration -5 degree C and an average ice thickness] 20% will be fair. It is my opinion.
       
 

RE: Conductor Tension

(OP)
Thanks for your reply. Could you please elaborate further that
-5degree C?I presume that it is the initial temperature applied which has an equivalent tension.How did you conclude that 20% is just fair? Please show computation to enlighten me on this.
Many Thanks
 

RE: Conductor Tension

There are many way to calculate the sag and the load [conductor tension].
Siemens Manual "Tables and Formulas for Electrical Engineer"
"Formel- und Tabellenbuch für Starkstrom-Ingenieure"
http://openlibrary.org/books/OL205294M/Formel-_und_Tabellenbuch_fu%CC%88r_Starkstrom-Ingenieure
use the following relationship:
sag=span^2*gammaZ/8/sigmaZ where:
gammaZ=conductor weight +supplimentary load[ice] divided by rated conductor cross-section [kgf/cm^3].
sigmaZ=the tension of conductor at -5 oC including supplementary load[ice].
Note: At -5 oC the ice thickness would be the maximum.
gammaZ=conductor weight [kgf/cm^3]+0.18*sqrt(d)/q    where:
d=conductor diameter [mm]   q=conductor cross section [mm^2].
sigmaZ=K%/100*CBL[kgf/cm^2].
From Manufacturer data:
d=7.5 mm ;  q= 34.36 mm^2;  CBL =72.39 kgf/mm^2 [recalculated].
CBL=72.39*100=7239 kg/cm^2.
Conductor weight =193 kg/km.
conductor weight/q= weight[kg/km]/10^3/q=193/10^3/34.36=0.005617 kgf/cm^3
gammaZ=0.005617+0.18*SQRT(7.5)/34.36=0.019964 kgf/cm^3.
If K%=20% then sag=(180*100)^2*0.019964/(20/100*7239)/8=558.46 cm.=5.58 m.
The wind is not taken into consideration.
In the same Manual it is indicated wind pressure 10.5 kg/m^2.
For wind load see[in the same forum]:
http://www.eng-tips.com/viewthread.cfm?qid=8018
Following the calculation indicated in these posts the conductor tension will be 7 kN [28%].If we intend to be not more than 20% we have to admit a bigger sag [8 m].
You may get information from:
http://ieee-tpc.org/ieee_tutorials/IEEETPCTutorial_Sag-tensionCalcs.pdf
http://www.ergon.com.au/__data/assets/pdf_file/0015/6612/P56M02R09-Ver-1-Guidelines-for-Overhead-Line-Design.pdf
 

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