ACSR Short Circuit Current Withstand 5

[ThePunisher]

I have on my old notes below formula for calculating the short circuit withstand of the ACSR conductors as:

I = S/(sqrt(t)) where: S = conductor cross sectional area, mm2; t = fault clearing time

I was not able to list the reference standard where I took this formulae, can anybody help me?

With the above formula, is this going to be compared with the calculated symmetrical short circuit current OR the asymmetrical short circuit current?

Your assistance is very much appreciated.

 

[magoo2]

All of the conductor burndown curves follow a constant i^2 t characteristic.

With i^2 t = constant,

you can work back to get the current values for a fixed duration which is what your equation looks like.

I'll have to look and see if I have the constant value for ACSR conductors.

 

[cuky2000]

Isc= K. A/SQRT(t)

K=4.368x10^-5 for conductor area in mm^2 , conductor temperature of 100oC (60oC rise over 40oC ambient) and maximum ACSR conductor temperature under 645oC during short circuit.

See also:

http://www.eng-tips.com/viewthread.cfm?qid=268033

 

[7anoter4]

The heat balance in a conductor is as follows:
i^2*R(T)*dt=Kext*Sext*(T-Ta)*dt+Qc*vol*dT
i^2*R(T)*dt=total heat produced by current flowing in conductor in dt seconds
Kext*Sext*(T-Ta)*dt=heat dissipated outside in dt seconds.
Qc*vol*dT=part of heat used to rise the conductor temperature by dT degrees.
R(T)=Ro*[1+ao*(T-To)] considering ao=constant from To up to T.
Ro=ro*length/Aria
Qc=conductor specific heat capacity [J/oC/mm^3]
vol=conductor volume=length*cross section Aria
In an adiabatic regime no heat will be dissipated outside.
That means all the heat will produce the conductor temperature rise.
Then:
i^2*R(T)*dt=Qc*vol*dT
i^2*Ro*[1+ao*(T-To)]*dt=Qc*vol*dT
integral(i^2*Ro*dt)|from t=0 to time|=integral[Qc*vol*dT/(1-ao*To+ao*T)|from T=Ti to Tf|
I[rms]^2=integral(i^2*dt)/time
I[rms]^2*Ro*time=Qc*length*cross section Aria/ao*ln[(Tf+1/ao-To)/(Ti+1/ao-To)]
I[rms]^2*ro*length/Aria=Qc*length*[cross section] Aria/ao/time*ln[(Tf+1/ao-To)/(Ti+1/ao-To)]
I[rms]^2=Qc* Aria^2/ao/ro/time*ln[(Tf+1/ao-To)/(Ti+1/ao-To)]
I=K*Aria*sqrt[ln(Tf+Ko)/(Ti+Ko)/time]
This corresponds to the formula from IEC 724
'Guide to the short-circuit temperature limits of electric cables with a
rated voltage not exceeding 0.6/1.0 kV' :

I^2=K^2*S^2/t*ln(theta1+betha)/(thetao+betha)
That means I it is actual rms current will flow through the conductor.
For aluminum if To=20 dgr.C ro=28.264/10^6 ohm.mm Qc=2.5/10^3 J/oC/mm^3 ao=0.00403 1/oC
K=2.5/10^3/28.264*10^6/0.00403= 148.15
Ko=228
ln((645+228)/(100+228))=0.9789
I=148.15*0.9789*A/sqrt(t)
I=145*A/sqrt(t)/10^3 kA
If one will take instead of d.c. resistance the a.c.the resistance
K will be less as skin effect could be high.
But this is an approximate calculation and for a homogeneous
metal in a conductor[not for ACSR conductor where K is variable
according to cross section al/st ratio.]
And the maximum temperature [proposed by Manufacturer] is
only 200 dgr.C and not 645 when the aluminum will fuse.

 

[cuky2000]

Hi 7anoter4,

It is very impressive how you re-developed the entire formula.
Regarding the maximum temperature for SCSR conductors, we need to consider the effect of the steel core. That is way the maximum temperature for ACSR is 645 0C at which the aluminum starts to melt.
For additional infrmation in this suggest, see the enclosed SCSR damage curve adapted from Aluminum Electrical Conductor Handbook. For additional reference see last item of table 1 in the link below.

http://www.skm.com/applicationguides4.html

 

[7anoter4]

Thank you Cuky for your kind appreciation, nevertheless it is a modest demonstration and I think electricpete or davidbeach or other may develop a similar [even improved] solution.
On the other hand the links indicated by you are very interesting.
I agree one could use the aluminum melting point as maximum temperature since
the heating process is not entirely adiabatic and actually the ACSR wire could withstand a more elevated current.
This could be similar with the copper grounding grid conductor- as per IEEE 80-
where the maximum 1089 dgr.C [copper melting point]could be taken into consideration but 250-400 dgr.C would be better if bolt connector- and even Cadweld- could be employed.

 

[ThePunisher]

Impressive derivation Zanoter. Thank you cuky2000 for the references. Stars to both of you.