Asymmetrical short-circuit current constant (r)

[rigidbar]

Hi!

I'm searching the equation for the short-circuit force on a substation busbar. I already find the equation in the IEEE Std 106-1998 (on page 16) (IEEE Guide for Design of Substation Rigid-Bus Structures) and I don't understand where does the constant (see Table 2), used in the equation, come from.

So, the second constant from table 2 (0.866) is (sqrt(3)/2) or cos (30), are the ieee editors use the line-line short-circuit current?

Pleas help!

rigidbar

 

[rigidbar]

Ok... maybe my post is not very clear... (I'm new) So I'll write you down the classic equation for a conductor fault current forces. It's taken from the IEEE Std 605.

For Metric units:

Fsc=(1.6*(r)*Isc^2)/((10^4)*(D))

For Imperial units:

Fsc=(43.2*(r)*Isc^2)/((10^7)*(D))

Where
Fsc=Fault current unit force, N/m [lbf/ft]
Isc=symmetrical rms fault current,A
D= conductor spacing center-to-center, cm [in]
(r)= constant base on type of fault and conductor location (see table)* ********THERE IS MY PROBLEM!!!!*********

type of short| CONFIG. | F. on cond. | ??(r)??| ____________________________________________
| Ph-to-ph | A---D---B | A or B | 1.00 | ?
___________________________________________
| 3 Ph | A -D- B -D- C | B | .866 | ?
___________________________________________
| 3 Ph | A -D- B -D- C | A or C | .808 | ?
___________________________________________

Where
A B C= are the 3 phases/conductors
D= distance between phase

I would like to know where does the (r) come from. Any documents or references would be appreciated!

So here is all the information I have, hope this can help.
Thanks a lot for your help!!!

rigidbar

 

[jghrist]

It has to do with the maximum instantaneous current difference between phases. For a Ø-Ø fault, the difference is 2 times the peak current because the currents are 180° out of phase. For 3-Ø faults, the currents between each phase are 120° apart. The force on phase A depends on the current in phase B over a distance of D and on the current in phase C over a distance of 2·D. For phase B, the distance to each other phase is D.

 

[cuky2000]

**************THERE IS YOUR SOLUTION!!!**************

type of SC| CONFIG. | F. on cond. | [Γ]|Inst Max F ________________________________________________________
| Ph-to-ph | A---D---B | A or B | 1.00 =43.2/43.2 | F=43.2K.I^2/10^7d _________________________________________________________________
| 3 Ph | A -D- B -D- C | B | .866 =37.4/43.2 | F=37.4K.I^2/10^7d
_________________________________________________________________
| 3 Ph | A -D- B -D- C | A or C | .808 =34.9/43.2| F=34.9K.I^2/10^7d
_________________________________________________________________
NOTE: F = lb/ft; d=Conductor Spacing (in), I = Current RMS Symmetrical

Additional inoformation could be found in the enclose link

http://www.cda.org.uk/megab2/elecapps/pub22/sec6.htm#6. Short-Circuit Effects

 

[rigidbar]

Thanks a lot!

Your info are very interesting and they confirm what I tought...

Thanks again!