×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Log In

Come Join Us!

Are you an
Engineering professional?
Join Eng-Tips Forums!
  • Talk With Other Members
  • Be Notified Of Responses
    To Your Posts
  • Keyword Search
  • One-Click Access To Your
    Favorite Forums
  • Automated Signatures
    On Your Posts
  • Best Of All, It's Free!
  • Students Click Here

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

Students Click Here

Jobs

IEEE 605 Short Circuit Force

IEEE 605 Short Circuit Force

IEEE 605 Short Circuit Force

(OP)
Can anyone explain the difference between Short Circuit force calculation in the 1998 standard and the 2008 standard of IEEE 605?

The 1998 equation (eq 11) can be re-worked into a similar form as the 2008 equation.

F_sc = (C x Gamma x (D_f x SQRT(2) x I_sc)^2) / D_in 1998

C = 5.4x10^-7 per the 1998 standard

Squaring the associated parameters of the 1998 equation gives

F_sc = (5.4 x Gamma x D_f^2 x 2 x I_sc^2) / (10^7 x D_in) 1998

Converting the distance parameter (D_in) to feet (D_ft) and re-arranging gives

F_sc = (5.4 x 2 x Gamma x I_sc^2) / (12 x 10^7 x D_ft) x D_f^2

= (0.90 x Gamma x I_sc^2) / (10^7 x D_ft) x D_f^2

= (3.6 x Gamma x I_sc^2) / (10^7 x D_ft) x (1/2 x D_f)^2 1998

The last expression is in a similar format to the 2008 equation. The only difference is the 1/2 included in the square of the Decrement factor.

F_sc = (3.6 x Gamma x I_sc^2) / (10^7 x D_ft) x D_f^2 2008

The difference between the 1998 & 2008 equations stems from the differing definitions of the decrement factor. Using a system frequency of 60 Hz and a fault current duration of a half second for the 1998 Decrement factor, the two equations are equal at a X/R ratio of zero.

DECREMENT FACTOR & SHORT CIRCUIT FORCE

I_sc = 40,000 A D_ft = 10 ft

1998 | 2008
60 Hz 0.5 sec* |
X/R T_a D_f F_sc | D_f F_sc 2008 / 1998
0 0.00000 1.0000 14.400 | 0.5000 14.400 1.0000
1 0.00265 1.0026 14.476 | 0.5216 15.671
2 0.00531 1.0053 14.553 | 0.6039 21.009
5 0.01326 1.0132 14.782 | 0.7667 33.863
10 0.02653 1.0262 15.164 | 0.8652 43.118 2.8434 Three times greater
20 0.05305 1.0517 15.928 | 0.9273 49.531 3.1097 at typical values of
30 0.07958 1.0766 16.692 | 0.9503 52.016 3.1162 X/R
50 0.13263 1.1248 18.218 | 0.9696 54.146
100 0.26526 1.2322 21.863 | 0.9845 55.832
300 0.79577 1.4624 30.796 | 0.9948 47.002 At the limit, the 2008
~ ~ 1.7320 43.200 | 1.0000 57.600 1.3333 equation is 4/3 times
greater than the 1998
* Fault Current Duration
~ Infinity

Notes:
1. The 1998 decrement factor approaches the square root of 3 as the X/R ratio goes to
infinity because the fault current duration is one half. The exponential
expression and the T_a value go to unity as X/R goes to infinity, so the 1998
equation reduces to square root of one plus the reciprocal of one half, or the
square root of three.

Also, Table 14 of IEEE 605 2008 makes me wonder if there is a typo. Table 14 shows the decrement factor and the square of the decrement factor as a function of X/R (& T_a) for both 50 Hz and 60 Hz systems. However, the decrement factor as defined in the standard is independent of the system frequency. Per eq 19, the decrement factor is a function of the product of the frequency and T_a (approximate period??):

T_a = X/R x (1 / (2 x pi x f))

f x T_a = X/R x (1 / (2 x pi))

which results in a the decrement factor being merely a function of the X/R ratio.

RE: IEEE 605 Short Circuit Force

(OP)
Can anyone explain the difference between Short Circuit force calculation in the 1998 standard and the 2008 standard of IEEE 605? For practical situations, the short circuit force is three times greater than previous standard. This causes various problems related to substation layout and insulator hardware selection.

The 1998 equation (eq 11) can be re-worked into a similar form as the 2008 equation.

.... F_sc = (C x Gamma x (D_f x SQRT(2) x I_sc)^2) / D_in .................... 1998

....... C = 5.4x10^-7 per the 1998 standard

Squaring the associated parameters of the 1998 equation gives

..... F_sc = (5.4 x Gamma x D_f^2 x 2 x I_sc^2) / (10^7 x D_in) .............. 1998

Converting the distance parameter (D_in) to feet (D_ft) and re-arranging gives

..... F_sc = (5.4 x 2 x Gamma x I_sc^2) / (12 x 10^7 x D_ft) x D_f^2

.......... = (0.90 x Gamma x I_sc^2) / (10^7 x D_ft) x D_f^2

.......... = (3.6 x Gamma x I_sc^2) / (10^7 x D_ft) x (1/2 x D_f)^2 .......... 1998

The last expression is in a similar format to the 2008 equation. The only difference is the 1/2 included in the square of the Decrement factor.

..... F_sc = (3.6 x Gamma x I_sc^2) / (10^7 x D_ft) x D_f^2 .................. 2008

The difference between the 1998 & 2008 equations stems from the differing definitions of the decrement factor. Using a system frequency of 60 Hz and a fault current duration of a half second for the 1998 Decrement factor, the two equations are equal at a X/R ratio of zero.

............... DECREMENT FACTOR & SHORT CIRCUIT FORCE

................ I_sc = 40,000 A ........ D_ft = 10 ft

...................... 1998 ......... | ..... 2008
................. 60 Hz .. 0.5 sec* . |
. X/R .... T_a ..... D_f ... F_sc ... | .. D_f ...... F_sc .. 2008 / 1998
.. 0 ... 0.00000 . 1.0000 . 14.400 .. | .. 0.5000 .. 14.400 ... 1.0000
.. 1 ... 0.00265 . 1.0026 . 14.476 .. | .. 0.5216 .. 15.671
.. 2 ... 0.00531 . 1.0053 . 14.553 .. | .. 0.6039 .. 21.009
.. 5 ... 0.01326 . 1.0132 . 14.782 .. | .. 0.7667 .. 33.863
. 10 ... 0.02653 . 1.0262 . 15.164 .. | .. 0.8652 .. 43.118 ... 2.8434 Three times greater
. 20 ... 0.05305 . 1.0517 . 15.928 .. | .. 0.9273 .. 49.531 ... 3.1097 at typical values of
. 30 ... 0.07958 . 1.0766 . 16.692 .. | .. 0.9503 .. 52.016 ... 3.1162 X/R
. 50 ... 0.13263 . 1.1248 . 18.218 .. | .. 0.9696 .. 54.146
100 ... 0.26526 . 1.2322 . 21.863 .. | .. 0.9845 .. 55.832
300 ... 0.79577 . 1.4624 . 30.796 .. | .. 0.9948 .. 47.002 ......... At the limit, the 2008
.. ~ ...... ~ .... 1.7320 . 43.200 .. | .. 1.0000 57.600 1.3333 equation is 4/3 times
..................................................................... greater than the 1998
..... * Fault Current Duration
..... ~ Infinity

.... Notes:
...... 1. The 1998 decrement factor approaches the square root of 3 as the X/R ratio goes to
......... infinity because the fault current duration is one half. The exponential
......... expression and the T_a value go to unity as X/R goes to infinity, so the 1998
......... equation reduces to square root of one plus the reciprocal of one half, or the
......... square root of three.

Also, Table 14 of IEEE 605 2008 makes me wonder if there is a typo. Table 14 shows the decrement factor and the square of the decrement factor as a function of X/R (& T_a) for both 50 Hz and 60 Hz systems. However, the decrement factor as defined in the standard is independent of the system frequency. Per eq 19, the decrement factor is a function of the product of the frequency and T_a (approximate period??):

.... T_a = X/R x (1 / (2 x pi x f))

... f x T_a = X/R x (1 / (2 x pi))

which results in a the decrement factor being merely a function of the X/R ratio.

Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Reply To This Thread

Posting in the Eng-Tips forums is a member-only feature.

Click Here to join Eng-Tips and talk with other members!


Resources