×
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

Questions about IEEE Std 551-2006

Questions about IEEE Std 551-2006

Questions about IEEE Std 551-2006

(OP)
I have some couple of confusions regarding IEEE Std. 551-2006 which I would appreciate much if anyone in this forum may enlighten me further:

1. Equation 9.8 in section 9.6.3 (page 216) of IEEE std 551-2006 indicates an equation to calculate the peak current for unfused low voltage circuit breakers. However, the peak equation in section 9.1.4.4.c) in ANSI C37.13-2008 indicates the same equation but with a 2.29 divisor. Which one is correct? Where did the 2.29 divisor came from?

2. Similarly, Equation 9.7 in section 9.6.3 (page 216) of IEEE std 551-2006 indicates an equation to calculate the asymmetrical current for fused low voltage circuit breakers. However, the asymmetrical equation in section 9.1.4.4.d) in ANSI C37.13-2008 indicates the same equation but with a 1.25 divisor. Which one is correct? Where did the 1.25 divisor came from?

3. Is the LV circuit breaker definition in sectionsection 9.6.3 (page 216) of IEEE std 551-2006 applies to both LV power circuit breakers and LV molded case circuit breakers?

RE: Questions about IEEE Std 551-2006

I think IEEE-Std 551-2006 is based on IEEE Std C37.13-1990 [ as mentioned in 9.6.3 First cycle currents Ch.]
The new one-IEEE/ANSI C37.13-2008-I have not this standard so I can't say more- may be an improved one.
But compared with IEC 60909-1/2001 - eq.54 :
ip=K*sqrt(2)*I"k   [I"k=Isym.]
K=1.02+0.98*exp(-3/(X/R))  for non-mashed networks [approx.the IEEE 551 value]
For mashed networks  ip=1.15*K*sqrt(2)*I"k [that means "more" not less!]
For Iasym formula IEC 60909-1 gives separately idc=sqrt(2)*I"k*exp(-4*pi()*f*t/(X/R))
if t=1/f -at the end of first cycle-  idc=sqrt(2)*I"k*exp(-2*pi()/(X/R))
so Iasym=sqrt(1+idc^2) as per IEEE 551
If t=1.5/f [1 and a half cycle] you have to divide by 1.25 [but it is beyond "first cycle"].
 

RE: Questions about IEEE Std 551-2006

If I understand well these sentences from  ch. 9.6.3 of IEEE-Std 551-2006 :
"Unfused low-voltage circuit breakers need to be evaluated on basis of first cycle peak currents.
Due to the fact, however, that these breakers are rated on a symmetrical basis according to IEEE Std C37.13,
there is already an embedded asymmetry assumed that rests on assumption of 15% test power factor,
equivalent to a test fault point X/R ratio of 6.6.This necessitates a further calculation for breaker duty only
 when power factors smaller than 15%(X/R ratio greater than 6.6) are encountered."
"Fused low-voltage circuit breakers are evaluated on basis of total asymmetrical rms first cycle currents.
Due to the fact, however, that these breakers are rated on a symmetrical basis according to IEEE Std C37.13,
there is already an embedded asymmetry assumed that rests on assumption of 20% test power factor,
equivalent to a test fault point X/R ratio of 4.9.This necessitates a further calculation for breaker duty only
 when power factors smaller than 20%(X/R ratio greater than 4.9) are encountered."
it seems to me that IEEE C37.13 takes into account the X/R =4.9 for fused breaker and 6.6 for unfused.
For any pf more than 0.15 [respectively 0.20] the breaker duty will be as per above standard, that means the derating factor
will be 1.For less one has to calculate it.
We can calculate X/R=tan(acos(pf))
In this case the maximum derating factor [for pf=0.05 X/R=19.97 ] will be for unfused 1.144 [Ipeak/1.15 approx.]
and for fused 1.258 [Iasymm/1.26].
The factor 2.29 seems to be 2*1.144=2.288 .But I cannot say why.
 

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