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CB Short Time Withstand Current selection
3

CB Short Time Withstand Current selection

CB Short Time Withstand Current selection

(OP)
Dear All,

  In our LV distribution system the MCCs have  Short Time Withstand Current "Icw" of 65kA/1sec however the offered incomer CBs for these MCCs (Located phisicaly in the Load Center) is 15kA/1sec with 65 kA service short circuit breaking capacity "Ics"; Does this work?
  
Is there  any solid reference regarding this issue rather than IEC 947-2 & 439-1?

What will be the case if the offered incomer CB is of a limitting current type?

RE: CB Short Time Withstand Current selection

What is the available fault current at the MCC, at the main breaker?

RE: CB Short Time Withstand Current selection

The Icw value is the ultimate breaking caapcity: after an interruption the breaker is not expected to be fit for service, or at least not without major overhaul. The Ics value is the service breaking capcity: after an interrruption it is expected to be capable of being put back in service.

What type of breaker is it - an ACB or an MCCB?
  

----------------------------------
  
If we learn from our mistakes I'm getting a great education!
 

RE: CB Short Time Withstand Current selection

2
No, Icw is not the ultimate breaking capacity. Icu is. Herewith some description of these terms:

Icu  =  rated ultimate short-circuit current
This is the maximum symmetrical short-circuit current the breaker can interrupt and is tested as follows, O – t – CO.

O    =  breaking operation
t    =  time separating two operations equal to 3 minutes or the length of time needed to reset the breaker, whichever is longer.
CO    =  making operation followed by a breaking operation
  
After the test, the breaker is tested to show that it is electrically safe. The dielectric test involves applying twice rated voltage, the minimum being 1000V.

Ics  =  rated service short-circuit current
It is tested as follows, O – t – CO – t – CO. Following the test a temperature rise, dielectric and tripping test is carried out. This is to verify that the breaker can be returned to service. In essence, Ics is the maximum current the breaker can interrupt three times and be returned to service with its operational integrity intact.

Ics is expressed as a % of Icu (25%, 50%, 75% or 100% of Icu).

Both Icu and Ics are threephase symmetrical (r.m.s.) current duties.

Icm  =  rated short-circuit making capacity
This is the half-cycle peak of the fault current waveform that the breaker has to close onto.

Icw  =  Rated short time withstand current
This is the steady state symmetrical fault current the breaker has to be able to carry for a duration of 1s or 3s without violating its thermal integrity. In order to verify that a breaker is capable of handling a given short-time current, the protection operating time of the breaker needs to be known. A more conservative estimate may see the back-up protection operating time being used.

Alsherbieny:
Your question of whether the breaker is suitable is not a simple as it seems. If you live in the IEC world, you need to have an IEC 60909 faultstudy done in order to determine the following:

I"k  =  initial symmetrical short-circuit current at the fault location. This is the maximum r.m.s. value of the symmetrical fault current. If there was no AC decay, then this current would be constant in that the peak-to-peak values would not change with time. If there is only DC offset (or DC decay) then the peak-to-peak distances remain the same but are not symmetrical w.r.t. the horizontal axis. With only AC decay, the peak-to-peak distances remain symmetrical w.r.t. the horizontal axis but they decrease with time to a steady state value.
  
Ik  =  steady state current. This is the conventional symmetrical fault current when the AC and DC decrements have disappeared. For far from generator faults, Ik  =  I"k.

IDC  =  DC Current. The exponentially decaying DC component of the fault current waveform due to the inductance of the system. It is determined by the point on the voltage waveform at which the fault occurs. Note this is not a current that can be measured, rather it is calculated. It is a mathematical model of a decaying transient inherent in the fault current waveform.
 
Ip  =  Peak Current. This is the maximum instantaneous peak of the fault current waveform i(t). Used to specify electrodynamic stresses of switchgear.

Ib  =  Breaking Current.  A current duty intended for the rating of the current interrupting device for when the contacts part. For far contributions, breaking duty equals initial symmetrical current, I"k. For near contributions special multipliers need to be applied to account for the AC decrement. Ib has no DC decrement but Ib(asym) considers both AC and DC decrement. Ib is a symmetrical r.m.s. current 50ms (for example) down the line from I"k.

With regards to current duties the following must be considered as a minimum:

1. Ics > Ik"
2. Icm > Ip  (this is a very important criteria which is often overlooked)
3. Icw > Ib (or even more concervatively Icw > Ik")

Some may suggest Icu > Ik". However, I deem it much safer to use Ics > Ik".

Regards.


  

RE: CB Short Time Withstand Current selection

(OP)
Davidbeach,
 The fault level at the MCC according to the ETAP is 20kA.
ScottyUK,
Your definitions may be not accurate, while the type of breaker is ACB.
Veritas,
 Thanks for your detailed reply. First of all, I am living in an IEC world (at least for the switchgears).
1. I still want to know if I can protect an MCC of Icw=65kA/1sec with an ACB (Loacted in the Load Center) of Icw=15kA/1sec. however both have Ics=65kA & larger.
2. What do you mean by current duties in "With regards to current duties the following must be considered as a minimum"?
3. What is the case if the ACB is of a current limiter type?
 

RE: CB Short Time Withstand Current selection

If the both have an interrupting rating of 65kA and the available fault current is 20kA, where's the problem?

RE: CB Short Time Withstand Current selection

"ScottyUK,
Your definitions may be not accurate"

May?? blush Thanks for the correction veritas. Seems I am suffering from brain fade. smile
  

----------------------------------
  
If we learn from our mistakes I'm getting a great education!
 

RE: CB Short Time Withstand Current selection

(OP)
davidbeach,
 The problem isn't the incomer ACB capability of clearing the fault, but it the best approach/practise of achieving a complete selective and reliable distribution system.
 Based on my understanding this incomer ACB will not sustain such MCC fault for more than 0.5 sec!
 

RE: CB Short Time Withstand Current selection

veritas,

Related to IEC 60909, I thought the Icw of the circuit breaker should be equal or greater than Ith.

Anyway, the calculation of Ith is always a topic of discussion within my workmates. IEC 60909 gives the "m" and "n" parameter calculation formulas, but as short circuit currents decrease with time until they reach the Ik value, some workmates suggest calculating the Ib value for different breaking times: 0.01, 0.02, 0.03, 0.05, 0.1, 0.2,..., 1.0 and the calculate the rms value.

 

RE: CB Short Time Withstand Current selection

Before I respond please could you define Ith for me?

Thanks.

RE: CB Short Time Withstand Current selection

veritas,

Ith is shown in point 4.8 of IEC 60909-0:

4.8 Joule integral and thermal equivalent short-circuit current

The joule integral li' dt is a measure of the energy generated in the resistive element of the system by the short-circuit current. In this standard it is calculated using a factor m for the timedependent heat effect of the d.c. component of the short-circuit current and a factor n for the time-dependent heat effect of the a.c. component of the short-circuit current (see figures 21 and 22)
The thermal equivalent short-circuit current is:
Ith=Ik''*SQRT(m+n)

For a series of i (i = 1, 2,....,r) three-phase successive individual short-circuit currents, the
following equation shall be used for the calculation of the Joule integral or the thermal
equivalent short-circuit current.

NOTE
The factors m and n first appeared as Figures 12a and 12b of IEC 60865-1 and are identical to them.
Up to now the thermal equivalent short-time current and the Joule integral are given in IEC 60865-1:1993.

RE: CB Short Time Withstand Current selection

radug

Sorry for the delay - been Christmas shopping - ho, ho, ho...

You raise an interesting point. I've never considered Ith before. Should I?

Quick question:
some workmates suggest calculating the Ib value for different breaking times: 0.01, 0.02, 0.03, 0.05, 0.1, 0.2,..., 1.0 and the calculate the rms value.

Did you mean Ith instead of Ib? Secondly Ith by definition must be a rms current as Ik" is an rms current. Thus do you mean to calculate the average for 0.01, 0.02, etc.?

I'll have to think some more about how to deploy the Ith term.

Regards.

 

RE: CB Short Time Withstand Current selection

radug

Considering the formula for Ith in IEC 60909 again, (Ith^2)*t  =  (Ik"^2)*(m+n)*tmax. Suppose t = tmax  =  1s. Then your formula Ith = Ik"*sqrt(m+n) is valid.

To get an idea of what we're looking at, suppose X/R  =  20. This will definitely cover transmission lines and cables and most small to medium sized transformers. From this get x  =  1.02 + 0.98*e^(-3R/X)  =  1.86 and from the table get m  = 0.1.

With regards to n, let Ik"/Ik  =  1.5. From tables get n  =  0.75. Thus sqrt(m+n)  =  0.92. Since this is less than Ik" I see no harm in just using Ik" to compare to Icw. Even Ib should be okay since for most ACB at the LV level operating time is about 20ms for the INST element.

Consider X >>> R so that x = 2. Then m = 0.3 and sqrt(m+n)  =  1.02. Using Ik" still fine.
Now suppose there is no AC decay such that n  =  1. With m  =  0.1 get sqrt(m+n)  =  1.05. This value of m was based on X/R  =  20. As X/R gets larger sqrt(m+n) approaches 1.14.

I guess for most cases using Ik" or Ib is sufficient.
I'll be interested to hear comments from you/others regarding this.

Then there is the case when t does not equal tmax...

Next time I'll address the original posting.

Regards.

RE: CB Short Time Withstand Current selection

ahelsherbieny

To get back to your original question:

In our LV distribution system the MCCs have  Short Time Withstand Current "Icw" of 65kA/1sec however the offered incomer CBs for these MCCs (Located phisicaly in the Load Center) is 15kA/1sec with 65 kA service short circuit breaking capacity "Ics"; Does this work?

1. So what you're saying is that you have an incomer with Ics  =  65kA and Icw = 15kA/1s? If that be the case then I find it very unusual. From what I've seen the Icw 1s value is very close or the same as the Ics value. The Icw 3s rating is substantially less than the Ics rating. So is your Icw 1s of 15kA obtained from manufacturer's specs?

2. If your Ik" fault level is 20kA, then whether the incomer breaker is suitable or not depends on the protection operating time. From a breaking current perspective, 65kA is more than enough. But using the I2t principle a 15kA/1s breaker can only handle a 20kA fault for at most 0.563s - and this is an estimation only, I would use 50% of this time as being my protection clearing time.

3. I have not heard of ACB's being current limiting - only MCCB's and MCB's. I may be wrong?

Regards.

RE: CB Short Time Withstand Current selection

Veritas,

I agree with your comments regarding Ith. The problem is that this is not clarified in IEC standards. Perhaps the upcoming IEC 62271-306 will clarify it.

Nonetheless, conceptualy it seems that Ith should be used for short-time withstand current, although from a practical point of view, using Ik'' will be enough.

Regarding your Ith examples, IEC 60909 says that in distribution networks n can be assumed to be equal to one. Doing so, any m calculation would produce a square root of a number a bit greater than 1, so Ith>Ik''.

Conceptually speaking I though that as Ith is the integral of current during say 1 second, as it takes into account both the symmetrical and asymmetrical components, its result should be greater than Ik''.
But, considering than motor contributions only last a few cycles, then fault current should decrease and for the most part of the 1 second duration motor contribution would not be present. And so, then perhaps Ith would be less than Ik''.
Problem here for me is that IEC 60909 formulas are based on Ik'', so it is difficult to obtain precise results, for example in Ith.
 

RE: CB Short Time Withstand Current selection

Hi radug

I think we are on the same page. Using Ith would be the ideal but there are practical limitations the biggest for me is that the engineering analysis software I use does not calculate Ith! I queried them about it and it appears that there was just never a need for it.

I will probably stick to Ik".

Regards.

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