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Time Constant of Circuit Breakers

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Siyou

Electrical
Jul 12, 2020
30

Dear All,

How can I know the time in milliseconds or the breaker or the time constant, if the breaker has 44% DC component,
25kA breaking current and 65kA making current?


Thank You
 
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The electrical system has a time constant at each fault point, not the breaker. It's calculated. Basically it's the X/R ratio at the point of the fault.
 
Manufacturers curves.

Bill
--------------------
"Why not the best?"
Jimmy Carter
 
@ waross is there any other way? the manufacturer didnt reply to me.
The manual only have opening time, breaking time, closing time and dead time in milliseconds.
It has also the operating duty cycle.

@dpc how about the X/R test ratio of CB, because it will affect the asymmetrical interrupting rating if the X/R calculated is higher than the X/R test ratio..
How can I know the X/R test ratio of the CB.

Thank You.
 
The manual only have opening time, total break time, closing time, and dead time..
are there any other way I can know the minimum time delay of CB?

Thank You
 
You are using several posts for the same question.What is the the opening time given in the manual?
 
Regarding test X/R of breaker - that is based on the standard the breaker was built and tested to. If you don't know or have access to the standard, I would ask the manufacturer.

In the ANSI world, if the calculated symmetrical fault current is less than 80% of the breaker's symmetrical rating, you don't have to worry about the dc offset or X/R.

 
According to IEC medium-voltage circuit-breaker interrupting ratings-unstated short-circuit considerations see: The breaker's opening time (Top) is the time from receipt of trip command to the first contact parting.
For breakers tripped by auxiliary power, a relay time (Tr) of 10 msec is added.
The 'standard' IEC breaker is designed for a system time constant not exceeding 45 msec.
idc=√(2)I"ke-2πftR/X [IEC 60909-0 ] idc%=idc/I”k%
where:
I"k is the initial symmetrical short-circuit current
f is the nominal frequency
t is the time[sec]
From here:
R/X=LN(idc/I"k/√(2))/(2*π*f*t)
idc%=idc/I”k% I”k=Ibreak
R/X= 0.082588 [or X/R=12.1[rated for this breaker].
The actual R/X it has to be calculated from the network data.
 
Thank You Everyone,

@Kiribanda the maximum closing time, total breaking time, opening time and deadtime stated in the manual are:
Opening time : 35ms
Closing time : 60 ms
Total breaktime: 55 ms
Deadtime : 300ms
Rated reclosing time : 20 cycles
RATED OPERATING Sequence : O – 0.3 sec – CO – 3 min – CO
ON-OFF time: 40 ms

The manual used Values complying with IEC 62271-100 and ANSI C37
 
What Will I do The manufacturer didnt reply?


Thank You everyone for sharing your knowledge and expertise
 
From the given operating time the minimum delay should be 2ms so that,
Opening time + Minimum delay time (Total relay time) = 35ms+2ms=37ms
to give 44% DC component as given in the bkr name plate.
I should say that 2ms is an extremely short delay time which is 1/8th of 60Hz wave
to use as the minimum delay for this type of CB because 2ms protection
is ONLY possible with fast semicincductor CL fuses. Normally we take half cycle (8.3ms on 690HZ)
as the minimum standard delay time. You are fee to use any reasonable delay time in your calculation.
Now if you put 37ms in your ETAP as the minimum delay, you should get the DC component as 44%.
As I mentioned before in my very first post, use your protection relay time and get the
corresponding DC component. It should be definitely lower than 44%.
 
1)Your Original Question: Since your bkr has opening time=35ms per its manual, the Minimum Delay time should be 2ms. At 37ms the breaker can break 44% DC component as per the name plate.
2)Minimum Delay of IEC Bkrs: I think you are trying to compare apples to oranges. If you select an IEEE C37.06 bkr having a 2Cy CPT, then it is capable of breaking an asym sc current having a 47.68% DC component. It is same as selecting an IEC bkr having a Minimum Time Delay of 33.2ms. There is no term call CPT for IEC bkrs. But it is the time that the bkr had been delayed to break an asym. sc current having a maximum 47.68% DC component.
3)Total Bus Fault Current & Max. Through Fault Current: Again I should mention that ETAP is just a tool and you have to use it as per your requirements. So you have to select one of these two options to compare it against the duty of the bkr.
 
Thank You @ Kiribanda
Thank You for sharing your knowledge and expertise.
The nameplate of the PCB used the IEC 62271-100. While using 0.037sec the asymmetrical current exceeds the device capability as well as Idc exceeded.

Anyways, the system where I did its short circuit calculation has 72.5kV PCB with 25kA breaking current and 65kA making current with 44% dc component rated in IEC standard which is used in the 69kV line, the 13.8 kV line used 15kV vacuum circuit breaker with rated short circuit current of 25kA, the rating used ANSI standard. Since the system used both ANSI and IEC standards. I'm confused about what standard I will use in my short circuit calculation. I tried using only ANSI standard calculation for the whole system. What will I do in my short circuit calculation? What Standard will I used in the computation?

Attached here is my result in ETAP using 0.037 sec as minimum time delay,.
Will I used only the symmetrical breaking current result or the asymmetrical breaking current to compare with the nameplate breaking current?
How can I evaluate the result based from the short circuit rating in the nameplate?
 
 https://files.engineering.com/getfile.aspx?folder=9efd8285-193a-42ec-8b19-3d4c720cc0fc&file=IEC_ETAP_result.docx
Will I used only the symmetrical breaking current result or the asymmetrical breaking current to compare with the nameplate breaking current?
How can I evaluate the result based from the short circuit rating in the nameplate?
The X/R ratio is greater than 17
 
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