DC circuit breaker capacity vs time constant 1

[Yuma]

Dear all,
I have a question on the breaking capacity of a DC circuit breaker.
If a DC circuit breaker is able to break 50 kA with a time constant of 15 ms, how do you calculate the new breaking capacity for other time constant, say 45 ms for example?

Searching the web I came across the following formula: I2 = I1 * sqrt (t1/t2), so it would be 50 *sqrt(15/45) = 28,9 kA

But the formula was given 'as is', with no explanation.

Can anybody confirm that this formula is correct, and if so, explain where it comes from, or point me to a good reference where it is explained?

Thanks

 

[Compositepro]

I guess because power dissipated is proportional to I squared, and energy is P times t.

 

[waross]

I am sure that that formula is valid for something.
I suspect that it takes more than fun with numbers to re-rate a DC circuit breaker.
It may be safer to use the manufacturer's type tested and verified interrupting ratings.

 

[Gr8blu]

Searching the web I came across the following formula: I2 = I1 * sqrt (t1/t2), so it would be 50 *sqrt(15/45) = 28,9 kA+

In general, the formula is referring to the amount of heat buildup because of electric current. A lower current for a longer time creates the same amount of heat (or thermal stress, if you prefer) as a higher current for a shorter time period. This is how the duration of applicable momentary or repetitive electrical loading is determined.

It is NOT applicable to the make/break performance of a circuit breaker - AC or DC.

 

[FacEngrPE]

Cataloged circuit breaker (AC and DC) interrupting ratings need to be respected, changing trip curve time constants does not change the interrupting rating.

IE if the circuit breaker is rated at 120V and 50,000A interrupting capacity, exceeding the interrupting rating can result in the breaker failing to clear the fault when it opens. This generally results in destroyed equipment.

An apparent exception to the above rule is encountered when there is a highly inductive load.
The problem here is that when a circuit with a highly inductive load is interrupted, the inductive energy will cause very high voltages to appear across the breaker terminals. This is sometimes solved by installing freewheeling diodes across the inductor.

 

[Yuma]

In general, the formula is referring to the amount of heat buildup because of electric current. A lower current for a longer time creates the same amount of heat (or thermal stress, if you prefer) as a higher current for a shorter time period. This is how the duration of applicable momentary or repetitive electrical loading is determined.

It is NOT applicable to the make/break performance of a circuit breaker - AC or DC.

Thanks Gr8blu!

 

[Yuma]

Hello to all, and sorry for not having answered before, I was on holidays.

FacEngrPE: yes, I agree with what you say, but I am still wondering how to assess (if possible) the validity of a breaker for a higher time constant. According to the nameplate, the breaker's rating in DC is 250 V, 50 kA and t= 15 ms. This happens to be an IEC breaker, but I also tried IEEE in case I could find some guidance.

So I read an ABB manual about DC circuit breakers, and also IEC 60947-2, IEEE 946, IEEE 1375 and IEEE C37.14

I am not reproducing here the specific wording because I don't want to infringe any copyright, but basically all I have found is that if the time constant exceeds the nameplate value, the manufacturer of the breaker should be consulted.

I was hoping I could find a way to calculate the new V and kA rating for a bigger time constant, but maybe I was wrong and it is impossible.

Thanks to all the responders

 

[Gr8blu]

DC switching happens FAST. Much faster than with AC circuits, usually.
If you're trying to protect rotating machines, think about how fast that switching really is.
In a typical DC motor, the direction of current flow in the armature changes from 1 per unit (forward) to 1 per unit (reverse) in approximately 0.066 MILLISECONDS.
That's roughly 227 times FASTER than the 15 ms breaker acts, and orders of magnitude faster still by the time all the delays (detect the signal, tell the breaker, activate the breaker, and clear the arc) are added up.

 

[waross]

I suspect that the time constant refers to the circuit being interrupted.
The higher the time constant, often results in a higher inductive kick voltage.
This may exceed the safe voltage to ground or to opposite polarity of the breaker.
I am not absolutely sure as I haven't run the numbers.
I suggest that you do some calculations as to the maximum possible inductive kick voltage at different currents and time constants..