Circuit Breaker settings
- Thread starter steve66
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- Wisconsin
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- PE (Retired) - Power Systems
Re: Circuit Breaker settings
It has to do with the device's trip curve.
If there is a sharp (right angle) transition from one trip band to another (usually "short time delay" to the instantaneous trip) then I^2t is OUT. If there is a slope (a gradual) transition then I^2t is IN.
For most applications OUT is the typical setting. The IN setting is used to allow an upstream setting to coordinate "closer" to a fuse or to a device with a similar trip curve.
I used this feature on my last two ground fault coordinations projects.
It has to do with the device's trip curve.
If there is a sharp (right angle) transition from one trip band to another (usually "short time delay" to the instantaneous trip) then I^2t is OUT. If there is a slope (a gradual) transition then I^2t is IN.
For most applications OUT is the typical setting. The IN setting is used to allow an upstream setting to coordinate "closer" to a fuse or to a device with a similar trip curve.
I used this feature on my last two ground fault coordinations projects.
- Location
- Lockport, IL
- Occupation
- Retired Electrical Engineer
Re: Circuit Breaker settings
OK. I'll admit to a long standing ignorance of something here.
The unit of measure for current (amps) breaks down to fundamental units of "coulombs per second." Thus, the units of I^2T would be "coulombs squared per second squared times seconds," or just "coulombs squared per second." I can't see any physical meaning to that unit of measure. What does it mean?
OK. I'll admit to a long standing ignorance of something here.
The unit of measure for current (amps) breaks down to fundamental units of "coulombs per second." Thus, the units of I^2T would be "coulombs squared per second squared times seconds," or just "coulombs squared per second." I can't see any physical meaning to that unit of measure. What does it mean?
petersonra
Senior Member
- Location
- Northern illinois
- Occupation
- engineer
Re: Circuit Breaker settings
The thermal trip curve typical follows I^2. T is time. The thing trips eventually when the area under the curve exceeds the trip setting. So it will trip much faster at higher currents than at much lower currents.
Roughly, if a breaker trips in 100 seconds at 30 amp current, it should trip in 9 seconds at 100 amps. This is fairly intuitive because heat is produced proportional to the square of the current, and the thermal side of the breaker trips when it gets warm enough.
It is not all that simple because it is a mechanical device, but that is a very basic view.
It is all about how much heat the breaker produces and how fast it can be dissipated.
The thermal trip curve typical follows I^2. T is time. The thing trips eventually when the area under the curve exceeds the trip setting. So it will trip much faster at higher currents than at much lower currents.
Roughly, if a breaker trips in 100 seconds at 30 amp current, it should trip in 9 seconds at 100 amps. This is fairly intuitive because heat is produced proportional to the square of the current, and the thermal side of the breaker trips when it gets warm enough.
It is not all that simple because it is a mechanical device, but that is a very basic view.
It is all about how much heat the breaker produces and how fast it can be dissipated.
- Location
- Lockport, IL
- Occupation
- Retired Electrical Engineer
Re: Circuit Breaker settings
Sorry, I wasn't clear about my question. It's hard to be clear about being confused.
I know how the curves are used, and I understand the relationship between higher currents and faster trip times. What I don't understand is the "unit of measure."
Let me give an example. I understand how multiplying voltage by current will give you a unit of power. Voltage is the energy needed to move an amount of charge from one point to another. Current is the rate of movement of charge. So the multiplication of "V" times "A" will give you units of (joules per coulomb) times (coulombs per second), with a final result of (joules per second). I know that power is the rate of use (or generation) of energy, so getting "joules per second" from VA makes sense.
Now, my question is, how do you get units of heat, or energy, or power, or anything else of physical significance, from "coulombs squared per second"?
Sorry, I wasn't clear about my question. It's hard to be clear about being confused.
I know how the curves are used, and I understand the relationship between higher currents and faster trip times. What I don't understand is the "unit of measure."
Let me give an example. I understand how multiplying voltage by current will give you a unit of power. Voltage is the energy needed to move an amount of charge from one point to another. Current is the rate of movement of charge. So the multiplication of "V" times "A" will give you units of (joules per coulomb) times (coulombs per second), with a final result of (joules per second). I know that power is the rate of use (or generation) of energy, so getting "joules per second" from VA makes sense.
Now, my question is, how do you get units of heat, or energy, or power, or anything else of physical significance, from "coulombs squared per second"?
- Location
- Wisconsin
- Occupation
- PE (Retired) - Power Systems
Re: Circuit Breaker settings
Charlie,
Because grandpa did?
IMO, to re-enforce the concept that we are dealing with a "standard" inverse time function the phrase "I squared t" is used rather than a resultant energy unit. This terminology allows a differentiation between typical trip curves like inverse, very inverse, and extremely inverse. the use of the resultant energy value would require additional explanation in order to compare one curve against the other.
Charlie,
Because grandpa did?
IMO, to re-enforce the concept that we are dealing with a "standard" inverse time function the phrase "I squared t" is used rather than a resultant energy unit. This terminology allows a differentiation between typical trip curves like inverse, very inverse, and extremely inverse. the use of the resultant energy value would require additional explanation in order to compare one curve against the other.
- Location
- Mission Viejo, CA
- Occupation
- Professional Electrical Engineer
Re: Circuit Breaker settings
Basically it is an empirical statement that, for short periods of time (usually 2 seconds or less), DAMAGE at a faulted location is proportional to I^2 t.
In other words, I would expect roughly the same amount of DAMAGE from a 5000A fault for 0.1 sec as I would from a 10000A fault for 0.025 sec.
Edit Add: BTW, Jim and the ?other? Bob?s statements are also true but, for longer periods of time, other heat dissipation phenomena begin to exert more influence.
[ February 14, 2006, 11:51 AM: Message edited by: rbalex ]
Basically it is an empirical statement that, for short periods of time (usually 2 seconds or less), DAMAGE at a faulted location is proportional to I^2 t.
In other words, I would expect roughly the same amount of DAMAGE from a 5000A fault for 0.1 sec as I would from a 10000A fault for 0.025 sec.
Edit Add: BTW, Jim and the ?other? Bob?s statements are also true but, for longer periods of time, other heat dissipation phenomena begin to exert more influence.
[ February 14, 2006, 11:51 AM: Message edited by: rbalex ]
- Location
- Lockport, IL
- Occupation
- Retired Electrical Engineer
Re: Circuit Breaker settings
OK. That I can buy. Many thanks.
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