Asymmetrical or symmetrical current?

[Grouch1980]

Hi all,
I attached a graph below that explains current limiting fuses. this is from the 'Selecting Protective Devices' Manual by Cooper Bussmann. I have a bit of confusion where the x-axis on the graph says 'Symmetrical RMS amps'. If you look on the current waveform on the right, it shows the first half cycle right after a fault condition... but this is the Asymmetrical current... this is not symmetrical. The values from the waveform are placed onto the graph on the left... how come the x-axis wouldn't be labeled "PROSPECTIVE SHORT-CIRCUIT CURRENT - ASYMMETRICAL RMS AMPS"?

 

[Grouch1980]

I think I can answer my own question... the waveform on the right is not showing the asymmetrical RMS current at all. I thought that was the asymmetrical RMS current.... it's actually showing symmetrical RMS current, so the 86,000 amps is actually the RMS current you will see once the DC component of the waveform decays and the symmetrical current is flowing. Is that correct?

 

[Ingenieur]

Looks like they are showing both

first few cycles sub-transient or asymmetrical
next cycles, transient, still asymmetric but tapering/settling
after those 2 stages steady-state or symmetrical

the fault current starts high and decays to steady state
most fault calcs are done steady state/symmetrical and a factor applied to get the sub-transient/asymmetrical, 1.5-1.7 depending on the x/r of the system

so although the cb must withstand the initial current most don't trip >3 cycles or in the steady state/symmetrical range

the cycles and peaking factor are obviously for example and will depend on the system

 

[Phil Corso]

Grouch...

The total current-wave is considered as Asymmetric because it is a function of time!

Isc, could include 3 Iac-components, the sub-transient I'', the transient I', Iss (steady-state), and Idc. I", I' and Idc decay to zero, with Idc being related to the X/R of the circuit to the point of fault!!

The RMS value at any point in time can be found using the Square-Rule:

Irms = Sqrt (Idc^2 + Iac^2).

Regards, Phil Corso

 

[Grouch1980]

Grouch...

The total current-wave is considered as Asymmetric because it is a function of time!

Isc, could include 3 Iac-components, the sub-transient I'', the transient I', Iss (steady-state), and Idc. I", I' and Idc decay to zero, with Idc being related to the X/R of the circuit to the point of fault!!

The RMS value at any point in time can be found using the Square-Rule:

Irms = Sqrt (Idc^2 + Iac^2).

Regards, Phil Corso

Thanks guys.... So the 86,000 RMS symmetrical amps (shown however on the asymmetrical waveform) is what you would see when the short circuit current stabilizes correct?

 

[topgone]

Thanks guys.... So the 86,000 RMS symmetrical amps (shown however on the asymmetrical waveform) is what you would see when the short circuit current stabilizes correct?

Nope, that's not what he meant. He's saying that the RMS value of the current (root-mean-squared) can be computed using the formula he just gave (sqrt(Idc^2 + Iac^2). During any energization, when the switch is closed at the instant the voltage wave is not zero, you will have a phenomenon called "DC-offset" current (due to asymmetry of resulting current). Take a close look at the formula; if there is no DC content, your RMS current will just be Iac.

 

[Grouch1980]

Nope, that's not what he meant. He's saying that the RMS value of the current (root-mean-squared) can be computed using the formula he just gave (sqrt(Idc^2 + Iac^2). During any energization, when the switch is closed at the instant the voltage wave is not zero, you will have a phenomenon called "DC-offset" current (due to asymmetry of resulting current). Take a close look at the formula; if there is no DC content, your RMS current will just be Iac.

Hi Topgone... That part i understand. If there's no DC component, then Irms = sqrt (Iac^2) = Iac. But my question is regarding the waveform... it's showing 86,000 amps RMS symmetrical current superimposed onto the asymmetrical cycle. Is this value of 86,000 amps without the Idc component?

 

[topgone]

Hi Topgone... That part i understand. If there's no DC component, then Irms = sqrt (Iac^2) = Iac. But my question is regarding the waveform... it's showing 86,000 amps RMS symmetrical current superimposed onto the asymmetrical cycle. Is this value of 86,000 amps without the Idc component?

Basically, what is shown is "symmetrical amps rms". The thing with rms values is that this is the current that will be generating heat to melt the fusible element.

 

[Ingenieur]

Hi Topgone... That part i understand. If there's no DC component, then Irms = sqrt (Iac^2) = Iac. But my question is regarding the waveform... it's showing 86,000 amps RMS symmetrical current superimposed onto the asymmetrical cycle. Is this value of 86,000 amps without the Idc component?

That is a representative/example curve of the fault current the fuse can handle
the fuse will open and clear well below that as indicated by
melt + arcing time = total clearing time

the rms is a calculated value based on the waveform ac/dc components as noted above

the peak let thru is 49 ka limited by the fuse
its rms value is 21 ka

the potential fault if uninterrupted is 198 peak/86 rms ka

 

[Sahib]

Hi all,
I attached a graph below that explains current limiting fuses. this is from the 'Selecting Protective Devices' Manual by Cooper Bussmann. I have a bit of confusion where the x-axis on the graph says 'Symmetrical RMS amps'. If you look on the current waveform on the right, it shows the first half cycle right after a fault condition... but this is the Asymmetrical current... this is not symmetrical. The values from the waveform are placed onto the graph on the left... how come the x-axis wouldn't be labeled "PROSPECTIVE SHORT-CIRCUIT CURRENT - ASYMMETRICAL RMS AMPS"?

A protective device is based on symmetrical rating taking into account any asymmetry by a suitable factor. That is why the graph's abscissa is marked in symmetrical current units.