Is There a Ratio Between Short Term and Continuous Current for Power Cables?

[rontruong]

Hi guys,

I'm looking to find out if there is a ratio between short term and continuous current for power cables. I know that people usually generalize around 5-10 x rated ampacity, but I'm just asking around to see if there is a certain #. On McMaster Carr, the highest rated cable is for 600 AMPS continuous under welding cables. I will be needing to short a 125 VDC capacitor module that is estimated to put out 8000-9000 amps for less than 1 second.

Let me know if you guys need any other information.

 

[negitron]

Should be fine with something like #2 AWG.

 

[m.s.j]

For determination of cable withstands short time current following relation shall be considered:

S =( Ia.√t) / k

Where :

S : Conductor cross section in mm2
Ia: Effective (rms/amount of DC current) value of current (A)
t: duration of hazardous shock current (sec.)
k: Cable Insulation Material Coefficient:

For PVC-insulated CU conductors : k=115 As/mm2-
For PVC-insulated Al conductors : k=76 As/mm2-
For rubber-insulated CU conductors : k=141 As/mm2-

For your concerned case : t = 1 s , Ia=9000 A , k=115 (e.g.) the suitable cable size is S=78 mm2 and rated size is S=95 mm2


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[negitron]

Capacitors don't have a constant-current discharge curve.

 

[Carl Pugh]

Capacitors don't like being shorted.

Discharge capacitor into a resistor.
A wire wound resistor is probably preferred for this application.

Probaly #16 wire will be Ok with any reasonably resistance resistor.

You can calculate the temperature rise of the wire, or much easier, just try it (being careful that if wire or resistor or capacitor explodes, no one is injured). If everything works OK, then maybe go to a higher wattage resistor and larger wire to have a safey factor.

Just out of curiosity, what are you using to short the capacitor?

 

[m.s.j]

For variable current functions, the rms value can be calculated as follow if we access to its current-time curve:

Ia = Irms = √ (1/T∫ i2 .dt)

 

[rontruong]

For determination of cable withstands short time current following relation shall be considered:

S =( Ia.√t) / k

Where :

S : Conductor cross section in mm2
Ia: Effective (rms/amount of DC current) value of current (A)
t: duration of hazardous shock current (sec.)
k: Cable Insulation Material Coefficient:

For PVC-insulated CU conductors : k=115 As/mm2-
For PVC-insulated Al conductors : k=76 As/mm2-
For rubber-insulated CU conductors : k=141 As/mm2-

For your concerned case : t = 1 s , Ia=9000 A , k=115 (e.g.) the suitable cable size is S=78 mm2 and rated size is S=95 mm2-----------------------------------------------------------------
Creative thinking is enjoyable, Then think about your surrounding things and other thought products. http://electrical-riddles.com

Okay, I think I'm going to listen to you for simplicity reasons. With the 95 mm^2 results, then a size of 4/0 AWG copper cable should be sufficient. Can you show me where you got the equation and k values from, I need it for a presentation. Thank you for your help.

Capacitors don't like being shorted.

Discharge capacitor into a resistor.
A wire wound resistor is probably preferred for this application.

Probaly #16 wire will be Ok with any reasonably resistance resistor.

You can calculate the temperature rise of the wire, or much easier, just try it (being careful that if wire or resistor or capacitor explodes, no one is injured). If everything works OK, then maybe go to a higher wattage resistor and larger wire to have a safey factor.

Just out of curiosity, what are you using to short the capacitor?

Actually, the purpose of the test is to short the capacitor by itself. I know its dangerous, but for this company, the experiment is to determine if the capacitor module can survive a short circuit along with all the balance boards + monitoring boards.

 

[rontruong]

For variable current functions, the rms value can be calculated as follow if we access to its current-time curve:

Ia = Irms = √ (1/T∫ i2 .dt)

I redid calculations with real RMS values and the cable size is much much smaller!

thanks for the help!

 

[negitron]

I redid calculations with real RMS values and the cable size is much much smaller!

How much smaller?

 

[rontruong]

~40 mm^2 cross section

mostly because the current decays so fast after 1 time constant, so this makes sense.

 

[negitron]

Can't beat 30 years electrical experience. 2 AWG wire has a cross-sectional area of ~34 mm²--close enough for government work.

 

[m.s.j]

Okay, Can you show me where you got the equation and k values from, I need it for a presentation. Thank you for your help.QUOTE]

References:

1- ICEA ( the Insulated Cable Engineers Association) P-45-482-1994

2- Electrical handbooks such as : " switching, protection and distribution in low voltage network" handbook of simense. Section 9.4.1.2

 

[rontruong]

ty :)

 

[Carl Pugh]

Easy way to calculate cable temperature rise is to
Calculate energy stored in capacitor.
Calculate weight of copper (or aluminum) in cables).
Assume all the energy stored in capacitor is transferred to the copper in the cables.

 

[Darryl]

I believe there has been a good amount of research on this subject, mainly in High power radio, High power pulsed RADAR and I expect also in high energy physics work.

I know in High power radio transmitters, even continuous ones, the peak's of very high current and voltage (power) causes local and transient IR losses on the High Tension power cables feeding the transmitting valves.

This variation of the High Tension Voltage supply results in spurrious and parasidic radiation of the transmittion, because of the modulation effects of the varying High Tension supply.

..._____
___l_____l____

I know its not very clear, but that kind of arangement is used for parsidic suppression, it does not increase the total cross sectional area, but it reductes (by shorting out) the instanious losses to to high speed loads.