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charz

Electrical
Jan 11, 2011
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My specification calls for a short circuit rating of 35kA rms and 77kA peak value for a junction box with a bus bar in it.
The vendor proposes a junction box with a short circuit tested at 90kA for 0.1s.

He justifies that peak mechanical effects (potential damage) to a product due to a short circuit fault occurs in no more than 5ms. Any time longer than 0.1s would only serve to heat the conductors hence longer testing is unnecessary to determine fault survival capability.

Then I converted the rating of 90kA @ 0.1s to 1s using the adiabatic equation I squared t
(90 kA for 0.1s = 8.1E+08) << (35kA for 1s = 12.25E+08)

but still it is less than what the spec calls for.

How would you go with this?
 
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You are asking 35kA sym and 77kA peak for the bus bar strength of the JB.
That means the test PF=41%. The bus bar strength should be tested for the peak
and not for the sym wave because during short circuit, the magnetic forces are
coming at the first peak. Thermal effects come later.
Therefore, the manufacturer has to set his SC current with PF=41% and Isym=35kA.
This 41% DC offset should be depicted in the sc test current oscillograph at the test lab.
If he has tested for 90kA and the tested wave is fully symmetrical, then your requirement is not met.
 
I am confused. Maybe I will learn something. 90 kA symmetrical is 127kA peak.
The spec is 70 kA peak.
Why would 90 kA symmetrical not be greater than 70 kA peak?

--------------------
Ohm's law
Not just a good idea;
It's the LAW!
 
@ waross.
We are talking about asymmetrical current during a fault. What you are trying to calculate is the relationship of rms current to the peak value of an AC sine wave (1.414 x rms current). The AC current during the first half-cycle is the peak being talked about(value of the fault current at 0.01 seconds from inception of fault). ANSI uses the factor 2.6 times Isym as the peak current at 0.01 seconds. It is the fault current most electrical equipment must withstand during asymmetrical faults. With 35 kA rms symmetrical amps, your equipment should be rated for 2.6 X 35 kA = 91 kA to be safe. 70 kA peak perhaps is the computed peak other than X/R of 17 (ANSI base X/R).
 
What I am talking about is the 90 kA test. It does not specify that this was 90 kA peak. If the 90 kA was symmetrical, the peak would have easily exceeded the 91 kA peak that is required.
When a current value is not qualified as peak, asymmetrical, peak to peak, or average, it is generally assumed to be an RMS value.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!
 
Dear charz
"... My specification calls for a short circuit rating of 35kA rms and 77kA peak value for a junction box with a bus bar in it. The vendor proposes a junction box with a short circuit tested at 90kA for 0.1s....".
I am of the opinion on the [dynamic] expect that:
1. factor k=77kApeak/35kArms = 2.2 . This factor (k=2.2) is rather on the high side; as it usually not exceeding 2.0. In this case [ R<<<X ].
2. If the box is tested:
a) to 90kArms 0.1s. For 90kArms test, the k factor would be say 2.4. The test value would be 216kA peak. It fulfilled the 35kArms requirement,
b) to 90kApeak 0.1s . It fulfilled the 77kApeck requirement.
Che Kuan Yau (Singapore)
 
90kA for 100mS, I doubt that is peak. If it was 10mS, then it could be. I also doubt it was fault tested into a resistive fault only. The test labs always use an inductive fault load.

Hence why I asked about the time. That's the only part that doesn't make sense to me. If this is low voltage, then generally you fault test long enough to trip a plain old thermal magnetic breaker.

 
How does the thermal capability assuming 90kA for 0.1s compare with a thermal withstand calculation based on the sizes of the conductors involved? The test the manufacturer performed may not have been intended to test both the thermal and physical withstand capabilities in the same test.
 
Dear Mr. charz
@bacon4life
1. I wish to point out the (dynamic)= [mechanical withstand strength] and the (thermal)=[temperature limitation] are two separate properties.
2. For Copper, the (final temperature) after the short-circuit is [usually limited to 200 degC ]. The initial temperature of the bar is dependent on the loading, before the short-circuit.
3. For example: the ambient temperature is 30 degC, but the copper bar had been on load. The temperature is 70 degC , when shot-circuit occurred. Taking the initial temperature of 70 degC and final temperature of 200 degC within 1s duration; the current density is estimated to be around 132 A/sqmm, with any conductive heat removal disregarded. With this information, it is possible to estimate the minimum copper bar size.
4. See IEC 60685-1 for detail.
Che Kuan Yau (Singapore)


 
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