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# Maximum let-through energy of fuse into cable - max at 5 seconds disconnect time

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 gilly1 (Electrical) 11 Jul 12 4:28
 Hi guys (and gals) You might be able to assist me in relation to a query I have on the maximum let-through energy of fuse versus the thermal withstand energy of cable. Basically my query revolves around the wording in the ETCI (Irish Regs) and IEE Wiring Regulations that state that fuse I2 t let through energy is to be less than the cable k2 S2 thermal withstand capacity, for fault disconnection times up to 5 seconds For example if you look up the time-current fuse characteristic for a 125A NH2 gG fuse and you calculate I2 t for a fault of 630A with associated fuse operating time of 5 seconds we get a higher let-through energy (1984500 A2sec) than for a fault of 1800A with a shorter fuse operating time of 0.02 seconds (64800 A2sec). Yet the fuse manufacturers state the maximum let through energy of 125000 A2sec for a fault of 20 x In = 2500A for the 125A fuse. But if the wording in the IEE Regulations is correct in that the above adiabatic formula is valid for fault durations up to 5 seconds (as they say that the equation is considered to remain adiabatic for fault current durations up to 5 seconds, i.e., does not lose heat through insulation up to 5 secs. After 5 secs some heat will dissipate through the cable insulation into the air and so formula no longer applies), then a high resistance fault of 630A for longer duration would result in a greater energy let-through into the cable than for a high fault current of 2500A for durations less than 1/4 cycle (i.e. less than 0.01 secs). So for example if we had a 10sq.mm Copper/PVC cable wired from above 125A fuse to a load where the fuse is providing short-circuit protection only (I know its not a practical example but it serves as a example of my above query). The cable has a thermal withstand capacity of 1152 102 = 1322500 A2sec, it seems that from a thermal capacity persepective this cable will heat up and reach it's insulation limit temperature for a 630A fault and not for a 1800A fault, as for a 630A fault the let through energy by fuse into cable (1984500 A2sec) is greater than cable withstand thermal energy (1322500 A2sec). The fuse manufacturer's only seem to publish the I2 t values for very high fault currents in the current limiting range, but as you can see from above example the largest thermal let-through energy into cable is for much lower currents. As sopmeone put it 'a long slow blow is much more damaging for a cable than a quick high current disconnection' I can't understand why fuse manufacturers dont publish the let through energy for current corresponding to a 5 second disconnection time rather than for very high fault current levels ?? Regards Gilly
 gilly1 (Electrical) 11 Jul 12 4:32
 Hi again ust noticed the 2 intending to be 'squared' did not come out when I pasted...so 'I2t' should be I squared x t and also '1152 102 = 1322500 A2sec' should be 115 squared x 10 squared A squared sec
 dpc (Electrical) 11 Jul 12 11:28
 It's just the area under the maximum clearing time curve for the fuse. A five-second clearing time is well beyond the "current-limiting" range of the fuse, which is 0.5 cycles at most. The I2t data provided by the manufacturer is to support their claims of let-thru energy limitation for very large levels of fault current. In this sub-cycle region, the fuse response is very non-linear and the let-thru energy data may provide a more accurate idea of the energy than just looking at the fuse time-current curve. At 5 seconds, you can compute your own I2t by looking at the fuse curve. As for cable protection, it is rare to see problems with cable damage during faults for normal power systems with typical overcurrent protection. You might want to double-check your cable fault withstand data.
 gilly1 (Electrical) 12 Jul 12 8:01
 Thanks dpc I realise that (i.e. at 5 seconds you comput your own I2t by looking at fuse curve), but I still think this could be made clearer by the fuse manufacturer's. I came across one fuse manufacturer, GEC Alsthom, that made this very clear and even gave the adiabatic formula I2t = k2S2 and stated "where I = current which causes fuse to operate in 5 seconds, t = 5 seconds...." Now that is clear (see http://www.weutscheck.com/atta...strial_Fuse_Links... ) Regards Barry
 cuky2000 (Electrical) 12 Jul 12 16:12
 See if the enclosed excerpt could help. http://files.engineering.com/getfile.aspx?folder=c8a0aaae-307a-4e23-a361-c5
 gilly1 (Electrical) 13 Jul 12 7:22
 Thanks Cuky for the pdf But this was for sizing CPC earth conductor. While it is related it is not exactly the same as sizing the phase conductors. But thanks for replying

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