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Maximum Short Circuit Current (FaultCurrent) before breakdown of cond 
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#1
Mar1714, 02:46 AM

P: 12

Here find some attached circuit of three phase cabinet which operation voltage is 400V /415V and have operated current 800A, fault current for 1 sec is 46 KA and have incoming and outgoing links with one phase Lighting and socket.
I want to know is there some method that one can calculate the short circuit current per second that the busbar can bear. Moreover I also want to calculate the short circuit current for 3 second. Is there any relation between the current and time in this case (directly, or Inversely Proportion) ? Please provide me the formula if exist to calculate short circuit current (faultcurrent) in any time. 


#2
Mar1714, 06:38 AM

Sci Advisor
PF Gold
P: 3,657

Product of I^{2} X t = constant think about that.... can you explain why? 


#3
Mar1714, 12:03 PM

P: 53

The shortcircuit current presents different values significant for withstand calculation:
ip[peak value] =k*sqrt(2)*I”k [IEC 609091 definition] k=1.71.9[max.2] depending on X/R of the shortcircuit grid. it is the highest possible instantaneous value of the shortcircuit current. I”k = Initial symmetrical shortcircuit current [rms]. The electromagnetic effect a circuit [physical] element has to withstand is defined as maximum force acting on this element. For threephase shortcircuit [according IEC 608651]: Fm3=miuo/(4 *pi())*sqrt(3)*ip^2/am pi()=3.14159... [in excel language] am=distance between elements[busbars]. As you can see this value does not depends on time and in any case the busbar has to withstand it. So you cannot allow a more elevated value of the shortcircuit current. The shortcircuit current in time will decrease from I”k up to Iksteady state shortcircuit and in this time the losses produced in busbars have to be limited so not to heat the circuit element more than permitted.This is the thermal effect which the busbar has to withstand. If Ithr =rated shorttime withstand current Tkr=rated shorttime If the new current Ith is LESS than Ithr then:Ith<=Ithr*sqrt(Tkr/Tk) Tk=new shorttime.Tk>Tkr If Tk<Tkr the new current has to be NOT MORE than Ikr in any case. So 46 kA for 1 sec [rated] could be 26.55 kA for 3 sec. 


#4
Mar1814, 10:55 AM

P: 53

Maximum Short Circuit Current (FaultCurrent) before breakdown of cond
The steady state load current [not for shorttime] depends on the downstream consumers [motors, other panels and so on].Shortcircuit current depends on upstream [the supply systemutility or generators, the transformer, the low voltage cables] impedance.
For instance, let’s say the transformer ratings will be 2500 kVA/ 415V. If the shortcircuit voltage will be 6% then the transformer impedance will be: 6/100*0.415^2/2.5=0.004133 ohm. Neglecting the resistance then Xtrf=Ztrf The supply system for medium voltage shortcircuit apparent power Ssys=500 MVA and system impedance will be: Zsys [at 415 V system]=0.415^2/500=0.0003445 ohm and neglecting the resistance Xsys=Zsys. A low voltage cable [2 parallel cables of 3*240 mm^2 copper conductors 90oC insulation for 2*400A rated current] presents a resistance approx.0.0947 ohm/km and a reactance of 0.0819 ohm/km. The length of cable [let’s say] =17 m and the cable impedance will be: Zcab=0.000805+j.0.000696 ohm. Total impedance will be: Xtot=0.0003445+0.004133+0.000696=0.00517 ohm Ztot=sqrt(0.00517^2+0.000805^2)=0.00523 ohm I”k3= three phases shortcircuit current [metallic contactno arcing]. I”K3=VoltL_L/sqrt(3)/Ztot=0.415/sqrt(3)/0.00523=45.8 kA. 


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