Maximum Short Circuit Current (Fault-Current) before breakdown of cond

  1. 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 bus-bar 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 (fault-current) in any time.

    Attached Files:

  2. jcsd
  3. jim hardy

    jim hardy 5,460
    Science Advisor
    Gold Member
    2014 Award

    Product of I2 X t = constant

    think about that.... can you explain why?
  4. The short-circuit current presents different values significant for withstand calculation:
    ip[peak value] =k*sqrt(2)*I”k [IEC 60909-1 definition]
    k=1.7-1.9[max.2] depending on X/R of the short-circuit grid.
    it is the highest possible instantaneous value of the short-circuit current.
    I”k = Initial symmetrical short-circuit current [rms].
    The electromagnetic effect a circuit [physical] element has to withstand is defined as maximum force acting on this element. For three-phase short-circuit [according IEC 60865-1]:
    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 short-circuit current.
    The short-circuit current in time will decrease from I”k up to Ik-steady state short-circuit 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 short-time withstand current
    Tkr=rated short-time
    If the new current Ith is LESS than Ithr then:Ith<=Ithr*sqrt(Tkr/Tk) Tk=new short-time.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.
  5. The steady state load current [not for short-time] depends on the downstream consumers [motors, other panels and so on].Short-circuit current depends on upstream [the supply system-utility or generators, the transformer, the low voltage cables] impedance.
    For instance, let’s say the transformer ratings will be 2500 kVA/ 415V. If the short-circuit 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 short-circuit 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 short-circuit current [metallic contact-no arcing].
    I”K3=VoltL_L/sqrt(3)/Ztot=0.415/sqrt(3)/0.00523=45.8 kA.
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