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

[tipu_sultan]

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.

 

[jim hardy]

Is there any relation between the current and time in this case (directly, or Inversely Proportion) ?

Product of I² X t = constant

think about that... can you explain why?

 

[Babadag]

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.

 

[Babadag]

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.

 

[alphy]

I can provide some information about calculating short circuit current in a three-phase cabinet. The maximum short circuit current, also known as fault current, is the highest level of current that can flow through a circuit in the event of a short circuit. This current can cause damage to the circuit and its components, and it is important to know the maximum short circuit current before breakdown of the circuit.

In this case, the operation voltage of the circuit is 400V/415V and the operated current is 800A. The fault current for 1 second is 46 KA, which means that in the event of a short circuit, the current can reach up to 46,000 amps in just 1 second. This is a very high level of current and can cause significant damage.

To calculate the short circuit current per second that the bus-bar can bear, one can use the formula I = V/Z, where I is the current, V is the voltage, and Z is the impedance of the circuit. The impedance of the circuit can be calculated using the formula Z = √(R^2 + X^2), where R is the resistance and X is the reactance of the circuit. By knowing the impedance, one can calculate the maximum short circuit current that the bus-bar can bear.

To calculate the short circuit current for 3 seconds, one can use the formula I = (V/Z)(e^(-t/τ)), where t is the time in seconds and τ is the time constant of the circuit. The time constant can be calculated using the formula τ = L/R, where L is the inductance and R is the resistance of the circuit. By plugging in the values, one can calculate the short circuit current for 3 seconds.

There is a direct relationship between the current and time in this case. As the time increases, the current also increases. This is because the impedance of the circuit decreases as the time passes, resulting in a higher current flow.

In conclusion, there are formulas and methods available to calculate the short circuit current in a three-phase cabinet. It is important to know the maximum short circuit current before breakdown of the circuit to ensure the safety and proper functioning of the circuit.