Let through Energy - Cable Sizing

AJJewsbury:

To be clear, I wanted to ask, When the cables have been already verified using adiabatic equation, Do I still need to check the cables for let through energy verification?

In principle they're both the same thing (S = √(I²t)/k is the same a S²k² = I²t, just re-arranged (divide through by k and square both sides)) - but in practice people often refer to different conditions - e.g. the adiabatic is often only considered when sizing c.p.c.s - i.e. for Earth Faults whereas L-N, L-L and L-L-L faults are more commonly thought of when thinking of energy let-though.  Of course faults between live conductors can produce quite different levels fault currents than L-PE faults - so a calculation based on an earth fault current might not be valid for the other kinds of faults.


As a first approximation if a conductor is "reduced" - i.e. it has a rating below that of its protective device (Iz < In) - then some check will likely be necessary.


Also keep in mind that the worst case isn't necessarily a fault at the far end of the circuit - that gives the lowest fault current and likely the longest disconnection time - and for fuses probably the highest energy let-through but for MCBs faults with higher currents (e.g. a fault next to the MCB) often don't disconnect significantly faster than a far-off fault - so I²t can often be higher. Unfortunatley some text book examples seem to have been written with fuses in mind and miss that point.


    - Andy.

Though both the equations are same (just rearrangement) but the I2t value comes from the manufacturer in case of verifying the let through energy. Now my question is can we use this let through energy value (received from the manufacturer in the form of curve) to apply in this adiabatic equation S=(I√t)/k, to find out the minimum required CSA of conductor

AJJewsbury:

To be clear, I wanted to ask, When the cables have been already verified using adiabatic equation, Do I still need to check the cables for let through energy verification?

In principle they're both the same thing (S = √(I²t)/k is the same a S²k² = I²t, just re-arranged (divide through by k and square both sides)) - but in practice people often refer to different conditions - e.g. the adiabatic is often only considered when sizing c.p.c.s - i.e. for Earth Faults whereas L-N, L-L and L-L-L faults are more commonly thought of when thinking of energy let-though.  Of course faults between live conductors can produce quite different levels fault currents than L-PE faults - so a calculation based on an earth fault current might not be valid for the other kinds of faults.


As a first approximation if a conductor is "reduced" - i.e. it has a rating below that of its protective device (Iz < In) - then some check will likely be necessary.


Also keep in mind that the worst case isn't necessarily a fault at the far end of the circuit - that gives the lowest fault current and likely the longest disconnection time - and for fuses probably the highest energy let-through but for MCBs faults with higher currents (e.g. a fault next to the MCB) often don't disconnect significantly faster than a far-off fault - so I²t can often be higher. Unfortunatley some text book examples seem to have been written with fuses in mind and miss that point.


    - Andy.

Though both the equations are same (just rearrangement) but the I2t value comes from the manufacturer in case of verifying the let through energy. Now my question is can we use this let through energy value (received from the manufacturer in the form of curve) to apply in this adiabatic equation S=(I√t)/k, to find out the minimum required CSA of conductor