How to determine Arc Flash level?

You can do it that way,  and to get back to simple formula, you just convert your PSSC to calories per electron-volt instead of amps.?

 You do however  end up with dazzlingly complex looking formulae with weird constants in if you mix units, and it rather obfuscates what is going on, especially in the improved formulae when folk use arc length in inches to estimate the fraction of the voltage drop that is not available for flash production.  Since my last posts on this a couple of months ago I have looked into it more deeply and the underlying ideas are becoming clearer.

But I stand by my comment that it is complex, and if in doubt use the Lee method and over- estimate the danger.

Mike

Edit 

Some metric numbers to conjure with - for the bare skin burn limit of 5j/cm2

at 10cm, (4 inches) from the event, the sphere area is 4pi r2 = 1256 cm2, so  a flash energy of 6000joules would be needed to give permanent skin damage, if omnidirectional. Perhaps 2000 joules if in a box and geometry unlucky.

If we have say a 0.1sec disconnection we need 60kW for the duration to illuminate that area to that intensity, and with an arc voltage of 120, and an arc current of  500A we get there. So under those conditions an open air arc from a  circuit PSSC of 1000A or so could be dangerous to unprotected skin at 4 inches or closer. In a box you should perhaps double that distance when in the firing line from the open side of the box.

If you wanted to be twice as far, perhaps 8 inches away, then you need 4 times the energy , as the sphere surface area has gone up by 4, to ~ 5000 square cm,  so 25000 joules. If again we assume 0.1 seconds and 120V arc voltage then arc current of   more like 2kA, so a PSSC of perhaps 4kA. Again double the distance for an arc in a box.

In practice then, down stream of MCBs that break in well under 0.1seconds,  and on PSSC of a few kA or less,  if you keep your head out of the boxes, then all that is needed is gloves and some forearm cover,, and maybe googles.

On the other hand, with a PSSC of 16kA and only a 100A cut out fuse to cover you,  the breaking time is more like 10msec, so  perhaps 8000A and 120V for 0.01 seconds is worst case ==   9600 joules - less than the let though of the MCB.  However, if there is a fault on the supply side of that incoming fuse, which is where the DNOs operate, it is a very different conclusion.

Low voltage ADS techniques on the load side of the company fuse in the UK and EU lead to disconnection times that are fast enough to limit the protection required to quite modest levels.

You can do it that way,  and to get back to simple formula, you just convert your PSSC to calories per electron-volt instead of amps.?

 You do however  end up with dazzlingly complex looking formulae with weird constants in if you mix units, and it rather obfuscates what is going on, especially in the improved formulae when folk use arc length in inches to estimate the fraction of the voltage drop that is not available for flash production.  Since my last posts on this a couple of months ago I have looked into it more deeply and the underlying ideas are becoming clearer.

But I stand by my comment that it is complex, and if in doubt use the Lee method and over- estimate the danger.

Mike

Edit 

Some metric numbers to conjure with - for the bare skin burn limit of 5j/cm2

at 10cm, (4 inches) from the event, the sphere area is 4pi r2 = 1256 cm2, so  a flash energy of 6000joules would be needed to give permanent skin damage, if omnidirectional. Perhaps 2000 joules if in a box and geometry unlucky.

If we have say a 0.1sec disconnection we need 60kW for the duration to illuminate that area to that intensity, and with an arc voltage of 120, and an arc current of  500A we get there. So under those conditions an open air arc from a  circuit PSSC of 1000A or so could be dangerous to unprotected skin at 4 inches or closer. In a box you should perhaps double that distance when in the firing line from the open side of the box.

If you wanted to be twice as far, perhaps 8 inches away, then you need 4 times the energy , as the sphere surface area has gone up by 4, to ~ 5000 square cm,  so 25000 joules. If again we assume 0.1 seconds and 120V arc voltage then arc current of   more like 2kA, so a PSSC of perhaps 4kA. Again double the distance for an arc in a box.

In practice then, down stream of MCBs that break in well under 0.1seconds,  and on PSSC of a few kA or less,  if you keep your head out of the boxes, then all that is needed is gloves and some forearm cover,, and maybe googles.

On the other hand, with a PSSC of 16kA and only a 100A cut out fuse to cover you,  the breaking time is more like 10msec, so  perhaps 8000A and 120V for 0.01 seconds is worst case ==   9600 joules - less than the let though of the MCB.  However, if there is a fault on the supply side of that incoming fuse, which is where the DNOs operate, it is a very different conclusion.

Low voltage ADS techniques on the load side of the company fuse in the UK and EU lead to disconnection times that are fast enough to limit the protection required to quite modest levels.