Discuss MCB tripping time for 153.37A Short circuit L-N fault for 16A Type C MCB?? in the Electrical Forum area at ElectriciansForums.net

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walberto

I'm a qualified spark but studying building services engineering at uni and just need a little bit of advice to see if I'm going wrong.

Trying to answer a question where we have a 16A Type B MCB, and a short circuit fault current of 153.37A. From fig 3.4 of regs we can see that for a fault current of 80A or over the breaker will activate in 0.1second or less. However we are asked if this could be changed to a 16A Type C MCB (in terms of the short circuit fault current). Putting the thermal constraints of the cable to one side, I am more concerned with working out exactly how long it would take for this type of breaker to activate with a fault current of 153.37A. From looking at the regs (Fig 3.5 of the 17th edition without the latest amendment) I would have thought you could plot this on the chart, giving a tripping time of more than 5 seconds. I would have thought this did not comply as any circuit of rating less than 32A needs to trip in under 0.4 seconds.

I could be totally wrong so I wouldn't mind a comment or two to set me straight! Thanks in advance.

 
A C curve MCB will trip instantaneously at anything above 5 x In. So 153.37 is > 5 times 16 so it should still trip in < 0.4 secs.....unless I'm missing something obvious...
 
Type B: 3-5 x In.
Type C: 5-10 x In.
Type D: 10-20 x In.

So it might be instantaneous but it cannot be guaranteed as the standard allows up to 10 times. Do you have manufacturer's data?
 
As the time/current chart for this 16A MCB (as well as all other values) can trip anywhere on the straight magnetic trip line, i wish you luck trying to calculate the exact time a 153.37A fault current will fall on that straight line. Besides if you read the question being given correctly, that's not what it's asking.....
 
As the time/current chart for this 16A MCB (as well as all other values) can trip anywhere on the straight magnetic trip line, i wish you luck trying to calculate the exact time a 153.37A fault current will fall on that straight line. Besides if you read the question being given correctly, that's not what it's asking.....
yep....it looks to me like a straight yes/no question....

he should be looking at the time/current curves in BS7671...
 
As the time/current chart for this 16A MCB (as well as all other values) can trip anywhere on the straight magnetic trip line, i wish you luck trying to calculate the exact time a 153.37A fault current will fall on that straight line. Besides if you read the question being given correctly, that's not what it's asking.....

For a 16A Type C MCB, the device will trip in 0.1s or less for a fault current of 160A. As the fault is actually less than that, I've plotted the prospective current on the time/current characteristic curve for this type of breaker. This gives a time in excess of 5 seconds. I just need an approximate time so I can discuss how this compares to the thermal constraints of the cable.
The thing that threw me was that when all of my data was inputted into some design software, it gave satisfactory tripping times.
This leads me to believe that I'm doing something wrong, hence the need for some advice from those in the know.
 
The thing that threw me was that when all of my data was inputted into some design software, it gave satisfactory tripping times.

That's because BS7671 only shows the slowest tripping/operating time curve and not the fastest. Remember that the actual tripping curve is between 5 and 10 times. (eg the actual magnetic tripping curve is between two straight lines, one for the X5 and one for the X10) Get hold of a manufacturers time/current curve for this size breaker and you'll see what i mean...
 
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