Electrical Services Design exam Q

ED.D

Hi any help with this would be greatly appreciated,

BrianDug

To solve this you need to use the adiabatic equation.

t = (k^2 * S^2) / I^2

where:
t = breaker disconnection time (seconds)
k = is a factor that derived from the specific heat capacity, temperature coefficient and conductivity of the cable. Its quiet complex to manually calculate so if you want to know how to derive it let me know and il show you.
S = conductor size in mm^2
I = Short circuit fault current.

So in your problem:
t = (k^2 * S^2) / I^2 becomes
t = (115^2 * 70^2) / 20000^2 = 28.17 seconds

So this tells us that it will take 28.17 seconds for the 70mmssq cable carrying 20kA to reach its limiting temperature. As your protective device operates within 0.2 seconds this is sufficient.

ED.D

Thanks a lot Brian this is a great help, always given the K factor so there's no need.
Cheers

BrianDug

Sorry Ed I hoped you checked them figures as there is a mistake, I just rechecked it now as I was thinking 28 seconds was a hell of a time for a cable to carry 20kA. The result should be 0.16200625 seconds. I don't know how the result ended up as 28 seconds, must of missed a figure somewhere.

The method is correct though.

So as the cable will reach its limit within 0.16 seconds and the breaker wont trip till 0.2 seconds the MCCB is not suitable and needs to be changed.

Apologies for the mistake.

2011

BrianDug wrote: »

The method is correct though.

That is the important part.
Most of the exam marks are for method / technique, almost none would be lost for mathematical errors.