Another exam question, energy let through from short circuit!

rob w

Another exam type question I'm not sure of and wondering if anyone here could help clear It up for me, it goes;

A short circuit current of 16ka flows under fault conditions. The circuit is protected by a 250a 'MCCB with a tripping time of 0.2 seconds and its I2t is 17x10^6 a2 secs.

(I) Calculate the time taken to raise the conductors to the final limiting temp of 130c if the CSA of t,he cables is 70mm2. Use 115 as the k factor for copper.

(II)Calculate the energy let through by the fault.



So my answer for part (I) was

t=(k^2)x(s^2) / I^2= 0.25 seconds


And answer for part (II) was

I2t for fault was (i^2)x 0.25 = 64x10^6

(This is assuming the fault would be allowed flow for the 0.25 seconds but as its only allowed flow for 0.2 before disconnection the actual I2t would be 51.2x10^6)


I think that's the correct way of working out the answers but not 100% sure, as the lecturer wasn't great and always a bit unclear, and I'm also unsure as to how the i2t given in the question was calculated and if its needed somewhere else in the question?

Any ideas?

Thanks

frankmul

I2t (AMPERES SQUARED SECONDS): an expression related to the circuit energy as a result of current flow. With respect to circuit breakers, the I2t [A2s] is expressed for the current flow between the initiation of the fault current and the clearing of the circuit.

I got this from http://www05.abb.com/global/scot/scot260.nsf/veritydisplay/9147d214ed83662e852575cf005ac7b5/$file/1sxu210170b0201.pdf

It seems to be a value determined by the manufacturer during tests.

I think that the calculation appear to be ok.

rob w

frankmul wrote: »

I2t (AMPERES SQUARED SECONDS): an expression related to the circuit energy as a result of current flow. With respect to circuit breakers, the I2t [A2s] is expressed for the current flow between the initiation of the fault current and the clearing of the circuit.

I got this from http://www05.abb.com/global/scot/scot260.nsf/veritydisplay/9147d214ed83662e852575cf005ac7b5/$file/1sxu210170b0201.pdf

It seems to be a value determined by the manufacturer during tests.

I think that the calculation appear to be ok.

Thanks for the reply frank,

Yeah i wondered was it just a manufacturers value, tried to figure out where it came from but couldnt, so I thought Id ask in here as the advice is always good!

Once my calculations are ok then Im happy!

Thanks

2011

rob w wrote: »

A short circuit current of 16ka flows under fault conditions. The circuit is protected by a 250a 'MCCB with a tripping time of 0.2 seconds and its I2t is 17x10^6 a2 secs.

(I) Calculate the time taken to raise the conductors to the final limiting temp of 130c if the CSA of t,he cables is 70mm2. Use 115 as the k factor for copper.

To be honest I have not done this in quite some time. I will come back to you on this, but my first thought is that surely you should have been given the temperature of the cable just prior to the fault?

If the temperature was 50 DegC then it would reach the limiting temperature quicker than if it was 5 DegC.

answer for part (II) was

I2t for fault was (i^2)x 0.25 = 64x10^6

(This is assuming the fault would be allowed flow for the 0.25 seconds but as its only allowed flow for 0.2 before disconnection the actual I2t would be 51.2x10^6)

I agree that this part of the question is correctly answered. However you will loose marks if you do not state the units in your answer.

rob w

I just presume that it's from an ambient temp of 30º, but yeah the questions never seem to be worded too well, or seem to be missing some info!

I'll make sure to include the units in the exam too, didn't realise I had left them out there!

2011

rob w wrote: »

I just presume that it's from an ambient temp of 30º, but yeah the questions never seem to be worded too well, or seem to be missing some info!

If you are going to assume 30 DegC you should state that, in which case your calculation should be based on a 𝝙t of 100 DegC

2011

Page 77 here provides the following formula that aligns nicely with your answer:

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

These are the values that I would use:
k = 115 (Perhaps this is Δt with the ambient temperature was 15 DegC and that is why the K value is given)
S = 70 mm^2
I = 16 x 10^3 A

frankmul

I think the k value of 115 is used when you assume the I initial temperature of the cable is 70 deg or greater, a figure of 143 is used for the k value when the initial temperature is assumed to be 30 deg. They are in the rules, chapter 5 I think

frankmul

2011 wrote: »

Page 77 here provides the following formula that aligns nicely with your

Looks like a good book

rob w

Thanks for the advice lads,helpful as always!