Unfused spur.

mapj1:
I am unclear about the energy let-through subject. I squared t, or A squared t. (Amp2 seconds).


I understand current and I understand time. Where did the current squared originate?


I suspect you are not alone, the choice of units is not the most descriptive.  Really it is a measure of the total energy dissipated in  every small resistance in the fault loop, in the time between start of fault, and the disconnection.  At the risk of a long post, and not one for those short of sleep, it may help to play some algebra games.

If need be print this out and take your time to digest it. (There is no exam at the end, except safe designs !)


V*I = watts. (1)  (by definition) 

V*I*t = joules (2)  (heat energy)

V= I*R           (3) (Mr Ohm)

so I*R*I*t = joules (merging statements 2 and 3)

regrouping


I*I*t*R = Joules

divide both sides by R, this remains a true statement

I²t = joules/ohms (5)

  Ta-da really these units are telling you that if you have say a switch contact of so many milliohms in series with a fuse of that let-through, it will be heated by that many joules per ohm  during the blowing time. If you know the contact heat capacity (so many degrees rise per so many joules of heat per certain mass or volume to be heated - watch you are using figures where the base  units are the same), you can then deduce a temperature rise, and decide if it will be damaged or not.


Take the example of a simple copper wire 1mm² in cross-section, and 1m long.

total resistance call it 16 milliohms per metre of length (or 19 if it is hot)

Total volume 1000mm³  

total mass

m=9 grams per metre of length

(it is 9 tonnes  per cubic metre, but rescaled... )

heat capacity per unit mass (S or Q in some texts), rescaled gives ~

S=  0.38 joules will raise one gram by 1 degree C   (in vacuum with no heat escape allowed)

temperature rise is more with more electrical power, but reduced by more mass of stuff to heat or higher heat capacity


dT=I²R*t/(S*m)          (6)


you can think of dT= difference in temperature


So for an example say a 100 degree temperature difference between before and after


100=I²R*t/(S*m)       (7)


re order to get


I²t=(S*m*dT)/R    (s)


= (0.38*9*100)/0.016

(note the per metre of length cancel as there is one on top one below.. )


 ~22 000 joules per ohm

or 22,000 amps² seconds will give a 100 degree rise in 1mm² copper cable.


for a 1000 degree rise, more like what it would take for copper to melt


more like 220,000 amps² seconds would be needed.


In reality there is always some cooling, so these figures will not quite match tabulated ones from real experiments, the faster it blows the less time there is for heat to escape, the better this approximation fits.

But it shows the technique.


Come back if you need walking through it in finer steps.



Mike.







 

Wow and thanks Mike. Thanks very much. I have never had it explained in the clear understandable terms that you have used. I am off to work now but will re-read your post later and retain it for future reference.


Z.

mapj1:
I am unclear about the energy let-through subject. I squared t, or A squared t. (Amp2 seconds).


I understand current and I understand time. Where did the current squared originate?


I suspect you are not alone, the choice of units is not the most descriptive.  Really it is a measure of the total energy dissipated in  every small resistance in the fault loop, in the time between start of fault, and the disconnection.  At the risk of a long post, and not one for those short of sleep, it may help to play some algebra games.

If need be print this out and take your time to digest it. (There is no exam at the end, except safe designs !)


V*I = watts. (1)  (by definition) 

V*I*t = joules (2)  (heat energy)

V= I*R           (3) (Mr Ohm)

so I*R*I*t = joules (merging statements 2 and 3)

regrouping


I*I*t*R = Joules

divide both sides by R, this remains a true statement

I²t = joules/ohms (5)

  Ta-da really these units are telling you that if you have say a switch contact of so many milliohms in series with a fuse of that let-through, it will be heated by that many joules per ohm  during the blowing time. If you know the contact heat capacity (so many degrees rise per so many joules of heat per certain mass or volume to be heated - watch you are using figures where the base  units are the same), you can then deduce a temperature rise, and decide if it will be damaged or not.


Take the example of a simple copper wire 1mm² in cross-section, and 1m long.

total resistance call it 16 milliohms per metre of length (or 19 if it is hot)

Total volume 1000mm³  

total mass

m=9 grams per metre of length

(it is 9 tonnes  per cubic metre, but rescaled... )

heat capacity per unit mass (S or Q in some texts), rescaled gives ~

S=  0.38 joules will raise one gram by 1 degree C   (in vacuum with no heat escape allowed)

temperature rise is more with more electrical power, but reduced by more mass of stuff to heat or higher heat capacity


dT=I²R*t/(S*m)          (6)


you can think of dT= difference in temperature


So for an example say a 100 degree temperature difference between before and after


100=I²R*t/(S*m)       (7)


re order to get


I²t=(S*m*dT)/R    (s)


= (0.38*9*100)/0.016

(note the per metre of length cancel as there is one on top one below.. )


 ~22 000 joules per ohm

or 22,000 amps² seconds will give a 100 degree rise in 1mm² copper cable.


for a 1000 degree rise, more like what it would take for copper to melt


more like 220,000 amps² seconds would be needed.


In reality there is always some cooling, so these figures will not quite match tabulated ones from real experiments, the faster it blows the less time there is for heat to escape, the better this approximation fits.

But it shows the technique.


Come back if you need walking through it in finer steps.



Mike.







 

Wow and thanks Mike. Thanks very much. I have never had it explained in the clear understandable terms that you have used. I am off to work now but will re-read your post later and retain it for future reference.


Z.