Cable Temperatures for Minimum fault current calc? 1

[healyx]

Hi All,

I'm trying to work out how to calculate minimum fault current manually in the IEC domain (usually phase to ground fault is minimum). I know the Cmin parameter needs to be set to 0.95 but what I need to know is what temperature the various cable resistances should be calculated at. Cables I need answers for are:
(1) Phase cable temperature for circuit under fault.
(2) Protective Earth cable temp for circuit under fault (TN-S segments).
(3) Phase cable temperature for upstream cables.
(4) Protective Earth cable temp for upstream cables (TN-S segments).
(5) Neutral cable temp for upstream cables (TN-C segments).
I also need to know if the temps are different for circuit breakers or fuses.
I can't find any reliable reference for this except the CENELEC document (UTE NFC-15-500). It has all of these individual segment temperatures covered but I don't know if this document complies with IEC60909. IEC60909 doesn't have anything this specific in it, it just mentions that the temperature should be set to the temperature of the cable at the end of the short circuit duration - which is kind of difficult to work out.
Can anyone help?

 

[cuky2000]

The maximum allowable temperature for cable is a function of the type of insulation (ie XLPE, EPR, etc.)
For bare cable, the limiting factor is the fusing temperature, annealing temperature or fire hazard in accordance with the design bases.

Check any cable manufacture in your region to determine the allowable temperature

 

[cuky2000]

The enclosed excerpt usually used in the ANSI marketplaces is a good reference to start. See if there is any similar information used in the IEC world.

Regarding your last question associated with protective devices such as circuit breakers,fuses,etc. each type has differents clearing time and will impact the cable temperature

I hope this help.

 

[jghrist]

If the standards do not specify what temperature to use, I would use the cable insulation maximum temperature or, for overhead lines, the maximum design temperature. I would use the same temperature for earth cables. You may get higher temperatures during a fault, depending on the clearing time of the fault, but it seems unnecessarily cumbersome to use the fault clearing time and cable temperature to calculate the fault current that is needed to determine the fault clearing time.

What fault resistance would you use to calculate minimum fault current? This might be a bigger factor than conductor resistance corrections and would be more variable.

 

[7anoter4]

The temperature can to be calculated according to:
IEC 60865-1 "Short-circuit currents –Calculation of effects" See [for instance]:

http://webstore.iec.ch/preview/info_iec60865-1{ed2.0}en_d.img.pdf

Section 3: The thermal effect on bare conductors and electrical equipment
Pp 57-61 fig.12 a pp 99, 12 b pp101, 13 pp. 103 and Annex A pp 111,113,
No indication for insulated cable, indeed. In IEC 60909 somewhere is noted: for minimum short-circuit current the conductor temperature will be 80oC. Now, first of all you have to choice the cable conductor cross section according to
steady state current –heat and voltage drop- and then if the short-circuit heating calculated for the fault clearing maximum time the temperature is less than maximum permitted [for instance PVC 150oC ] you may state the cable is suitable and if it is not less you have to take another bigger.
Let's take an example:
System Ssys=500 MVA, Transformer Strf=1 MVA zk=6% Rated low voltage 600 V.
Itrf= Strf/Vtrf/sqrt(3)=962.25 A. 1 km cable [two single core cables per phase ] 3*2*300 sqr.mm Copper.
In a duct bank of 2 rows 3 columns the ampacity [current carrying capacity]=450 A per cable [70oC maximum].
The voltage drop is o.k. Cable resistance at 20oC [d.c.] will be 0.0601 ohm/km skin effect+proximity effect=0.08.
The reactance [for 250 mm apart center-line to center-line, 23 mm conductor diameter] will be 0.195 ohm/km.
In my opinion you may neglect the resistance even in order to calculate minimum short-current.
For short-circuit regime- as for short time the heat dissipation to ambient is negligible-it does not important if the conductor
is bare or insulated. Only the final permitted temperature is different.
From Annex A ch.A.9 Figure 13 from formula Sth=K/sqrt(Tkr) the rated short-time withstand current density:
Sth=Isc/Scopper, by transforming the K formula we can get
Tfinal=(exp(K^2/const)*(1+a20*(Tinitial-20)-1)/a20+20
K=sqrt(time)*Isc/Scopper
Const=k20*c*ro/a20
For copper c=390 J/kg/oC; ro=8900 kg/m^3; k20=56/10^6 1/(ohm.m) a20=0.0039 1/oC.
Const= 49459.542
First of all you have to calculate the I"k let's say for phase to ground short-circuit.
System Zsys=Vtrf^2/Strf=0.00072 ohm
Ztrf=zk%/100*Vtrf^2/Strf= 0.0216 ohm
Let's state at the beginning Rcab at 20 oC will be 0.0601*1.08=0.0649 ohm
Zcab=0.0649+j*0.195 ohm
Neglecting Rcab and considering all Z=X[System and transformer]
Zosys=0 Zotrf=Z1trf=Z2trf Zocab=0
I"k1=sqrt(3)*1.1*Vtrf/(0.00072*3+0.0216*3+0.195)=4.36 Ka
So K=sqrt(time)*I"k/Scopper=4.34*1000/300=14.4
Tfinal=(exp(14.4^2/49459.54)*(1+0.039*(Tinitial-20)-1)/0.039+20
For ground conductor Tinitial=Tambiant [40 oC] for power cable 70oC.
Tfinal[power cable]= 70.2 oC. Now you have to modify Rcab(1+0.039*(70.2-20))=0.0776 ohm
and recalculate the I"k1 [negligible]

 

[7anoter4]

Conclusion:
My opinion here: it is not a short-circuit current calculation method problem but a protection policy.
If do you think the supply voltage will appear short-time after the first fault was cleared and will follow a second fault
then you could expect a higher temperature and recalculate the minimum short-circuit correcting the resistance for this temperaturei[if it will be necessary].

 

[healyx]

Thanks Guys.

Thanks in particular to 7anoter4! Very detailed. I'll look into your calc and post back if I have any other questions.

 

[healyx]

OK, with thanks to 7antorter4 I think I have come up with a generalized method to produce the worst case minimum fault current (e.g. lowest). It would be great to get some feedback to see if you all agree. By the way, this calc is for insulated LV cables only.
(1) Determine the minimum falut current using the highest operational temperature for the cable under consideration. In Australia, we use 75oC for PVC insulated cables. So, for example, say this yields a fault current of 1.5kA for a 2.5mm2 cable at 75oC. You then need to check the clearing time for the protective device - so assuming a breaker, clearing time might be 0.02 seconds.
(2) Calculate the final worst case temperature of the cable. To do this, assume the initial temp is the temp used in step 1 (75oC).

If the device is NOT fault current limiting, calculate the temp using:

Tfinal = e^((k^2)/(226^2))(234.5 + Tinitial) - 234.5

This is for Copper cables where e is Eulers Number (2.71) and k is the familar heating constant given by:

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

where S is cross sectional area of cable in mm^2, I is fault current and t is clearing time.

So for the example above, assuming the device in NOT fault current limiting, we get:

k^2 = (1500^2)0.02/(2.5^2) = 7200

So T final is:

Tfinal = e^(7200/51076)(234.5 + 75) - 234.5 = 122oC

(3) Redo the calculation for minimum fault current using the cable at 122oC instead of 75oC. This will give a current smaller than the 1.5kA used previously. The calculated higher cable temperature is worst case as it assumes the cable remains at the 75oC resistance level even though it heats up to 122oC. This would not be the case in practice, but this at least guarantees you haven't under estimated it.

Finally, for the case where the breaker is a fault current limiting type. E.g. Assume a Schneider 20A C60 MCB. At 1.5kA, this device will limit I^2t to about 6000 A^2S. So:

K^2 = 6000/(2.5^2) = 960

Tfinal becomes:

Tfinal = e^(960/51076)(234.5 + 75) - 234.5 = 81oC

This is a significant reduction from 122oC (10% lower resistance), so this is a good way to increase the minimum fault current.

To do this properly, you might need to perform a similar check on upstream feeder cables - but just calculating their resistance at the maximum operating temp is likely to be all that is required (as cable size increases, temperature effects become more negligible).