Current rating for parallel single core cables in duct back (thread238-237460)

[ghetuflorin]

Hi everyone

Recently we had to calculate using IEC 60287 part 2-1 and 1-1 the cable derated ampacity for the following:
3 X (3X630 sqmm) cable, single core, copper conductor, armored, XLPE. (Prysmian) from Transformer (11/0.69 kV) to PC (0,69 kV).
The length is 70 m routed in air, but for 4 m is laid in 3 ducts embedded in concrete (to pass from substation to the transformer ).
Most of IEC is clear and all formula we applied accordingly, but the problem came for T’’’4 (thermal resistance of surrounding medium).

As per IEC-2-1 /2.2.7.3 : “ When the ducts are embedded in concrete, the calculation of the thermal resistance outside the ducts is first of all made assuming a uniform medium outside the ducts having a thermal resistivity equal to the concrete. A correction is then added algebraically to take account of the difference, if any, between the thermal resistivities of concrete and soil for that part of the thermal circuit exterior to the duct bank.”
Correction factor is clear defined, but what about the T4’’’? One could say that this should take the value of concrete thermal resistivity, then calculated correction factor (outside the duct back the filling medium is sand with thermal resistivity of 1,09 km/W) will be added and final result for thermal T’’’4 (total) will be obtained .But is it correct?

The duct bank (750x250 mm) undergorung (-230 mm) has 3 pipes of 225 diameter , wouldn’t be necessary to calculate the mutual heating of the pipe in the middle(the hottest pipe) using IEC 60287-2-1/2.2.3.2 and then to this value should be added the Correction factor?

As a last question, the formula for permissible current rating flowing in one conductor (IEC 60287-1-1/1.4.2.1) has the “n” defined as: number of load-carrying conductors in the cable (conductor of equal size and carrying the same load). Considering that we have 3 single cable in parallel per phase (3 in each pipe X 3 pipes) , what is the correct value for “n” : 1 or 3?

I can send or attach the handmade calculation if any one of you gentlemen would like to help and needs additional information. I mention also that the result obtained considering T4’’’ as rho concrete + correction factor and value of n=1 is 778 A but calculation using ETAP is 403A


Thank you

Florin

 

[7anoter4]

First of all you have to calculate T4 according to 2.2.3.2 "Equally loaded identical cables" taking ro=roconcrete [0.85-1.00] and second, calculate the supplement as per 2.2.7.3 External thermal resistance of the duct (or pipe) T4"'. This supplement is to be added to above calculated T4.
You cannot use 60287-1-1 1.4.2.1 since is not recommended for multi-duct :
"The following method shall be applied to a single isolated cable or circuit only"
Therefore, you have to employ the 1.4.1.1 formula considering maximum earth thermal resistance [dry].
I don't know what are the cable characteristics except 630 sqr.mm CU. [1 KV XLPE/PVC, I guess]
I don't know the surrounding earth temperature and the actual concrete depth but for 20 dgr.C and 0.23 m depth the cable ampacity could be 658 A.
If we shall follow 1.4.1.1. and 2.2.3.2 with ro=1.09 then ampacity will be 585 A.
All calculation is based on "equilateral triangle" cable configuration in duct [and not cradle].

 

[ghetuflorin]

Thanks 7anoter4, I was hoping to get your attention with this topic and get your opinion.

First of all you have to calculate T4 according to 2.2.3.2 "Equally loaded identical cables" taking ro=roconcrete [0.85-1.00] and second, calculate the supplement as per 2.2.7.3 External thermal resistance of the duct (or pipe) T4"'. This supplement is to be added to above calculated T4.

Confirmed

You cannot use 60287-1-1 1.4.2.1 since is not recommended for multi-duct :
"The following method shall be applied to a single isolated cable or circuit only"
Therefore, you have to employ the 1.4.1.1 formula considering maximum earth thermal resistance [dry].

Confirmed (it was a typing mistake)

I don't know what are the cable characteristics except 630 sqr.mm CU. [1 KV XLPE/PVC, I guess]

Phi conductor= 30 mm annealed stranded copper wires
Phi Insulation =34.9 mm; thickness=2.4 mm, XLPE
Phi Bedding 37 mm; thickness=1 mm ; PVC
Phi Armor=41 mm; phi wire=2 mm; N.53 aluminum wires
Phi over sheath =46 mm; thickens=2.3 mm; PVC

I don't know the surrounding earth temperature and the actual concrete depth but for 20 dgr.C and 0.23 m depth the cable ampacity could be 658 A.

Maximum earth temperature is 35 degrees Celsius. The top of duct bank is at -0.23 m, bottom of DB is -1.270 m.

If we shall follow 1.4.1.1. and 2.2.3.2 with ro=1.09 then ampacity will be 585 A.
All calculation is based on "equilateral triangle" cable configuration in duct [and not cradle].

The calculation considering n=3 in the formula 1.4.1.1 and with above assumption is 458 A. I have attached a scan of the hand calculation:

https://www.wetransfer.com/download...b8f9240d17eb970b013cbf4f20130304071723/302f8f

[i]Concluding, the problems I am facing:

1.) IEC 2.2.7.3 mentions “single way” ducts. We have 3 cables in each duct, is it applicable?
2.) “n” in formula 1.4.1.1 is n=3 or n=1?

Thank you[/i]

 

[7anoter4]

First, I have to congratulate you for your work and your English. I am Romania from and I am abroad for more than 30 years. My Romanian is still good but my English is still not.
Second, I did the calculation using a software developed by myself on Visual Basic 6 10 years ago and introducing your last data I get 455 A, very close to yours.
The 1.4.1.1 formula is valid for a multi-core cable. In a duct [or conduit] case, the T4 will include all 3 cables present in the duct/conduit:
DT= (I2R + ½ Wd) T1 + [I2R (1 + l1) + Wd] n T2 + [I2R (1 + l1 + l2) + Wd] n (T3 +N/n T4)
In our case N=3 n=1.
I=[DT- Wd[0.5T1+n(T2+T3+N/nT4)]/[RT1+n(1+l1)T2+nR(1 + l1 + l2)(T3+N/nT4)]
Greek letters are not available, I guess.
2.2.7.3 it is suitable for any number of ducts. But in the correction formula
"N is the number of loaded cables in the duct bank" that means 9[!].
For the time being, I am busy a little - but I intend a close analyzing your calculation, of course.
I have no ETAP knowledge but I think this kind of calculation is based on IEC 60364-5-52 (nevertheless the maximum conductor cross section is 300 sqr.mm in these tables).

 

[ghetuflorin]

Engleza ta e ok si cu siguranta sunt alte lucruri la care excelezi printre care si faptul ca impartasesti din cunostintele tale. Am refacut calculele cu formula descrisa de tine pentru T4 si am obtinut 452 de amperi. Cum varsta mea e aceeasi cu anii tai in afara tarii, nu reusesc sa ii conving pe colegii mei care urmaresc postul, drept urmare am sa continui in engleza.

7anoter4, someone could disagree with your statement as all assumptions are mainly based on a ” personal” interpretation of the IEC.

1. How can be demonstrated that para. of IEC 60287.2.1 for the cable in pipes is the 2.2.3.2 and not “only” 2.2.7 (with rho concrete as only value for T4) ?.
2. How can be demonstrated that the IEC 60287-1-1 para. 1.1.4.1 it is also applicable for multicore cables and not only for a single “cable” either single core or multicore and “n” is the number of the conductors in the cable.
3. Where is IEC 60287-1-1 stating the formula that does include any “N/n” that is present in your calculation ? .

Since the IEC is not clear, it leaves room for personal interpretation; the fact is that ” 3 single core cable/phase in duct bank” is a typical installation and the correct formulas to be applied needs a justification back-up (like references to existing studies as Neher-McGrath or similar experiments).

Thanks again for your interest in our subject

 

[7anoter4]

Nu-mi prea place sa ma laud de aceea am sa spun in romaneste: trebue sa spun ca am proiectat-mai précis,am participat la proiectarea a 8 centrale electrice de puteri 350-650 MW si eu am condus echipa care a proiectat-printre altele-si traseele de cabluri.Traseele subterane erau "duct-banks"cu tevi din p.v.c. 6" inglobate in beton-intre 20 si 50 de tevi.Cabluri de forta incarcate diferit-cablurile mai mici introduse cite 3-5 in aceasi conducta.
Nu retin ca au fost reclamatii vre-o-data.
Now, as we have to speak only English I'll try to answer to your question as follows:
"1. How can be demonstrated that para. of IEC 60287.2.1 for the cable in pipes is the 2.2.3.2 and not “only” 2.2.7 (with rho concrete as only value for T4) ?."
From 2.2.7.3:
"This shall be determined for single-way duct(s) not embedded in concrete in the same way as for cable, using the appropriate formulae from 2.2.1, 2.2.2, 2.2.3 or 2.2.4.
That means: 2.2.3 is the appropriate article since we have "a group of buried ducts (non touching)" :
2.2.3 Groups of buried cables (not touching)
We have also:
2.2.3.2 Equally loaded identical cables
"2. How can be demonstrated that the IEC 60287-1-1 para. 1.1.4.1 it is also
applicable for multicore cables and not only for a single “cable” either single core
or multicore and “n” is the number of the conductors in the cable."
3. Where is IEC 60287-1-1 stating the formula that does include any “N/n”
that is present in your calculation ? "
The calculation way is based on temperature drop DT=total losses passing through a layer*thermal resistance.
The differential form of Fourier's Law of thermal conduction:
q=-lambda*grad(T) where q=heat flow ;lambda=thermal conductivity
for linear [or radial]heat transfer grad(T)=dT/dx
or DT= integral(q*dx)/lambda q(x)=losses/2/pi/x then integrating DT=losses/lambda*ln(rinsulation/rconductor)
q= total losses passing through a layer/layer exterior area.
For the insulation layer[T1 thermal resistance] the losses are I^2*R[single-core] and for 3 cores cable also[or using G factor for combined T1 and T2].For bedding and jacket will be 3 conductors losses[if any], including shield losses[lambda1] and armor losses[lambda2] and dielectric loss[Wd].For a duct or conduit with N(3) single-core cable the losses are N(3) time single-core losses. In order to calculate a temperature drop through a concrete duct bank the total losses of all loaded conductors, dielectric losses, shields, armors and metallic ducts present in this duct bank have to be taken into consideration.
Therefore ,T4 –duct thermal resistance has to be multiplied by total losses produced in this duct [all cables= N(9)].

 

[7anoter4]

Sorry for the delay. I did not hand calculation for 25 years.
I think there are very correct calculations- most of them-except hand calculation- some slight differences vs. excel calculation. Nevertheless, the final formula is not correct.
Since it is a low-voltage cable Wd [dielectric losses] =0[approx.] the numerator [upper part] has to be only 90-35=55
In addition, De [in T4' formula] has to be 2.15*46 therefore T4'=0.316.
The Denominator [lower part] has to be:
0.0000405*0.082+0.0000405*(1+0.0725)*(0.044+0.084+3*0.316+3*0.065+3*0.977+3*0.05
For T3 n=1 as under the single-core cable outer jacket it is only one conductor.
For T4"a n=3 but for Correction Factor n=9.
Therefore I= 534 A

 

[7anoter4]

We have still two problems:
1) You have to be sure if your transformer and breaker terminals are designed for 90 dgr.C.
IEC 60364-5-52 limits the conductor maximum temperature at 70 dgr.C if you are not sure.
[NEC limits to 75 dgr.C].In this case the ampacity will be 455 A [as per my previous post]
2) The build-up induced voltage measured at the second end-not grounded armor end-has to be limited to 25 V. [or to maximum permitted by locally standards]