Ampacity of Buried Cable

You don't say where you are.

There's so much you don't say.

Try here:NFPA 70: National Electrical Code®

Well, this calculations has to be very lengthy and rigorous. I appreciate if anybody has done it before or can give me any suggestion on the ampacity of the cable with this buried configuaration. It was not possible for our case to maintain the spacing. ETAP gives some derated calculation based on ambient temp. and rho (thermal resistivity), but it doesn't give any calculation if spacing cannot be maintained. NFPA 70 doesn't actually say anything about it.

Indian Electrical Codes - Free PDF downloads

See previous threads on CR4 as we have answered this common question to death over the years.

Hint - this derating information is available from the cable manufacturers.

Yep. British Standard 7671. Again <sigh>.

If the spacings between conductors are unknown, assuming it was a triplex cable would serve as a worst-case scenario.

Don't know if this will help, but this company has some excellent resources and even apps for smart phones that help you calculate ampacity, wire size, etc.

http://www.cerrowire.com/ampacity-charts

Good luck

Since no voltage is mentioned I suppose it is low voltage.

From NEC [low voltage cable] 310.15 (B) Tables.

Table 310.15(B)(16) Allowable Ampacities of Insulated Conductors Rated Up to and Including 2000 Volts, 60°C Through 90°C (140°F Through 194°F), Not More Than Three Current-Carrying Conductors in Raceway, Cable, or Earth (Directly Buried), Based on Ambient Temperature of 30°C (86°F)*

No distance between these 3 cables it is required.

This is an "official" ampacity and AHJ will refer to this, I guess.

310.15 (C) Engineering Supervision [in Excel language]:

I=sqrt((Tc-Ta)/(Rdc*(1+Yc)*Rca)

where:

Tc = conductor temperature in degrees Celsius (°C)

Ta = ambient temperature in degrees Celsius (°C)

Rdc= dc resistance of conductor at temperature Tc

Yc = component ac resistance resulting from skin

effect and proximity effect

Rca= effective thermal resistance between

conductor and surrounding ambient

See:

AIEE Paper 57-660, The Calculation of the Temperature

Rise and Load Capability of Cable Systems, by J.

H. Neher and M. H. McGrath.

Rdc=12.9/CI copper conductor 75dgr.C CI=circular inches -expression(10)

Yc=Rac/Rdc=1+Ycs+Ycp Ycs=skin effect factor Ycp=proximity effect factor (20)

Rdc=12.9/CI copper conductor 75dgr.C CI=circular inches -expression(10)

Yc=Rac/Rdc=1+Ycs+Ycp Ycs=skin effect factor Ycp=proximity effect factor (20)

Rca=Re'=0.012*roe*n*[log(Dx/De+LF*log[(4*L/Dx)*F]] thermal ohm/ft (44)

Dx=8.3"[approx.211 mm] L=900-1000 mm[cable centerline depth]

F=(d21'/d21)*(d23'/d23) for the middle cable as hottest.

let's say for 900mm depth d21'=900*2=1800 mm [aprox]

let's say d21=cable o/d=50 mm

correct d21'=sqrt(1800^2+50^2)~1800 mm

d23=50 d23'=1800

F=1800/50*1800/50=1296

LF=load factor[usually=0.7-1]

You may calculate the ampacity following IEC 60287-1,2 also.

Hi 7anoter4:

Many thanks for your info.

I must say you are only one here who have understood the problem and given a close solution. There are many "Gurus" here who have a tendency to jump to all sorts of questions and try to answer those. I have clearly requested if someone has done it before.

Now, let me give you the full picture if you are able to help me. The cable is an URD cable 25kV 3-1C 250MCM buried underground at a depth of 1000mm. It is supposed to be placed with a spacing of 190mm in flat configuaration between conductors or in triplex form (we are not considering this triplex form here). But the spacing could not be maintained due to some reason and they are placed flat with no spacing in between and looks like "ooo". According to CEC, the ampacity should be 361 amps if spacings are maintained. My question is how do we calculate the derated ampacity since no spacing is maintained?

Thanks.

361 x the de-rating factor for touching cables. 303 Amps. Max, for the route length.

Your de rating factor is that, where the cables are closest, and that you can apply to the whole route. Decide over what distance you have the cables touching or in very close proximity to each other and simply use the manufacturers calculation which is simple. Without checking, your de rating factor is around 0.81 to 0.84. If you had maintained the spacing, I am sure the factor is 0.9. Not a great deal is gained.

The calc's in IEC and IEEE are sufficient for use. Easiest, call the cable manufacturer or designer or consultant for the calc's. You are obviously not either of them.

First of all, I agree with IQ. In order to calculate the ampacity of this cable I need to know cable details as insulation thickness ,shield details, jacket details and the Manufacturer knows these better.

You may send an inquiry to a few manufacturers with no more details like cable voltage, the 100% load current, the soil average temperature, thermal resistivity[RHO] , the depth of cable run and the shield grounding system. They will send you a proposal.

This cable it is a medium voltage cable and in this case you have to take into consideration shield losses and dielectric losses. The expression is showing in NEC 310.15 (C) Engineering Supervision I=sqrt((Tc-Ta)/(Rdc*(1+Yc)*Rca) it is not suitable for this cable.

I=sqrt((Tc-Ta-DTd)/(Rdc*(1+Yc)*Rca') [expression no.9].

Rca'=Ri+qs*Rse+qe*Re' thermal-ohm feet [exp. 8]

I think there is not significant difference between flat arrangement without distances and triplexed.

According to NEC Figure 310.60 "Cable Installation Dimensions" maximum depth to the top of direct buried cables shall be 900 mm (36 in.).

Table 310.60(C)(85) Triplexed single RHO 90 cond.temp. 90 dgr.C ambient 20 dgr.C.

for copper conductor 250 kcmil permissible 390 A[Detail 7].

For 30 dgr.C ambient using equation: I'=I*SQRT(Tc-Ta')/(Tc-Ta))=I*k k=sqrt((90-30)/(90-20))=0.9258

See: 310.60 Conductors Rated 2001 to 35,000 Volts (4) Ambient Temperature Correction.

Then I=390*.925=361 A.

According to (2) Burial Depth of Underground Circuits. (b) "Where burial depths are deeper than shown in a specific underground ampacity table or figure, an ampacity derating factor of 6 percent for each 300-mm (1-ft) increase in depth for all values of rho shall be permitted."

So for only 100 mm more than 900 mm the derating factor will be 6/3=2%

361*(1-2/100)=353.8 A

Calculated for flat arrangement and no clearance between cable-as the proximity effect will be a little more-the permissible is 353 A.

Remark: the shield shall be grounded in one single point only.