Cable ampacity change due to RHO 1

[guiyermo]

Hi everyone,

If I know the ampacity of a given cable directly buried, provided the conductor's temperature, thermal resistivity (RHO), soil temperature, etc, is it possible to calculate a factor in order to determine the new ampacity when only the RHO has changed? The rest of the parameters remain the same.

I've seen equations for determining changes in the ampacity due to a variations in ambient or conductor temperature, but not for variations in RHO.

Regards,

 

[WhiteyWhitey]

guiyermo,

Are you refering to the internal or external thermal resistance of the cable?

internal thermal resistance is the resistance of heat flux from the source (conductor and Sheath) to the cables outer surface,

the external thermal resistance is the resistance between the cables outer surface and the surface of the ground or the almighty heatsink.


If you are after a derating factor for the external thermal resistance then i cant really help but I have noticed that it is notoriously hard to quantify changes in ampacity over short lengths of thermally unfavourable surrounds such as service crossings, deep burial depths etc.

If you want to try the parametric analysis, the formula you are working with is this

I^2 = DeltaT - Wd(0.5T1 + n(T2+T3+T4))
--------------------------------
R(T1 + n(1+Theta1)T2 + n(1+theta1+theta2)(T3+T4))

where
I = ampacity
T4 = external Ther Res
T1 = Ther Res between Con & Sheath
T2 = Ther res between Cond & outer sheath
(steel conduit / Armouring if present)
T3 = Thermal resistance of sheath covering
theta1 = metallic sheath loss factor
theta2 = armouring loss factor
Wd = Dialectric loss per unit length in each phase
R = AC resistance at maximum tempurature of conductor.
n = number of conductors.


If anyone can explain a way to calculate the external thermal resistance of a conductor in Thermal backfill of trench width D that would be great! I am relying on results from an Finite Element Analysis Program run by someone else and I would love to be able to do some cross checking.

Cheers,
Andrew

 

[7anoter4]

The formula presented by WhiteyWhitey it is according to IEC 60287.
where DeltaT=Tmax-Tinitial[usually the ambient]
For PVC insulation Tmax=70 degrees C and for XLPE= 90.
For a single [one] buried cable:
T4=1/2/pi()*RHO*ln(u+sqrt(u^2-1)) [Excel spelling]
u=2h/odia odia=overall cable diameter h=cable center line depth[up to ground surface].
Let's say we have RHOa and RHOb and we know Ia. According to above formula:
Ia^2 = DeltaT - Wd(0.5T1 + n(T2+T3+T4))/(R(T1 + n(1+Theta1)T2 + n(1+theta1+theta2)(T3+T4a))
Ib^2 = DeltaT - Wd(0.5T1 + n(T2+T3+T4))/(R(T1 + n(1+Theta1)T2 + n(1+theta1+theta2)(T3+T4b))
For only one buried cable, up to 15 kV and shield grounded only one end[Wd[approx]=0,Theta1=Theta2=0.We also may neglect T1 and T2 as T4 >>T1,T2] :
Ia^2=DeltaT+R*n*T4a
Ib^2=DeltaT+R*n*T4b
Ib^2-Ia^2=R*n*(T4b-T4a)
R*n=(Ia^2-DeltaT)/T4a
Ib^2=Ia^2-(Ia^2-DeltaT)/T4a*(T4b-T4a)
If there are more buried cables we have to involve the mutual heating between cables so this formula is not applicable.

 

[jghrist]

I don't think that there is a simple way to adjust for RHO because the ampacity depends on the heat flow through the insulation and the earth. Heat flow through the thermal resistance is analogous to current flowing through two resistances in series. You can't determine the voltage increase across the series resulting from increasing one resistance without knowing the other resistance.

 

[guiyermo]

Hi,

Thanks for your response.

From the NEC 2008 Table B.310.8 "Ampacities of Three Triplexed Single Insulated Conductors, Rated 0 Through 2000 Volts, Directly Buried in Earth Based on Ambient Earth Temperature of 20°C (68°F), Arrangement per Figure B.310.2, 100 Percent Load Factor, Thermal Resistance (Rho) of 90", one can find the amapcities for copper and aluminum cables @60 °C and 75 °C for four different cable arrangements.

The arrangment I am interested in is a single 3-conductor cable with no other cables surrounding it (Figure B.310.2 Detail 5), directly burried in earth and the same conditions depicted in Table B.310.8 title with the only difference that the Thermal Resistance (Rho) is 120 instead of 90.

Is there a way to calculate the ampacity of a cable under these conditions (Rho = 120) based on the ampacities shown in Table B.310.8?

(Sorry I did not upload any print screen of the NEC since I think I am not allowed to because of Copyright issues)

Sorry for my bad English,

Best Regards,

 

[7anoter4]

Thank you jghrist for your remarks. First of all I did a big mistake –small algebra big mistake- as I took somewhere at a slash as * and the final formula was totally wrong.
In the same time I checked the T1-T2 to T4 ratio.
For a cable of 600 V [Okonite] 3*500 MCM O/D=2.44"=62 mm 60 mils insulation and 75 mils jacket RHO insulation 3.5 K.m/w.
For a non shielded cable T1 and T2 are together in a single formula:
T1=Rhoins/(2*PI())*G
G=1.242 as per Mie formula [IEC 60287]
Cable underground depth 2.5 ft =.762 m and Earth RHO =0.9 K.m/w
T1=0.692 and T4 =0.558.Surprising for me but very conclusive: You are right, we can't neglect T1.
In order to answer to guiyermo about NEC I'll use the J.H. Neher & M.H. McGrath method as it is indicated in NEC 310-15 (b) article, soon.
Best Regards

 

[dpc]

The NEC tables are based on the Neher-McGrath paper and calculation methods. You can apply the rho adjustment if you know what all the other factors used were. And I think there is sufficient data to do that.

 

[rcw retired EE]

Doesn't Figure B310.1 in Appendix B of the NEC show ampacity adjustments for various values of rho?

 

[guiyermo]

Hi everyone,

I've read a little more about this and as many of you replied, there is no “easy” factor to be applied against the cable’s ampacity.

I should study every single cable configuration to determine which is the maximum current that can flow through each cable without harming its insulation.

Thanks to 7anoter4, WhiteyWhitey, Jghrist, Dpc and rcwilson for your support.

Also, I want to reply rcwilson that I checked figure B310.1 and only applies for duct banks.

Best Regards,

 

[7anoter4]

You are right, dpc I have all the data to calculate the ampacity according to Neher & McGrath. Unfortunately, NEC is more conservative for large cross sections. As I know also VDE/DIN 298 does not follow IEC 60287 entirely and I think the above VDE takes into consideration soil water loss and increasing RHO around the cable. Also it takes the Load percent 70% only as usually.
After the correction, for 120 oC.cm/w and according to Neher and McGrath I propose the following[for 3 copper conductor 75 degrees C maximum]:
Size I[A]
AWG/MCM]
8 63
6 82
4 108
2 143
1 160
1/0 180
2/0 210
3/0 240
4/0 272
250 297
350 360
500 435
750 535
1000 581