Underground cables in Concrete Encased PVC ducts 1

[Adam1980]

Dear members,
for the above mentioned installation conditions for underground cables a thermal calculation is usually recommended. However in case we want to make a rough estimation of the cable amapcity derating what is the recommended factor to be used? According to available references from textbooks and manufacturer catalouge this factor varies between 0,8 and 0,9.
Any recommendations or suggestions based on experience?

Please notice that the subject here is a factor which is additional for the other factors of ground temperature, grouping, soil thermal resistivity.

Thank you.
Adam

 

[7anoter4]

I think a rough estimation could be NEC Art. 310 and Annex B. The cable ampacity depends on a lot of factors:
concrete and earth thermal resistivity[RHO],concrete duct bank dimensions, duct diameter, duct depth-from the surface-earth temperature, number of loaded ducts in the duct bank and other.
An average concrete thermal resistivity could be 100 oC.cm/w [1 K.m/w] as per IEC 60287-2.NEC uses 55oC.m/w and IEEE 835 -60.RHO[earth] from 60 to 120 [IEEE-835] and up to 250 according to IEC.[it could be more].

 

[Adam1980]

Hello 7anoter4,
thank you for the information.

My question is in case we have two cable arrangments which are at similar installation conditions (earth resistivity, grouping, depth, temperature) and one of those arrangements is in PVC duct and the other is directly buried what would be a rough estimation of the de-rating factor. In one catalouge from ABB it is mentioned that this factor is 0,9 while a textbook "Transmission and Distribution Electrical Engineering" states that the factor is 0,8?
i started to do the calculations according to IEC 287-1-1 and 287-2-1 hoping that i arrive to a satisfactory answer but any help is appreciated!
Thanks.

 

[7anoter4]

As I know-See ABB Switchgear Manual ed.11 ch.13.2.2 Current carrying capacity –according to DIN/VDE 0276-603-for PVC or XLPE insulated low-voltage cable it is only for directly laid underground. A reducing factor of 0.85 is
recommended when laying cable in conduits.
However, if you see, the factor f1[Table 13-54] for ground temperature depends also on earth thermal resistivity[from 0.7 to 2.5] and on load factor[0.5-1].This factor could be from 1.24 to 0.68 but it is referring to cable laying underground at 20 dgr.C 0.8-1.2 m depth.
If you would be more specific I could –may be-more helpful.

 

[Adam1980]

i already calculated the ampacity of three cables layed in 9-way PVC encased ductbank using thermal calculations however for some reason the customer had their own calculations which depend simply on factors and the are considering the de-rating of a PVC encased duct bank as 0.8 which is resulting in a difference from the output of the above mentioned calculations so i am trying to know how this 0.8 (not mentioned in IEC 287 or 520-2) was reached or how to approximate this value in order to explain to them that they did it wrong.

The system is 60 Hz 13,8 kV 300 mm.sq. single-core unarmoured cable. The cables are laid in trefoil in 160 mm PVC duct in a 9-way duct bank. the center of the bottom ducts (worst case) is 1500 mm underground with 40°C ground temperature and 2 K.W/m soil thermal resistivity.

Thanks

 

[7anoter4]

There are some data still missing. For instance shield losses-eddy current and circulating current if the shield is grounded both ends. The distance between two adjacent ducts and number of columns and the rows[usually eddy current will be 7% of main conductor losses and circulating current=0 and the ducts are arranged in 3*3 order].
The distance between ducts[vertically and horizontally]will be 250 mm[center-to-center].In this case the ampacity will be 166 A/cable. If the cables are laid directly in the ground keeping the same distances and arrangement the ampacity will be 184 A [It is equal to the ABB tables also:
3*300 sqr.mm copper XLPE insulated 12/20 kV-buried cables in trefoil in 40 dgr.C and 2 K.m/w will be 599*0.68=407.3 A and for 9 groups will be:
407.32*0.45=183.3 A/each].

 

[Adam1980]

Thanks!

so if the rating for cable in ducts is 166 A and for directly buried is 184 A the factor in between is around 0,9 (assuming all other conditions are the same). for a rough estimation is this valid?
for the duct in question it is a 3 x 3 configuration vertically and horizontally there is 50 mm spacing between the ducts (165 mm center to center)and 75 mm from the concrete encasing border to the duct order also vertically and horizontally.
i believe i have all data i need to calculate the ampacity using the equations in 287-1-1 and 287-2-1 but i am not getting a logical result. I am doing calcualtions only for one cable (3x1 300 mm.sq. in PVC duct) to see what is the ampacity and compare with the one given by the manufacturer for laid direclty in ground.
Finally, you said that you are using information from ABB for the 9 cables arrangement but is it valid to use the factors from IEC 502 when considering 3 x 3 configurations, the factors are for horizontal arrangement only! correct?
Thanks

 

[7anoter4]

You are right! But ,as ABB considers a dry-out factor of 1.5, for 9 groups in line and I neglected this factor since IEC 60287-1-1ch.1.4.2.1 does not permit to use it for more than one group and I got the same result for 3*3 arrangement[coincidence!].However, if concrete thermal resistivity is less than surrounding earth the concrete will "cool" the duct and rises the ratio between ampacities of the two laying systems from 0.8 to 0.9.

 

[Adam1980]

7anoter4 , Thanks for your input it gave me some confidence in the assumptions i was making, additinally i saw table B.6 in IEC 502-2 where they have the ampacities for laid in duct and directly buried mutli-core cables the factor varies from 0,86 to 0,9 depending on the cable cross-section. i used the multi-core cable since in that standard they are considering single-core cables each phase in separate duct.

Finally is it possible to explain more how to calculate the 3 x 3 arragnement from the IEC equations? we can do that on private email if you prefer?

Thank you.

 

[7anoter4]

I am sorry, my e-mail it is of reduced size and always full-I neglected to treat it, I agree, of course. The software I used to calculate it is a Visual Basic 6 –"hand made" and it is impossible to transfer through an e-mail and even attached here. But I intend to "translate" it-only for our case-into an excel file soon.

 

[7anoter4]

It seems to me even an ordinary excel file I cannot send it so only jpg images of this file. I am really sorry!

 

[7anoter4]

The first jpg actually is here!

 

[Adam1980]

Thank you very much for sharing this. I will go over the standard and try to remake the calculations to get the results.
Thanks

 

[Adam1980]

Hello,

is there a way according to IEC 287 to calculate the maximum capacity of a cable in a concrete duct bank while other cables are loaded equally. For example, if i have same to the above calculations 9 cables each with 150 A when all cables are operating, for 8 cable operating can i have 7 cables operating at 150 A and the 8th at 175 A?

Thanks

 

[7anoter4]

We calculated the ampacity for 27 single core cables equally loaded following IEC 60287-2-1 ch.2.2.7.3 External thermal resistance of the duct (or pipe) T4”’ connected with 2.2.3.2 Equally loaded identical cables. This way uses global formulae simplifying the calculation. If we intend to load differently one or more cables then we have to follow ch.2.2.3.1 Unequally loaded cables. This is not a relation but a procedure in order to find the final temperature distribution and iterate the way until reach the maximum permissible. This way is difficult to transfer in Excel. There are slight differences between results.
The first way result was 166 A [exactly 165.5 A] and the second-considering also equally loaded cables-163.5 A. If the 3 cables in one duct of the upper row –first or last [not this in the middle] will be loaded with 175 A the ampacity of the rest of cables will drop to 161.5 A [reaching maximum temperature at the duct number 5- in the middle of the duct bank].

 

[Adam1980]

Thanks again for your very precious reply. I will be doing some calculations and would like to consult you on them later.

 

[Adam1980]

Hello 7anoter4,
thank you for youe previous assistance. the sheets you sent helped me a lot in understanding the calculations. I made the calculations and have some questions. I hope it is not a problem to ask you.
1. when i am considering non-armoured cables lambda 2 in 1.4.1.1 from 287-1-1 and T2 in 2.1.2.1 from 287-2-1 are zero since there is no armour

2. when considering single-core cables in duct banks:
2.a we have the parameter n in 1.4.1.1 from 287-1-1 which is number of load-carrying conductors. in this case n should be 27 in case i have 9 ducts which have cables each having 3 loaded conductors for which we calcualted T1, T2, T3, R and all the other variables.
2.b we have the parameter N in in 2.2.7.3 from 287-2-1 this should be the number of ducts containing acables in service, example for a fully loaded 3 x 3 duct bank i will have N = 9

3. should i always consider Wd in 2.2 from 287-1-1 in table 3 it is mentioned that this could be ignored for operating voltage under 63 kV for XLPE! for 13,8 kV and 24,5 kV operating voltage this could be neglected correct?

4. in your calculations you considered the eddy currents calculations as in 2.3.6.1 from 287-1-1 for single bonded or cross-bonded systems. However isn't it more critical to calcualte current capacity for bonding at both ends? using 2.3.1 the eddy currents could be ignored. However it wasnt clear for me the process of iterative calculation in 2.3.1 and i assumed losses of 10% coming from circulating current.

5. just to clarify a final point if i am calculating for the same cable and same ductbank for different number of loaded cables i only have to change the N in 2.2.7.3 from 287-2-1 and n in 1.4.1.1 from 287-1-1 and the d'pk/dpk product in 2.2.3.2 from 287-2-1 to correspond to the distances between the expected worst condition cable and the other cables.

Thank you very much!

 

[7anoter4]

1)Correct. Lambda 2=0
2)Correct. If in 1.4.1.1 formula you’ll take n=3 then in 2.2.7.3 N=9.
3)Wd may be neglected, indeed.
4)I considered only eddy current but if the shield is grounded both ends then a circulating current will flow through it. In this case you may neglect eddy current, of course.
5)Since the current are different you have to follow 2.2.3.1 Unequally loaded cables and fk factor will not be useful. But you may split the temperature drops in two groups :
Dp=1/2/pi().W1.LN(d’p1/dp1)+1/2/pi().W2.LN(d’pk/dpk*…*d’pn/dpn)|k=2 to n-except k=p[k<>p]
New Fk= d’pk/dpk.….d’pn/dpn|k<>p,k<>1
W1=total losses in cable no.1 [main conductor losses+shield losses+wd(if any)+armor losses(if any) and W2 is the same sum of losses in the rest of cables.

 

[Adam1980]

Thanks!
one last question:
above you indicate n = 3 and N = 9 is there any place where we will use 27? that is 9 ducts and in each duct a cable with 3 conductors.

 

[7anoter4]

The total temperature drop from a loaded conductor up to environmental [ambient] medium it is as follows: [see 1.4.1.1 AC cables]
DT= (I^2*R + ½ Wd).T1 + [I2R (1 +lambda1) + Wd].n.T2 + [I2R (1 +lambda1 +lambda2) + Wd].n.(T3 + T4)
This is the sum of thermal resistances multiplied by losses-thermal flow passing through.
T1=Ri insulation thermal resistance of one cable only I^2*R=main conductor losses
T2=Rj jacket thermal resistance [for single-core cable one cable losses = main +shield]
T3 = external serving th.res.[for single-core cable one cable losses = main +shield+armor]
T’4=temperature drop from bunch of 3 cables to inner part of the duct [n=3 cables total losses]
T”4=temperature drop through the duct thickness [n=3 cables total losses]
T’”4=Tconcrete+Tearth
Tconcrete summarizes the thermal flux from all around ducts reaching the “one” then will be from all 8 remaining ducts. But you have to multiply by n=3 since each duct contains 3 cables.
Tearth –it refers to all cable together that means thermal flow = total losses of 3[n]x9[N] =27 cables.