What is the RHO value of SAND in an electrical duct as the result affects the AMPACITY of the cable?

[bdn2004]

We need a total of 885 amps of cable ampacity to feed a new AC motor that is replacing a large DC motor.
The new motor is to be fed from a VFD with special 3/C, 350kcmil, 1000V, shielded cable.

The easiest way to do it would be to pull out the old cables and run the new in the existing conduits. One in 3/C in each conduit.
The existing raceway is under the floor in a makeshift duct bank that has 3 PVC pipes and is filled in between the pipes with sand.

The closest detail is Detail 2 as shown in the attached. As you can tell using the 60 RHO column gives an ampacity of 340 x 3 = 1020A, but a 90 RHO 267 x 3 = 801A, not enough. How do you better determine the correct value of RHO ?

 

[dpc]

There are many different types of sand. That's your dilemma. Unless your trench geometry matches the NEC cartoon, your ampacity will not be exactly equal, even with the correct rho value. Check out AmpCalc software if you want a more precise calculation.

Maybe a start here:

https://www.tdworld.com/intelligent...66169/underground-cables-need-a-proper-burial

Good luck,

Dave

 

[7anoter4]

Because the ampacity is not so far from 300 A you have 3 possibilities:
1)use IEEE 835/1994 as recommended by NEC Annex B :
For additional information concerning the application of these ampacities, see IEEE STD 835-1994, Standard Power Cable Ampacity Tables.
In IEEE 835/1994 page 4 for in Underground Duct Bank - Triplexed - Three Circuits 25ºC Earth Ambient 60 Rho, 90 Rho ,120 Rho
75ºC - Copper Conductor - Concentric Strand
for RHO 90 100% LF 350 kmil ampacity= 269 A.
Further, using formula from IEEE 835 3.4.1 Adjust for changes in ambient temperature
I'=SQRT((75-20)/(75-25))*269=282.13 A at 20 oC ambient.
However, the cable is here 0.681 bare and 0.871 insulated.
2) Calculating -as permitted in NEC 310.15 Ampacities for Conductors Rated 0–2000 Volts. (C) Engineering Supervision.
for cable of 3*350 kmil 0.681 in. diameter and 0.811 in. insulated core the ampacity is 299 A.
3)you may reduce the sand depth to 1.5 ft instead of 2.5 ft standard [and permitted by art.300.5] and for the same cable as per IEEE 835 recalculated the ampacity will be 303 A.

 

[cuky2000]

Pure sand does not retain water resulting in has high thermal resistivity. In a duct bank surrounded by other types of soil the sand will have the soil migrating and over time improve the average thermal resistivity. Typical concrete in dut bank have much better thermal resistivity. So for the compound sand-silk-concrete, we typically use RHO=150 to 200.

See if the graph below from the NEC can help to do sensitivity analysis with different RHO and load factors.

>>>

 

[bdn2004]

Thanks for the answers. Awesome.

Can someone explain the NEC definition of the Load Factor - and exactly how it can be determined most accurately?
I understand it's a percentage of the total load.

 

[7anoter4]

The loss factor LF was introduced by Neher & McGrath in their publication
Temperature and Load Capability of Cable System October 1957
in order to calculate the temperature drop from loaded conductor to ambient.
ΔTc=Wc[Ri+qsRse+qe(Rex+(LF)Rxa]
where (LF)=0.3*lf+0.7*lf^2 per unit
and lf=load factor.
if lf [the load factor for a least one hour or daily load] lf=1 then LF=1 too.

 

[cuky2000]

The definition of load factor is the average power in a period of time in a representative load cycles divided by the time and maximum demand.
One way of determine the LF is taking a reading every hour then LF= ∑Amp/24.Amp_max.
Other way is using a recorder over longer representative time cycle obtain the average and divide it by the peak demand.
This later method usually used by utilities and other researchers to estimate typical load factor for residential, commercial and industrial load.
For cable sizing, usually the LF is assumed or based in typical database from previous load estudies.