Eddy Current losses on rebar buried in concrete
Eddy Current losses on rebar buried in concrete
(OP)
Hi
For a HV single conductor cable passing thru foundation with rebars around ,how much should be the clreance from rebar to cable to make sure that induced heating will not damage the PVC conduit and eventually the cable ?
Thanks
For a HV single conductor cable passing thru foundation with rebars around ,how much should be the clreance from rebar to cable to make sure that induced heating will not damage the PVC conduit and eventually the cable ?
Thanks






RE: Eddy Current losses on rebar buried in concrete
Bill
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"Why not the best?"
Jimmy Carter
RE: Eddy Current losses on rebar buried in concrete
RE: Eddy Current losses on rebar buried in concrete
"A single conductor cable carrying a current over 200A shall be run and supported in such a manner that the cable is not encircled by ferrous material.
Add to that field experience with (non code compliant) installations where ferrous material encircled not all of the service conductors and the subsequent heating. I have seen a 400 amp service disconnect switch destroyed by the heat conducted through the copper cables. The current was only about 200 amps but magnetic encirclement was a factor in producing the heat.
I consider 100 amps to be a little on the high side of conservative. I strive mightily to avoid ferrous encirclement at any current level.
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Eddy Current losses on rebar buried in concrete
RE: Eddy Current losses on rebar buried in concrete
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Eddy Current losses on rebar buried in concrete
RE: Eddy Current losses on rebar buried in concrete
length of the rebar around the conduit [felength]and the total current [intensity] which passes through the concrete hole.
I'll take a rebar of 1/4", a PVC conduit of 6" and a rebar around the conduit in 8" diameter circle [1" concrete around the conduit]. If H [magnetic field intensity]*felength[m] =Icable [A] and I=300 A felength=pi()*8*2.54/100= 0.63856 m
Then H=300/.63856= 469.8 A/m.
From a B_H diagram for a mild steel I get B[magnetic flux density]=0.12 T[or Wb/m^2] or 1200 Gs.
Neglecting hysteresis losses for eddy current losses in on the circular rebar I use the formula for a steel plate of a thickness
equal to the rebar diameter Pe=(pi()*B*t*f)^2/(6*10^7*d*ro) w/kg where f=60 Hz d=7.8 g/cm^3 ro= 10 micro ohm
Pe==26.4 w/kg. The rebar circle weights 157.7 gr=0.1577 kg.
The losses in one rebar will be 0.1577*26.4 =4.16 W. If we have 12 parallel circular rebars per one feet we get 12*4.16
=50 w/ft. Using formula q[heat flow= w/m^2 ]=k*dT/dx and k=1 K.m/w thermal conductivity of the concrete T=50/.3048^2*6*.3048 for 6 feet concrete depth from the surface. T=294.3 degree C. Adding the soil temperature of 25 degrees we'll get 319 oC rebar temperature. Don't touch it!
RE: Eddy Current losses on rebar buried in concrete
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Eddy Current losses on rebar buried in concrete
RE: Eddy Current losses on rebar buried in concrete
First of all the formula DT=q/lambda*depth is good for a definite way of the thermal flow-for instance through
[lambda=medium thermal conductivity=1/k] let's say a vertical wall of 1 ft depth or though an insulation or a through an other round solid. But the thermal flow in the soil is spreading during the rising to the surface so I'll take better the formula for round [cable or conduit] to ground surface from IEC 60287-2-1 -[taking the 12 rings of rebar as a conduit]:
The depth of 6 ft is also exaggerate as NEC will require only 750 cm[2.5 ft, only] and k=0.9[90 oC.cm/w]
So , ThermalRes=1/2/pi*k*ln(4*depth/Dia)=1/2/pi*0.9*ln(4*.75/0.63856) =0.246 oC.m/w
and DT=ThermalRes*losses==0.22166*164=36.34 oC losses=50 w/ft=164 w/m
Let's take a low voltage cable of 1*350 MCM copper 0.038855 ohm/1000 ft a.c resistance at 90 oC.
The conductor losses at 300 A will be 0.038855/1000*300^2=3.5 w/ft or 3.5/.3048=11.48 w/m
If we neglect the reciprocal heating of 3 phases the temperature drop from "rebar_ conduit" position up to surface will be ThermalRes*11.48=0.246*11.48=2.8oC. According to NEC the ampacity of 350 mcm copper cable will be 315 A. So 90oC-20=70oC temp.drop [20oC is the surrounding temperature -as per NEC 310.60(77).So the inner drop temperature
Up to "rebar conduit" will be 70-2.8=67.2 deg.C.
Subtracting the 36.4 degree for "rebar conduit" from 67.2 we'll get 30.8 remaining .That means in order to limit the
Cable conductor temperature to 90 deg.C we have to derate the ampacity by sqrt(67.2/30.8)=1.475.
The initial ampacity was 315A now will be 315/1.475= 213 A.
If the rebar will be bonded so that 1/8" gap will be provided then used the formula:
Hfe*LengthFe+Hgap*Lgap=I considering Bfe=Bgap=B then:
B/miufe*lengthFe+B/miuo*Lgap=I where miufe=Bfe/Hfe=0.0002554 T/(A/m)=2554/10^7 H/m and miuo=4*pi()/10^7=12.57/10^7 H/m lengthFe= 0.8379 m Lgap=25.4/1000/8=0.003175 m
B=0.0517 Wb/m^2=517 Gs. Losses= 0.17 w per rebar ring and 12*0.17=2.03 w/ft.
using the same ThermalRes=0.246 DT=0.246*2.03/.3048=1.6 deg.C= negligible