Derating LV cables installed in same trench as HV 600mm apart 2

[DrDrreeeaaa]

Hi All

Can anyone direct me to the applicable standard?

I am working in Australia, the Australian standards do not address this case.

I have a 4x1c95mm2 (430V 200A) Cu XLPE in conduit 600mm away from a 3x1c300mm2 (3.3kV 300A) XLPE in conduit.

LV at 500mm below ground, HV at 750.

Thanks

Michael

 

[DrDrreeeaaa]

Is it a case of needing computer-calculations (CYMCAP) etc.?

 

[7anoter4]

Hi Michael
As this cable is metric units I should recommend IEC 60287 .This is not a standard like NEC [indicated values on tables] but is a way of calculation.
According to my calculation based on this IEC 60287 no derating is required if:
earth temperature=25
Conductor max. temp.=90[in order to calculate the resistance]
earth thermal resist.=90
ins.+jacket thermal resist=350
Current freq.50 HZ
Max.cond.temperature will be no more than 60 degree C
Regards

 

[DrDrreeeaaa]

7anoter4, thankyou very much for the calculation.

Which volume of IEC 60287 did you use as I would like to purchase it. Also do you have the page/clause reference.

Thanks again

Michael

 

[sberbece]

Michael,
The applicable Australian standard is AS/NZS 3008.1.1/1998 - Electrical installations-Selection of cables Part1.1 - Cables for AC up to and including 0.6/1kV - Typical Australian installation conditions.
You can use the table 26(2)/page 62. This standard recommends a derating factor of 0.96.

Regards,
Stefan

 

[DrDrreeeaaa]

Yes, but, the standard is for cables 0.6/1kV and I have a 3.3kV cable on one side and a 0.4 kV cable on the other.

Won't the 3.3kV cable's current have more energy than the 415? precluding my appliation of AS 3008?

 

[sberbece]

The issue here is the temperature influence that cables will have against each other.
The voltage level do not count (at your voltage level, LV & 3.3 kV), only the carrying currents that will heat the conductors, count.
Therefore, I recommend you to use the AS3008 standard, which I found very conservative.

Regards,
Stefan

 

[DrDrreeeaaa]

Stefan, thanks alot for your comments.

Saying a 3.3kV cable and a 400V cable operating at the same current will have the same thermal emissions seems counter intuitive.

For the same cable size, installation method, and current, the current in the HV cable would be under a higher potential difference and therefore have more energy. For the same type of cable this would result in more thermal emissions??

For the same size cable and installation method, ERA 69-30 has about 50 Amps less CCC for the HV cable (at around 400A). This tells me there is a difference but it is minor.

Practically it seems that As3008 can be applied.

I would still like to hear other's opinions.

 

[jghrist]

The difference in current carrying capacity of HV vs. LV cable is not because of the voltage or energy carried. It is because the HV cable has thicker insulation and has shield wire with induced currents.

 

[jghrist]

It's the energy dissipated in losses that reduces current carrying capacity, not the energy transmitted.

 

[davidbeach]

An the energy dissipated is a function of I²*R; neither of which is related to voltage.

 

[7anoter4]

Hi Michael
I have the 1982 edition modified in 1988.Then all the volumes were united in a single volume: IEC 287.Now I see there are many volumes including parallel cables and other. What you need is:
1-Cable theoretical construction [rebuilding the cable in order to calculate the conductor dia, insulated core dia, inner jacket thickness, outer jacket thickness and dia, and so on. This may be IEC 502.
2- Further you need losses calculation as resistance loss, dielectric losses, shield losses [if any].This one could be found in this volume 1-1.
3. You need thermal resistance of cable itself, of conduits and of the earth. This may be in the volume 3-1.
I'll try to give you only for your application the usually used formulae.

First of all as thickness of insulation for 0.6/1 KV 95 sqr.mm is 1.1 nominal and for 3.3 is 2.2 nominal we try to "rebuild" the cables according IEC 502:
cond.dia core assy o/d
4*95 0.6/1 kV CU xlpe/PVC 12.9 16 38.5 44
3*300 3.3 kV CU xlpe/PVC 23.1 28 59 65
AC Resistance of conductor
R=R'(1+ys+yp)
R= ac current resistance of conductor at maximum operated temperature [for XLPE insulation for 90 oC](ohm/m)
R'=d.c. resistance of conductor at maximum operated temperature(ohm/m)
ys=the skin effect factor
yp=the proximity effect factor
D.C. Resistance of conductor
R'=Ro[1+a20*(T-20)]
Ro=d.c. resistance of the conductor at 20 oC[ohm/m]
Ro from IEC Publication 228
For copper conductors:
for 95 sqr.mm Ro= 0.193 stranded class 2[concentric stranded]
for 300 sqr.mm Ro= 0.0601 stranded class 2[concentric stranded]
a20=3.93.10^-3 Table I
T=90 oC for XLPE [Crosslink Polyethylene]
ys=the skin effect factor
ys=xs^4/(192+0.8*xs^4)
xs^2=8*pi*f/R'/10^7*ks
f=supply frequency in Herz [50 or 60]
for your application ks=1
yp=the proximity effect factor
xp^2=8*pi*f/R'/10^7*kp
for your application kp=1 then xp=xs
dc=diameter of conductor [mm]
s=distance between conductor axes[mm]
for your application s=dia.insulated.core
for three-core cables circular conductor :
yp=xp^4/(192+0.8*xp^4)*(dc/s)^2*[0.312*(dc/s)^2+1.18/(xp^4/(192+.8*xp^4)+.27)]
for your application:
yp=ys*(dc/s)^2*[0.312*(dc/s)^2+1.18/(ys+0.27)]
If the cables are in steel conduit :
R=R'(1+1.7*(ys+yp))
If the cables are in plastic conduit :
R=R'(1+ys+yp)
Resistance losses= 3*I^2*R[w/m]
If the cables are unshielded or the shield is not shortcircuited both ends for U<15 kV no other loss could be considered.
Calculation of thermal resistances [on the further post,soon]

 

[7anoter4]

Resume
R'=Ro/1000*(1+3.93/1000*(90-20))
R'1=0.193/1000*(1+3.93/1000*(90-20)
R'2=0.0601/1000*(1+3.93/1000*(90-20)
R'1= 0.000246
R'2= 0.0000766
xs^2=0.5106
xs^2=1.6398
ys1=0.0013566
ys2= 0.01385
dc/dins1=0.80625
dc/dins2= 0.825
yp1=0.00401
yp2=0.04119
Rac1=0.000247ohm/m
Rac2=0.00008085
Loss1= 29.69w/m
Loss2= 21.83 w/m
Thermal resistances calculation
Thermal resistance between one conductor and sheath T1
T1=roT/2/PI*G
roT=thermal resistance of insulation[for XLPE=3.50 K.m/w PVC=5 average=4.25]
G=geometric factor
From fig.3 Geometric factor for three-core belted cables with circular conductors
t=thickness of insulation between conductors[3.1 for 4*95;4.6 for 3*300 (3.3 kv)]
t1=thickness of insulation between conductor and sheath[3.525;4.15]
dc=diameter of circular conductor[12.9;23.1]
From fig.3 Geometric factor for three-core belted cables with circular conductors
G=0.6;0.6
T1[4*95]= 0.4058451
T1[3*300]= 0.4058451
As these cable are built only of insulation and jacket no other thermal resistance will be involved.
Thermal resistance T'4 between cable and conduit
T'4=U/(1+0.1*(V+Y*thm)*De)
From Table VII U=5.2 V=0.91 Y=0.006 [in fiberduct in concrete]
De=O/Dia = 44;65
thm=the mean temperature of filling medium[air]
thm=40 [appreciated]
T'4=5.2/(1+0.1*(0.91+0.006*50)*De)
T'4=0.822264;0.586576
DT4=Temperature drop from cable to conduit
DT41=20.694[o C]
DT42= 11.04
Now we have to calculate mutual heating from one conduit to another
dthp1-2=the temperature rise at surface of the conduit2 produced by the losses of cable 1
dthp1-2=1/2/pi()*roearth*Losses1*ln(d'12/d12)
d1-2 is the distance measured from the center of conduit 1 to the center of conduit 2
d'1-2 is the distance measured from the center of conduit 1 to the center of reflection of conduit 2 in the ground-air surface respectively
d1-2 = SQRT( 0.6 ^2+0.25^2)=0.65 m
d'1-2 =sqrt(0.6^2+1.25^2)= 1.3865m
d2-1=d1-2=0.65m
d'2-1=sqrt(0.6^2+1^2)= 1.1662
Roearth=0.9
Tearth=25 degree C
dthp1-2=3.22degree C
dthp2-1=1.828degree C
Tcable1=63.45degree C
Tcable2=48.58degree C
Regards

 

[DrDrreeeaaa]

I2R is not related to voltage?

i & V

Thermal & V^2
Thermal & 1/R

??

 

[davidbeach]

Consider 1m of conductor; same size conductor, same current in the conductor. Therefore the same, very small, voltage drop across your 1m of conductor. That is the only voltage of concern for the problem at hand. Your 400V and 3.3kV are externalities that along with the load determine the current flow in the conductor. But once the current is determined voltage ceases to be a factor.

Third or fourth order effects of leakage through the insulation may be related to voltage to ground of the conductor but you won't know your soil thermal resistivity well enough to determine the difference in heat due to insulation leakage.

The presence of a shield on the 3.3kV cable will increase the heat produced by the 3.3kV, but only as a factor of the construction of the cable. Use the 3.3kV cable for both the 3.3kV circuit and the 400V circuit and you will have similar heating if you have the same current in each cable.

 

[DrDrreeeaaa]

Great explanation mate, thanks for that.

 

[7anoter4]

I agree with davidbeach that the voltage here is not significant. The only change is the insulation thickness and as you see is not so important too. For more than 10-15 KV the dielectric losses will increase the total losses and has to be considered.
Up to 5 KV usually the cable is not shielded. But if is a shield and is grounded only at one end the losses are insignificants. In this case the shield will reduce the cable temperature [slightly] but for buried cables in duct [or conduits] is not important too, as you see. I did not want to complicate you with the shield [as I did for other cable is totally not important for your application].My computer [Visual Basic] is based on Neher theory and the results are slightly different [the temperature is 1-2% less]
Regards

 

[DrDrreeeaaa]

Thankyou 7anoter4