Max load on a Hochstadter type of cable?

[UHEngineer]

Hi all!

I've got a really old 36 kV copper cable of Hochstadter-type, and no data sheets available.

Anyone knows a decent table that can be used for max load of this cable? Or some resources as to which approximation can be made?
The closest I have is looking at a load table for a papper-insulated type of cable. Google search didn't find anything useful other than description of the cable model.

Thanks in advance!

 

[Desrod2]

If you can get details (type and thickness) of the various layers and core, it could be computed by software like CYMCAP.
You can even include the operating conditions like installation type and loading.

 

[UHEngineer]

Thanks for the tip Desrod2!

Unfortunately I don't have access to such software.
Basically, I'm in a need of a rough (but well grounded) guide of what I could load on that cable given a given conductor area.

UHEngineer

 

[7anoter4]

UHEngineer,
I know this type of cable, I think, a paper insulated lead covered cable but for not more than 15 kV. If you can inform us more like number of conductors-it has to be single-core, the conductor material-copper, I guess- and the overall cable diameter, we can try to “rebuild” the cable and calculate the ampacity. The cable ambient as underground or open free air or cable duct or cable tray and the ambient temperature would facilitate the calculation.

 

[7anoter4]

If we’ll take the information from “High Voltage Engineering” -by J R Lucas, 2001
High Voltage Cables - ch.5- 5.2.8 Hochstadter or "H" type Cable:
“In this type of cable (Figure 5.16), there is no belt insulation. The screening of individual cores is generally thin and flexible so that there is not much power dissipation in them. All the individual screens are earthed so that the potential at these sheaths are all zero and thus the stress lines between the cores and screens would be now radial. The screens are thin so that there is hardly any current induced. The sheaths surrounding the insulation of the cores consist of metallized perforated paper. These are wrapped round with copper woven fabric (cotton tape into which are woven copper wire). This outer screen is in contact with the inner screens and is earthed. The cable has the additional advantage that the separation of the cores by thermal expansion or mechanical displacement cannot introduce stresses in the dielectric. The metallized screens help to dissipate the heat. These are used up to 66 kV. In the H-type cable, the individual cores contain no lead covering. The three cores are laid up with fillers in the ordinary way. If the cable is to be buried, then the cable is armored with steel wire and tape. The wormings of the H-type cable are full of oil.”
and we’ll build the cable-following IEC 60502-2 way for 150 sqr.mm we get this sketch [see attachment].
Let’s calculate the ampacity according IEC 60287-1-1/2-1.
At first let's calculate the thermal resistance.
The specific thermal resistance of paper -as per Neher&McGrath- will be 700 oC.cm/w[7 K.m/w] nevertheless IEC indicates 6,and will take the paper worming thermal resistance the same [according to IEC could be less].
In this conditions T1=roT/2/pi*G [G from fig.3 for t1=8 mm t=8 mm dc=14.2 mm] t1/dc=~0.6 t/dc=~0.6 G=1.6
Since the shield is not metallic but metallized paper K=1
T1=7/2/pi*1.6=1.7825
Logically no resistance will be from the assembly of cores and the lead sheath since no prisoner air should be here [no gap].However let’s take as if there is a gap.
T'4= U/(1+0.1*(V+Y*m)*De) U=5.2 V=1.4 Y=0.011 and m=50 oC De=60.6 mm.T’4=0.406
Let’s say now there is a serving-a pvc sheath of 3 mm thick-outside the lead then
T3=roJ/2/pi*ln(1+tj/Dlead) roJ=6 Dlead=66 mm
T3=0.0424
In free open air
Tair=1/pi/Dout/h/^1/4
h=Z/(Dout)^d+E =0.21/(0.072)^0.6+3.94=4.958
Dout=(Dlead+2*3)/1000=72/1000=0.072 m
If KA=pi*Dout*h*(T1/3+T3+T’4) =1.17 then by iteration from formula:
(^1/4)n+1= (DTS+Wd/2*T1)/[1+KA*(^1/4)n]
where DTS=TempMax-Ta=65-40=25 oC
Uo=36/sqrt(3)=20784 V.
Wd = ω*C*Uo^2*tan δ (W/m)=2*pi*50*3.239/10^10*20784^2*0.01=0.4396 w/m
C=epsr/18/ln(Di/dc)/10^9=4/18/LN(28.2/14.2)/10^9=3.239E-10 F/m
tan δ=0.01
^1/4=4.25 oC.
Then Tair=1/pi/0.072/4.958/4.25=0.2098
Rdc[20 oC]=1.02/58/150*1=0.00011724 ohm/m
Rdc[65 oC]= Rdc[20 oC]*[1+3.93/10^3*(65-20)]=0.00013798 ohm/m
skin effect+proximity effect=0.004305+0.00478=0.009085
Rac= Rdc[65 oC]*(1+0.009085)=0.00013923 ohm/m
I={[DTS-Wd*(0.5*T1+n*(T3+T’4+Tair)]/Rac/[(T1+n*(T3+T’4+Tair)]}^0.5
I={[25-0.4396*(0.5*1.7825+3*(0.0424+0.406+0.2098))]/0.00013923/[1.7825+3*(0.0424+0.406+0.2098]} ^0.5=213 A

 

[7anoter4]

Corrections:
T'4= U/(1+0.1*(V+Y*Θm)*De) U=5.2 V=1.4 Y=0.011 and Θm=50 oC De=60.6
Tair=1/pi/Dout/h/ΔΘ^1/4
(ΔΘ^1/4)n+1=(DTS+Wd/2*T1)/[1+KA*(ΔΘ^1/4)n]

 

[7anoter4]

If we’ll neglect T’4 –the thermal resistance between core assembly and lead-, we’ll take the cooling effect of the screen [K=0.63] and we calculate the losses factor in the lead sheath according to IEC 60287-1-1 2.3.8 Three-core unarmoured cables with common sheath for Rs>100 micro ohm/m [λ2=0.0814] then we’ll get I=311A [in IEEE 835/1994 for 300 MCM [=150 sqr.mm] copper paper insulated 35 kV lead sheathed cable in air [still air]-no sun I=318 A [since the shield is copper tape here a eddy-current losses has to be added].
I think you may use IEEE-835/1994.However since the cable is very old a factor of 0.8 has to be employed.