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Default Resistance per Meter for copper cable (R1+R2)

Hi

Can someone help out with this Electrical problem.
(I'm doing a CG2330 course)

Table 9A of the IEE On site guide only goes upto 50mm csa cable for the
values (R1+R2)

Eg
50mm Phase+CPC = 0.774 mOhms/m

What would you do if the conductor was say 70mm ???

Is there a formula I could use ?


Thanks

Leigh

  #2   Report Post  
dennis@home
 
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Default Resistance per Meter for copper cable (R1+R2)


wrote in message
oups.com...
Hi

Can someone help out with this Electrical problem.
(I'm doing a CG2330 course)

Table 9A of the IEE On site guide only goes upto 50mm csa cable for the
values (R1+R2)

Eg
50mm Phase+CPC = 0.774 mOhms/m

What would you do if the conductor was say 70mm ???

Is there a formula I could use ?



50/70 x 0.774 mOhms/m?


  #3   Report Post  
Slurp
 
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Default Resistance per Meter for copper cable (R1+R2)


"dennis@home" wrote in message
o.uk...

wrote in message
oups.com...
Hi

Can someone help out with this Electrical problem.
(I'm doing a CG2330 course)

Table 9A of the IEE On site guide only goes upto 50mm csa cable for the
values (R1+R2)

Eg
50mm Phase+CPC = 0.774 mOhms/m

What would you do if the conductor was say 70mm ???

Is there a formula I could use ?



50/70 x 0.774 mOhms/m?


Nope, nothing like.
Its cross sectional area you need to use (pi * r^2)

So for 50mm = approx 1962 sq mm, 70 mm = approx 3846 sq mm.

i.e. it will be arount HALF the resistrance or about 385milliOhms/metre.

Slurp


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dennis@home
 
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Default Resistance per Meter for copper cable (R1+R2)


"Slurp" wrote in message
...

"dennis@home" wrote in message
o.uk...

wrote in message
oups.com...
Hi

Can someone help out with this Electrical problem.
(I'm doing a CG2330 course)

Table 9A of the IEE On site guide only goes upto 50mm csa cable for the
values (R1+R2)

Eg
50mm Phase+CPC = 0.774 mOhms/m

What would you do if the conductor was say 70mm ???

Is there a formula I could use ?



50/70 x 0.774 mOhms/m?


Nope, nothing like.
Its cross sectional area you need to use (pi * r^2)

So for 50mm = approx 1962 sq mm, 70 mm = approx 3846 sq mm.

i.e. it will be arount HALF the resistrance or about 385milliOhms/metre.


Cables are usually quoted in mm2.
It correct for 50mm2 and 70mm2 is it not.


  #5   Report Post  
Andy Wade
 
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Default Resistance per Meter for copper cable (R1+R2)

Slurp wrote:

"dennis@home" wrote in message
o.uk...

50/70 x 0.774 mOhms/m?


Nope, nothing like.
Its cross sectional area you need to use (pi * r^2)


No, the figures mentioned were clearly conductor CSAs in mm^2, so the
answer was right, at least for DC.

An easier way would to use the tabulated values for 35mm^2 and halve
them, giving 0.524 mohm/m for R1+R2 for 70+70 mm^2.

However the OP should be aware that for these large sizes used on AC the
impedance will be higher then the DC resistance because (a) skin
effect increases the resistance and (b) the inductive reactance is not
necessarily negligible. The OSG is only relevant to installation work
up to 100 A per phase, in which you would not usually need to use 70
mm^2 conductors!

--
Andy


  #6   Report Post  
Slurp
 
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Default Resistance per Meter for copper cable (R1+R2)


"dennis@home" wrote in message
.uk...

"Slurp" wrote in message
...

"dennis@home" wrote in message
o.uk...

wrote in message
oups.com...
Hi

Can someone help out with this Electrical problem.
(I'm doing a CG2330 course)

Table 9A of the IEE On site guide only goes upto 50mm csa cable for the
values (R1+R2)

Eg
50mm Phase+CPC = 0.774 mOhms/m

What would you do if the conductor was say 70mm ???

Is there a formula I could use ?


50/70 x 0.774 mOhms/m?


Nope, nothing like.
Its cross sectional area you need to use (pi * r^2)

So for 50mm = approx 1962 sq mm, 70 mm = approx 3846 sq mm.

i.e. it will be arount HALF the resistrance or about 385milliOhms/metre.


Cables are usually quoted in mm2.
It correct for 50mm2 and 70mm2 is it not.


Yes you are right. I was going by the OP's spec of 50mm/70mm, not thinking
it was mm2!

Have a team point and gold star.

Slurp


  #7   Report Post  
Mr Fizzion
 
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Default Resistance per Meter for copper cable (R1+R2)

On Tue, 25 Oct 2005 17:16:15 GMT, "dennis@home"
wrote:


"Slurp" wrote in message
...

"dennis@home" wrote in message
o.uk...

wrote in message
oups.com...
Hi

Can someone help out with this Electrical problem.
(I'm doing a CG2330 course)

Table 9A of the IEE On site guide only goes upto 50mm csa cable for the
values (R1+R2)

Eg
50mm Phase+CPC = 0.774 mOhms/m

What would you do if the conductor was say 70mm ???

Is there a formula I could use ?


50/70 x 0.774 mOhms/m?


Nope, nothing like.
Its cross sectional area you need to use (pi * r^2)

So for 50mm = approx 1962 sq mm, 70 mm = approx 3846 sq mm.

i.e. it will be arount HALF the resistrance or about 385milliOhms/metre.


Cables are usually quoted in mm2.
It correct for 50mm2 and 70mm2 is it not.


The OP stated 50mm csa - csa = cross sectional area. For a direct
current, resistance is inversely proportional to the cross sectional
area of a material so doubling the area halves the resistance.

Mr F.

  #8   Report Post  
 
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Default Resistance per Meter for copper cable (R1+R2)

Hmm, looks like I've opened up a can of worms here !

Here's the full example question and the the bits I've got to.

Consumer unit, 190V A.C., with BS3036 fuse, Ze=0.49Ohms
45m PVC Armoured cable, clipped directly to wall with 5 other circuits
50 Degree C ambient temp, 9Kw load
Cable runs through 100mm Insulation

IB = 9*103/190 = 47.36Amps

In = 60A (From Page 195 Fig3.2A Next biggest to 47A)

Correction Factors

1). Cf - A BS3036 device 0.725
2). Cg - From Table 4A1, (Page 210) Method 1 used
From Table 4B1, 1+5=6 Circuits, Bunched = 0.57
3). Ca - From Table 4C2 50°C, Armoured PVC = 0.87
4). Ci - From Table 52A, page 97 100mm = 0.81

Iz = 60 = 206Amps
(0.725*0.57*0.87*0.81)

From Table 4D4A, Armoured PVC, Method 1, Single Phase, 222A nearest


70mm2 csa conductor

Then the problem, table 9A O/S/G only goes upto 50mm

Table 9A (Page 158) of the O/S/G gives mO per meter values to
calculate R1 + R2

R1+R2 = mO/m * Length (m) * Temperature multiplier (Table 9C)
1000

So I'm a bit confused as to who's right.

Leigh

  #9   Report Post  
Andy Wade
 
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Default Resistance per Meter for copper cable (R1+R2)

wrote:

Hmm, looks like I've opened up a can of worms here !

Here's the full example question and the the bits I've got to.

Consumer unit, 190V A.C., with BS3036 fuse, Ze=0.49Ohms
45m PVC Armoured cable, clipped directly to wall with 5 other circuits
50 Degree C ambient temp, 9Kw load
Cable runs through 100mm Insulation

IB = 9*103/190 = 47.36Amps

In = 60A (From Page 195 Fig3.2A Next biggest to 47A)


OK so far, assuming single-phase, which is not explicitly stated
(although the words "consumer unit" do tend to suggest it's 1-ph).

Correction Factors

1). Cf - A BS3036 device 0.725


Only applies if overload protection is required. Also if the six
circuits of the group are not liable to simultaneous overload you can
use Appendix 4, 6.2.2, equations (7) & (8) which will almost certainly
help towards a smaller cable.

2). Cg - From Table 4A1, (Page 210) Method 1 used
From Table 4B1, 1+5=6 Circuits, Bunched = 0.57


Six armoured cables clipped to a wall will not be bunched, they'll be in
a single layer and spaced by more than one diameter. (Otherwise there
won't be room to get the cleats in!) This gives Cg = 0.9. (If spaced
by more than 2 diameters you can used Cg = 1.0 - no derating.)

3). Ca - From Table 4C2 50°C, Armoured PVC = 0.87
4). Ci - From Table 52A, page 97 100mm = 0.81


Do all six cables go through the insulation, or just this one? The
question isn't very clear. If only this cable, then Cg and Ci won't
both apply at the same time. Then you have to consider the installation
conditions for each section of the circuit in turn and take the highest
It rating obtained.

Iz = 60 = 206Amps
(0.725*0.57*0.87*0.81)


This is getting silly - no-one would install like that, on grounds of
cost if nothing else. You'd adjust the method of installation so that
not all factors applied simultaneously, and not bunch the cables as I
said above.

R1+R2 = mO/m * Length (m) * Temperature multiplier (Table 9C)
1000


Since it's SWA Table 9C won't help with R2 (assuming armour as CPC). Or
if a third core is used as CPC then R2 will equal R1.

So I'm a bit confused as to who's right.


The question seems to leave a lot of necessary detail out. Also what's
the ultimate aim of the problem - to verify that Zs is low enough, or to
verify compliance on all grounds?

--
Andy
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Default Resistance per Meter for copper cable (R1+R2)

Hi Andy

The values were just picked out of the air by the lecturer, it was so
that the class could work through correction factors, finding out Zs,
fault currents and selecting cable sizes.
I think he left out all the details so that our poor little brains
don't overload :-), but the values he used cause more problem in the
end.

Thanks for information, helps move class room knowledge to the real
world (Even if the above example would never happen).

Leigh



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