Old cable data sheets.

[Gnasher]

Hello

I understand that this may be an impossible one to answer, but I have been surprised on many occasions with what info is out there if you only ask. So here goes.

I am trying to locate historical cable data sheets for paper lead cables that were probably installed in the 1960s

I think that the manufacture was probably BICC who are no longer in existence. I have done a google search and found a company in Cairo who carry the BICC logo. I have emailed them but as yet had no reply.

The cables are imperial 2 core 19/0.64 and 19/1.53 lead sheathed,

The specific info I am looking for is the ohmic values of the conductors and lead sheath. Similar to the Table 11 in the onsite guide R1 + R2 values.

Thanks

Steve

 

[7029 dave]

Would it not be best to get the cable replaced.

 

[Vortigern]

You could work out the resistance

 

[Gnasher]

Would it not be best to get the cable replaced.

Unfortunately that's not an option at the moment. These cables need to be incorporated into an existing system, with designs demonstrating compliance.

All cables have had insulation resistance tests carried out on them and have perfect readings.
The cost would prohibit replacement at the moment.

 

[Gnasher]

You could work out the resistance

Unfortunately I am unable to find resistance values for old imperial conductors. 35mm sq and 95mmsq equivalents.

Additionally I an unable to locate the resistance values of the lead sheath .

One option is to measure the longest length cable and do some maths. Problem is it is not possible to isolate the circuits without a lot of planning and disruption. This will take a few weeks to arrange.

I might be missing something hear if I am please let me know.

 

[Gnasher]

Unfortunately I am unable to find resistance values for old imperial conductors. 35mm sq and 95mmsq equivalents.

Additionally I an unable to locate the resistance values of the lead sheath .

One option is to measure the longest length cable and do some maths. Problem is it is not possible to isolate the circuits without a lot of planning and disruption. This will take a few weeks to arrange.

I might be missing something hear if I am please let me know.

just re-read my reply might sound a bit abrupt. Not intentional.

 

[Lucien Nunes]

I have the original BICC data books (called 'Cables and Tables') which will probably list your specific cables. It might be a few days before I am near the books though, as they are in the museum library.

The conductor resistance can be calculated easily to reasonable accuracy (there will be a minor influence from the layup, drawing tolerance etc.) The sheath will be more difficult as we don't know its CSA but TBH it might not be in the tables anyway. Please can you check the sizes though:

19/.064 (not 19/0.64) is a standard imperial size equal to 39mm². Approx 0.49mΩ per metre of conductor.

19/.153 (not 19/1.53) would be 225mm² so I don't think that's right because you mention 95mm².

IIRC, the nearest 19-strand imperial size for 95mm² would be 19/.103 which is 102mm². Could that be what was meant? If so about 0.18mΩ per metre of conductor.

Would it not be best to get the cable replaced.

Not sure why, if it's sound. Much of the UK still runs on PILC much older than this.

 

[pc1966]

Based on this quick web-search the resistance of lead is about 13 times higher than copper:

Resistivities for common metals

<b>The following table shows the resistivities of several materials. The values are correct at 20 degrees Celsius.</b>

cleanroom.byu.edu

So if you can measure the sheath outer diameter and thickness you ought to be able to compute the equivalent CSA, and from that get an idea of the R2 value to use. However, if Lucian can get you his book values soon enough that would be the best answer.

 

[Gnasher]

I have the original BICC data books (called 'Cables and Tables') which will probably list your specific cables. It might be a few days before I am near the books though, as they are in the museum library.

The conductor resistance can be calculated easily to reasonable accuracy (there will be a minor influence from the layup, drawing tolerance etc.) The sheath will be more difficult as we don't know its CSA but TBH it might not be in the tables anyway. Please can you check the sizes though:

19/.064 (not 19/0.64) is a standard imperial size equal to 39mm². Approx 0.49mΩ per metre of conductor.

19/.153 (not 19/1.53) would be 225mm² so I don't think that's right because you mention 95mm².

IIRC, the nearest 19-strand imperial size for 95mm² would be 19/.103 which is 102mm². Could that be what was meant? If so about 0.18mΩ per metre of conductor.



Not sure why, if it's sound. Much of the UK still runs on PILC much older than this.

Hello Lucien

Thanks for your reply. I never cease to be amazed what info is available if you ask.
More than happy to wait a few days. It will probably take that long to dust the cobwebs off.

You are correct with your observations on the cable sizes. Decimal point in the wrong place.

As you say the sheath may be the difficult one to find out.

Thanks again.

Steve.

 

[Lucien Nunes]

If the cables are currently terminated it might be impossible to establish the sheath thickness by measurement. If ithe resistance or CSA of the lead is not in the book, I guess it will be a case of isolating a good length (hopefully it's on insulated cleats or served) and putting a Ducter on it.

We could probably make a good estimate by weight because the weight per unit length will certainly be tabulated, and copper and lead make up most of the cable weight. Subtracting the known weight per unit length of copper will probably give an answer accurate to 10% for the lead. That could perhaps be tweaked a little by calculating the CSA of paper by subtracting lead and copper from total, and then re-calculating the CSA of lead with the paper weight included. The error will then be reduced by the ratio of (lead density-1) / paper density and probably as close as you would get by measuring a chunk of PILC with a vernier.

 

[7029 dave]

I have the original BICC data books (called 'Cables and Tables') which will probably list your specific cables. It might be a few days before I am near the books though, as they are in the museum library.

The conductor resistance can be calculated easily to reasonable accuracy (there will be a minor influence from the layup, drawing tolerance etc.) The sheath will be more difficult as we don't know its CSA but TBH it might not be in the tables anyway. Please can you check the sizes though:

19/.064 (not 19/0.64) is a standard imperial size equal to 39mm². Approx 0.49mΩ per metre of conductor.

19/.153 (not 19/1.53) would be 225mm² so I don't think that's right because you mention 95mm².

IIRC, the nearest 19-strand imperial size for 95mm² would be 19/.103 which is 102mm². Could that be what was meant? If so about 0.18mΩ per metre of conductor.



Not sure why, if it's sound. Much of the UK still runs on PILC much older than this.

Yeah, did not realize it was MU type cable.

 

[Gnasher]

Based on this quick web-search the resistance of lead is about 13 times higher than copper:

Resistivities for common metals

<b>The following table shows the resistivities of several materials. The values are correct at 20 degrees Celsius.</b>

cleanroom.byu.edu

So if you can measure the sheath outer diameter and thickness you ought to be able to compute the equivalent CSA, and from that get an idea of the R2 value to use. However, if Lucian can get you his book values soon enough that would be the best answer.

Based on this quick web-search the resistance of lead is about 13 times higher than copper:

Resistivities for common metals

<b>The following table shows the resistivities of several materials. The values are correct at 20 degrees Celsius.</b>

cleanroom.byu.edu

So if you can measure the sheath outer diameter and thickness you ought to be able to compute the equivalent CSA, and from that get an idea of the R2 value to use. However, if Lucian can get you his book values soon enough that would be the best answer.

Thanks pc1966

Good suggestion. The outer sheath should be easy enough, the thickness will be a bit more difficult though as the cables are in service.
I can ask if there is an old section about to measure. Fingers crossed.

Steve

 

[Lister1987]

John Ward has CCC sheets for various PILC and MICC if memory serves, night be worth dropping him a bell

 

[Gnasher]

John Ward has CCC sheets for various PILC and MICC if memory serves, night be worth dropping him a bell

Thanks Lister.

How do I contact John.

Steve.

 

[richy3333]

John Ward Flameport Electrical

Flameport Electrical, an electrical company owned by John Ward, Electrician in Dorset, UK. Electrical content including the John Ward Youtube channel

www.flameport.com

 

[Gnasher]

John Ward Flameport Electrical

Flameport Electrical, an electrical company owned by John Ward, Electrician in Dorset, UK. Electrical content including the John Ward Youtube channel

www.flameport.com

Thanks.

 

[Lucien Nunes]

As I had to swing by the stores yesterday, I picked up some relevant info. There wasn't time to search for the books nearest to the date of your cable, as some of it is still boxed up from the move, so what I ended up with is all older than your assumed 1960s cable.

The first thing I noticed is the lack of 19/.103 conductors in the tables. There's 19/.101 hard-drawn copper but not annealed, so not relevant IMO, the largest 19-strand annealed conductor being 19/.083 at 0.1 sq. ins. So I think it is worth just double-checking your reference for the larger cable; i.e. where did the 95mm² and 19/something figures come from? Could it be a 0.15 sq. ins. cable of different stranding such as 37/.072?

Also, do you have the voltage rating or BS for the cables?

Looking only at the smaller cable for now:

BICC 'Cables & Tables' 1953 ed.
Quoting BS7 for 19/.064 annealed copper conductor at 20°C:
Standard resistance per 1000 yds. 0.4074Ω
Maximum for tinned conductors per 1000 yds. 0.4195Ω
Weight per 1000 yds: 720.3 lbs
Table 8, quoting IEE 12th ed, for one or two 2-core PILC 0.15 sq. ins. cable at 32°C ambient:
CCC: 128A DC or 1-phase AC
Voltage drop: one volt per 24 feet DC or 1-phase AC
No info on the lead sheath.

The first reference I have found to the lead thickness for a 660V cable is actually from before the BICC merger:
BIC handbook 4th. ed. of 1927.
For 660V twin plain class A PILC with either circular or segmental 0.06 sq. ins. conductor:
Lead thickness 0.080 ins
Diameter overall: 0.85 ins.
Weight per 1000 yds. 43.75 cwt.

For comparison, from a different manufacturer nearer in date to your cable and probably to the correct BS:
Enfield Standard Power Cable Handbook ES101 of August 1962, paper insulated cables to to BS480 : 1954
Table 2.2, twin-core belted 1100V cable with segmental 19/.064 conductors, type PLY (bare sheath)
Minimum thickness of sheath 0.05 ins.
Nominal diameter overall: 0.78 ins.
Approx. weight per 1000 yds: 30cwt.

Next tea break I will bash the numbers into the calculator and see what we get for the lead resistances.

 

[Lucien Nunes]

1927 handbook

 

[Lucien Nunes]

Let's try to validate the lead CSA for the older BIC 19/.064 by calculating it both from the stated dimensions and from the linear weights.

Lead O/D = 0.85 ins. = 21.59mm
Lead I/D = 0.85 - (2 * 0.08) = 0.69 ins. = 17.53mm
Lead CSA from dims = ((21.59 / 2)² - (17.53 / 2)²) * Π = 125mm²

Weight of twin cable = 43.75cwt / 1000 yds. = 2.431kg/m
Weight of 19/.064 copper = 720.3 lbs / 1000 yds. = 0.3573kg/m
Weight of lead (ignoring paper) = 2.431 - (2 * 0.3573) = 1.7164kg/m
Lead density (ignoring alloying elements) = 11.34g/cm³
Lead volume per metre = 1716.4 / 11.34 = 151.4cm³
Lead CSA from weight = 151mm²

Lead resistivity 2.2E-7 Ωm @ 20°C ignoring alloying.
Resistance of 125mm² lead = 2.2E-7 / 125E-6 = 1.76mΩ/m
Resistance of 151mm² lead = 2.2E-7 / 151E-6 = 1.46mΩ/m
Resistance of 19/.064 TC = 0.4195Ω / 1000 yds. = 0.459mΩ/m

I'll do the later Enfield cable in the morning unless someone wants to save me the effort.

 

[freddo]

While we are mentioning BICC here is a little promotional copper ashtray I have here.

 

[Lucien Nunes]

Nice! Sitting on a DAC90 unless I'm mistaken.

 

[freddo]

Nice! Sitting on a DAC90 unless I'm mistaken.

Well spotted!

 

[pc1966]

Resistance of 125mm² lead = 2.2E-7 / 125E-6 = 1.76mΩ/m
Resistance of 151mm² lead = 2.2E-7 / 151E-6 = 1.46mΩ/m
Resistance of 19/.064 TC = 0.4195Ω / 1000 yds. = 0.459mΩ/m

Comparing that to 2-core 35mm SWA which looks closes today I see R1 as 0.524 mOhm/m and the armour R2 as 2.6mOhm/m, so the lead sheathed cable is better!

For a given definition based on Zs (or more specifically R2/R1 as about 5 for SWA and about 3.8 for old stuff) and using the Prysmian data sheet for BS5467 cable

 

[Lucien Nunes]

The Enfield cable has a thinner sheath so might be closer to SWA. Obviously some PILC has wire armour in parallel, some has tape which is less beneficial as a CPC.

Hopefully the OP will be able to confirm the larger cable data and if I go to the warehouse tomorrow I will have a look for 1960s data books.

In the meantime, while we're doing electrical-industry-promotional-items-on-radios here's a pocket electrical slide-rule from Hackbridge and Hewittic (the transformer and rectifier people)

 

[freddo]

That looks in nice clean condition! Sat on a Pye MM?

 

[Lucien Nunes]

Yes, one of the MMs that won't be working any time soon due to disintegrated Mazak castings in the tuning gang.

 

[freddo]

A friend had one, the mazak was also in a terrible state and crumbling, the moving parts seized.

 

[Gnasher]

Let's try to validate the lead CSA for the older BIC 19/.064 by calculating it both from the stated dimensions and from the linear weights.

Lead O/D = 0.85 ins. = 21.59mm
Lead I/D = 0.85 - (2 * 0.08) = 0.69 ins. = 17.53mm
Lead CSA from dims = ((21.59 / 2)² - (17.53 / 2)²) * Π = 125mm²

Weight of twin cable = 43.75cwt / 1000 yds. = 2.431kg/m
Weight of 19/.064 copper = 720.3 lbs / 1000 yds. = 0.3573kg/m
Weight of lead (ignoring paper) = 2.431 - (2 * 0.3573) = 1.7164kg/m
Lead density (ignoring alloying elements) = 11.34g/cm³
Lead volume per metre = 1716.4 / 11.34 = 151.4cm³
Lead CSA from weight = 151mm²

Lead resistivity 2.2E-7 Ωm @ 20°C ignoring alloying.
Resistance of 125mm² lead = 2.2E-7 / 125E-6 = 1.76mΩ/m
Resistance of 151mm² lead = 2.2E-7 / 151E-6 = 1.46mΩ/m
Resistance of 19/.064 TC = 0.4195Ω / 1000 yds. = 0.459mΩ/m

I'll do the later Enfield cable in the morning unless someone wants to save me the effort.

Lucian

This brilliant. I cant believe you have gone to this amount of effort to help.

Perhaps the best way would take an average of the 125mm² and the 151mm². That's probably a reasonable assumption.

You did mention the lack of 19/.103, this should be 19/153.

if you could assist with the larger cable that would be much appreciated.

Steve

 

[Lucien Nunes]

Feel free to mash the winner button if you like the service!

I didn't get round to the later Enfield 19/.064 cable yet, that is more likely to be of similar construction to your 1960s cable as they were probably both made to the later British Standard. The lead is significantly thinner so we should calculate that before making a final call on the sheath resistance.

Re the larger cable, I can't make sense of the '19/153' or '19/1.53' as given earlier.

• 19/1.53 is a valid metric cable stranding for 35mm²
• Trying to read it as an imperial size '19/.153' would make it 225mm²
• My reference to 19/.103 was a top-of head attempt to match it to your comment that it was 95mm² equivalent. 19/.103 did exist as a stranding and is just over 95mm², but does not seem to have been used as a core of PILC.
• 95mm² equivalent imperial size as a PILC core invariably seems to have been 37/.072

So we really need to clarify that number and gather any other possible information before we can make headway with the larger cable.