hertzal123 said:

.Have worked out R1+R2 for the cores as 50x3.66x1.2/1000 =0.0216 ohms

I think I get that at 0.2196 ohms?

The figures on the left hand side look correct for copper DC resistance 1.83 mΩ/m per core at 20 deg C, and 1.2 factor to correct to 70 deg C.

hertzal123 said:

Can the armour be included in the calcs to lower ZS and how would it be done?

There isn't a published formula, at least that I'm aware of, to use SWA in parallel with an internal copper conductor (there is with an external conductor). However, the following formula for an external cpc, from NA.4.5 of PD IEC/TR 50480, is likely to give you an over-estimate of the impedance for the internal copper cpc in parallel with the armour (because the loop reactance for the portion of current returning down the other copper internal conductor ought to be zero, but the formula below accounts for external conductor separated from internal copper conductor by armour):

Resistive component of (Z₁+Z₂),  r = R₁ + R₂ //1.1R_2a

where

R₁ and R₂ are the DC copper conductor resistances impedance we have used already ({1.83 × 1.2} mΩ/m)
R_2a is the DC resistance of the armour (I think {4 × 1.225} mΩ/m for 3 c 10 mm² SWA corrected to 70 deg C)

Reactive component of (Z₁+Z₂), x = 0.4 (you'd have to take reactance into account even for small csa armour, as for copper csa 2.5 sq mm and above, armour csa > 16 sq mm - in the case of 10 sq mm 3 c SWA, the armour csa is 39 sq mm).

and overall impedance is then (Z₁+Z₂) = √(r²+x²)

If you work this through, for 3 c 10 sq mm you might get a loop impedance per metre around 85 % of that assuming copper conductors only.

This broadly supports what Mike says about the fact you won't get huge reductions taking the armour into account.

hertzal123 said:

.Have worked out R1+R2 for the cores as 50x3.66x1.2/1000 =0.0216 ohms

I think I get that at 0.2196 ohms?

The figures on the left hand side look correct for copper DC resistance 1.83 mΩ/m per core at 20 deg C, and 1.2 factor to correct to 70 deg C.

hertzal123 said:

Can the armour be included in the calcs to lower ZS and how would it be done?

There isn't a published formula, at least that I'm aware of, to use SWA in parallel with an internal copper conductor (there is with an external conductor). However, the following formula for an external cpc, from NA.4.5 of PD IEC/TR 50480, is likely to give you an over-estimate of the impedance for the internal copper cpc in parallel with the armour (because the loop reactance for the portion of current returning down the other copper internal conductor ought to be zero, but the formula below accounts for external conductor separated from internal copper conductor by armour):

Resistive component of (Z₁+Z₂),  r = R₁ + R₂ //1.1R_2a

where

R₁ and R₂ are the DC copper conductor resistances impedance we have used already ({1.83 × 1.2} mΩ/m)
R_2a is the DC resistance of the armour (I think {4 × 1.225} mΩ/m for 3 c 10 mm² SWA corrected to 70 deg C)

Reactive component of (Z₁+Z₂), x = 0.4 (you'd have to take reactance into account even for small csa armour, as for copper csa 2.5 sq mm and above, armour csa > 16 sq mm - in the case of 10 sq mm 3 c SWA, the armour csa is 39 sq mm).

and overall impedance is then (Z₁+Z₂) = √(r²+x²)

If you work this through, for 3 c 10 sq mm you might get a loop impedance per metre around 85 % of that assuming copper conductors only.

This broadly supports what Mike says about the fact you won't get huge reductions taking the armour into account.