Working out the length of a ring finals circuit

[Tetris1985]

r1: 1.15
r2: 1.80
rn: 1.15
r1r2 0.69

To work out the length of a ring in 2.5mm twin would this be correct

0.69/(19.51/1000) = 35m

or would you use the r1r2 before you divide it by 4?

 

[pc1966]

The r1 and r2 values you have are both measured end-end on the RFC, so if using tables of R1+R2 (such as Table I1 in the OSG) then you would add them and divide by the table value:

length = (r1 + r2) / R1plusR2perLength = (1.15 + 1.80) / (19.51/1000)

 

[Tetris1985]

The r1 and r2 values you have are both measured end-end on the RFC, so if using tables of R1+R2 (such as Table I1 in the OSG) then you would add them and divide by the table value:

length = (r1 + r2) / R1plusR2perLength = (1.15 + 1.80) / (19.51/1000)

So it would go something like this?

(1.15 + 1.80) / (19.51/1000)
2.95 / 0.01951 = 151 meters

so this to work out the voltage drop it would then be

Voltage Drop = ( mV/A/m ) x L x Ib / 1000

18 x 151 x 32 / 4 = 21744 / 1000 = 21 volts

does not comply with voltage drop

 

[Rockingit]

To be pedantric, you can't work out the length of a ring, only the circumference,

 

[Pretty Mouth]

When calculating the length of an existing ring, for simplicity and accuracy I prefer to use the resistance of just one of the live conductors for the calculation.

r2 is not always accurately measured, due to different sized CPCs, and parallel paths, so including it in the calculation can lead to inaccuracies. The live conductors shouldn't have parallel paths (assuming equipment is not in circuit), so if the measured r1 and rN are identical, then you can be fairly confident that both live conductors are continuous and without loose connections.

From OSG table I1, the resistance of a 2.5mm2 conductor is 0.00741 ohms/m at 20 deg C.

1.15 / 0.00741 = 155.2 m

Note that I prefer to use ohms rather than milliohms for the resistance/m part of the calculation, I find it simpler that way.

 

[pc1966]

So it would go something like this?

(1.15 + 1.80) / (19.51/1000)
2.95 / 0.01951 = 151 meters

Yes. Though see @Pretty Mouth comment about accuracy on typical measured r2

so this to work out the voltage drop it would then be

Voltage Drop = ( mV/A/m ) x L x Ib / 1000

18 x 151 x 32 / 4 = 21744 / 1000 = 21 volts

does not comply with voltage drop

Not quite, but a quick check in the OSG Table 7.1(ii) shows the max length is 106m for 2.5mm so you are right it will fail.

The drop for a RFC is not end-end as it is for a radial.

Starting with the Zs situation, in the case of the RFC the worst-case R1+R2 seen at a socket is half-way around the ring, closer to either end and you have a shorter cable and less drop. Note that you might not actually have a socket at the half-way point!

But if you did, then you have two legs feeding it, each of L/2 so they have voltage (r1+r2)/2 and as those two are in parallel your R1+R2 would be computed from (r1+r2)/4 and adding your Ze you get Zs

Volt drop follows a similar logic, in effect you have two parallel paths so your L & N drop is from end-end divided by four to account for being half-way, and those in parallel. You seem to be doing this but the way you write out the equation is slightly odd, as you divide the current by 4 which achieves the same, but is not really reflecting the L/4 reasoning.

But the VD for a RFC and similar has the further complication that your load is not usually in one place but spread out over multiple sockets, similar to a light circuit, but for normal computation you can assume the worst-case is 26A at the mid-point from two 13A sockets at full load (which also ties in with OSG table 7.1(i) of assumed loads).

If your loads are all small compared to the OCPD as in the lighting case then you typically assume they are evenly spread out so the total load can be assigned to half-way on the cable for VD (i.e. half the VD compared to an end-of-circuit radial load like a cooker, immersion heater, etc).

For Zs it is always worst-case at end of radial, or RFC mid-point, that you use for checking circuit protection.

 

[Pretty Mouth]

So it would go something like this?

(1.15 + 1.80) / (19.51/1000)
2.95 / 0.01951 = 151 meters

so this to work out the voltage drop it would then be

Voltage Drop = ( mV/A/m ) x L x Ib / 1000

18 x 151 x 32 / 4 = 21744 / 1000 = 21 volts

does not comply with voltage drop

Your voltage drop calculation is about right. A slightly different way to go about it:

The most onerous conditions for voltage drop on the ring, is the full 32A load at the mid-point of the ring. As the resistance of both legs between this mid-point and the CU is identical, half of this current will flow through one leg, half down the other.

So, 16A flows from the CU, through the line conductor of one leg to the mid-point, then through the N of that leg back to the CU. Using V=IR, we can calculate the voltage drop for this current flowing through this resistance.

Resistance of one leg = (r1 /2) + (rN /2) = 1.15 ohms.

At the full 70 deg C operating temperature, this resistance would be: 1.15 x 1.2 = 1.38 ohms

V = IR = 16 x 1.38 = 22.34 V

For comparison, let's use your way of calculating voltage drop, with the 155.2m that I arrived at previously:

18 x 155.2 x 32 / 4 / 1000 = 22.3488 V

Same result, slightly different way of getting there.


Note that the standard ring final in the OSG uses a design current of 26A. I believe it also uses the correction factor from p429 of the regs to arrive at a 106m long circuit.

 

[Tetris1985]

Yes. Though see @Pretty Mouth comment about accuracy on typical measured r2

Not quite, but a quick check in the OSG Table 7.1(ii) shows the max length is 106m for 2.5mm so you are right it will fail.

The drop for a RFC is not end-end as it is for a radial.

Starting with the Zs situation, in the case of the RFC the worst-case R1+R2 seen at a socket is half-way around the ring, closer to either end and you have a shorter cable and less drop. Note that you might not actually have a socket at the half-way point!

But if you did, then you have two legs feeding it, each of L/2 so they have voltage (r1+r2)/2 and as those two are in parallel your R1+R2 would be computed from (r1+r2)/4 and adding your Ze you get Zs

Volt drop follows a similar logic, in effect you have two parallel paths so your L & N drop is from end-end divided by four to account for being half-way, and those in parallel. You seem to be doing this but the way you write out the equation is slightly odd, as you divide the current by 4 which achieves the same, but is not really reflecting the L/4 reasoning.

But the VD for a RFC and similar has the further complication that your load is not usually in one place but spread out over multiple sockets, similar to a light circuit, but for normal computation you can assume the worst-case is 26A at the mid-point from two 13A sockets at full load (which also ties in with OSG table 7.1(i) of assumed loads).

If your loads are all small compared to the OCPD as in the lighting case then you typically assume they are evenly spread out so the total load can be assigned to half-way on the cable for VD (i.e. half the VD compared to an end-of-circuit radial load like a cooker, immersion heater, etc).

For Zs it is always worst-case at end of radial, or RFC mid-point, that you use for checking circuit protection.

I'm getting there with it but a drop of 22 volts on a rfc is a fail of vd?

So if you used a 16amp mcb on a rfc the formula would stay the same? But in the above case the vd would then comply

What would happen if you exceed VD than say less than 10%

 

[pc1966]

I'm getting there with it but a drop of 22 volts on a rfc is a fail of vd?

Yes, the regs say 5% max which is 11.5V

So if you used a 16amp mcb on a rfc the formula would stay the same? But in the above case the vd would then comply

Yes and no.

If it was designed for 16A of typical load then it would meet the VD requirement, and without any special design information you would assume that a 16A MCB means it is designed for a 16A load.

However, if you need more than 16A because it covers a lot of rooms with various significant loads then you will simply trip the MCB. So in that case you have failed to design for the required use.

For something like a set of sockets what constitutes the correct design load is not hard and fast. If you look in the OSG it has typical coverage area for 20A radial, 32A radial, and 32A RFC circuits and for a lot of cases that is your reasonable usage.

However, if you have a non-domestic system it can go far in the other directions:

• For example you might have 6 * 32A sockets in a factory but only one mobile item that uses them so you can design for 32A.
• Or you might find you have 6 * 13A sockets in a commercial kitchen and they all get loaded to pretty close max at the same time so a 32A RFC is not enough and you need two 32A radials or similar.

What would happen if you exceed VD than say less than 10%

It depends, and in many cases not a lot. The ESQCR states that the nominal supply voltage should be 230V -6% +10% which is 216V to 253V and if your combination of supply voltage and circuit VD keeps the appliance voltage within that range then all equipment should work normally.

If the voltage drops very low then some equipment will malfunction and in unusual cases even be damaged by out of range operation.

In reality if you have found an unacceptably large RFC as this appears to be you need to make a judgement call about what to do. The first aspect is if it is safe or not, and that comes down the protection being able to act quickly. In today's "RCD everything" situation that should be simple but many folks like myself like to see a sporting chance that it could disconnect on the OCPD side safely, which comes down to the worst-case Zs and the MCB choice. That depends on your R1+R2 as well as Ze, so it is more than just the VD value (but they are closely related as VD depends on R1 of course).

The second aspect of the judgement is if the circuit is likely to cause trouble, and if it has been used for years without complaint then you have a reasonable answer "no". For example, you might have a very long RFC feeding a handful of sockets in an attic conversion, in which case are you likely to need 26A for any normal activity?

So while it is not compliant, if it is unlikely to lead to problems then I would leave it with a note to that effect. The only way to serve more current is to rewire some or all in 4mm or similar, dropping the MCB to 20A or 16A might make it design-consistent but it does not help the person who needs any more current.

 

[Tetris1985]

Yes, the regs say 5% max which is 11.5V

Yes and no.

If it was designed for 16A of typical load then it would meet the VD requirement, and without any special design information you would assume that a 16A MCB means it is designed for a 16A load.

However, if you need more than 16A because it covers a lot of rooms with various significant loads then you will simply trip the MCB. So in that case you have failed to design for the required use.

For something like a set of sockets what constitutes the correct design load is not hard and fast. If you look in the OSG it has typical coverage area for 20A radial, 32A radial, and 32A RFC circuits and for a lot of cases that is your reasonable usage.

However, if you have a non-domestic system it can go far in the other directions:

• For example you might have 6 * 32A sockets in a factory but only one mobile item that uses them so you can design for 32A.
• Or you might find you have 6 * 13A sockets in a commercial kitchen and they all get loaded to pretty close max at the same time so a 32A RFC is not enough and you need two 32A radials or similar.

It depends, and in many cases not a lot. The ESQCR states that the nominal supply voltage should be 230V -6% +10% which is 216V to 253V and if your combination of supply voltage and circuit VD keeps the appliance voltage within that range then all equipment should work normally.

If the voltage drops very low then some equipment will malfunction and in unusual cases even be damaged by out of range operation.

In reality if you have found an unacceptably large RFC as this appears to be you need to make a judgement call about what to do. The first aspect is if it is safe or not, and that comes down the protection being able to act quickly. In today's "RCD everything" situation that should be simple but many folks like myself like to see a sporting chance that it could disconnect on the OCPD side safely, which comes down to the worst-case Zs and the MCB choice. That depends on your R1+R2 as well as Ze, so it is more than just the VD value (but they are closely related as VD depends on R1 of course).

The second aspect of the judgement is if the circuit is likely to cause trouble, and if it has been used for years without complaint then you have a reasonable answer "no". For example, you might have a very long RFC feeding a handful of sockets in an attic conversion, in which case are you likely to need 26A for any normal activity?

So while it is not compliant, if it is unlikely to lead to problems then I would leave it with a note to that effect. The only way to serve more current is to rewire some or all in 4mm or similar, dropping the MCB to 20A or 16A might make it design-consistent but it does not help the person who needs any more current.

Thanks for the reply its educational to say the least.

I found a ring of 151 meters and one of 135 meters so they are both over the range for VD

The longest one feeds two bedrooms and the shorter one feeds a front room and hallway

Leaving them on 32 amps does not sit well with me.

They meet the Zs requirement.

With a ring on 32 amps you have a design current of 26 amps

If the ring was on a 25amp breaker or 20amp breaker would the design current be 25amps and 20 amps or would it reduced like the 32amps is reduced to 26amps

 

[SparkyChick]

If I were you I would get some realistic loading figures. It's unlikely a circuit feeding two bedrooms will get anywhere close to 26A in this day and age. It's more likely going to be less then 2kW which is 8.3A which conveniently (using PrettyMouths approach) is a voltage drop of 11.5v I believe which is the absolute maximum.

So you need to establish the loads that are used in the rooms, how much and how they overlap to establish an actual predicted maximum. Not an absolute design maximum.

 

[westward10]

What is it you are actually doing?

 

[timhoward]

If the ring was on a 25amp breaker or 20amp breaker

If strictly complying with regs, a ring final circuit can only have a 30 amp or 32 amp protective device.
You are allowed to increase the cable CSA beyond 2.5 sq mm though, to reduce your resistance and hence volt drop.

In your case if Zs is met, you have to ask how much is likely to be drawn by two bedrooms or a front room.

If it's a freezing 14th century mansion with no central heating and you are expecting a fan heater, a hair dryer and an electric blanket in both bedrooms to be on simultaneously then you might have a point, or e.g. student rooms with a microwave and air fryer in each, but chances are it will be an LED bedside light and a phone charger?

I'm not saying ignore volt drop requirements, but you do have to be realistic.

I can't tell if this is a real life example, or a college exam questions, as I first assumed? Do tell us the whole story!

 

[Pretty Mouth]

Thanks for the reply its educational to say the least.

I found a ring of 151 meters and one of 135 meters so they are both over the range for VD

The longest one feeds two bedrooms and the shorter one feeds a front room and hallway

Leaving them on 32 amps does not sit well with me.

They meet the Zs requirement.

With a ring on 32 amps you have a design current of 26 amps

If the ring was on a 25amp breaker or 20amp breaker would the design current be 25amps and 20 amps or would it reduced like the 32amps is reduced to 26amps

The breaker is there for overload and fault protection. The only relation the breaker has with design current, is that the design current should be no greater than its rating. Lowering the breaker rating doesn't lower the design current however - you design the circuit and protective device(s) based on the design current, not the other way round.

In this case, you're dealing with existing circuits. @pc1966 and @SparkyChick have covered this nicely in the last few posts, the only things I'll add are:

It is unlikely that the circuits will be heavily loaded for any length of time, if ever.
Even if they are, the supply voltage, in my experience, usually measures 240V or more, which is a large margin of error over the lower acceptable limit of 216V.

Personally, I wouldn't be concerned with these circuits.

 

[Tetris1985]

What is it you are actually doing?

Learning I have done a level 2 and 3 and working on my NVQ.

I have started working with my dad who is QS and old spark and he brought a new property and has let me test it to get some real world experience before he tests its. ( its empty I cant hurt anyone )

why cant you comply with the regs if a ring is not on 30 or 32 amps seems strange

Its a fairly new building

so at 32 amps the volt drop would be

18x26x151/4 = 17667 / 1000
= 17 volts which is about 7.5% volts dropped in the bedrooms

18x26x133/4 = 15561/1000
=15.5 volts dropped which is about 6.75% volts dropped in the front room

im just asking realistically in the real world is that a problem where you would rewire in 4mm