Should I use 4mm or 6mm cable? Or which size MCB?

[HappyHippyDad]

Hello all,

I have a 2 way CU in my shed with a 16A MCB for the radial (3 double sockets) and 6A MCB for the lights (four of them, max usage 330w). I have 20 metres of SWA 4mm[SUP]2 [/SUP]which I will take to the house from the shed and then use 4mm[SUP]2 [/SUP]in trunking to the CU.

I then presumed I would have to use a 32A MCB (in the house CU) as it would have to be greater or equal to the total of the 2 way CU in the shed, but...

The OSG states that 4mm[SUP]2[/SUP] (in trunking) can take up to 30A which is absolutely fine for my shed but the MCB obviously has to trip at less than the current carrying capacity of the cable so it should be between 22A (max of the 2 way consumer unit in shed) and 30A.

It seems crazy to use 6mm[SUP]2[/SUP] when I dont need that many amps and already have 20metres of 4mm[SUP]2[/SUP] SWA.

Am I missing something very simple? What MCB should I use in main CU?

Thanks..

 

[telectrix]

25A . simple as that.

 

[Sintra]

Use a 25A MCB

 

[Paul.M]

Just use a 25A mcb.

- - - Updated - - -

E3 of us posted the same thing at the same time lol.

 

[spark 68]

I then presumed I would have to use a 32A MCB (in the house CU) as it would have to be greater or equal to the total of the 2 way CU in the shed, but...

Why ?, it should be to your design current Ib also taking diversity into account.

Obviously it is better to use a 32A MCB at the supply end if possible, as this is twice the In of the 16A in the shed giving partial discrimination.
But I see no reason not to use a 20A at the supply end depending on your anticipated load at the shed (power tools etc.), but you know your requirements rather better than us on a forum

Voltage drop needs taking into account too, depending on your Ib.

 

[JUD]

Hello all,

I have a 2 way CU in my shed with a 16A MCB for the radial (3 double sockets) and 6A MCB for the lights (four of them, max usage 330w). I have 20 metres of SWA 4mm[SUP]2 [/SUP]which I will take to the house from the shed and then use 4mm[SUP]2 [/SUP]in trunking to the CU.

I then presumed I would have to use a 32A MCB (in the house CU) as it would have to be greater or equal to the total of the 2 way CU in the shed, but...

The OSG states that 4mm[SUP]2[/SUP] (in trunking) can take up to 30A which is absolutely fine for my shed but the MCB obviously has to trip at less than the current carrying capacity of the cable so it should be between 22A (max of the 2 way consumer unit in shed) and 30A.

It seems crazy to use 6mm[SUP]2[/SUP] when I dont need that many amps and already have 20metres of 4mm[SUP]2[/SUP] SWA.

Am I missing something very simple? What MCB should I use in main CU?

Thanks..

Base your calculations on design current NOT the size of your MCBs i.e. total current = 16A for the sockets and 1.43A for the lights (330 ÷ 230) for a total of 17.43A so would be fine on a 20A breaker.

Voltage drop on 20m of 4mm² is 3.8V at 17.43A. How far is it to the CU after you enter the house?

 

[HappyHippyDad]

Thanks all,

Obviously I feel a bit silly now as I have only seen 6, 16 and 32A MCB and didn't realise there are sizes in between. We have done very little work on MCB's so far on the course, but thats still no excuse.

I'd be really interested to know more about voltage drop and how that affects your decision on cable/MCB's? If you have voltage drop does that actually mean there is more current running through the cable?

I would think I'll be using another 8m's 4mm2 cable in the house.

 

[telectrix]

volt drop..... any cable has a finite resistance. therefore , according to ohms law. V=IR, (V being the volt drop over the length of cable), the volt drop is dependant on both the resistance ( an therefore the length) and the current. tables in BS7671 and OSG give us the VD in mV/A/m for different cable CSA, so you can calculate by mA/V/m x.I x L.

 

[Mr.Ian]

Voltage drop is due to the resistance in the cable over a length, smaller cable higher resistance hence greater drop, it's all in the regs.

 

[JUD]

So 28m for your distribution circuit gives you a voltage drop of 5.37V which will be included in your total voltage drop for the final circuits.

The maximum voltage drop for your lighting circuit is 6.9V (3% of the supply voltage). This leaves you with around 1.5V to play with. Do you know what the lengths of the final circuits in the shed are going to be?

 

[JUD]

As others have said V = I x R (Voltage Drop = Current x Resistance)

Resistance decreases as csa increases and increases as csa decreases.

Resistance increases as length increases.

Therefore the more resistance there is in a circuit the higher the voltage drop will be which is why you may need larger cables for long runs.

 

[HappyHippyDad]

There is half a meter (1mm) to a 2 gang light switch, then 2m's (from switch) to 2 lights and 2 metres (from switch) to the other light. Also 2m's from shed CU to PIR light on shed front.

As for sockets, around 5m's of 2.5mm cable inside shed.

Thanks Jud for using figures and a real life example to understand voltage drop, I'm finding it easier to understand.

 

[JUD]

There is half a meter (1mm) to a 2 gang light switch, then 2m's (from switch) to 2 lights and 2 metres (from switch) to the other light. Also 2m's from shed CU to PIR light on shed front.

As for sockets, around 5m's of 2.5mm cable inside shed.

Thanks Jud for using figures and a real life example to understand voltage drop, I'm finding it easier to understand.

So to work out the voltage drop for the lights...

mV/A/m for 1mm² is 44.

Let's say we work it out on 100W per light fitting. 2m to 2 lights (200W or 0.87A) would be (44 x 2 x 0.87) ÷ 1000 = 0.08V (hardly worth worrying about). Add this to the 5.37V on the distribution circuit and you get 5.45V which is fine as it's less than the allowed 6.9V.

The maximum volt drop for sockets is 11.5V(5% of the supply voltage) so...

mV/A/m for 2.5mm² is 18.

(18 x 5 x 16) ÷ 1000 = 1.44V. Added to your 5.37V gives you 6.81V so well under the maximum.

 

[HappyHippyDad]

I'm almost there, I understand the maths but the equation is confusing. Telectrix has said that mV/A/m is how VD is expressed for a given CSA in the OSG, but you cant express VD just with CSA you need length and amps. V = I x R so perhaps this mV/A/m is resistance as VD = I x mV/A/m x L which looks very similar?

What is the figure given for mV/A/m (i.e 18 for 2.5mm[SUP]2[/SUP]) expressed in. 18 what? ohms? Or is it just an arbitory figure for a given CSA which is then used in the equation mV/A/m x I x L to find out VD (given length and ampage)?

I reckon this might be my last question!

 

[JUD]

Yes the mV/A/m is a simplified way to work out voltage drop and is expressed in mΩ. It's worked out by using the resistance of 1m of a given size of cable carrying 1amp at 70°C.

So the formula for working out the mV/A/m values is ((R1+Rn) x 1.2) x I. The 1.2 is the correction factor for the operating temperature of the cable. This is because the temperature of a conductor when testing resistance is usually given as 20°C. Resistance increases with temperature by 0.4% per °C. Therefore with the 50°C difference between 20°C and 70°C the resistance increases by 20% (50 x 0.4% = 20%).