Volt drop to summer house heating

[HandySparks]

If you do run 35ml or ,25 ml is the house supplied by 16 ml

Would it make any difference if it is? (Assuming that the maximum total load was within the rating of the 16mm² cable.)

 

[dingledong]

No but depending on what he puts in the house then working out the house load and this shed then I would say it will matter

 

[telectrix]

i'd be walking away. 6kW heaters. does she think it's siberia? looks like a load of trouble for whoever takes it on.

 

[Silly Sausage]

65m of 35mm SWA, get them to run a gas pipe and install a small gas fired heater instead!

 

[brucelee]

sounds like she needs a carpenter to add a front to the shed lol instead of changing the climate

 

[Chr!s]

Hi. First time poster here so please go easy on me.


A customer of mine has had a summer house built at the rear of her garden which has been supplied with two 6Kw heaters! Yep, 12Kw!! She would also like 2 switched double sockets and about 4 LED spots installed. Now the problem, as if a 12Kw load isn't enough, is that the summer house will be on the end of roughly a 65m run from a switch fuse inside the house. The obvious problem is volt drop and whether enough diversity can be applied to make this install possible. As a note, the heaters very basic instructions state that they DO each need a 26A supply.


Should I even apply diversity to the heaters knowing full well that the customer is intending to use them to full capacity? The customer explained that they would like to eat Christmas dinner in the summer house. This set the obvious alarm bells ringing - heaters on, cooker on, kettle on, hair dryers on....Boom....darkness.


At 52A I can get away with 16mm SWA if I have calculated correctly. But chuck in 10A + 50% of a power circuit and 66% of a lighting circuit and I think it all begins to fall apart. I have tried talking her into more efficient heating methods but she remains hell bent on what she already has. Surely other people have bought these summer houses and heaters. And at £20k a pop I'm thinking a fair few of them will be at the end of big gardens! Surely not all of these people can be on 3 phase supplies?


Any advice, even if it's 'say no to the customer' greatly appreciated.

Just give her what she wants, she'll be happy that way. 60 amp MCB and 16mm cable, job done. Get paid and off you go.

 

[telectrix]

Just give her what she wants, she'll be happy that way. 60 amp MCB and 16mm cable, job done. Get paid and off you go.

top of my head that'll give around a 10V VD.

 

[Chr!s]

top of my head that'll give around a 10V VD.

What do you get from the bottom?

 

[telectrix]

a plan view of my arse.

 

[KFH]

Stuck the figure into Cable Mate:

Supply Voltage = 230 Volts
Power factor = 1
Ib - Design current = 52 Amps
Protective Device Type = MCB type B (BS EN 60898)
In - Protective Device Rating = 63 Amps

Cable Type : General purpose PVC ARMOURED - Multicore
Length of run of cable = 65 metres
Maximum permissable Voltdrop: 3% (Lighting) = 6.9 volts
Maximum selected voltdrop for this calculation = 6.9 volts

Installation Method : Multicore unarmoured cable in conduit or in cable ducting in the ground.
An installation depth of 0.7 Mtr, A soil thermal resistivity of 2.5 K.m/W
(method D)

Ambient temp = 20 °C
Number of circuits including this one = 1
Length of cable in thermal insulation = none

Apply Correction factors:
From TABLE 4C3 : Cg = 1 (Grouping)
From TABLE 4B2 : Ca = 1 (Ambient temp) - Ground Temperature : 20 °C
From TABLE 52.2 : Ci = 1 (Insulation)
Protective device factor for Buried cables : Cc = 0.9 (Burried direct)
For an installation depth of 0.7 Mtr : TABLE 4B4: Cd = 1
For soil thermal resistivity of 2.5 K.m/W : TABLE 4B3: Cs = 1
Protective device factor : Cf = 1

It = tabulated current carrying capacity
It = In / (Cg x Ci x Ca x Cf x Cc x Cs x Cd)
It = 63 / (1 x 1 x 1 x 1 x 0.9 x 1 x 1 )
It = 70.00 Amps
From TABLE 4D4A Cable selected = 16 mm²

TABLE 4D4B For 16 mm²: mV/A/m = 2.8
mV/A/m corrected for power factor = mV/A/m x Power Factor = 2.8 x 1 = 2.8

Voltdrop = (mV/A/m x Length x Design current) / 1000
Voltdrop = ( 2.8 x 65 x 52 ) / 1000
Voltdrop = 9.46 Volts
(Maximum permissible voltdrop (regulation - 525) = 6.9 Volts)
This exceeds the maximum voltdrop, so we have to Increase cable size and recalculate

TABLE 4D4B For 25 mm² : mV/A/m r = 1.75 and mV/A/m x = 0.17
mV/A/m Corrected for Power Factor = (cosØ mV/A/m r)+(sinØ mV/A/m x) = 1.75
Voltdrop = (mV/A/m x Length x Design current) / 1000
Voltdrop = ( 1.75 x 65 x 52 ) / 1000
Voltdrop = 5.92 Volts

Calculated Cable size = 25 mm², Minimum Earth conductor size = 16 mm² (Table 54.7)
Current capacity of cable selected = 99 Amps
Maximum Cable Length = 75.8 Metres

 

[D Skelton]

Stuck the figure into Cable Mate:

Supply Voltage = 230 Volts
Power factor = 1
Ib - Design current = 52 Amps
Protective Device Type = MCB type B (BS EN 60898)
In - Protective Device Rating = 63 Amps

Cable Type : General purpose PVC ARMOURED - Multicore
Length of run of cable = 65 metres
Maximum permissable Voltdrop: 3% (Lighting) = 6.9 volts
Maximum selected voltdrop for this calculation = 6.9 volts

Installation Method : Multicore unarmoured cable in conduit or in cable ducting in the ground.
An installation depth of 0.7 Mtr, A soil thermal resistivity of 2.5 K.m/W
(method D)

Ambient temp = 20 °C
Number of circuits including this one = 1
Length of cable in thermal insulation = none

Apply Correction factors:
From TABLE 4C3 : Cg = 1 (Grouping)
From TABLE 4B2 : Ca = 1 (Ambient temp) - Ground Temperature : 20 °C
From TABLE 52.2 : Ci = 1 (Insulation)
Protective device factor for Buried cables : Cc = 0.9 (Burried direct)
For an installation depth of 0.7 Mtr : TABLE 4B4: Cd = 1
For soil thermal resistivity of 2.5 K.m/W : TABLE 4B3: Cs = 1
Protective device factor : Cf = 1

It = tabulated current carrying capacity
It = In / (Cg x Ci x Ca x Cf x Cc x Cs x Cd)
It = 63 / (1 x 1 x 1 x 1 x 0.9 x 1 x 1 )
It = 70.00 Amps
From TABLE 4D4A Cable selected = 16 mm²

TABLE 4D4B For 16 mm²: mV/A/m = 2.8
mV/A/m corrected for power factor = mV/A/m x Power Factor = 2.8 x 1 = 2.8

Voltdrop = (mV/A/m x Length x Design current) / 1000
Voltdrop = ( 2.8 x 65 x 52 ) / 1000
Voltdrop = 9.46 Volts
(Maximum permissible voltdrop (regulation - 525) = 6.9 Volts)
This exceeds the maximum voltdrop, so we have to Increase cable size and recalculate

TABLE 4D4B For 25 mm² : mV/A/m r = 1.75 and mV/A/m x = 0.17
mV/A/m Corrected for Power Factor = (cosØ mV/A/m r)+(sinØ mV/A/m x) = 1.75
Voltdrop = (mV/A/m x Length x Design current) / 1000
Voltdrop = ( 1.75 x 65 x 52 ) / 1000
Voltdrop = 5.92 Volts

Calculated Cable size = 25 mm², Minimum Earth conductor size = 16 mm² (Table 54.7)
Current capacity of cable selected = 99 Amps
Maximum Cable Length = 75.8 Metres

All of that for something that you could work out yourself in a few seconds with a flick through the regs book!?

You've also not taken into account any extra demand from the two double socket outlets and LED lighting. The design current is higher than 52A if you allow for these too, albeit not by much, but these have to be taken account of. Your maximum permissable volt drop is too high also. You need to take into account the volt drop up to and including the installation within the outbuilding and not just use 6.9V as a value up to the outbuilding.

Shame your computer program can't seem to transpose a simple calculation to tell you the minimum size needed without having to perform a seperate calculation for every cable size selected.

35mm is what would be needed if you wished to stick religiously to the guidance given in appendix 4 regarding maximum suggested volt drop for consumers installations.

The devil is in the detail

 

[HandySparks]

I've not done the calcs, but in this sort of case where the majority of the load is heating, you may be better off laying two cables; one big one for heating (with the larger permissible voltage drop allowed) and a much smaller one for lighting (and maybe the other small loads).

Yes you'd have two supplies to the same 'building', but I'm sure that it's possible with the right warning labels.

Just a thought.

 

[dingledong]

Also no diversity on the heating as no thermoststic control just on/off and usually when you turn it on its left on especially with no front to the building

 

[davesparks]

Also no diversity on the heating as no thermoststic control just on/off and usually when you turn it on its left on especially with no front to the building

No diversity if there was a thermostat controlling it either.
It would be all on at the same time even with the thermostat.

 

[KFH]

All of that for something that you could work out yourself in a few seconds with a flick through the regs book!?

You've also not taken into account any extra demand from the two double socket outlets and LED lighting.

All I had done is stick in the figures the OP gave for his original 16mm calc. I had not tried to do the full design for him.

 

[D Skelton]

All I had done is stick in the figures the OP gave for his original 16mm calc. I had not tried to do the full design for him.

I know what you did, but the OP will read it and say "It's ok boss, some fella on a forum said to stick 25mm in", which if like I say he wants to stick to the guidance given, won't be big enough.

And you put in the wrong volt drop value.

 

[davesparks]

I've not done the calcs, but in this sort of case where the majority of the load is heating, you may be better off laying two cables; one big one for heating (with the larger permissible voltage drop allowed) and a much smaller one for lighting (and maybe the other small loads).

Yes you'd have two supplies to the same 'building', but I'm sure that it's possible with the right warning labels.

Just a thought.

Why on earth would you need a warning label to run two supplies in to a building?

 

[D Skelton]

Why on earth would you need a warning label to run two supplies in to a building?

More importantly, there would be no warning label warning of the warning label warning of the two supplies in to the building!

C1 that shiznit all day long!

 

[HandySparks]

Why on earth would you need a warning label to run two supplies in to a building?

Well, maybe you don't, especially if it's obvious from the arrangement of the equipment. Just being overcautious about two supplies when most people would expect to find only one.

 

[davesparks]

What exactly would a warning label add to the situation? Apart from making the installer look amateurish to the next electrician to work there?

Warning labels achieve nothing as nobody ever reads them!