Voltage drop workout

[PiousJim]

Hi guys. Those who don't know me I have recently started on my own.

I am testing my understanding, re Volt drop. And trying to stick to one formula. But I'm getting mixed results.

My example scenario would be to feed a CU in an outhouse / shed.
The CU will feed sockets, lights and some form of machinery of 2200 watts. Cable buried.

Length 25m
Design Current 27 amps (6200)

1.Using OSG P124 formula
VD=(mV/A/m) x lb x L ÷ 1000

I got
11.5v (5% single phase) x 27a x 25m
= 7.6
According to BS7176 p339 should use 4mm

2.On another thread someone suggested the following

11.5x1000 /25/27 = 17.03

According to BS7176 p339 should use 2.5

To be honest I would rather follow no 1 as this is how we were taught.
But no 2 is the result I got from tlc calculator and a few others.

Have I got something wrong or missed something?

In your opinion which would you go for and why?

Any advice is helpful.

 

[Andy78]

Where the formula states mV/A/m, then you need to be putting in the mV/A/m value for the selected cable size and type from appendix 4 to work out your actual volt drop rather than using the max VD in volts.


EDIT: Having said that, it depends if you are trying to work out the actual volt drop of your selected cable, or trying to work out the minimum cable size for volt drop permitted.

 

[telectrix]

i calculated a VD of 7.425 with 2.5mm. OK for non-lighting, but just outside the 3% for lighting. as the actual supply voltage will be well above 230V, i'd go with 2.5mm. assuming the ccc of 2.5mm is OK depending on installation method.

edit: in the real world, however, i'd be inclined to use 6mm. this would allow for any future increase in load up to 40+A.

 

[PiousJim]

Thanks for the reply Andy78 and Telectrix.

Andy78 to be honest I was trying to workout which cable to use.

I should also keep a note of how to work out the voltage drop of a circuit.

Telectrix did you use a formula similar to my no 1?

 

[telectrix]

VD = (mV/A/m x L x A)/1000

 

[Andy78]

Thanks for the reply Andy78 and Telectrix.

Andy78 to be honest I was trying to workout which cable to use.

I should also keep a note of how to work out the voltage drop of a circuit.

Telectrix did you use a formula similar to my no 1?

If you are working out which cable size to be used for volt drop alone the the formula is
mV/A/m = (MaxVD x 1000) / (Ib x L)
If you are working out the volt drop on your selected cable the formula is
VD = (mV/A/m x Ib x L) / 1000

You have of course got to size your cable for CCC before any volt drop stage.

 

[PiousJim]

Ok so my no 1 is correct?
(11.5×1000)/ (20*25)
=11500 / 675
≠17.03
Page 339 table 4D4B would suggest 2.5mm

I'm sorry for being paranoid 2.5 feels to small. Could I use 4mm as a precaution?

 

[PiousJim]

I think Telectrix answered the question in the edit on post 3

 

[PiousJim]

To be honest I would still have went 6mm as well they told us at college to not go lower than 6 as a precaution.

 

[Andy78]

Well you are sizing your entire cable on the requirements for volt drop. You should be sizing your cable on requirements for CCC taking into account load and corection factors and then checking for compliance with VD, not the other way round.

 

[PiousJim]

Sorry Andy78 you lost me ccc refers to correct carrying capacity?
Load is 27amps

That's why I wanted to go for 6mm.

The voltage drop formula was to ensure it would be compliant.

Is this not the correct way?

 

[Andy78]

Sorry Andy78 you lost me ccc refers to correct carrying capacity?
Load is 27amps

That's why I wanted to go for 6mm.

The voltage drop formula was to ensure it would be compliant.

Is this not the correct way?

No not really. This thread has been concerned only with VD and you wanting to work out the minimum size cable for VD.
If you want to make sure your selected cable for CCC complies with VD, then it's a different process and a different formula.
Standard cable design process has you sizing the cable appropriately for CCC requirements taking into account derating factors, then you check for compliance with volt drop.

 

[PiousJim]

Cheers Andy78

 

[Andy78]

And you will need to know the actual volts dropped over your distribution cable so you can work out the remainder of VD left for your final circuits from your outbuilding CU.

 

[Leesparkykent]

Just to add, you may need to size the cable accordingly if bonding is required in the outhouse.

 

[Deleted member 72666]

Out of interest what is this 'some form of machinery of 2200 watts' you speak of?

 

[PiousJim]

I just picked a random machine from screw fix.

http://m.NoLinkingToThis/p/record-power-pt107-265mm-planer-thicknesser-240v/48940

 

[PiousJim]

I've created a essay/document to test my knowledge and that I could refer back to.

I was designing the circuit. This thread is in relation to powering a shed.

 

[Richard Burns]

I know this is veering slightly from the plan.
Load of 27A on a buried SWA cable run for 25m from the origin of the installation.

Load is 27A so a 32A breaker required, cable must be able to take 32A.
Current carrying capacity of SWA method D no other rating factors in table 4D4A:
2.5mm² has 29A not enough for 32A breaker.
4.0mm² has 37A OK so the minimum cable size so far is 4.0mm².

So for a 4.0mm² cable the volt drop is 11 mV/A/m.

Using the values we now have in VD = (mV/A/m x L x A)/1000
Volt drop = (11 mV/A/m * 25 m * 27 A)/1000
Volt drop = 7.425 V

maximum volt drop for power is 11.5 V (5% of 230 V) so this is OK for power.
maximum volt drop for lighting is 6.9 V (3% of 230V) so this is not OK for lighting

OK so move up to 6.0 mm² cable with a volt drop of 7.3 mV/A/m as it is a bigger cable the current carrying capacity is going to be OK.

Using the values we now have in VD = (mV/A/m x L x A)/1000
Volt drop = (7.3 mV/A/m * 25 m * 27 A)/1000
Volt drop = 4.93 V

maximum volt drop for power is 11.5 V (5% of 230 V) so this is OK for power.
maximum volt drop for lighting is 6.9 V (3% of 230V) so this is OK for lighting.

Therefore for this run of cable, assuming the cable comes from the origin and does not have a significant circuit length to supply the sockets and lights and machinery in the shed, and for this installation method carrying a design curent of 27A, the minimum size cable that can be used to supply lighting is 6 mm².

 

[PiousJim]

Thanks Richard I was struggling with the calculation it self.

I wasn't happy with myself for getting the answer but not understanding the calculation.


Thank you for clearing this up for me. I think I owe you a few by now.
Thanks again Richard

 

[Gavin A]

Also, watch for the surge current on start up of that machinery. Put a big lathe or something in there and the start up current could well be 3 times the nominal rated current, so you could end up with a 15% VD on start up.

I remember one fault report with an inverter on one of the forums that turned out to be caused by the inverter tripping out due to the massive voltage swing when a big bit of machinery started up in the shed it was connected in to.

Also when doing these calcs, if you could double check them to be sure to meet the 1% VD limit for a 16A solar PV supply that'd be much appreciated. Nowt worse* than having to dig the driveway up to install solar on the garage because the existing supply to the garage doesn't quite cut it for our purposes.


*ok there are actually quite a few things worse than that.