This thread's a prime example of why you have to view all postings with a certain amount of suspicion and then make up your own mind.
Outspoken is putting forward a method of calculating voltage drop, based on the total load and total circuit length. This will always give a pessimistic value of voltage drop, and therefore will be 'safe', but will, in some cases lead to installing a larger conductor than necessary.
Richard Burns is describing the accurate analysis, calculating the actual voltage drop for each section of cable based on the current and length for each section and then adding up the voltage drops for the supply route to each load (and then take the maximum drop as the stated value for the circuit). I think most others here agree with this method, and it is the method mentioned by the OP as taught to him previously.
Here's another thread where Richard describes the correct method of calculation, although not a branched circuit, it shows the process:
http://www.electriciansforums.net/e...lectrical-regulations/46784-voltage-drop.html
Handy, to some degree I agree with your comments and those of Richard, however as the calculations show, simply adding volt drops together gets a lower reading than the calculations do in the method I gave examples of, which is the method I was taught at college some 30 years ago, taught by responsible contractor and advocated by all the governing bodies, including the IET and British Standards. The reason being is the at the section of cable nearest to the MCB is subject to the total volt drop along the circuit and the total load on the circuit will have an impact on the calculations and alter them from the method you ascribe to and Richard advocates.
In the grand scheme of things I do not say any method is inherently wrong, so long as some form of consideration is given to accounting for voltage drop on a given circuit, even if you 1-3V out it will not make a major impact on the circuit as we all know the actual voltage of the mains is more like the average of 239.6V that the harmonised voltage of 230V.
Following on from this discussion with Monkeyblain regarding volt drop in circuits that are not simple straight line circuits I have received a number of PM’s regarding this subject whereby the message has requested further information because they either feel I have explained it wrong, or that they have misunderstood.
Obviously I like to think that over the years I have got this aspect of my job correct, having never had a serious volt drop issue on a cable I think I am certainly doing something right, but I am loathe to imply or state others are wrong, just that perhaps the way they are thinking about this problem needs tweaking or perhaps they have simply been steered down the wrong road in the past.
I believe there is a misconception of the impact that Volt drop can have, and how it actually operates in practice in a given circuit, but it can have a radical and critical impact on the efficient and the safe operation of circuit, the classic lights going out on a 110V site temporary supply when the 9” Angle grinder starts is an extreme example of this, one that should never be seen on permanent services.
We all know that on a simple circuit, calculating the volt drop is a simple matter; See Fig 1 below:
Figure 1: Simple Lighting circuit
![[ElectriciansForums.net] Volt drop cont.](https://electricians.s3.eu-west-1.amazonaws.com/data/attachments/12/12438-09f545ab2f7ac9883bfd1be87aa96133.jpg "[ElectriciansForums.net] Volt drop cont.")
Now we can see this is a typical type of lighting circuit, nothing complicated but branched into two sections from the first switch as a two plated system in conduit (for arguments sake).
Assuming that all lights are on at the same time we can calculate the voltage dropped along the length of the circuit thus:
Total length of cable: 7.25m +8m +12m+7m = 34.25m
Total load on the circuit = (500W*2) + (92W*4) = 1368W or 5.95A @ 230V
The circuit has cables that pass through and buried in insulating walls, so from Table 4D2 of BS7671 we can see the volt drop of 1.5mm² is 29mV per Ampere per meter, thus:
29 * 5.95 * 34.25
1000
= 5.909V
Now we may wish to consider what the voltage drop would be on each section of this circuit because in reality not every light may be on at the same time.
So let’s deal with the halogens first:
Circuit length: 7.25m + 8m = 15.25m
Maximum demand = 500*2 = 1000W = 4.34A @ 230V
So:
29 * 4.34 * 15.25
1000
= 1.91V
Now let’s calculate the voltage dropped on the section containing the Fluorescent lights.
Circuit length: 7.25m + 12m = 7m = 26.25m
Maximum demand = 92*4 = 368W = 1.6A @ 230V
So:
29 * 1.6 * 26.25
1000
= 1.21V
Now we can see that the combined voltage drop of both sections is 1.21V + 1.91V = 3.12V
However our calculations earlier showed us that the total voltage drop across the entire circuit is 5.9V, so why the discrepancy?
Well this a combination of two circuits, that are effectively running in parallel, being combined into one circuit, the increased cable length of the circuit and the combined effect on the section of cable that will carry the total load of this circuit causing a heavier impact on the voltage drop than the two sections would have individually.
I know that many might think that
“surely we can treat these as two circuits and only worry about the individual volt drops for each section and simply add them together”, but this would be wrong because you have a common neutral that makes it one circuit, remember, current flows the opposite way to voltage, neutral to positive, and this is the same whether the circuit has a direct or alternating current.
Now let’s look at something a little more complicated, along the lines of a message I received last night.
The scenario:
Mechanical Services circuit, fed from a sub main served DB on a 63A single phase MCB. Circuit length is 68m fed via 10mm² SWA, on tray and basket. The Equipment at the end of this is a HVAC unit with a 7.5kW motor and a 3.5kW heater element and an associated control panel that has a demand itself of some 4.24A.
This circuit has now been broken into and some new offices have had their lights and small power (radials) taken from this circuit via a small consumer unit, however this is only 12m along the circuit from the point of origin, potentially installing an issue caused by volt drop when the HVAC starts up.
Figure 2: Layout of the described circuit.
![[ElectriciansForums.net] Volt drop cont.](https://electricians.s3.eu-west-1.amazonaws.com/data/attachments/12/12439-74d1ba8018748784ead0db3f45eb979f.jpg "[ElectriciansForums.net] Volt drop cont.")
Before we do any calculations on this you can see that this is a poorly designed circuit, the original circuit is likely not compliant, although calculations will prove this one way or another, but the addition of the circuits for the two offices simply adds to the problems, not least of which is the issue of adequate isolation of the two sections of the main circuit.
Let’s do some basic calculations for the original circuit, we know the following.
Maximum length: 68m
Maximum Demand: 11kW + 4.24A = 52.06A
Cable Type: 10mm² 3 Core Steel Wire Armoured (3[SUP]rd[/SUP] core providing earth path)
Protective device: 63A Type D BS 60898 MCB (max Zs 0.82Ω) [BS 7671:2008:2011.pp56: Table 41.4]
Table 4D4A: (BS7671:2008:2011. pp398/399)
Maximum CCC for 10mm² SWA = 72A
Volt Drop (mV/A/m) = 4.4mV
Now let’s calculate the volt drop on this circuit;
4.4 * 52.06 * 68
1000
= 15.57V
Clearly this does not comply with the regulations which allow a maximum of 11.5V (5%) (BS7671:2008:2011. Appendix 4: sec 6.4 table 4Ab: pp314).
Straight away we can all see there would be some serious problems here. In the summer the circuit likely complies as the heater element is unlikely to be used, but in colder months the unit will run at full demand for periods and thus the circuits fed from this will be impacted.
Assuming the Original sub-main was calculated correctly, and assuming the demand on this does not cause severe volt drop we have a supply voltage at source of 236.5V, with the volt drop calculated we have a voltage of 220.93V, which is thankfully within the minimum allowed of 6% of supply voltage (Assumed to be 230V) of 216.2V
We do not know what equipment will be fed from the sockets in these two offices, but being offices we can make a reasonable assumption of IT equipment and as such we can assume a maximum demand, including lighting, from both offices, of 12A (likely a lot lower).
Now we get into some territory that appears to cause confusion for many. To calculate the true volt drop on this circuit we have to calculate the volt drop on the sub circuits back to their respective MCB.
The lights are wired in 1.5mm² Twin and Earth, let’s assume these are twin 36W 1500mm T5 types thus each fitting is pulling (Philips Technical Guide 2012) some 92W from the supply, we have 4 lights and a total cable length of 15m.
Thus:
29 * 1.67 * 15
1000
= 0.72V
No problem there.
The radials are a different proposition but if we make an educated guess that each will have a load of 5A placed upon it we can do the following (I do not know if they are 2.5mm² T&E on a 16A MCB or 4.0mm² T&E on a 32A, but I will assume the former as 32A would be OTT)
From BS7671:2008:2011 Table 4D2A pp 334/335 we have the following information.
Max capacity: 18.5A (Ref Method A: Column 2)
Volt Drop: 18mV/A/m (Volt drop Column 3)
Radial 1 is 7m long, so we get;
18 * 5.00 * 5
1000
= 0.45V
OK so far:
Radial 2 is 11m long, so we get;
18 * 5.00 * 11
1000
= 0.99V
Again, this circuit is compliant.
These are separate circuits and thus we are allowed to add the combined volt drops together to have a guide as to the total on this sub-main.
Thus: 0.72 + 0.45 + 0.99 = 2.16V
I do not know if the cable feeding this small DB is 10mm² SWA or a different cable, or how this has been split off the main feed, so I will simply assume the volt drop for this part of the circuit is the 2.16V above (will not be more if the cable is 10mm²)
Now we have determined that the main volt drop on the supply cable is some 15.57V, if we add the 2.16V to this we have a volt drop of 17.73V.
Clearly, as this is an existing circuit a measurement of the volt drop should be completed to ensure that the calculations are correct, the final volt drop will be more or less than the 17.73V depending on the total demand on this first section of the circuit and the method used to split the mains, whether this has an impact on the final volt drop of this sub-section or not is open to debate at this point due to the lack of information, however we definitely know for a fact that the absolute minimum volt drop will be some 15.57V and thus non-compliant.
Here we have another issue, if the offices are occupied and in use when the HVAC turns on the demand at start-up will likely cause a significant drop in the supply voltage, this will have a potentially damaging impact on the light fittings and any IT equipment in the offices. The lights may dim or fail momentarily, but continued impact of this will damage the control gear and the lamps. The PSU’s in the computer equipment could well be damaged by this spike that would be induced due to the sudden drop in voltage.
There is not enough space here to address every aspect of circuit configuration we may encounter in our professional lives, but we have to apply common sense and ensure that the method of calculation is sound. Whether we use the method ascribed by Richard and Hants, or the method I ascribe to is less of an issue as both give reasonably accurate answers to this thorny question, and I am certainly not going to get into an argument over a procedural issue that at times may be the only method left open to the spark faced with an unconventional circuit arrangement.
What this does open up is the can of worms surrounding proper design of circuits in the first place so that such situations are not faced in the future, but that is for another discussion altogether.
