OP
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Not just what the max VD is for a circuit. Surely it's what voltage is left at end so lights/equipment work correctly.
-
Please checkout the new poll we will be pushing hard, and want some good stories to come out of this
result are public - you can change your vote over time, which might genuinely be the case from time to time.
New Zealand Volt drop cont.
Discuss Volt drop cont. in the UK Electrical Forum area at ElectriciansForums.net
O
Outspoken
Monkey, FFS, the calculation of the TOTAL voltage drop across the entire circuit tells you the voltage at the END of the circuit FFS....
If the Volt drop is calculated at 4V (Thus dropping to 226V) the the voltage at the end of the circuit is ...Yep, you guessed it 226V ...it is not Rocket Science!
OP
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I'm not stupid and I do get what you are saying. I know how to calculate the entire load of a circuit then use VD calcs to obtain total VD. I was always taught that you had to work out each leg individually. As the way your explaining it is assuming the total load is at the end of the circuit rather than it branching off at various points. Do you not see my point. But I will go with everything you say as you clearly have worlds more knowledge/experience. And if in your examples is correct way to
Determine total VD for a branch circuit, then I will leave it as that. As its a simpler way to calculate.
Determine total VD for a branch circuit, then I will leave it as that. As its a simpler way to calculate.
O
Outspoken
I had considered pointing out that if you measure voltages in the UK the averages for single and three phase are 236.7V and 415.8V respectively...but realised he was confused enough as it is!! 
O
Outspoken
Monkey, FFS, whoever taught you that is a frigging idiot as you will always end up with incorrect figures, that is why i gave you the example I did, to PROVE you MUST calculate the total maximum demand on the maximum length of the circuit to establish the true maximum volt drop you may experience. Obviously in the example I gave this will be variable as lights are turned on and off on various branches, thus you need to calculate the worst case scenario, if this is within the permitted levels with BS7671 then everything else is a pointless waste of your time!
OP
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Thank you. I do understand. And sorry for all the frustration. Bad training.
OP
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Haha this could of been over a lot earlier if I'd seen this sentence at the beginning.
you MUST calculate the total maximum demand on the maximum length of the circuit to establish the true maximum vd
you MUST calculate the total maximum demand on the maximum length of the circuit to establish the true maximum vd
4
44ALLAN
Monkey, FFS, the calculation of the TOTAL voltage drop across the entire circuit tells you the voltage at the END of the circuit FFS....
If the Volt drop is calculated at 4V (Thus dropping to 226V) the the voltage at the end of the circuit is ...Yep, you guessed it 226V ...it is not Rocket Science!
Well put and better than a thousand diagrams
If the Volt drop is calculated at 4V (Thus dropping to 226V) the the voltage at the end of the circuit is ...Yep, you guessed it 226V ...it is not Rocket Science!
Well put and better than a thousand diagrams
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Couldn't be spelled out to folk any clearer than that! And yet still it continues.......
OP
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Alright guys. Your all much smarter than me. I was taught wrong in the beginning. Which explains my confusion. If someone had told me to work it out from entire load and entire length then there wouldn't be a problem. But I was trying to make things too complicated calculating each leg rather than worse case scenario. Your smart remarks are very constructive though. Thanks.
K
KAS1
I think some people are getting themselves confused with volt drop calculations, in that they are not treating a radial circuit with branches within the circuit as a parallel circuit ie with the total load of the circuit in mind and that the voltage is the same within the circuit. It's as though they are thinking on the lines of a series circuit with volt drop across different parts of the circuit, if that makes any sense!!!!
OP
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I'm gonna ask a stupid again obviously. On the lengths, when you mention lengths from MCB why aren't they they the leg length plus the 5m from cu to joint as mentioned on diagram. They're all different. And on the final calc when you have the lengths in brackets. Where does the 11.5 come from. I see the others but I can't see where that's come from. I would of thought it would of been all legs plus 5m from Ccu. I can't work out what's what here.
E
Engineer54
I don't think missed anything completely, ...not from the two posts that i quoted anyway. Now that you have explained and shown what you meant, all becomes clear. That can't really be said of those original two posts, when all you were basically quoting, was origin of supply....
O
Outspoken
Look at the picture carefully, read the information and you will understand...
OP
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I'm sorry. But leg 1 is 6m from joint to last fitting. And from ccu to joint it's 5m. So that's 11m. But then it states that from mcb to last fitting its 8m??? Same with all of them. Then like I said before the final calc. (11.5+6+3+5+3)??? Am I really that dumb that I can't understand anything today?
O
Outspoken
MB, it's a different drawing to the first one, work it out, all the information is there if you READ it. Don't worry about drawing 1, just look at drawing 2
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OK I will respond, but I think it will not be worth it, in the information you give above you are jumping from measurement to calculation.
I agree that if you were to measure volt drop at the junction box with with the loads disconnected you would measure zero.
Then you go on to say that this section with suffer max volt drop, despite it reading zero in measurement.
However I agree that the calculated volt drop would be under the most onerous conditions (i.e. maximum current flow) along that length.
I do not agree that it is the maximum volt drop because if it was a 1m length of cable taking 10A, and the rest of the circuit was a 100m length of the same cable taking 0.5A (assuming the 9.5A remainder were used at the JB) the volt drop proportion would be five times greater over the long part of the circuit.
Could you let me know how you would calculate volt drop where the circuit has differing sizes of cable.
I also notice in your calculations in later posts that on summing the individual volt drops (that you state as incorrect) you miss out the volt drop on the supply cable, presumably you are assigning this as zero?
(This would not affect your "proof", the calculation would still come out with a higher volt drop by your method)
Anyway an interesting diversion, I will not go further as the horse is at the water.
(Sorry you were subjected to this Monkeyblaine)
OP
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Outspoken. I am looking at drawing 2 and it says 5m to Ccu and the info I gave previously is all I have to go on and it doesn't add up.
O
Outspoken
Richard, your not thinking about it correctly, perhaps how I worded it has not helped, On measurement we both agree that the Volt drop at the JB would be zero as there would be no load, however you're questioning my comment about this section also seeing the largest volt drop, well it will do under operational conditions because the full load of the circuit will pass along it thus the full amount of volt drop would be measured here, between the JB and the MCB. Do you understand that now...measurement with no load = zero volt drop, measurement at full demand = maximum volt drop.
That would simply be a badly designed circuit and proof positive of why this calculation should be completed before you install cables. However the maximum volt drop would still be on section 1 because if the load on section 2 caused say a 5V drop, this would impact the voltage on section 1, causing a higher load to be drawn, thus causing higher volt drop and the first section would still see the highest volt drop because the full load of the circuit would be passing along it.....regardless of where the load on the circuit was.
Richard, I think you should be able to figure that out yourself.
This is why in picture 2 I reverted to the MCB instead of a JB because for all intents and purposes the supply can be rightly assumed to be at peak voltage because even if this were a board fed by a sub main the calculations should have been done to ensure the voltage on the output side of all MCB's would be 230V and so the sub-circuit calculations can be based on this figure....you will be well within because unless you get ridiculous loading the real voltage will be ~ 238V so even a drop along the run will compensate for this.
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Oh dear.
Fair enough, let sleeping dogs lie.
Fair enough, let sleeping dogs lie.
O
Outspoken
OK lets take this one step at a time:
Leg 1 is 6m from the joint to the last fitting, but this section is 8m long, therefore the leg from the MCB to the joint is 2m long, making 8 m
Leg 2 is 3m from the joint to the last fitting, but this section has a total length of 6m, thus 3m + 3m from the MCB (1m further than leg 1)
Does this make sense to you know...
The length of the cable to the first joint is 2m, then from joint 1 to joint 2 = 1m, from Joint 1 to Joint 3 = 2m, Joint 1 to joint 4 = 3.5m, Joint 1 to joint 5 = 4.5m. You then need to add the initial 2m to each, plus the length of the each leg...
The information is on the drawing and self explanatory.
OP
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Ok but on your picture it clearly states 5m to ccu. Where I would imagine the MCBs are and also they look switch symbols not joint symbols. So I couldn't the reference point of your lengths.
O
Outspoken
Apologies, I failed to delete the JB comment in the bottom corner. The clue should have been that is stated JB still not MCB, but that is my fault, i now see why you got confused by that MB...my humble apologies!!
OP
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Ah ok. Thank you. That makes sense now. Haha. Thanks for taking the time to see what I was talking about.
OP
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Just one more thing haha. I's the final calc correct? Shouldn't it be (2+6+3+5+3+5) rather than (11.5+6+3+5+3) as stated in your diagram?
O
Outspoken
Highly likely, I've been typing this and being attacked by my 3 year old who thinks he is a Monster robot for the last few hours...attacking me with pillows, blankets, stuffed toys and generally being mad!
It should be the total length of all the cable, so 11.5m for the longest single section, then the length of each section from the joint...will look at the diagram in a mo when...I have fought him off again!
O
Outspoken
Just double checked, I did do the total correct...
OP
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Haha know the feeling have a crazy 4yr old princess myself. As I sent last message I realised I was wrong as well but. From CU to joint 5 I see the length as 6.5m. (2m from cu to 1st joint then 4.5m from joint 1 to 5). Then add the branch of legs. Where is the 11.5 from?
OP
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Once again I apologise haha just worked it out. 11.5 from cu to last light on last leg. Then all other lengths. Got it!!!
4
44ALLAN
Go to c/unit................do voltage check.
Go to last fitting..........do voltage check.
Difference is volt drop............voila
Only joking guys ffs lol
Go to last fitting..........do voltage check.
Difference is volt drop............voila
Only joking guys ffs lol
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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
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
O
Outspoken
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
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.1000
= 5.909V
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
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
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.
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).1000
= 15.57V
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
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
1000
= 0.45V
OK so far:
Radial 2 is 11m long, so we get;
18 * 5.00 * 11
1000
= 0.99V
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.
4
44ALLAN
Well..............my way was easier :laugh:
OP
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- 5
Wow... I'm Outta my depth.
S
Silly Sausage
Is that the Ben-Hur of forum answers?
OP
- Reaction score
- 5
handyspark where your mention, "and then adding up the voltage drops for the supply route to each load" if you can picture a cable route from CU going down to 5 joints with lights coming off each joint, do you mean add up the VD on the cables from joint to joint (NOT including cables to lights) as we can assume that If the VD is satisfactory to each joint then we can assume its fine to the lights branched off. This goes back to something I was trying to get my head round earlier. Say for example at one particular joint the voltage is at 220v then that's the supply voltage to the branched off lights at that particular joint AND the supply voltage to the next joint. What I'm trying to say is, you wouldn't calculate VD from CU to first joint, to the lights on that branch, then to the next joint (in that order). You would calculate CU, to first joint, to second joint and so on (obviously to get max VD of entire circuit. NOT individual leg!) Because at the joint it doesn't include the VD to the lights, where it carries on to the next joint. (Due to the voltage behaving in a parallel way not series) do you follow me at all?
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- 2,602
To use the above example of a feed from a CU with spurs off to individual lights. If the supply to the first light branches off at JB1, second at JB2, etc:
For light 1, total VD = VD from CU to JB1 + VD from JB1 to light 1.
For light 2, total VD = VD from CU to JB1 + VD from JB1 to JB2 + VD from JB2 to light 2.
For light 3, total VD = VD from CU to JB1 + VD from JB1 to JB2 + VD from JB2 to JB3 + VD from JB3 to light 3.
and so on.
If the CU is any significant distance from the origin of the installation, add the VD for the submain (at the expected maximum loading) to each distribution value.
The VD in each section of cable is calculated on the mV/A/m, actual current and length of that particular section.
When you've calculated the total VD at each load point, the VD for the whole circuit is just the maximum value of those figures. It's likely that you can omit calculating the VD for some of the loads by 'guessing' which will have the highest VD. ie one of the loads at the far end is likely to have the highest VD, unless one of the other spurs is particularly long or heavily loaded.
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- 3,768
how come people are really strugling with this?
the smaller the csa of the cable the higher the resistance.
in other words if you have a run of 250m you can't run the sockets in 2.5 but what you could do is increase the size of the cable.
why are you trying to work out volt drop at each section? your wasting time in this case. the supply to either the jb or anothet dist board needs to be big enough to not be affected by the leghth of the run.
if you think of it like a length of pipe, you can only put so much water through it you cant increase this amount, at every junction the amount reduces.
now if the start of the pipe is a lot larger than the offcoming shoots then you wont get a reduction.
this is the reason people run in circuits in 2.5 to a fused spur
the smaller the csa of the cable the higher the resistance.
in other words if you have a run of 250m you can't run the sockets in 2.5 but what you could do is increase the size of the cable.
why are you trying to work out volt drop at each section? your wasting time in this case. the supply to either the jb or anothet dist board needs to be big enough to not be affected by the leghth of the run.
if you think of it like a length of pipe, you can only put so much water through it you cant increase this amount, at every junction the amount reduces.
now if the start of the pipe is a lot larger than the offcoming shoots then you wont get a reduction.
this is the reason people run in circuits in 2.5 to a fused spur
OP
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- 5
Yes. So I'm right in saying, you take worse case scenario for the circuit. Ie end of line. Or heavily loaded/long spur. Not an accimulation of every spur and supply cable. Just worse case route. And of its sarisfactory there then we can assume all other lengths/loads are fine.
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- 2,602
Yes, but you need to be sure that your are calculating for the 'worst case'. If there's doubt, you need to calculate the VD at more than one (or all) the loads and compare results. Remember, it may not be the far end load that has the greatest VD, depending on length and load on each spur.
Also remember that, for each section calculated, you need to take into account the whole current in that section (based on the loads supplied through that section), not just the current due to the load that you've chosen to calculate the VD for.
O
Outspoken
So please explain why you're making a simply task so darned complicated when it need not be, one simple calculation covers all that and you are then compliant and safe...
OP
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- 5
Handy spark. I understand completely. Sections will contain load to its particular light/fitting and to all loads downstream of it. Outspoken I will definitely be doing it your way because of the simplicity and you know that you are completely safe. Unless I have some crazy designing where by I need to install smaller CSA cables in sections to fine tune installs.
OP
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- 5
I think the point were trying to make is that the whole load of the entire circuit is only upon the first section. After that it changes at each point as a load is dropped. So you could possibly get away with Using a smaller cable in places, which is obviously good for costings and ease of installs. When you say the first section will have the total VD across it, it won't because it will keep falling from that point. So where it could be at 227v at first section with the total load upon it, a drop of three volts. Over the next 3 sections (for example) it could drop another 1v per section. Taking it to 224v, the Max VD for the circuit. But like I said before outspoken your way is the way I will aim to do it.
OP
- Reaction score
- 5
Ok another question. When you turn on a10kW shower on in some houses and the lights dim, is that due to a poorly designed supply cable from the DNO. In other words in a home with lights and power under a standard load everything is fine so no major drop on incomer, then when shower is under full load, that extra current the supply cable is dealing with causes a volt drop on the entire installation, for every circuit. Including the shower. Because the supply (DNO
Supply) is dropping to a Lower voltage than it should be?
Supply) is dropping to a Lower voltage than it should be?
O
Outspoken
MB, sorry chap, whilst you have grasped one part of this thorny issue, you have totally missed the most important part...the MAXIMUM voltage drop, the biggest drop in electrical potential, occurs where the circuit is under the most pressure, and this occurs where the circuit has the heaviest load placed upon it, thus it will always be at the point nearest to the source of energy for that circuit, such as the MCB/Fuse/RCBO/RCD.
You are correct to think that different parts of the circuit will have differing voltage drops along their length because this is wholly dependent on the load on that section, but if there is a common connection/section for all of the circuit loads on that section, then you will suffer the highest significant drop at this point.
OP
- Reaction score
- 5
Outspoken please don't thnk I'm trying to teach my grandmother how to suck eggs (excuse the expression). Like I said before you are worlds apart from me in both knowledge and experience. I was just wondering if we were all on the same page and as I had created this post and people joined along the way, I was just checking we all were up to date. And yes I do understand that. The biggest significant drop will be at that point, and whatever's left after that point continuing to further loads downstream will drop further equal to the factors of length and load. Perhaps we should put this to bed now. I think I've annoyed everyone enough (you mainly) haha.
Reply to Volt drop cont. in the UK Electrical Forum area at ElectriciansForums.net
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