New Zealand Volt drop cont.

[Monkeyblaine]

Ok I have another VD question. I'm installing 9 LED floods around my friends garden which totals at just over 100m in length. Obviously for each branch the VD varies so for argument sake I worked It out as if all the cable would be taking all the load (rather than dropping relevant current after each light, worse case scenario) the load is minimal being LEDS but to be safe I'm putting 2.5mm 3c hightuf in to cope with the distance and to help for the zs reading. Now I am also using a wise box control box (wireless switching) now am I right in thinking I can supply this box from the CU with a length of 1mm t+e as this is a very short distance and is my thinking correct that as long as no voltage has been dropped through the 1mm to the wise box then I have a full 230v (give or take) for the garden circuit and as its run in 2.5mm it will stop any VD issues. Basically what I'm asking is, whatever voltage you have at a certain point regardless of cable size, then that's what you have from that point onwards. I.e if I lost 3volts through initial 1mm cable then I'm down to 227v at wise box (for example)

 

[Monkeyblaine]

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.

 

[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!!!!

 

[Monkeyblaine]

Monkey, Look at this FFS..


View attachment 19415

The maximum voltage dropped it at the START of the circuit because this has the HIGHEST load placed upon it.

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.

 

[Engineer54]

You have missed it completely. Monkeyblaine is discussing calculating the volt drop on different parts of a circuit (lighting) and then applying these values to those parts of the circuit and sizing the cable accordingly, a clear nonsense because the volt drop must be calculated across the length of the circuit as the maximum VD will be on the leg from the protective device to the first accessory/equipment.

I don't get what you're not getting?

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....

 

[Outspoken]

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.

Look at the picture carefully, read the information and you will understand...

 

[Monkeyblaine]

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?

 

[Outspoken]

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?

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

 

[Richard Burns]

Richard, in the example you have given you are wrong i am afraid to say.

Voltage drop can only be calculated along the entire length of a circuit for design purposes. In your example the Voltage drop on the supply to the JB is ZERO because to measure this properly you need to disconnect from the branch to avoid loading by the installed fittings, and as there would be no load then there would be no volt drop.

This section of the circuit will suffer the maximum voltage drop on the circuit, the voltage drop on the branch sections is irrelevant in this example because the circuit should be designed to take into account all the volt drop along it's length.

This is why we use the equation VD=Mv/A/m

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)

 

[Monkeyblaine]

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.

 

[Outspoken]

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.

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.

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.

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.

Could you let me know how you would calculate volt drop where the circuit has differing sizes of cable.

Richard, I think you should be able to figure that out yourself.

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)

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.

 

[Richard Burns]

Oh dear.

Fair enough, let sleeping dogs lie.

 

[Outspoken]

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.

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.

 

[Monkeyblaine]

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.

 

[Outspoken]

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.

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!!

 

[Monkeyblaine]

Ah ok. Thank you. That makes sense now. Haha. Thanks for taking the time to see what I was talking about.