Voltage drop/2391

[Markelec]

Hello im currently doing the 2391 online I have failed 4 times but I realise there are variations to the voltage drop calculation could someone help me to know what the variations are to the voltage drop calculation I know the basic 1 thats fine but its the other ways for instance multiplying or dividing by 4 I have seen

 

[Markelec]

I'll try to explain as I understand it:

At the origin of a single phase installation, you have a Line at 230V, and a Neutral at 0V - a potential difference of 230V.

Let's build a simple circuit - 10m of 2.5mm² T+E feeding a heater pulling 13A. That 13A will flow from the origin, through 10m of line conductor, through the heater, through 10m of neutral conductor back to the origin.

The bulk of the circuit resistance is in the heater, so the bulk of the potential difference will be across the heater. But the circuit conductors have a small resistance too, so there will be a small potential difference across the 10m of line conductor, and across the 10m of neutral conductor. Added together, this potential difference is the 'voltage drop'.

From the OSG, p196;
1m of 2.5mm² copper conductor has a resistance of 0.00741ohms at 20 deg C.
Therefore 20m would be 0.00741 X 20 = 0.1482ohms at 20 deg C.

The resistance of a conductor increases as it gets warmer, so we want to calculate the voltage drop at the maximum operating temperature of the conductor, when the resistance, and therefore voltage drop would be at its greatest. For T+E this is 70 deg C. A conductor at 70 deg C will have roughly 1.2 times the resistance of a conductor at 20 deg C.

So, 0.1482ohms X 1.2 = 0.17784ohms at 70 deg C.

V=IR
V = 13 X 0.17784
V = 2.31192V This is your voltage drop for the circuit.

Let's check that it tallies with the other method. From OSG P161:

1m 2.5mm² T+E has a VD of 0.018V/A/m. That's the VD for both line and neutral conductors, at 70 deg C.

0.018 X 13 X 10 = 2.34V/A/m

More or less the same, a very slight difference I expect due to rounding.

Note that I prefer to always work with volts rather than millivolts, which i think can confuse calculations sometimes.

So if the rn is not available the resistance and you have the r1 how do you find it or vice versa say you have to size of r1 but not rn say

 

[Markelec]

So if the rn is not available the resistance and you have the r1 how do you find it or vice versa say you have to size of r1 but not rn say

So meaning when you go to look at the table for copper resistance and you have one but not the other how do you find it to complete resistance in calculation

 

[Pretty Mouth]

So meaning when you go to look at the table for copper resistance and you have one but not the other how do you find it to complete resistance in calculation

For the example I gave, I used the data from the row saying "line conductor 2.5mm², protective conductor - ". This just gives you the resistance of 1m of a single 2.5mm² conductor. We were using 10m of twin and earth, which is 20m of conductor. Hence why I used 20m for the calculation.

Another way would be to use the data from the column "line conductor 2.5mm², protective conductor 2.5mm²" . This is the same as the resistance of 1m of 2 X 2.5mm² conductors in series, ie 1m of line conductor plus 1m of neutral conductor. Had I used this, I would have used 10m for the calculation instead.

 

[Markelec]

So this would be as if both cpc and line conductors are the same Size but sometimes its not the case

 

[Markelec]

And I'm seeing formulas talking about using ratio to find resistance if you are missing or sizes are different

 

[Pretty Mouth]

So this would be as if both cpc and line conductors are the same Size but sometimes its not the case

CPC, line conductor, neutral conductor, it doesn't matter what they're called in the table. They're all copper, and they all have the same resistance per meter for a given cross sectional area. 1 meter of 2.5mm² twin and earth is 2 meters of live conductor: 1 meter of 2.5mm² line and 1 meter of 2.5mm² neutral.

Read back over my post carefully, it should explain things

 

[Zerax]

To the OP: Focus on understanding the principle.. NOT a formula. The OSG & BS7671 & all the IET books that I've looked at... try to make things easier by giving you the same thing in a different way. But IMHO this just makes things far more confusing...

I think all the standard publications could be made half the size they are if they removed all the variations. Oh, and added page numbers to the index !!

 

[Pretty Mouth]

To the OP: Focus on understanding the principle.. NOT a formula. The OSG & BS7671 & all the IET books that I've looked at... try to make things easier by giving you the same thing in a different way. But IMHO this just makes things far more confusing...

Completely agree with this. It's messy, and a nightmare when trying to learn. Eg the table on p196 should have just 3 columns: conductor CSA, resistance/m copper, resistance/m Aluminium. No need to mention line conductor, protective conductor, R1 or R2.

 

[Lister1987]

VD itself is as we know mV/A/m * Ib * L

Max VD is either 6.9v (3%) or 11.5v (5%) for lighting & power respectively

We can rearrange the formula (using Max & Actual VD) to find Max circuit length for the calculated VD



Let's say we have a 6 mm² T&E with pvc insulated and sheathed cable feeding a a lighting circuit which has a mV/A/m value of 7.3. The actual circuit current is 12.5 A, and the length of run is 14 m.
Volt drop = 7.3 x 12.5 x 14 / 1000
= 1.28 V


If a 14 m run gives a volt drop of 1.28 V,
the length of run for a 11.5 V drop will be:
11.5 x 14m = 105m

Completely agree with this. It's messy, and a nightmare when trying to learn. Eg the table on p196 should have just 3 columns: conductor CSA, resistance/m copper, resistance/m Aluminium. No need to mention line conductor, protective conductor, R1 or R2.

I'd drop the individual rows for line and CPC, as you say you just want a value for 2.5 and 1.5, doesn't matter if it's line neutral or CPC.

I disagree on dropping R1+R2, I find that useful for determining the length of a circuit (or an implausibly low/high reading), dividing my R1+R2 value by the listed value

 

[Pretty Mouth]

I'd drop the individual rows for line and CPC, as you say you just want a value for 2.5 and 1.5, doesn't matter if it's line neutral or CPC.

I disagree on dropping R1+R2, I find that useful for determining the length of a circuit (or an implausibly low/high reading), dividing my R1+R2 value by the listed value

Problem is, you need both line and CPC columns present in order to have an R1+R2 column. Can't drop one without dropping the other!

Personally I think it clutters the mind when learning. And it's not difficult to calculate (R1+R2)/m if you have the resistance/m of each of the conductors: just add them together.

I would have the resistances in ohms/m as well. It would remove the annoying step of dividing by 1000 to convert from milli ohms to ohms.

 

[AJshep]

The other basic method of working out the voltdrop they tend to use on the 2391 is using ohms law, I believe they use this to test the candidates basic knowledge

You would need to use your R1+RN mutipled by the design current and then mutilpying by 1.2 to allow for the circuit to be running at 70°c

A basic question would be
"Whats the volt drop on a 13 amp immersion circuit with an R1+RN of 0.30 ?"

A=4.68v
B=3.59v
C=4.45v
D=5.68v

Answer
13x0.3x(1.2) = 4.68v