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

 

[radiohead]

5% with the exception of lighting which is 3%

 

[Pretty Mouth]

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

Hi Mark. When you say the basic one, which one do you mean?

 

[Lucien Nunes]

For circuits smaller than 25mm², there really is only one equation:

V(drop) = [tabulated VD] * L * Ib / 1000

The tabulated voltage drop is in millivolts per amp per metre, which is a fixed value for a particular size and type of cable. Multiply by the length L to get the total drop per amp along the length of the cable, then multiply by the design current Ib which is the maximum current that will flow under normal operating conditions. Then, because the tabulated value is in millivolts per amp per metre, the result would be in millivolts, so divide by 1000 to get volts.

If you consider that the tabulated voltage drop in mV/A/m is more or less a representation of its resistance in ohms per metre (R=V/I) then the voltage drop equation is an application of ohm's law: V (drop) = I (load) * R (cable). So it follows that there can really only be one equation at least where resistance is the main cause of drop, as it is in smaller cables. Once you get into very high current AC circuits, the reactance starts to become a significant factor and you have to consider that separately to the resistance, but I believe that is beyond the scope of your exam questions.

I am not sure whether your questions involve a mixture of 3-phase and single-phase circuits, but care is needed when calculating the VD to the end of a single-phase circuit supplied by a 3-phase submain. For a balanced 3-phase load, the VD relates to the line-line voltage, not the line-neutral voltage. The easiest way to deal with this is to convert the drop in each section into a percentage and add. So a 3-phase submain with 6V drop contributes 6/400 * 100 = 1.5%, while a single-phase final with 5.75V drop contributes 5.75/230 * 100 = 2.5% These can be added to give 4% total drop.

 

[moose]

mV/A/m / Length / Amps div 1000

volt drop like 9.5v max. if its change let me know its all from memory.

not sure what you mean by 4. 4 is for ring mains.

R1 + R2 / 4 = ring/ all same at sockets.

 

[Lucien Nunes]

I was about to complain that @moose was breaking a number of laws of physics there, when I noticed some gobbledegook in my own post #4. Perhaps a mod can change it or let me edit the post: para 1, 'tabulated value is in millivolts per amp per metre,'

 

[westward10]

I was about to complain that @moose was breaking a number of laws of physics there, when I noticed some gobbledegook in my own post #4. Perhaps a mod can change it or let me edit the post: para 1, 'tabulated value is in millivolts per amp per metre,'

Edited.

 

[Markelec]

5% with the exception of lighting which is

For circuits smaller than 25mm², there really is only one equation:

V(drop) = [tabulated VD] * L * Ib / 1000

The tabulated voltage drop is in millivolts per amp per metre, which is a fixed value for a particular size and type of cable. Multiply by the length L to get the total drop per amp along the length of the cable, then multiply by the design current Ib which is the maximum current that will flow under normal operating conditions. Then, because the tabulated value is in millivolts per amp per metre, the result would be in millivolts, so divide by 1000 to get volts.

If you consider that the tabulated voltage drop in mV/A/m is more or less a representation of its resistance in ohms per metre (R=V/I) then the voltage drop equation is an application of ohm's law: V (drop) = I (load) * R (cable). So it follows that there can really only be one equation at least where resistance is the main cause of drop, as it is in smaller cables. Once you get into very high current AC circuits, the reactance starts to become a significant factor and you have to consider that separately to the resistance, but I believe that is beyond the scope of your exam questions.

I am not sure whether your questions involve a mixture of 3-phase and single-phase circuits, but care is needed when calculating the VD to the end of a single-phase circuit supplied by a 3-phase submain. For a balanced 3-phase load, the VD relates to the line-line voltage, not the line-neutral voltage. The easiest way to deal with this is to convert the drop in each section into a percentage and add. So a 3-phase submain with 6V drop contributes 6/400 * 100 = 1.5%, while a single-phase final with 5.75V drop contributes 5.75/230 * 100 = 2.5% These can be added to give 4% total drop.

I understand the millivolts times length times ib I get that but some calculations im looking at you apply the 1.2 factor some is just ohms law thats needed and I'm struggling to differciate between the 2 and aswell the transposing is a bit confusing

 

[Strima]

im looking at you apply the 1.2 factor

Edit: I'm talking out of my bum again, ignore me...

 

[Pretty Mouth]

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.

 

[Lucien Nunes]

I think @Pretty Mouth has hit on what the OP means by different formulas, i.e. calculating the VD from the tables at design stage, vs. verifying it during testing.

The method we've all given is convenient for calculating it at design stage where you know the length of the cable, so can easily use the VD figures from the table which apply to a cable under operating conditions.

If you are testing an installation and want to confirm the VD is in limits, you probably won't know the length of the circuit but you can measure its resistance, R1 + Rn. When you measure it, it's not energised, so the conductor will be at room temperature (typically 20°C) instead of maximum working temperature (typically 70°). If you apply ohms law to the measured resistance: VD = (R1 + Rn) * Ib you will get a figure for drop at room temperature, whereas it is necessary to check it would be OK at working temperature because the resistance of copper rises by about 0.39% per degree centigrade. This equates to a factor of 1.2 between room temp and working temp.

Multiplying the resistance measurement by 1.2 gives VD = (R1+Rn) * 1.2 * Ib

In a sense it is really the same calculation, you are just correcting the measurement for the fact that you cannot measure using a normal resistance meter while the circuit is loaded. Of course there might be situations in which the temperature correction factor of 1.2 does not apply. For example, a cable run in a cool location carrying only a small fraction of its rated Iz, might never exceed 30°C. A cable operating at a high temperature (MI in a blast furnace?) might need a greater correction factor.

 

[brianmoooore]

For the divide by 4 thing on a ring: You've measured the total resistance of the wire in the ring.
Taking a socket exactly half way around the ring, it has only half the total length of this wire going to it, so half the resistance = divide by two.
But, you also have two wires in parallel going to this socket, so to find the combined resistance of them, you divide by two again.
Divide by two, then divide by two = divide by four.

 

[telectrix]

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

I'm the opposite. I use mV/m x A x L to get VD in mV, e.g. 2340mV. then just add a dec.pt. after the "2". saves a fat finger error when using a calculator.

and i disagree with the 1.2. factor when the magnitude of the cable load will not take it anywhere near 70 deg. in some cases this could be a major saving in cable cost.

 

[Markelec]

I think @Pretty Mouth has hit on what the OP means by different formulas, i.e. calculating the VD from the tables at design stage, vs. verifying it during testing.

The method we've all given is convenient for calculating it at design stage where you know the length of the cable, so can easily use the VD figures from the table which apply to a cable under operating conditions.

If you are testing an installation and want to confirm the VD is in limits, you probably won't know the length of the circuit but you can measure its resistance, R1 + Rn. When you measure it, it's not energised, so the conductor will be at room temperature (typically 20°C) instead of maximum working temperature (typically 70°). If you apply ohms law to the measured resistance: VD = (R1 + Rn) * Ib you will get a figure for drop at room temperature, whereas it is necessary to check it would be OK at working temperature because the resistance of copper rises by about 0.39% per degree centigrade. This equates to a factor of 1.2 between room temp and working temp.

Multiplying the resistance measurement by 1.2 gives VD = (R1+Rn) * 1.2 * Ib

In a sense it is really the same calculation, you are just correcting the measurement for the fact that you cannot measure using a normal resistance meter while the circuit is loaded. Of course there might be situations in which the temperature correction factor of 1.2 does not apply. For example, a cable run in a cool location carrying only a small fraction of its rated Iz, might never exceed 30°C. A cable operating at a high temperature (MI in a blast furnace?) might need a greater correction factor.

Thank you very much what I'm gonna do im gonna go through past papers and apply this and see how I get on I will give feed back thanks

 

[Markelec]

I think @Pretty Mouth has hit on what the OP means by different formulas, i.e. calculating the VD from the tables at design stage, vs. verifying it during testing.

The method we've all given is convenient for calculating it at design stage where you know the length of the cable, so can easily use the VD figures from the table which apply to a cable under operating conditions.

If you are testing an installation and want to confirm the VD is in limits, you probably won't know the length of the circuit but you can measure its resistance, R1 + Rn. When you measure it, it's not energised, so the conductor will be at room temperature (typically 20°C) instead of maximum working temperature (typically 70°). If you apply ohms law to the measured resistance: VD = (R1 + Rn) * Ib you will get a figure for drop at room temperature, whereas it is necessary to check it would be OK at working temperature because the resistance of copper rises by about 0.39% per degree centigrade. This equates to a factor of 1.2 between room temp and working temp.

Multiplying the resistance measurement by 1.2 gives VD = (R1+Rn) * 1.2 * Ib

In a sense it is really the same calculation, you are just correcting the measurement for the fact that you cannot measure using a normal resistance meter while the circuit is loaded. Of course there might be situations in which the temperature correction factor of 1.2 does not apply. For example, a cable run in a cool location carrying only a small fraction of its rated Iz, might never exceed 30°C. A cable operating at a high temperature (MI in a blast furnace?) might need a greater correction factor.

Thanks another I will apply this through my revision and give you feed back I've got 10 days till test so fingers crossed

 

[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