N
Nigel
Interesting thread but can I ask if they went through any of this during the actual course?
Briefly, but I’m just asking others to check what I’ve done.
Here is Question 4. Below are some of the circuits that have already been defined in the specification:
Pool Lighting: PVC Conduit 90°C thermosetting singles
We had some notes:
Pool Lighting: PVC Conduit 90°C thermosetting singles
- Circuit 1 - Main Pool Lights
- Circuit 2 - Main Pool Lights
- Circuit 3 - Changing Rooms, Pump Room and Plunge Pool
- Circuit 1 - Cafe Lights
- Circuit 2 - Cafe Lights
- Circuit 3 - Kitchen Lights
- Circuits 1 - Changing Room Sockets
- Circuit 2 - Pump Room Equipment ( 5-core 90°C SWA Clipped Direct)
- Circuit 1 - Dining Room Sockets
- Circuit 2 - Kitchen Fridges - 1 ( 2x 16A Circuits)
- Circuit 3 - Kitchen Fridges - 1 ( 2x 16A Circuits)
- Circuit 4 - General Kitchen Sockets
- Circuit 5 - Dishwasher ( 3-core 70°C SWA Clipped Direct)
- Circuit 6 - Food Holding Equipment ( 5-core 70°C SWA Clipped Direct)
- Circuit 7 - Fire Alarm 13A unswitched fuse spur connection unit will be required in the pool/cafe block to supply a repeater panel. To be installed by a specialist contractor.
- Circuit 8 - Vending Machines ( 2x 230v 1.2kW each)
We had some notes:
- Circuits must be grouped together.
- Pool - 3 Max circuits grouped together
- Cafe - 2 Max circuits grouped together
- Kitchen 4 Max circuits grouped together
- Pool - 40°C
- Cafe - 25°C
- Kitchen - 35°C
Main Pool Lights - Circuit 1
- Ib = 1.51A
- In = 6A
- Installation Method = 4B
- RF = Ca 0.91 Cg = 0.80
- Iz = 8.24A
- csa = 1mm² 13.5A VD 44
- Actual VD = 1.99v
- Max Vd = 6.9v
- Max Disconnection Time: 0.4 seconds
- Earth Fault Loop Impedance = 7.28Ω
- Maximum Earth Fault Loop Impedance
- Ze-Ze = R1+R2 = 7.28 - 0.11 = 7.17Ω
- Calculated Zs = 18.10 + 18.10 x 30 x 1.28/1000 = 0.71 +0.11 = 0.82Ω
Made a Mistake:
Changing Rooms, Pump Room, Plunge Pool Lights
- csa = 1mm² 17A VD 46
- Actual VD = 2.08v
- Ib = 1.51A
- In = 6A Type B breaker
- Installation Method = 4B
- RF = Ca 0.91 Cg = 0.80
- Iz = 8.24A
- csa = 1mm² 17A VD 46
- Actual VD = 2.08v
- Max Vd = 6.9v
- Max Disconnection Time: 0.4 seconds
- Earth Fault Loop Impedance = 7.28Ω
- Maximum Earth Fault Loop Impedance
- Zs-Ze = R1+R2 = 7.28 - 0.11 = 7.17Ω
- Calculated Zs = 18.10 + 18.10 x 30 x 1.28/1000 = 0.71 +0.11 = 0.82Ω
Changing Rooms, Pump Room, Plunge Pool Lights
- Ib = 0.45A
- In = 6A Type B breaker
- Installation Method = 4B
- RF = Ca 0.91 Cg = 0.80
- Iz = 8.24A
- csa = 1mm² 17A VD 46
- Actual VD = 2.08v
- Max Vd = 6.9v
- Max Disconnection Time: 0.4 seconds
- Earth Fault Loop Impedance = 7.28Ω
- Maximum Earth Fault Loop Impedance
- Zs-Ze = R1+R2 = 7.28 - 0.11 = 7.17Ω
- Calculated Zs = 18.10 + 18.10 x 30 x 1.28/1000 = 0.71 +0.11 = 0.82Ω
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Circuit 1 & 2 - Cafe Lights
- Ib = 0.27A
- In = 6A Type B breaker
- Installation Method = 4B
- RF = Ca 1.03 Cg = 0.80
- Iz = 7.28A
- csa = 1mm² 13.5A VD 44
- Actual VD = 0.35v
- Max Vd = 6.9v
- Max Disconnection Time: 0.4 seconds
- Earth Fault Loop Impedance = 7.28Ω
- Maximum Earth Fault Loop Impedance BS7671: 7.28Ω
- Zs-Ze = R1+R2 = 7.28 - 0.11 = 7.17Ω
- Calculated Zs = 18.10 + 18.10 x 30 x 1.20/1000 = 0.67 +0.11 = 0.78Ω
Kitchen Lights
- Ib = 3.91A
- In = 6A Type B breaker
- Installation Method = 4B
- RF = Ca 0.94 Cg = 0.80
- Iz = 7.97A
- csa = 1mm² 13.5A VD 44
- Actual VD = 5.16v
- Max Vd = 6.9v
- Max Disconnection Time: 0.4 seconds
- Earth Fault Loop Impedance = 7.28Ω
- Maximum Earth Fault Loop Impedance BS7671: 7.28Ω
- Zs-Ze = R1+R2 = 7.28 - 0.11 = 7.17Ω
- Calculated Zs = 18.10 + 18.10 x 30 x 1.20/1000 = 0.67 +0.11 = 0.78Ω
Pool Power PVC Conduit 90°C thermosetting singles
Circuits 1 - Changing Room Sockets - not sure whether to do a 20A radial.
Pump Room Equipment ( 32A 400v 5-core 90°C SWA Clipped Direct)
Circuits 1 - Changing Room Sockets - not sure whether to do a 20A radial.
Pump Room Equipment ( 32A 400v 5-core 90°C SWA Clipped Direct)
- Ib = 32A
- In = 32A Type B breaker
- Installation Method = 20C
- RF = Ca 0.91 Cg = 0.73
- Iz = 48.17A
- csa = 6mm² 53A VD 6.8
- Actual VD = 6.528v
- Max Vd = 6.9v
- Max Disconnection Time: 0.4 seconds
- Earth Fault Loop Impedance = 1.36Ω
- Maximum Earth Fault Loop Impedance BS7671: 1.37Ω
- Zs-Ze = R1+R2 = 1.37 - 0.11 = 1.26Ω
- Calculated Zs = 3.08 + 3.08 x 30 x 1.28/1000 = 1.18 +0.11 = 1.29Ω
It takes a long time to go through multiple questions and it can be hard to identify what numbers you have used and correlating the information that you have used with the results is another difficulty.
Note that when using 90°C cables that all the cable terminations and enclosures should be rated to 90°C as well, I am sure this is covered in your training and is taken account of for this project, but it is important to note that if it is not taken into account the cables should be downrated to 70°C.
Also note that 70°C cables are thermoplastic not thermosetting (probably just a typing error there)
Based on your current circuits I notice that every Zs calculation shows the circuit length as 30m, but then the results do not match with your answers.
For the lighting circuits not knowing the circuit length makes it impossible to calculate Volt drop, however you have three lighting circuits with different design currents and presumably different lengths and yet the calculated volt drop is the same in each case. The values also seem high unless you have very long circuits perhaps you could show your working there.
The pump room equipment appears to be a single circuit and presumably not grouped as it is clipped direct SWA yet you have a grouping factor of 0.73, I cannot work out where this is derived or why but you may have more information about it.
Check your maximum permitted volt drop for this circuit as well.
For installation method it is common convention just to use the letter e.g. C rather than including the installation method number e.g.20, however including the number does give more information so could well be useful.
Note that when using 90°C cables that all the cable terminations and enclosures should be rated to 90°C as well, I am sure this is covered in your training and is taken account of for this project, but it is important to note that if it is not taken into account the cables should be downrated to 70°C.
Also note that 70°C cables are thermoplastic not thermosetting (probably just a typing error there)
Based on your current circuits I notice that every Zs calculation shows the circuit length as 30m, but then the results do not match with your answers.
For the lighting circuits not knowing the circuit length makes it impossible to calculate Volt drop, however you have three lighting circuits with different design currents and presumably different lengths and yet the calculated volt drop is the same in each case. The values also seem high unless you have very long circuits perhaps you could show your working there.
The pump room equipment appears to be a single circuit and presumably not grouped as it is clipped direct SWA yet you have a grouping factor of 0.73, I cannot work out where this is derived or why but you may have more information about it.
Check your maximum permitted volt drop for this circuit as well.
For installation method it is common convention just to use the letter e.g. C rather than including the installation method number e.g.20, however including the number does give more information so could well be useful.
How we are to do the length of the circuit, is by measuring each circuit, we have a scale drawing 1:50, on A4, however that has been submitted for critique marking, As that was Q2 were we needed to draw the lighting circuits for each room, as I am only allowed to take certain questions home. I thought it best to use the maximum allowed which was 30M As we have to fill in a table at the very end in pencil, I can just change some of the calculations once Q2 which is the drawings has been marked and given back to me for any changes.
I will show you my workings for the Pool, Kitchen and Cafe Lighting, my apologies. My mistake with the SWA, I will adjust that accordingly. As its a single cable, there would be no grouping. I will also check the maximum permitted volt drop for the circuit also.
For the sockets in the changing rooms, I was thinking of putting a 20A radial socket as it seems that it would not require a 32A ring circuit, as it would only have items such as hair dryers etc being plugged in.
I will add my circuits now with corrections and also the workings out.
The scale of 1:50 means for every 1 it is 50 full scale. So 1cm on your plan equates to 50cm in reality.
Okay, I am just trying to work out what I could use as the maximum circuit length is 30m. As the spec states All circuits lengths to be estimated from drawing scales. For the purpose of the project, it no circuit should not exceed 30m.
Are we taking this course or are you?
I am just explaining where the 30M comes from. But I need a Triangle Ruler in order to measure the circuits, as you can't do it with a ruler. Sorry if I caused offence.
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2 cm is 100 mm is a workable way of doing it.
Westward10 has explained how to use a normal ruler to get your measurements.
If you have printed the diagrams at full size on A3 paper then they will be scaled 1:50 so if you measure 9 cm as the width of a room on the page then in reality the room will be 50 x 9 cm i.e. 450 cm or 4500 mm.
The picture is 50 times smaller than the real thing.
To get you started I have drawn out approximate dimensions on what I believe is the diagram you are using for the pool. If you can measure on your printed versions values to give similar results you should be able to work on from there for the rest of the building.
Because my measurements are manual on rough copies they could be a few mm out in measurement, for instance the front to back of the pool building is about 230 mm, but on measuring again it might be 231 mm this would change the size from the given 11500 mm to 11550 mm.
The purpose of the project is allow you the chance to learn how to do things like scale measurements and accurate circuits so that you can use this confidently in work at a later date; so it is good to practice now and get things correct, much better than getting things wrong on a real installation!
Westward10 has explained how to use a normal ruler to get your measurements.
If you have printed the diagrams at full size on A3 paper then they will be scaled 1:50 so if you measure 9 cm as the width of a room on the page then in reality the room will be 50 x 9 cm i.e. 450 cm or 4500 mm.
The picture is 50 times smaller than the real thing.
To get you started I have drawn out approximate dimensions on what I believe is the diagram you are using for the pool. If you can measure on your printed versions values to give similar results you should be able to work on from there for the rest of the building.
Because my measurements are manual on rough copies they could be a few mm out in measurement, for instance the front to back of the pool building is about 230 mm, but on measuring again it might be 231 mm this would change the size from the given 11500 mm to 11550 mm.
The purpose of the project is allow you the chance to learn how to do things like scale measurements and accurate circuits so that you can use this confidently in work at a later date; so it is good to practice now and get things correct, much better than getting things wrong on a real installation!
Bought a scaled ruler. I spoke to the tutor who said that I could use the ruler, or just wait until we transfer the drawing over to A3. I will have a go with the ruler. I did show him my Q5, but he said that I should just use 0.1 seconds in the equation. He also mentioned that my R1+R2 didn't seem quite right.
I need to use the one shown in the image.
I need to use the one shown in the image.
Attachments
The first equation I cannot follow. Surely it should be maximum R1+R2 (not r1+r2) per metre in milliohms.
The calculation is working backward to give you the figure to cross reference in table I1. Zs - Ze is the total circuit resistance, Ω, at normal operating temperature, dividing by length gives you resistance,Ω, per metre, dividing by the table I3 multiplier gives you resistance, Ω, per metre at 20°C and multiplying by 1000 gives the maximum resistance, mΩ, per metre at 20°C. So any value in the table lower than your result should be compliant.
The second equation is what you have already been doing to get your calculated Zs.
I have just realised your print outs you say are on A4 paper which if the original that fits on an A3 page has been scaled down to fit on an A4 page then the scale of the diagram will be smaller at 1:70.7, not a standard value!
If your tutor says use 0.1s then that is what you do. Did he specify what current to use?
Just be aware that the results will be incorrect for when you are calculating in real life, From your calculations with a 2090 A PEFC the minimum csa for the cpc would be 6mm², this would not go down well in designed a lighting circuit with a 6mm² cpc.
The calculation is working backward to give you the figure to cross reference in table I1. Zs - Ze is the total circuit resistance, Ω, at normal operating temperature, dividing by length gives you resistance,Ω, per metre, dividing by the table I3 multiplier gives you resistance, Ω, per metre at 20°C and multiplying by 1000 gives the maximum resistance, mΩ, per metre at 20°C. So any value in the table lower than your result should be compliant.
The second equation is what you have already been doing to get your calculated Zs.
I have just realised your print outs you say are on A4 paper which if the original that fits on an A3 page has been scaled down to fit on an A4 page then the scale of the diagram will be smaller at 1:70.7, not a standard value!
If your tutor says use 0.1s then that is what you do. Did he specify what current to use?
Just be aware that the results will be incorrect for when you are calculating in real life, From your calculations with a 2090 A PEFC the minimum csa for the cpc would be 6mm², this would not go down well in designed a lighting circuit with a 6mm² cpc.
So I had a go, based on those calculations:
- Ib 16A
- In 16A
- Method of Installation: C
- Rating Factor: 16/0.94x0.80 = 21.27A
- It 2.5mm2 24A VD 18
- Actual VD: 18X12X16/1000 = 3.456V
- Max VD: 11.5V
- Max R1+R2 2.73-0.11/1.2x12 = 181.94Ω
- Calculated R1+R2 7.41+7.41x1.2x12/1000 = 0.11Ω
- Calculated Zs 2.73x0.8 = 2.18+0.11 = 2.2Ω
- Fault Current: 230/2.2 = 104A
- S = Square Root: 104² x 0.1/115 = 0.28mm²
See my responses in red above
Thanks Richard. The maximum r1+r2 is a strange one. There is a calculation in the book
Method is B I must of read something else.
1.37-0.24/1.2*18* 1000 = 52.31m ohms/metre.
Mine for some reason is very high. Also the Calculated Zs should that be Ze 0.11 and R1 and R2 are 7.41 total is 14.93
Would that be correct?
Method is B I must of read something else.
1.37-0.24/1.2*18* 1000 = 52.31m ohms/metre.
Mine for some reason is very high. Also the Calculated Zs should that be Ze 0.11 and R1 and R2 are 7.41 total is 14.93
Would that be correct?
Come on, remember what units you are working in at each stage.
The calculation you show but do not explain seems to show
Zs - Ze (1.37-0.24)
Adjust for temperature (1.2) as a divisor goes from 70°C to 20°C
Circuit length m (18) as a divisor give a value per m
Ω to mΩ (1000) as a multiplier to give a result in mΩ/m @20°C
So take the BS7671 max Zs for your breaker subtract Ze divide by 1.2 divide by circuit length multiply by 1000 to give mΩ/m @20°C.
i.e. the number you would look up in table I1 for the 2.5 line 2.5 cpc row value to check if the cable you want to use has a lower resistance.
In your case the circuit is short and the cable is very much lower resistance than the 182mΩ/m that you would get so in that instance a cable could be 12 times longer (144m) before you came close to exceeding the maximum Zs, though the volt drop would limit the length of the circuit more.
For the calculated Zs=Ze +(R1+R2)
Your Ze is 0.11Ω, your calculated value of R1+R2 for the cable you have chosen is in your calculation list "Calculated R1+R2 = 7.41+7.41x1.2x12/1000 = 0.11Ω"
(The 7.41 value you give is the resistance in mΩ/m for a single of 2.5mm² not the total R1+R2 of the circuit)
You do need to look at what you are doing and understand why you are doing these things, what do the values mean, what units are they in, is this a sensible calculation that I am trying to do?
If you understand why and how your should do something then when faced with the unexpected you can use this understanding to apply the correct methods of calculating things.
The calculation you show but do not explain seems to show
Zs - Ze (1.37-0.24)
Adjust for temperature (1.2) as a divisor goes from 70°C to 20°C
Circuit length m (18) as a divisor give a value per m
Ω to mΩ (1000) as a multiplier to give a result in mΩ/m @20°C
So take the BS7671 max Zs for your breaker subtract Ze divide by 1.2 divide by circuit length multiply by 1000 to give mΩ/m @20°C.
i.e. the number you would look up in table I1 for the 2.5 line 2.5 cpc row value to check if the cable you want to use has a lower resistance.
In your case the circuit is short and the cable is very much lower resistance than the 182mΩ/m that you would get so in that instance a cable could be 12 times longer (144m) before you came close to exceeding the maximum Zs, though the volt drop would limit the length of the circuit more.
For the calculated Zs=Ze +(R1+R2)
Your Ze is 0.11Ω, your calculated value of R1+R2 for the cable you have chosen is in your calculation list "Calculated R1+R2 = 7.41+7.41x1.2x12/1000 = 0.11Ω"
(The 7.41 value you give is the resistance in mΩ/m for a single of 2.5mm² not the total R1+R2 of the circuit)
You do need to look at what you are doing and understand why you are doing these things, what do the values mean, what units are they in, is this a sensible calculation that I am trying to do?
If you understand why and how your should do something then when faced with the unexpected you can use this understanding to apply the correct methods of calculating things.
Thanks I will have another go. I understand what you mean.
The calculation above is out of a book we were given it’s to find the maximum R1+ R2 to determine the size of the CPC.
Calculation are
Maximum Zs -Ze / I3 factor x length overall x 1000
So you then add your Ze to that value to get your Zs.
Then use the adiabatic equation.
Fault Current: 230/Calculated Zs that will give you a result in A
Then apply the equation used
S = Square Root of I2 x t / k
I will have another go as I seem to have got mixed up somewhere with my calculations.
The calculation above is out of a book we were given it’s to find the maximum R1+ R2 to determine the size of the CPC.
Calculation are
Maximum Zs -Ze / I3 factor x length overall x 1000
So you then add your Ze to that value to get your Zs.
Then use the adiabatic equation.
Fault Current: 230/Calculated Zs that will give you a result in A
Then apply the equation used
S = Square Root of I2 x t / k
I will have another go as I seem to have got mixed up somewhere with my calculations.
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I’ve looked at the calculation again. It’s definitely correct as that’s what’s shown in the book. I don’t think you could break it down using BODMAS as it comes out the same. For the total R1+R2 with the line and CPC the same in table I1 shows 14.86.
I understand what’s going into the calculation, it just seems that it’s a very high number.
I will write the steps out again just so I’m doing it in the correct order the the correct values. Thanks for the help and advice.
I understand what’s going into the calculation, it just seems that it’s a very high number.
I will write the steps out again just so I’m doing it in the correct order the the correct values. Thanks for the help and advice.
I think you are missing a key point
Look at this calculation you typed
1.37-0.24/1.2*18* 1000 = 52.31m ohms/metre.
Maximum Zs -Ze / I3 factor x length overall x 1000
What is the (small but important) difference?
I can only work from what you post and if there are assumptions or specifics that are not mentioned or are changed I cannot work from these.
I have been trying to provide enough information to allow you to calculate the answers based on the normal approaches, however the training approach is not quite the same. Where things appear to vary from what I expect I have queried them or tried to work out what you have meant, not necessarily very successfully.
I have prepared the attached calculations that I think should provide the answers that you have been asked to find.
I am now falling behind on my paperwork so will have to reduce the responses I am making.
Look at this calculation you typed
- Max R1+R2 2.73-0.11/1.2x12 = 181.94Ω
1.37-0.24/1.2*18* 1000 = 52.31m ohms/metre.
Maximum Zs -Ze / I3 factor x length overall x 1000
What is the (small but important) difference?
I can only work from what you post and if there are assumptions or specifics that are not mentioned or are changed I cannot work from these.
I have been trying to provide enough information to allow you to calculate the answers based on the normal approaches, however the training approach is not quite the same. Where things appear to vary from what I expect I have queried them or tried to work out what you have meant, not necessarily very successfully.
I have prepared the attached calculations that I think should provide the answers that you have been asked to find.
I am now falling behind on my paperwork so will have to reduce the responses I am making.
Thanks Richard. The above has helped. In the calculation I posted above I forgot to include the x 1000 to get the m ohms per meter.
I also see how it’s broken. That’s much better and easier to follow. Is that a program you are using? Or is it something you have put together?
I also see how it’s broken. That’s much better and easier to follow. Is that a program you are using? Or is it something you have put together?
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Well done on the error, it is vital to be able to spot these (and correct them)
The diagram and equations are something I have just written/drawn out to provide a total overview.
The diagram and equations are something I have just written/drawn out to provide a total overview.
So we got our project back for some changes.
- I have said that I will use Type B MCB for most of my Lighting and Power Circuits. I need to say what MCB type to be used for the Fridges. As they are given in the spec as 16A so I would assume they would be on an isolator? As the feedback just says what types for the Fridges.
- I need to say where Functional Earthing would be found in the complex, and also the Minimum permissible CSA in BS7671. Looking at the spec the existing hotel is a TN-C-S System
- I also need to find the Main Protective Bonding where PME conditions apply.
From my point of view you have specified Type B for the MCBs for all circuits except the Fridges, therefore you need to specify what type MCB you will use for the Fridges.
The Fridges are quite large, it appears; as I said in an earlier post most commercial fridges would not exceed 500W (2.1A). However a fridge, since it is running a compressor motor, when started does have a high peak current. This current might be 10X the normal running current but for a fraction of a second. If you only had one fridge on each 16A radial circuit then the peak current of, say 21A, would not trip the 16A type B circuit breaker. If you had six fridges on one 16A circuit and they were all started simultaneously then it is possible the peak current could trip a 16A type B breaker and you may wish to select a type C breaker to ensure this high starting current did not trip the breaker.
Functional earthing is generally only found in things like PELV systems or RCDs with functional earthing conductors. Functional earths are an earth reference that is not intended to provide a safety function. If the conductor is both functional and protective then the CSA is determined by the protective part, if just functional then by the minimum CSA given in table 52.3 in section 524.
Main protective bonding is covered in section 544.
The Fridges are quite large, it appears; as I said in an earlier post most commercial fridges would not exceed 500W (2.1A). However a fridge, since it is running a compressor motor, when started does have a high peak current. This current might be 10X the normal running current but for a fraction of a second. If you only had one fridge on each 16A radial circuit then the peak current of, say 21A, would not trip the 16A type B circuit breaker. If you had six fridges on one 16A circuit and they were all started simultaneously then it is possible the peak current could trip a 16A type B breaker and you may wish to select a type C breaker to ensure this high starting current did not trip the breaker.
Functional earthing is generally only found in things like PELV systems or RCDs with functional earthing conductors. Functional earths are an earth reference that is not intended to provide a safety function. If the conductor is both functional and protective then the CSA is determined by the protective part, if just functional then by the minimum CSA given in table 52.3 in section 524.
Main protective bonding is covered in section 544.
Thanks for your help on that Richard. Much appreciated.
I have looked at the Main Protective Bonding, Supplementary Bonding, Earthing Conductor and CPC in the regs, I am just wondering if I am on the right lines:
- PME Conditions Apply - 25mm2
- Non PME Applications - 10mm2
- Without Mechanical Protection - 16mm2
- With Mechanical Protection - 16mm2
- TT Installation - 2.5mm2
- TN Installation - 6mm2
- Without Mechanical Protection - 4mm2
- With Mechanical Protection - 2.5mm2
BS7671 states that every (non protective bonding) protective conductor must be calculated from the adiabatic equation or selected from table 54.7. This is the basic requirement to meet.
Then there are also other limits applied for various circumstances:
For unenclosed single core copper cables: 2.5mm², or 4mm² if not mechanically protected (sheathed)
For buried earthing conductors a lot of different scenarios; the smallest csa permitted being 2.5mm².
For supplementary bonding conductors generally a minimum of 2.5mm², or 4mm² if not mechanically protected (sheathed) however where exposed conductive parts are connected the csa of the cpc(s) of the circuit supplying those parts must be taken into consideration when determining the minimum value.
Then for main protective bonding conductors the csa is determined by table 54.8 and reference to the csa of the incoming supply PEN conductor, where the supply is PME, the minimum csa being 10mm². Note that the earthing conductor on PME supplies also needs to meet these conditions.
Where the supply is not PME then the minimum csa is 6mm² and >= half the minimum required csa of the earthing conductor (not the actual installed csa of the earthing conductor).
Therefore I would suggest that you look again at the values you have given for the main protective bonding conductors.
Consider the practicality of using 16mm² cable for supplementary bonding and review the sizes you have given.
For the earthing conductor the TT size is correct.
Consider the requirements of 542.3.1 when giving the minimum csa for the earthing conductor on a TN system in general. Perhaps provide values for TNS and TNCS separately.
For cpc csa if the minimum is 2.5mm² then how can we use 1.0mm² twin and earth with a cpc size of 1.0mm² and 1.5 - 4 mm² twin and earth with a cpc size of 1.5mm²?
What conditions for the cpc allow this to occur?
The minimum sizes you have given would correctly apply for single core copper cables.
Then there are also other limits applied for various circumstances:
For unenclosed single core copper cables: 2.5mm², or 4mm² if not mechanically protected (sheathed)
For buried earthing conductors a lot of different scenarios; the smallest csa permitted being 2.5mm².
For supplementary bonding conductors generally a minimum of 2.5mm², or 4mm² if not mechanically protected (sheathed) however where exposed conductive parts are connected the csa of the cpc(s) of the circuit supplying those parts must be taken into consideration when determining the minimum value.
Then for main protective bonding conductors the csa is determined by table 54.8 and reference to the csa of the incoming supply PEN conductor, where the supply is PME, the minimum csa being 10mm². Note that the earthing conductor on PME supplies also needs to meet these conditions.
Where the supply is not PME then the minimum csa is 6mm² and >= half the minimum required csa of the earthing conductor (not the actual installed csa of the earthing conductor).
Therefore I would suggest that you look again at the values you have given for the main protective bonding conductors.
Consider the practicality of using 16mm² cable for supplementary bonding and review the sizes you have given.
For the earthing conductor the TT size is correct.
Consider the requirements of 542.3.1 when giving the minimum csa for the earthing conductor on a TN system in general. Perhaps provide values for TNS and TNCS separately.
For cpc csa if the minimum is 2.5mm² then how can we use 1.0mm² twin and earth with a cpc size of 1.0mm² and 1.5 - 4 mm² twin and earth with a cpc size of 1.5mm²?
What conditions for the cpc allow this to occur?
The minimum sizes you have given would correctly apply for single core copper cables.
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