2365 Design Project - May need some advice.

[RDB85]

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.

 

[Richard Burns]

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.

 

[RDB85]

He said to use 16A current as that’s what’s specified in the question. I will have another go with the new information. I will also measure the length. So it’s accurate.

 

[RDB85]

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

2.5mm² CPC is suitable.

 

[Richard Burns]

So I had a go, based on those calculations:

• Ib 16A YES
• In 16A YES
• Method of Installation: C Not known I thought cafe power was PVC singles in conduit (B) but fridges may be different.
• Rating Factor: 16/0.94x0.80 = 21.27A YES 35°C two grouped circuits
• It 2.5mm2 24A VD 18 Suitable YES
• Actual VD: 18X12X16/1000 = 3.456V For a 12m circuit YES
• Max VD: 11.5V Correct, non lighting circuit
• Max R1+R2 2.73-0.11/1.2x12 = 181.94Ω NO Maximum permissible R1+R2 for circuit at 20°C = Zs-Ze/1.2 =2.18Ω, useful to do a "sanity check" on your answers, you are working in the mΩ and low single figure Ω range for circuit resistances is 182Ω a likely result?
• Calculated R1+R2 7.41+7.41x1.2x12/1000 = 0.11Ω YES at normal operating temperature
• Calculated Zs 2.73x0.8 = 2.18+0.11 = 2.2Ω I would say 2.3Ω (round up) However Zs = Ze +R1+R2, you have used tabulated maximum Zs corrected to 20°C not the given Ze which I think was 0.11Ω, confusingly the same as your R1+R2.
• Fault Current: 230/2.2 = 104A Maths correct but these results are affected by the above paragraph
• S = Square Root: 104² x 0.1/115 = 0.28mm² Maths correct but these results are affected by the above paragraph

2.5mm² CPC is suitable.

See my responses in red above

 

[RDB85]

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?

 

[Richard Burns]

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.

 

[RDB85]

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.

 

[RDB85]

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.

 

[RDB85]

Having another go:

It just doesn’t seem right when doing that calculation. As you say once you get that value you then look up in Table I1 which for 2.5 line and 2.5 CPC is 14.86

Which then would be

14.86x1.2x12/1000 = 0.213 ohms.

 

[Richard Burns]

I think you are missing a key point
Look at this calculation you typed

• Max R1+R2 2.73-0.11/1.2x12 = 181.94Ω

And compare it to these that 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.

 

[RDB85]

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?

 

[Richard Burns]

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.

 

[RDB85]

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.

 

[Richard Burns]

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.