2396 style question, 2400, short circuit, RCBO,

[Pete999]

You basically need to work out your R1, R2 and Rn. Cable resistances are in the on site guide.

For short circuit protection you need to work out the PSSC using R1 and Rn. You have a PFC of 2.1Ka and you know the voltage so can work out the external Zn. Add them up, then look up the trip curves for the device and see if it will disconnect in time.
You mean r1,r2 and rn don't you Strima?

Bear in mind the final answer may not need to be correct, just your method of working it out.

 

[Pete999]

You basically need to work out your R1, R2 and Rn. Cable resistances are in the on site guide.

For short circuit protection you need to work out the PSSC using R1 and Rn. You have a PFC of 2.1Ka and you know the voltage so can work out the external Zn. Add them up, then look up the trip curves for the device and see if it will disconnect in time.
You mean r1,r2 and rn don't you Strima? may get marked down using the wrong terminoogy, or am I barking up the wrong tree?

Bear in mind the final answer may not need to be correct, just your method of working it out.

 

[Richard Burns]

You mean r1,r2 and rn don't you Strima? may get marked down using the wrong terminology, or am I barking up the wrong tree?

I think here the calculations are for a radial circuit but going into more detail than usual; so using the capital letters for the conductor resistances, as we are not considering a ring final circuit.

 

[Pete999]

I think here the calculations are for a radial circuit but going into more detail than usual; so using the capital letters for the conductor resistances, as we are not considering a ring final circuit.

I see, still somewhat confusing imo, not doubting your thoughts, just find them confusing. My reason being that r1,r2 and rn are end to end resistance readings, I know on the test certificates the lower case r designation figures are for RFCs only, if you were testing a radial you would use the terms R1+R2, or R1+RN etc wouldn't you? perhaps someone with a better insight could clarify.

 

[Strima]

You're correct Pete, a small oversight on my behalf coupled with a skin full of beer...

And TBH I still get confused what to use unless I'm working on a ring final.

I've now gone cross eyed thinking about it...

 

[Richard Burns]

Determine Earth Fault Loop Impedance (Zs = Ze + (R1 + R2) × Cr)

R1 = 0.387 (mΩ/m) From the On-Site Guide table I1,
R2 = From BS7671 table 54.7 Minimum protective conductor can be selected as the line conductor is 50mm² then the equation is S ∕ 2
∴ 50mm² ∕ 2 = 25mm² the next greater cable CSA is 35mm²
∴ R2 = 0.524(Ω/km) From the On-Site Guide table I1.

Ze = 0.08Ω, Cr = 1.28 From the On-Site Guide table I3.

∴ Zs = 0.08 + ((0.387 × 0.5) + (0.524 × 1) ∕ 1000) × 1.28 = 0.081

The maximum Earth Loop Fault Impedance for the Over Current Protective Device BS-88-2 160A is 0.27Ω From BS7671 table 41.4
so satisfactory
Determine line Fault Loop Impedance (Zn = Zne + (R1 + Rn)

Zne = Uo × Cmax ∕ Pscc ∴ Zne = 230 × 1.1 ∕ 5000 = 0.0506

Determine thermal compliance for short circuit

I²t ≤ K²S². From BS7671 table 43.1 K = 143, from the manufactures data t = 0.01

∴ 4962² × 0.01 = 246217 A²s ≤ 143² × 50² = 51122500

My question is this,
how should i end a cable calculation?
Do i need to work out short circuit thermal compliance or do i stop once i have ensured that the maximum fault current for the device has been confirmed not to exceed the rated?
Have i correctly work out my I²t or should i quote the manufactures data?

Regards

Dan

Dan
I have reduced your post in the quote to refer to some items on the calculations.
For the EFLI you have a given Zs as Zs = (Ze + (R1 + R2)) × Cr however you have not included Cmin.
The correction factor of 1.28 is for 90°C cables which also require that all terminations of those cables are suitable for that temperature, which is often not the case, so the limiting temperature would be 70°C in most cases and so a 1.20 correction factor. You have also used k as 187 elsewhere which is for 90°C as well. This may well be right but it is worth checking.
You have checked S/2 and got 25mm², so 25mm² cable would be OK, there is no need to go up a cable size unless the value exceeds 25mm².
I do not recognise Zn and Zne though they are identified by context but just seem odd in referring only to perhaps Neutral (n) and earth (ne).
You presumably have the PSCC (5000A) from your information.
A²s seems to jump in to the equation looking like a paste error.

Right now to the questions
I²t should be taken from manufactures data as the devices are current limiting and so the actual let through energy may be lower than a calculated value.
In practical terms if the fault current does not exceed the rating of the protective device then this is usually enough for installation calculations unless there is a reason to expect problems.

Check Volt drop, Zs, PSCC and if all is OK then installation can proceed.

 

[Daniel Jones]

Richard thanks for your response, i have revised the calculation and included the full calc

this is at the origin of the installation, and the PSCC in 5KA, as the devices are 88 at 80KA and 60898 at 6KA do i need to determine short circuti current?

Main Switch to CCU

the design current of the circuit is 100A

As the circuit is greater than 32A and is a Distribution circuit the maximum disconnection time will be 5 secs as regulation 411.3.2.3 permits

we are installing to a Busbar chamber with fused switches feeding the existing Distribution circuit’s, so the installation method is to be single core cable installed from Main switch to bus bar chamber hence reference method B.

The cable will be group with no other circuits and in an ambient temperature of 20 degrees, The Over current protective device will be an BS-88-2

Over Current Protective Device Selection (In ≥ Ib) ∴ In = 125A

Cable Selection (It = In∕ Correction Factors) ∴ no correction factors ∴ It = 125A

From BS7671 table 4D1A column 5 A 50mm² has a current carrying capacity of 134A so satisfactory

Volt Drop (Vd = mV/A/m × Ib × Length)

From BS7671 table 4D1B column 6 a 50mm2 single core cable has an mV/A/m of 0.85

∴ 0.85 × 100.41 × 0.5 ∕ 1000 = 0.043V ∴ Phase Vd = 0.043 ∕ 1.732 = 0.025V

Volt Drop Percentage 0.043 ∕ 400 × 100 = 0.010% so this below the 5% allowed so satisfactory

Determine Earth Fault Loop Impedance (Zs = Ze + (R1 + R2) × Cf)

R1 = 0.387 (mΩ/m) From the On-Site Guide table I1, as the R2 has not been determined I have selected it from BS7671 table 54.7 Minimum protective conductor can be selected as the line conductor is 50mm² then the equation is S ∕ 2 ∴ 50mm² ∕ 2 = 25mm² ∴ R2 = 0.727(Ω/km) From the On-Site Guide table I1.

Ze = 0.08Ω From Technical Specification, Cf = 1.20 From the On-Site Guide table I3.

∴ Zs = 0.08 + ((0.387 × 0.5) + (0.727 × 1) ∕ 1000) × 1.20 = 0.081

The maximum Earth Loop Fault Impedance for the Over Current Protective Device BS-88-2 160A is 0.27Ω From BS7671 table 41.4 so satisfactory

Determine Earth Current, Ief = (Uo × Cmin ∕ Zs) ∴ 230 × 0.95 ∕ 0.081 = 2698A

Determine Minimum Circuit Protective Conductor, (S= √ (Ief² × t) ∕ K)

t = the time it will take the device to disconnect at the fault current from manufacture’s time current curves

K = a factor taking in account of the resistivity, temperature coefficient and conductor material from BS7671 table 54.4

S= √ (2698² × 0.013) ∕ 143 = 2.15mm² As the CPC cable is a separate 25mm² single core cable this is adequate.

Determine line Fault Loop Impedance (Z neutral = Z neutral external + (R1 + Rn)

∴ Z neutral external = Uo × Cmax ∕ Pscc ∴ Z neutral external = 230 × 1.1 ∕ 5000 = 0.0506

As the neutral is the same size conductor as the line conductor the R1 + Rn = 0.774(mΩ/m)

∴ Z neutral = 0.0506 + (0.774) × 0.5 ∕ 1000) = 0.050987

Determine Short Circuit Current (Uo × Cmax ∕ Z neutral) ∴ 230 × 1.1 ∕ 0.050987 = 4962A

Determine thermal compliance for short circuit

I²t ≤ K²S², From BS7671 table 43.1 K = 115, from the manufactures data the I²t = 168000 (A²s)

∴ 168000 (A²s) ≤ 115² × 50² = 33062500 (A²s)

Therefore as the energy let through of the devices is let than that the energy withstand of the conductor compliance is met.

The maximum breaking capacity of the device is 80Ka so satisfactory and the circuit meets the requirements of BS7671-2015

Regards

Dan

 

[Richard Burns]

Dan
Right this is looking much better.
These calculations are the calculations that you would perform when designing a new electrical installation and cover all the calculations required.
For a simpler installation then some calculations could be omitted, because of the section you are designing here then consideration of short circuit current would be relevant to ensure that the cable can withstand this, though in practice most circuits after the main incomer will be able to withstand the fault current and in your case as the fault current is only 5kA this should not present any problems.

In all cases I have not actually checked the mathematical numerical calculations are correct only the method used.

Cable selection: correct no problems there you have considered the possible issues and selected the correct cable for the situation described.

Volt drop: correct method of calculation, though I have not checked the maths itself. Note that whilst the volt drop is compliant at this point it is important to remember that the volt drop that will be relevant will be the drop from supply to the end of the final circuit so compliance at this point in the installation is almost a given (or should be!).
I would not normally calculate the volt drop at this point expect to add it to the following distribution and final circuits to ensure their compliance.

Earth fault loop impedance: the formula is correct, the selection of R2 from 54.7 is OK (you can also do the calculation instead), the correction factor is correct assuming the single core cables are bunched together.
I am not sure why you have halved the 0.387Ω the value in I3 is for a single conductor which is what you have.
Zs is compliant. (though again more relevant further on in the installation)

Earth fault current: Correct method, subject to Zs being correct.

Minimum pc size: formula is correct, the value of K is correct but would come from table 54.2, as a single conductor not bunched.

Line fault loop impedance (thank you for clarifying the n ne references): the formula is correct, again I do not know why you are halving the resistance.

Short circuit current: maximum short circuit current will be at origin so no need to use R1 + Rn in the impedance value just 0.0506Ω, this would back confirm your 5000A from the technical specification. (circular calculation)

Thermal compliance: No problems there.
Notation wise you are suddenly unexpectedly including the units, A²s.

Good work.

 

[Daniel Jones]

Thank you for all your help Richard.

The reason i have halved the 0.387Ω is because the circuit conductor is only 0.5m long.
A²s i think is measurement of let through energy.

Dan

 

[Daniel Jones]

Sorry another New Question,

I am trying to work out if the CPC is adequate but get some conflicting information.

I was always taught that if the fault current causes instantaneous operation then use 0.1 in the adiabatic equation.

However, I’ve been reading up and regulation 434.5.2 states that for a fault of very short duration less than 0.1 the I²t for that class of device should be less than that of the K²S² of the cable.

so....

Option 1 S = √ (I² × t) ∕ K)

S= √ (96.7² × 0.1) ∕ 115 = 0.26mm²

Therefore, the circuit has a 1mm² CPC so is adequate.

Option 2 S = √ (let through energy (I²t)) ∕ K

So, you go to manufacture’s data (Schneider electrical Acti 9 pg 335/ (11/21)) and I can’t understand the chart, so I went to Beama guide and they give you a max value of 35000A²s

S = √(35000A²s) ∕ 115 = 1.62mm²

Therefore, the circuit has a 1mm² so is NOT adequate.

In the electrical installation design guide (pg 94) other manufactures are quote as having a lower let through energy which would make it compliant for instance.

Hager MCB

S = √(13000A²s) ∕ 115 = 0.99mm²

Question: which equation should I be following? or does anyone know the let through energy for a 6A Acti 9 iC60H MCB Type B or can explain the charts.

Thanks

Dan

 

[Pete999]

Sorry another New Question,

I am trying to work out if the CPC is adequate but get some conflicting information.

I was always taught that if the fault current causes instantaneous operation then use 0.1 in the adiabatic equation.

However, I’ve been reading up and regulation 434.5.2 states that for a fault of very short duration less than 0.1 the I²t for that class of device should be less than that of the K²S² of the cable.

so....

Option 1 S = √ (I² × t) ∕ K)

S= √ (96.7² × 0.1) ∕ 115 = 0.26mm²

Therefore, the circuit has a 1mm² CPC so is adequate.

Option 2 S = √ (let through energy (I²t)) ∕ K

So, you go to manufacture’s data (Schneider electrical Acti 9 pg 335/ (11/21)) and I can’t understand the chart, so I went to Beama guide and they give you a max value of 35000A²s

S = √(35000A²s) ∕ 115 = 1.62mm²

Therefore, the circuit has a 1mm² so is NOT adequate.

In the electrical installation design guide (pg 94) other manufactures are quote as having a lower let through energy which would make it compliant for instance.

Hager MCB

S = √(13000A²s) ∕ 115 = 0.99mm²

Question: which equation should I be following? or does anyone know the let through energy for a 6A Acti 9 iC60H MCB Type B or can explain the charts.

Thanks

Dan

In the equation marked in red where is the t element? could be barking up the wrong tree here as it's Saturday and the bar is open.

 

[Daniel Jones]

In the equation marked in red where is the t element? could be barking up the wrong tree here as it's Saturday and the bar is open.

35000 is the I2t (let through energy) from the Beama guide to low voltage switches for a BS60898

 

[Daniel Jones]

Ok, so been reading all week and still can’t get my head around this.

can someone please confirm my thinking on this.

Sizing of CPC with an MCB where the disconnection time is less than 0.1 sec.

When the disconnection times for an MCB is less than 0.1 sec (instantaneous) you can no longer use the time/current characteristics charts.

So you calculate that the device meets the maximum earth loop fault impedance are the end of the circuit.

Determine Earth Fault Loop Impedance (Zs = Zdb + (R1 + R2) × Cf)

From the On-Site Guide table I1 combined R1 + R2 for a 1.5mm² with a 1mm² circuit protective conductor has a mΩ/m of 30.20

Zdb = 0.094 From DB Submain Calculation, Cf = 1.20 From the On-Site Guide table I3

∴ Zs = 0.094 + ((30.20) × 50m ∕ 1000) × 1.20 = 1.90Ω

The maximum Earth Loop Fault Impedance for the Over Current Protective Device is 7.28Ω From BS7671 table 41.3 so compliant

To determine the minimum circuit protective conductor, (S= √ (Ief² × t) ∕ K)

t = the time it will take the device to disconnect at the fault current from BS7671 Fig 3A4 But because the fault current at the end of the circuit will cause the device to operate in less than 0.1 sec you have to use the energy let through for that specific manufacture

K = a factor taking in account of the resistivity, temperature coefficient and conductor material from BS7671 table 54.3

To calculate the energy let through for Schneider you need the need the minimum earth fault current. Which will be at the origin of the circuit e.g. the sub DB. So using the Zs at the DB (Zdb) you calculate the fault current,

Ief = Uo × Cmin ∕ Zdb ∴ Ief = 230 × 0.95 ∕ 0.094 = 2324A

You then go to the Schneider data sheets and look at the charts, from there a 6A MCB with a fault current of 2324A has a let through energy of 1500A²s

Check if k²s² ≥ I²t

The energy withstand of the CPC is k²s² = 115² × 1² = 133225A²s

Therefore as the CPC withstand is less than that of the let through energy the 1mm² is adequate.

regards

Dan

 

[Pete999]

Ok, so been reading all week and still can’t get my head around this.

can someone please confirm my thinking on this.

Sizing of CPC with an MCB where the disconnection time is less than 0.1 sec.

When the disconnection times for an MCB is less than 0.1 sec (instantaneous) you can no longer use the time/current characteristics charts.

So you calculate that the device meets the maximum earth loop fault impedance are the end of the circuit.

Determine Earth Fault Loop Impedance (Zs = Zdb + (R1 + R2) × Cf)

From the On-Site Guide table I1 combined R1 + R2 for a 1.5mm² with a 1mm² circuit protective conductor has a mΩ/m of 30.20

Zdb = 0.094 From DB Submain Calculation, Cf = 1.20 From the On-Site Guide table I3

∴ Zs = 0.094 + ((30.20) × 50m ∕ 1000) × 1.20 = 1.90Ω

The maximum Earth Loop Fault Impedance for the Over Current Protective Device is 7.28Ω From BS7671 table 41.3 so compliant

To determine the minimum circuit protective conductor, (S= √ (Ief² × t) ∕ K)

t = the time it will take the device to disconnect at the fault current from BS7671 Fig 3A4 But because the fault current at the end of the circuit will cause the device to operate in less than 0.1 sec you have to use the energy let through for that specific manufacture

K = a factor taking in account of the resistivity, temperature coefficient and conductor material from BS7671 table 54.3

To calculate the energy let through for Schneider you need the need the minimum earth fault current. Which will be at the origin of the circuit e.g. the sub DB. So using the Zs at the DB (Zdb) you calculate the fault current,

Ief = Uo × Cmin ∕ Zdb ∴ Ief = 230 × 0.95 ∕ 0.094 = 2324A

You then go to the Schneider data sheets and look at the charts, from there a 6A MCB with a fault current of 2324A has a let through energy of 1500A²s

Check if k²s² ≥ I²t

The energy withstand of the CPC is k²s² = 115² × 1² = 133225A²s

Therefore as the CPC withstand is less than that of the let through energy the 1mm² is adequate.

regards

Dan

Ok, so been reading all week and still can’t get my head around this.

can someone please confirm my thinking on this.

Sizing of CPC with an MCB where the disconnection time is less than 0.1 sec.

When the disconnection times for an MCB is less than 0.1 sec (instantaneous) you can no longer use the time/current characteristics charts.

So you calculate that the device meets the maximum earth loop fault impedance are the end of the circuit.

Determine Earth Fault Loop Impedance (Zs = Zdb + (R1 + R2) × Cf)

From the On-Site Guide table I1 combined R1 + R2 for a 1.5mm² with a 1mm² circuit protective conductor has a mΩ/m of 30.20

Zdb = 0.094 From DB Submain Calculation, Cf = 1.20 From the On-Site Guide table I3

∴ Zs = 0.094 + ((30.20) × 50m ∕ 1000) × 1.20 = 1.90Ω

The maximum Earth Loop Fault Impedance for the Over Current Protective Device is 7.28Ω From BS7671 table 41.3 so compliant

To determine the minimum circuit protective conductor, (S= √ (Ief² × t) ∕ K)

t = the time it will take the device to disconnect at the fault current from BS7671 Fig 3A4 But because the fault current at the end of the circuit will cause the device to operate in less than 0.1 sec you have to use the energy let through for that specific manufacture

K = a factor taking in account of the resistivity, temperature coefficient and conductor material from BS7671 table 54.3

To calculate the energy let through for Schneider you need the need the minimum earth fault current. Which will be at the origin of the circuit e.g. the sub DB. So using the Zs at the DB (Zdb) you calculate the fault current,

Ief = Uo × Cmin ∕ Zdb ∴ Ief = 230 × 0.95 ∕ 0.094 = 2324A

You then go to the Schneider data sheets and look at the charts, from there a 6A MCB with a fault current of 2324A has a let through energy of 1500A²s

Check if k²s² ≥ I²t

The energy withstand of the CPC is k²s² = 115² × 1² = 133225A²s

Therefore as the CPC withstand is less than that of the let through energy the 1mm² is adequate.

regards

Dan

Dan, thanks for your post, but it's Friday for goodness sake, you can't really expect a sensible anwser

 

[Daniel Jones]

Dan, thanks for your post, but it's Friday for goodness sake, you can't really expect a sensible anwser

lol. I thought i might have been a big ask, only just recovered from last weekend myself.