2396 style question, 2400, short circuit, RCBO,

[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.

 

[Richard Burns]

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

This is a non valid short cut route to checking, often used but not really giving a reasonable answer.

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.

You have this correct as far as I am aware.
The class of device has limits in the product standard but manufacturers try to get better and more useful products so the actual device will likely be (much) better than the standard limit.

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.

Follow the second equation using I²t. You seem to have worked out the charts, very similar in design to the tripping times charts, logarithmic scale.

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

Yes no problems there. Easier to follow now that you are using the units.

To determine the minimum circuit protective conductor, (S= √ (Ief² × t) ∕ k)
k = a factor taking in account of the resistivity, temperature coefficient and conductor material from BS7671 table 54.3

remember that k and K may be different items. When writing always make sure you use the correct case.
Similarly below with S (csa) and s (seconds).
Even though they are the same number, technically you are calculating the fault current withstand and should use table 43.1 for the value of k rather than 54.3, do not worry about it.

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
Dan

Obviously for calculating the energy withstand you need the worst case which is the highest possible value of fault current, not the minimum. This is, as you have stated, at the protective device at the origin of the circuit. You would then use Cmax in the equation rather than Cmin.
I have different charts from Schneider that that give an energy let through value of about 2400 for a 6A IC60 acti9 MCB at an Ief of 2300A. But this is five times less than the energy withstand of the cable so well within the limits.

The end result however you have correctly determined.
The cpc is protected from the fault current because of the energy limiting of the protective device, which may not be the case for other manufactures devices if they had less energy limiting designs.

It sounds like it is all coming together well.
(You will probably never use this after the learning but it is good to know!)
Lost the post there at the end so I hope it comes out OK.

 

[Daniel Jones]

Thank you,

That makes everything crystal clearer.

still have a problem with the logarithmic scale for schneider.

I have downloaded the brochure.

but there are 4 charts for the IC60H 2 on page 335 and 2 on page 338. i have assumed i am to use the one with the A2s but what are the others for?

am i using the correct scale?

Dan

 

[Daniel Jones]

expanded version of the charts

 

[Richard Burns]

OK the header indicates that the second set of charts includes 3P+N, not sure why this is added, on the document I have only the first two charts.
The two thermal stress charts are the same the two peak current charts just have a different origin.
I would use the first set of charts.
The first chart just tells you the maximum current that can flow in a fault situation with that circuit breaker.
The second chart is the one you have correctly identified as the I²t chart that you need to identify the energy let through.
From the expanded chart you seem to be using it correctly.
Going along the bottom row each block of bold lines is subdivided by the finer lines in equal range of values (not equal spread across the graph).
So for 0.1 to 1 each finer line is a range of 0.1, so 0.1-0.2, 0.2-0.3, etc.
At the next line from 1 to 10 each finer line is a range of 1 and so on.
I have attached the graph with arrows showing the values mentioned in your answer in post #28 and how I derived 2400 A²s as the I²t value that we want.
Fault current of 2324A is 2.3kA so one third (ish) of the way from 2 to 3 on the bottom line. Move up from there to intersect the line labelled 6 (a 6A MCB). Track horizontally along to the left hand edge and read off the value. As it is about half way between 2000 and 3000 I estimated this at 2400.

 

[Daniel Jones]

This forum has some awesome members. The speed that my questions have been answered is amazing.

Following on from the cpc sizing when the ocpd is a mcb. I now have a situation where I have a 16A fuse to bs88 with a fault current of over 700A all the manufactures chart only go down to 0.01 so again should I use the let through energy.

Will this equation give me the minimum cpc

S = square root of (I2t)(energy let through)/k.

Regards

Dan

 

[Richard Burns]

Now you are thinking!
Yes, you are outside the range of the tripping time charts so you can use the let through energy charts instead.
As it states in BS7671 for the adiabatic equation, I is the current that will flow in the case of a fault taking due account of the current limiting (I²t)of the device.
This is what you are doing when you use the I²t values.

On a practical basis I would not push too hard against the limits of csa as there are usually other factors that require a larger csa, such as simple mechanical strength.

 

[Daniel Jones]

Thank you Richard,

As you are an expert i have another question for you regarding volt drop.

I came across a volt drop calculation whilst reading one of the many books i now own. This was a lighting circuits and the equation was
Volt drop = mV/A/m × (Ib ∕ 2) × Length
with an evenly distributed load.

Question is when is it appropriate to use?
and what constitutes an evenly distributed load?

Dan

 

[Strima]

Thank you Richard,

As you are an expert i have another question for you regarding volt drop.

I came across a volt drop calculation whilst reading one of the many books i now own. This was a lighting circuits and the equation was
Volt drop = mV/A/m × (Ib ∕ 2) × Length
with an evenly distributed load.

Question is when is it appropriate to use?
and what constitutes an evenly distributed load?

Dan

That calculation is the only one you would really use unless you transpose the formula to get the maximum allowable length of a circuit.

Also you need to divide by 1000 to get the correct answer: Volt drop = (mV/A/m × (Ib ∕ 2) × Length)/1000

Think about a long lighting circuit will several different loads attached an how the might affect the voltage drop along the circuit. For example a couple of 5000 watt lamps near the origin of the circuit and a few 3 watt LED at the end would have a different VD if the light loads were near the origin and heavys at the end.

 

[Daniel Jones]

Sorry. Someone shoot me.

why do i need to work out the fault current at the end of the final circuit?

 

[Strima]

I never mentioned fault current.