2396 style question, 2400, short circuit, RCBO,

[Daniel Jones]

hello,

studying for my 2396 exam and i need some help!

Been banging my head against a wall for hours trying to get this one straight in my mind.

The question is as follows:
An electrical installation within a small retail outlet, located in a farm building is to be designed with the protective measure of ADS. The supply and installation form a 230V singles phase TT system. All wiring systems are surface mounted Ra(Ze) 12 ohm and PFC 2.1Ka
A) Answered this
B) one of the circuits is protected by a 32A 30mA RCBO and the circuit is wired using 4mm2/1.5mm2 flat profile cable to a length of 17m.
Determine if this circuit has adequate,
I) Earth fault protection
work out Zs and then a 32A has an max Zs of 1.37 assuming Type b but with the RCBO should it be 1667? just so confused
II) Short circuit protection
Dont know where to start

Any help would be greatly appreciated. just point me i a direction.

Dan

 

[Richard Burns]

I) earth fault protection
Will the specified cable and and installation method mean that in the case of an earth fault the protective device will operate within 0.2s and will the cable be able to handle the fault current.
Calculate the Zs and see if it complies, calculate the fault current to earth and see is the cable can handle this current for 0.2s i.e. use the adiabatic equation. All fairly overkill as the fault current is 19A.
II) short circuit protection
see if the cable meets the requirement in 434.5.2 (t<=((k²S²/I²) assuming the RCBO is operating correctly.

 

[Strima]

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.

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

 

[Daniel Jones]

wow amazed how quick that was answered thought i would leave it a day or 2 and i had replies within hours

one last thing. what do you think to these answers

An electrical installation within a small retail outlet, located in a farm building is to be designed with the protective measure of ADS.

The supply and installation form a 230V singles phase TT system. All wiring systems are surface mounted Ra(Ze) 12 ohm and PFC 2.1Ka


A) Explain the conditions that must to be met if this installation is to comply with the disconnection times applicable to TN system? (8 Marks)

See note below Table 41.1 the maximum disconnection times

The disconnection times of a TN system may be used where in a TT system disconnection is achieved by an overcurrent protective device and equipotential bonding is connected to all the extraneous conductive parts within the installation in accordance with regulation 411.3.1.2

Regulation 411.3.1.2 requires that main protective bonding conductor comply with chapter 54 and shall connect the MET to all extraneous conductive parts including. Water Pipe, Gas Pipe, Structural Steel, Central Heating, Air conditioning and other installation pipe work and ducting

B) one of the circuits is protected by a 32A 30mA RCBO and the circuit is wired using 4mm2/1.5mm2 flat profile cable to a length of 17m.
Determine if this circuit has adequate,

I) Earth fault protection? (4 Marks)

Earth Fault loop impedance calculation

Zs = Ze + ((R1 + R2) Cf)

Zs = 12 + ((16.71 x 17/1000) x 1.20) = 12.34Ω

Assuming 32A 30mA type B RCBO the maximum earth loop fault impedance of 1667Ω

So satisfactory

II) Short circuit protection (8 Marks)

Zn = (R2 + RN)

Zn = (9.22 x 17/1000) = 0.15

Ipf = V x Cmin/Zn

Ipf = 230 x 1.1 / 0.15 = 1.686

For a 32A 30mA type B RCBO the current causing instant operation is 160A therefore the devices will disconnect with in 0.2s

Using the equations t=k²s²/I²

t must equal less than 0.2s

Total Ipf = 1.686 + 2.1 = 3.786kA

115² x 4² / 3786² = 0.014s

So satisfactory

 

[Strima]

Looks like you've got the idea. Would also be worthwhile on the earth fault referencing maximum touch voltages, all additional points.

I haven't checked your results, TBH the method you use is the main point of the questions and how you understand the fundamental principles.

IIRC one question on my exam was based on Cmin and why it is used. Take a look at the chief examiners response to each exam and you'll see where people go wrong.

To make life easier during the exam I actually wrote out all the relevant formulas in the notes pages in the back of my regs book. This is allowed as the book already contains these, it just makes it easier and quicker to use them. Also don't have too many flags on pages, just the important ones such as tripping curves, CCC for cables, adiabatic, etc etc etc.

One flag on the first tripping curve chart just so you can find the tables quickly.

And a spare calculator!!!

 

[Daniel Jones]

II) Short circuit protection (8 Marks)

Zn = (R2 + RN)

Zn = (9.22 x 17/1000) = 0.15

Ipf = V x Cmin/Zn

Ipf = 230 x 1.1 / 0.15 = 1.686

Total Ipf = 1.686 + 2.1 = 3.786kA

Using the equations t=k²s²/I²

115² x 4² / 3786² = 0.014s

For a 32A 30mA type B RCBO 3786A will take 0.02s to disconnect

This is not So satisfactory the cable would be damaged before the device would disconnect

 

[Richard Burns]

I) Earth fault protection? (4 Marks)
Earth Fault loop impedance calculation
Zs = Ze + ((R1 + R2) Cf)
Zs = 12 + ((16.71 x 17/1000) x 1.20) = 12.34Ω
Assuming 32A 30mA type B RCBO the maximum earth loop fault impedance of 1667Ω So satisfactory.

II) Short circuit protection (8 Marks)
Zn = (R2 + RN)
Zn = (9.22 x 17/1000) = 0.15
Ipf = V x Cmin/Zn
Ipf = 230 x 1.1 / 0.15 = 1.686
For a 32A 30mA type B RCBO the current causing instant operation is 160A therefore the devices will disconnect with in 0.2s
Using the equations t=k²s²/I²
t must equal less than 0.2s
Total Ipf = 1.686 + 2.1 = 3.786kA
115² x 4² / 3786² = 0.014s
So satisfactory

II) Short circuit protection (8 Marks)

Zn = (R2 + RN) Zn = (9.22 x 17/1000) = 0.15 Ipf = V x Cmin/Zn

Ipf = 230 x 1.1 / 0.15 = 1.686 Total Ipf = 1.686 + 2.1 = 3.786kA Using the equations t=k²s²/I²

115² x 4² / 3786² = 0.014s For a 32A 30mA type B RCBO 3786A will take 0.02s to disconnect
This is not So satisfactory the cable would be damaged before the device would disconnect

You have made good attempts at answering the questions there and have taken account of the required factors and values.
It is a good idea to check your terminology as this will be noted in an exam.
For I) you have given a correction factor Cf which is (according to BS7671) the correction factor for the use of a semi enclosed fuse instead of a circuit breaker. You are meaning that you are correcting the tabulated resistance of the cable at 20°C to the maximum permitted operating temperature in this case 70°C. It might be better to state correcting to maximum permitted operating temperature using the factor 1.20.
The maximum earth fault loop impedance for a 30mA RCD is 1667Ω and is unaffected by the type of the over current protection.
The final answer of satisfactory is correct.

For II) You have given Zn as the resistance of the line neutral path from the origin along the final circuit.
You have given the values of R2 + RN which should be R1+RN.
You have also stated Cmin and then used (Accurately) the value for Cmax.
However on the rest of the calculation your approach is somewhat flawed you cannot add the fault currents or calculate the fault current like that.
The maximum fault current will be seen at origin, thereafter as you move further out into the installation the added resistance of the conductors will reduce the fault current seen.
As an aside which is not relevant to the question you could say that at the extremity of the circuit the total resistance of the short circuit fault path would be the resistance at origin plus the resistance of the live conductors as follows: Resistance at origin can be calculated from the given fault current (2.1kA) as R=V/I =230/2100 =0.011Ω.
The live conductor resistance would be as you have calculated 0.15Ω at 20°C, this is the (almost) worst case scenario so does not need to be corrected for temperature, adding the two values would give 0.16Ω and therefore a fault current of 1438 A.

However you are planning for the worst short circuit fault, this can occur anywhere on the circuit but the highest current would be seen just as the circuit starts at origin and so the fault current would be 2100A.
This value is beyond the graphs given in appendix three for the disconnection time of an RCBO, using the equation t=k²s²/I² gives a time of 0.04s but this value cannot be used with the graphs.
It is possible that you could consider the k²S² > I²t for a class 3 RCBO type B where I²t is 52,000 as given in the Beama guide to low voltage circuit breakers and k²S² is 211,600 and so greater than I²t and therefore safe against short circuit.
Alternatively in case I am going too far into the realms of complexity the RCBO is rated to withstand 6000A as a short circuit and so would be protected against short circuit at 2100A but this would not appear to afford 4 marks.


For your answer to question A it may be an idea to say that the earth fault loop impedance of the TT system should be low enough that the earth fault current is high enough to allow the use of an over current protective device. in addition to what you have already correct stated.

 

[Strima]

You have made good attempts at answering the questions there and have taken account of the required factors and values.
It is a good idea to check your terminology as this will be noted in an exam.
For I) you have given a correction factor Cf which is (according to BS7671) the correction factor for the use of a semi enclosed fuse instead of a circuit breaker. You are meaning that you are correcting the tabulated resistance of the cable at 20°C to the maximum permitted operating temperature in this case 70°C. It might be better to state correcting to maximum permitted operating temperature using the factor 1.20.
The maximum earth fault loop impedance for a 30mA RCD is 1667Ω and is unaffected by the type of the over current protection.
The final answer of satisfactory is correct.

For II) You have given Zn as the resistance of the line neutral path from the origin along the final circuit.
You have given the values of R2 + RN which should be R1+RN.
You have also stated Cmin and then used (Accurately) the value for Cmax.
However on the rest of the calculation your approach is somewhat flawed you cannot add the fault currents or calculate the fault current like that.
The maximum fault current will be seen at origin, thereafter as you move further out into the installation the added resistance of the conductors will reduce the fault current seen.
As an aside which is not relevant to the question you could say that at the extremity of the circuit the total resistance of the short circuit fault path would be the resistance at origin plus the resistance of the live conductors as follows: Resistance at origin can be calculated from the given fault current (2.1kA) as R=V/I =230/2100 =0.011Ω.
The live conductor resistance would be as you have calculated 0.15Ω at 20°C, this is the (almost) worst case scenario so does not need to be corrected for temperature, adding the two values would give 0.16Ω and therefore a fault current of 1438 A.

However you are planning for the worst short circuit fault, this can occur anywhere on the circuit but the highest current would be seen just as the circuit starts at origin and so the fault current would be 2100A.
This value is beyond the graphs given in appendix three for the disconnection time of an RCBO, using the equation t=k²s²/I² gives a time of 0.04s but this value cannot be used with the graphs.
It is possible that you could consider the k²S² > I²t for a class 3 RCBO type B where I²t is 52,000 as given in the Beama guide to low voltage circuit breakers and k²S² is 211,600 and so greater than I²t and therefore safe against short circuit.
Alternatively in case I am going too far into the realms of complexity the RCBO is rated to withstand 6000A as a short circuit and so would be protected against short circuit at 2100A but this would not appear to afford 4 marks.


For your answer to question A it may be an idea to say that the earth fault loop impedance of the TT system should be low enough that the earth fault current is high enough to allow the use of an over current protective device. in addition to what you have already correct stated.

What more can be said? Bang on.

 

[timbobelfast]

The Short Circuit question has been in the 2396 Exam on the two ocassions I had the joy of sitting them. Carrying big marks as well, so if you can bat that out of the park you will be in with a good chance

 

[Daniel Jones]

Thanks for all your replies. Just going back through notes now and its something that we covered but had just forgotten. Got the exam next Thursday. So 5 days left of rereading books and notes. Thanks for the advice on the examiners reports. Got the last six. The same things seem to come up. Not enough explanation of how the regulations are implemented on certain installation. Need to give examples. Not just quote regulations. Show that you understand the regulation.

Any further thoughts and help will be greatly appreciated.

Dan

 

[timbobelfast]

One more bit of advice, know how to use a calculator inside out.
I had my scientific calculator taken off me in the exam and was given one from toys r us. Hence no squared root button so had to work everything out manually! Was going to grab the examier but just bit my tongue!

 

[Strima]

One more bit of advice, know how to use a calculator inside out.
I had my scientific calculator taken off me in the exam and was given one from toys r us. Hence no squared root button so had to work everything out manually! Was going to grab the examier but just bit my tongue!

What was his reason for removing your calculator?

Have you contacted C&G to confirm this was the correct course of action for the invigilator?

 

[timbobelfast]

The centre said it was because it could be 'programmed', even though if offered to let her default the calculator.

Toughest exam I have sat, glad to say overcame that hurdle. C&G are as much use as a motorbike ashtray tbh!

 

[Strima]

The centre said it was because it could be 'programmed', even though if offered to let her default the calculator.

What a load of old cock...

Toughest exam I have sat, glad to say overcame that hurdle. C&G are as much use as a motorbike ashtray tbh!

Is still think I was done out of 2 1/2 hours on mine.

 

[Daniel Jones]

Hi everyone,

Finished my exam last Thursday and it has taken till now to get over it. how is it only 3 hours. I think it all went well, but didn't managed to complete hopefully scored enough on the questions i did answer thou.

New question for you experts, still on the same topic short circuit.

just a little something for my own piece of mind.

so i have determined the cable size for current carrying capacity and volt drop.

and then i progressed to here

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 Earth Current, (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) ∕ 176 = 1.74mm²

As the CPC cable is a separate cable of 35mm² so satisfactory

Determine line Fault Loop Impedance (Zn = Zne + (R1 + Rn)

Zne = Uo × Cmax ∕ Pscc ∴ Zne = 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)

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

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

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

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

The maximum fault current of the device is 80Ka so satisfactory

Circuit meets the requirements of BS7671-2015

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