2396 style question, 2400, short circuit, RCBO,

[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.

 

[Richard Burns]

I am certainly not an expert, just generally able to apply the requirements of the regs.
To expand on Strimas post, if you have an evenly distributed load on a long lighting circuit, say 100W every 20m then on average you can use the formula you describe as the volt drop will be high because all lights are taking current at the start of the circuit and drop down as you get to the end where only one light is taking current but it will also rise at the end because of the length of circuit, so on average your volt drop would be that for when the load is only half the total design current.
Not an equation I would usually use as you normally need to know the maximum volt drop. but for a rough estimate and a likely successful outcome it would work.

Additionally it is possible on long radial circuits to have a large csa at the start where the load current is high, so as to reduce volt drop and reduce the csa as you get further out and the load current, and so volt drop, is lower, in this case it may be necessary to know the fault current at the points where the csa changes in case the cable does not have the energy withstand of the fault current at origin.
I cannot see why you would ever need to know fault current at the end of the circuit though.

I attach a diagram from a previous post about calculating volt drop that is not totally in line with this post but may help explain the idea.

 

[Daniel Jones]

Sorry guys went off on a bit of a tangent asking about the volt drop. But thank you for your responses.

Sorry Strima, we must have posted those replies at the same time.

The reason I was asking about the fault current was in all the books you calculate the fault current at the extremity of the circuit then use that to calculate the minimum CPC. How ever that is not the largest current that will flow in a fault. A fault close to the source would be the largest. so shouldn’t we calculate the CPC size based on that?. Why do we need to calculate the fault current at the extremity? Is it just to check disconnection times are less than that of table 41.1.

Dan

 

[Strima]

Calculate the fault at the end of the circuit. If you did it at the source the current would be much higher therefore you would need a larger CPC which would in turn increase the fault current meaning you would need a larger CPC... And breathe...

 

[Richard Burns]

I am sorry, I was in error, I was still thinking energy withstand and maximum fault currents. You do not need to know the maximum fault current at the end of circuit but you do need to know the minimum.
You need to know the lowest fault current that can flow to ensure that the circuit will meet the disconnection times in all cases.
As the current will be at a minimum at the end of the circuit (or mid point of a ring) as the resistance of the cables will be highest. You also use Cmin to ensure that even if the supply voltage drops to (just above) the minimum permitted level you will still meet disconnection times.

 

[Daniel Jones]

ok,

a little confused now. sorry for throwing in the volt drop calculate that hasn't helped to flow.

I am trying the establish the minimum circuit protective conductor.
i have a 50mm² line conductor with a 25mm² CPC protected by a bs88 125A. i have determined the fault current at the extremity of the circuit.

Determine Earth Current,
Ief = (Uo × Cmin ∕ Zs)
therefore 230 × 0.95 ∕ 0.081 = 2698A
therefor the device will disconnect in less than 0.01 seconds

Determine Minimum Circuit Protective Conductor,
(S= √ (Ief² × t) ∕ K)
as t is less than 0.01 i use the energy let through = 78400(A²s)
therefore.
√(78400) ∕ 115 = 2.43mm²
as the CPC is 25mm² this is adequate

or does that equation not work and should i just be checking that the energy let through does not exceed the with stand of the cable. I.E
I²t ≤ K²S²
therefore
78400 ≤(115²25²)8'265'625
as the CPC is 25mm² this is adequate.

Dan