Adiabatic Equation help

[Sam1995]

Hi was wondering if im doing this right. The question in: Determine for a chosen circuit "the minimum possible csa of cpc which will satisfy the requirements disconnection as ADS under earth fault conditions and the adiabatic equation as reg 543.1.3."

Circuit: Single core pvc 70* 2.5mm2 + 2.5mm2. 16A Type B 61009-1
r1+r2 = 0.22
Ze = 0.11
Zs = 0.23
Pfc = 230/0.23 = 1000A

Square root of 1000 squared x 0.1 = 316.227

316.227/k(115) = 2.749

So 2.5mm is undersized? Should I be calculating a different way with let through energy or leave it as and use a 4mm?

Thanks!

 

[Pretty Mouth]

@Sam1995 , mind if I ask what this is for? College project/exam?

Real world, this would likely need no more than a 1mm2 CPC, but the limited info you have means the adiabatic equation requires an unrealistically thick CPC

 

[Sam1995]

@Sam1995 , mind if I ask what this is for? College project/exam?

Real world, this would likely need no more than a 1mm2 CPC, but the limited info you have means the adiabatic equation requires an unrealistically thick CPC

For the 305 design assignment.
I'd have thought you would come across for example a socket next to a CU using 2.5+1.5 t+e with similar pfcs... in the real world?

What I've done is 2.5 with a 4mm cpc. Zs works out at .21 pfc=1095A I got S=3.011mm2

When Ive calculated my circuit resistance I've used the multiplier for max operating temp/thought this would be best for worse case scenario although design current isnt that high.

Is the equation all that is needed to show the cpc satisfies ADS?

 

[Pretty Mouth]

For the 305 design assignment.
I'd have thought you would come across for example a socket next to a CU using 2.5+1.5 t+e with similar pfcs... in the real world?

What I've done is 2.5 with a 4mm cpc. Zs works out at .21 pfc=1095A I got S=3.011mm2

When Ive calculated my circuit resistance I've used the multiplier for max operating temp/thought this would be best for worse case scenario although design current isnt that high.

Is the equation all that is needed to show the cpc satisfies ADS?

434.5.2
A fault occurring at any point in a circuit shall be interrupted within a time such that the fault current does not cause the permitted limiting temperature of any conductor or cable to be exceeded.

For a B-type breaker, let through energy (I2t) tends to increase as fault current increases. This means that a fault immediately downstream of the protective device will cause the greatest increase in temperature of the conductors. If the CPC can withstand the temperature rise from a fault occurring there, then it can withstand it from a fault anywhere else in the circuit.

So, you should actually use the fault current at the origin of the circuit to calculate the adiabatic for B-type breakers. In this case, I make that 2091A. Run that through the adiabatic, using t=0.1s, and you get a ridiculously large CPC of 5.75mm2!

The problem is, using data from appendix 3 of BS7671, the lowest value of t is 0.1s, whereas manufacturers data will go lower, down to 0.001s or lower. Also, manufacturers supply data for I2t which can be used in the calculation.

For example, this is for Hager MCBs:

 

[Pretty Mouth]

Using the above, for a fault current of 2091A on a 16A breaker, you get approx 6000A²s. Run that through the adiabatic:

S=(√6000)/115 = 0.67mm²

, a much more realistic sized CPC.

Note also table B7 of the OSG, that allows you to select a CPC for circuits protected by breakers, without calculation, based on fault current.

 

[buzzlightyear]

 

[timhoward]

First things first, I think there's a simple mistake right at the beginning, otherwise the I'd say the method is right.

Circuit: Single core pvc 70* 2.5mm2 + 2.5mm2. 16A Type B 61009-1
r1+r2 = 0.22
Ze = 0.11
Zs = 0.23 ??
Pfc = 230/0.23 = 1000A

Zs will be Ze+(R1+R2) = 0.33
So PFC will be 696A

We know that with a 16A type B RCBO it will disconnect in 0.1s if current exceeds 80A from table on right of graph, page 417 of regs.
As @Pretty Mouth said, using manufacturers data you could get the 0.1 to an even lower value, but it becomes a bit acedmic for reasons you will see.

So your calc's need adjusting a bit:

Square root of 1000 squared x 0.1 = 316.227

316.227/k(115) = 2.749

Becomes:
Square root of 696 squared X 0.1 = 220.0945
220.0945 / k(115) = 1.91

This is under the minimum allowed separate CPC size. So getting a lower number using manufacturers data on their RCBO is a bit pointless in this case. (If it were 25 sq mm SWA then it's a different story!)

Assuming mechanical protection the minimum size is 2.5 sq mm. Otherwise it would be 4 sq mm.
In this case using reg 543.1.4 is simpler.

I believe this is one of the clearest videos out there on the subject:

 

[Sam1995]

Using the above, for a fault current of 2091A on a 16A breaker, you get approx 6000A²s. Run that through the adiabatic:

S=(√6000)/115 = 0.67mm²

, a much more realistic sized CPC.

Note also table B7 of the OSG, that allows you to select a CPC for circuits protected by breakers, without calculation, based on fault current.

I wasnt sure how to use let through energy data. So fault current is at the bottom of the graph which is 2091A at origin and let through energy is left hand of graph being 6000A squared (is the s 1 second or 0.1s??) and in the equation how comes it isnt the square root of 6000squared x t divided by k?
Thankyou!
Also theyve asked for the adiabatic equation to be used

 

[Sam1995]

First things first, I think there's a simple mistake right at the beginning, otherwise the I'd say the method is right.

Zs will be Ze+(R1+R2) = 0.33
So PFC will be 696A

We know that with a 16A type B RCBO it will disconnect in 0.1s if current exceeds 80A from table on right of graph, page 417 of regs.
As @Pretty Mouth said, using manufacturers data you could get the 0.1 to an even lower value, but it becomes a bit acedmic for reasons you will see.

So your calc's need adjusting a bit:

Becomes:
Square root of 696 squared X 0.1 = 220.0945
220.0945 / k(115) = 1.91

This is under the minimum allowed separate CPC size. So getting a lower number using manufacturers data on their RCBO is a bit pointless in this case. (If it were 25 sq mm SWA then it's a different story!)

Assuming mechanical protection the minimum size is 2.5 sq mm. Otherwise it would be 4 sq mm.
In this case using reg 543.1.4 is simpler.

I believe this is one of the clearest videos out there on the subject:

When i see that mistake I really kicked my self! I was shocked so i double checked and I wrote the post wrong
My circuit was 70* pvc 2.5mm2 + 2.5mm2 at 7 meters so I worked out my r1+r2 to be 14.82 x 7 divided by 1000 x by the max operating factor 1.2 = 0.124488 rounded down to .12

Does that seem correct? And that would make S = 2.749 with the info from graph in 7671

Thankyou

 

[timhoward]

My circuit was 70* pvc 2.5mm2 + 2.5mm2 at 7 meters so I worked out my r1+r2 to be 14.82 x 7 divided by 1000 x by the max operating factor 1.2 = 0.124488 rounded down to .12

I concur.

And that would make S = 2.749 with the info from graph in 7671

It would, but the point is that the graph stops at 0.1s when in fact a device would trip quicker than that.
Here's a Wylex graph:


You have 1000A fault current. (10 to power 3)
The I squared t is on the left and would be 1250

So square root of 1250 / 115 gives S = 0.31 sq mm
Then back to minimum sizes, and you end up with 2.5 sq mm

 

[Pretty Mouth]

I wasnt sure how to use let through energy data. So fault current is at the bottom of the graph which is 2091A at origin and let through energy is left hand of graph being 6000A squared (is the s 1 second or 0.1s??) and in the equation how comes it isnt the square root of 6000squared x t divided by k?
Thankyou!
Also theyve asked for the adiabatic equation to be used

Fault current is horizontal axis (bottom of graph)
Let through energy is vertical axis (left hand of graph).

The equation I posted is the adiabatic, but not quite as you were expecting.

Note that the let-through energy is measured in A²s, and replaces the (I²t) in the adiabatic equation. The (I²t) part of the equation has already been done by hager, if you like.

 

[Pretty Mouth]

I concur.


It would, but the point is that the graph stops at 0.1s when in fact a device would trip quicker than that.
Here's a Wylex graph:
View attachment 107471

You have 1000A fault current. (10 to power 3)
The I squared t is on the left and would be 1250

So square root of 1250 / 115 gives S = 0.31 sq mm
Then back to minimum sizes, and you end up with 2.5 sq mm

@timhoward , I believe that, for B and C type breakers, the fault current used for the adiabatic equation should be taken at the point of the circuit that it would be highest, ie the origin. The reasons for this are in my above post, also the note to fig3A4 ('The application of the rules of Chapter 43 should take into account both minimum and maximum fault current conditions.').

To me, it seems that using the fault current at the far end of the circuit leads to a CPC being selected on a best case scenario, and a fault anywhere else in the circuit would raise the temperature of the conductors to a higher temperature than a fault here would.

However, pretty much everyone else, all books, videos etc on the matter seem to suggest using the fault current at the end of the circuit for this. Also, the info included with questions that students pose on the forum (such as the OP's) kind of make me suspect that the examiners are expecting the fault current to be taken from the end of the circuit too.

Can you offer any insight into this? Whilst I think that I'm right on this, I don't want to give out info that may get someone a fail in their exams.

(note, this is only for B and C type breakers, different for other protective devices IMO)

 

[timhoward]

@timhoward , I believe that, for B and C type breakers, the fault current used for the adiabatic equation should be taken at the point of the circuit that it would be highest, ie the origin. The reasons for this are in my above post, also the note to fig3A4 ('The application of the rules of Chapter 43 should take into account both minimum and maximum fault current conditions.').

To me, it seems that using the fault current at the far end of the circuit leads to a CPC being selected on a best case scenario, and a fault anywhere else in the circuit would raise the temperature of the conductors to a higher temperature than a fault here would.

However, pretty much everyone else, all books, videos etc on the matter seem to suggest using the fault current at the end of the circuit for this. Also, the info included with questions that students pose on the forum (such as the OP's) kind of make me suspect that the examiners are expecting the fault current to be taken from the end of the circuit too.

Can you offer any insight into this? Whilst I think that I'm right on this, I don't want to give out info that may get someone a fail in their exams.

(note, this is only for B and C type breakers, different free for other protective devices IMO)

I only have time for a quick comment. There’s two concerns
1 - will the thing trip quickly enough. Clearly for a B16 it will, using min or max fault current
2 - how hot will the thing get during the time it takes to trip. This is where the higher fault current nearer the origin could be an issue, and using the equation in its other form to find how long it takes to get to 70 degrees would be required. This should be > answer 1.

I agree with your interpretation of the regs. Rightly or wrongly I suspect the point of the original question in this case is that whatever we do we end up back at 2.5 sq mm as the minimum mech. protected size.
I’m also on the fence regarding whether a trainee should focus on the min fault current at end of circuit which is what their tutor is probably expecting. Maybe!

(I’ve also seen one example that says a B16 will trip instantaneously at whatever the box on the right of the graph says (80A?) so 80 and 0.1 should be used in the equation as as soon as it reaches 80A it will trip. I’m still musing that one!)

 

[buzzlightyear]

I’ve also seen one example that says a B16 will trip instantaneously at whatever the box on the right of the graph says (80A?) so 80 and 0.1 should be used in the equation as as soon as it reaches 80A it will trip. I’m still musing that one!)

But the box also says. 0.1 to 5 instant trip time, you need to know exactly which one to use.

 

[pc1966]

Can you offer any insight into this? Whilst I think that I'm right on this, I don't want to give out info that may get someone a fail in their exams.

That would also be my approach, though it is worth noting that very high PFC (several kA) are brought down quickly by even a couple of meters of T&E

However, I don't know what the examiners in this situation expect. Basically there are two concerns:

• At origin the PFC is highest and so demands on OCPD break limit and let-though (for MCB, etc) usually highest.
• At end of circuit PFC is (usually) lowest so you need to determine it is high enough to meet ADS times.

(note, this is only for B and C type breakers, different for other protective devices IMO)

Type D are similar, but usually used where 5s is acceptable for sub-main and so you might well see a very much higer I2t value as t is up to 5s and not the 10-20ms that a typical MCB goes at.

Fuses are the opposite: normally I2t drops as PFC increases until it plateaus to a fairly constant level as it limits the let-through, usually to much lower than similar breaker (MCB / MCCB).

 

[Lister1987]

I only have time for a quick comment. There’s two concerns
1 - will the thing trip quickly enough. Clearly for a B16 it will, using min or max fault current
2 - how hot will the thing get during the time it takes to trip. This is where the higher fault current nearer the origin could be an issue, and using the equation in its other form to find how long it takes to get to 70 degrees would be required. This should be > answer 1.

I agree with your interpretation of the regs. Rightly or wrongly I suspect the point of the original question in this case is that whatever we do we end up back at 2.5 sq mm as the minimum mech. protected size.
I’m also on the fence regarding whether a trainee should focus on the min fault current at end of circuit which is what their tutor is probably expecting. Maybe!

(I’ve also seen one example that says a B16 will trip instantaneously at whatever the box on the right of the graph says (80A?) so 80 and 0.1 should be used in the equation as as soon as it reaches 80A it will trip. I’m still musing that one!)

The whole point of the design project is to demonstrate you understand the process. Ask yourself why they'd wanting a specific manufacturer (as opposed to generic 7671) data ["....with the aid of manufacturer data...."] and the purpose and reasoning is clear; so can work the equations and extrapolate the required information, even if that ends up just back at the 2.5mm minimum.

You can, in theory, run the equation and steps, get them wrong and STILL get positive marks because while the end result may be wrong, you are demonstrating knowledge of the process, equations and expected results.

 

[Sam1995]

I assumed I'd do the calcs with the pfcs at end of circuit but what you say makes sense. Although they have stated to show it will satisfy requirements for ADS.

I did find on page 409 of BS7671 "the impedance values given in Tables 41.3 and 41.6 can be used for BS EN 60898 circuit-breakers. These values are far more onerous and in some cases may be difficult to achieve without installing larger sized cpcs.

Above this "Where ever possible designers should use the manufacturer's specific data"

so hager 16A 60898 type B will disconnect in 0.01s at 719A. Is this correct?

So with this I could go 2.5mm2 + 1mm2. r1+r2 = 0.21
Zs = .11 + .21 = .32
pfc = 230/.32 = 719A

so square root of 719 squared x 0.01/115 = 0.62 = S

 

[Sam1995]

The whole point of the design project is to demonstrate you understand the process. Ask yourself why they'd wanting a specific manufacturer (as opposed to generic 7671) data ["....with the aid of manufacturer data...."] and the purpose and reasoning is clear; so can work the equations and extrapolate the required information, even if that ends up just back at the 2.5mm minimum.

You can, in theory, run the equation and steps, get them wrong and STILL get positive marks because while the end result may be wrong, you are demonstrating knowledge of the process, equations and expected results.

Yes to be fair i could have maybe avoided this as I haven't done the previous question which is selecting the devices. First time doing all this so didnt realise how important skipping that question would have been.
Thankyou!

 

[Sam1995]

On this graph would you multiply the breaker rating for example 16 by the bottom number of this graph?

 

[buzzlightyear]

so hager 16A 60898 type B will disconnect in 0.01s at 719A. Is this correct?

16x5=80amps
Square root (80x80x0.1)÷115=answer.