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