Adiabatic equation

[Scott F]

Recent pir i did last week ring main 2.5mm and cpc 1mm so using the equation

If =Uo/Zs=230/0.82=78673.183 so this is my fault current sq root of =15734 x time(type b32 amp)0.1s divided by 115

88.69/115=0.77 so 1mm is cool

looking at app 3 pg 249 in bs7671 cannot seem to see the time current curve (i can see it ) but not how you get the 0.1 what am i missing

 

[62patrick]

Hi, I hope this helps Uo/Zs 230/0.82= 280.49A

see reg 543.1.3 page 128 the big red book

so, S = sq root 280.49 x 280.49 x 0.4 / 115 = 1.55mm

the type B breaker will operate with in 0.04 at 160A

cheers patrick

 

[DanBrown]

62patrick, not sure about your answer mate maybe someone else can confirm and also the adiabatic equation isnt there to tell you if the breaker will operate intime - the Zs is for that...,In response to the question here is what i got.
The adiabatic equation is there to tell you the required minumum earth CSA required..
My working's are as follows;

S = I squared x Time, square-root / k factor.
Uo / Zs = 230/0.82 = 280.5 Amps Fault Current
280.5 A curve = 0.1 seconds
I squared = 78680.3 A
Time = 0.1 Seconds
So 78680.3 x 0.1 = 7868
Square root 7868 = 88.7
K factor for copper conductor upto 300mm2 = 115
So, 88.7 / 115 = 0.77mm2

So, the 1mm2 copper conductor for you ring main, is adequate enough to withstand a fault without causing damage to the conductor itself...

Hope this helps..

 

[Scott F]

Thanks guys its not the equaton it self its how you arrrive at the time "t"part

By doing a bit of research it seems t is 0.1 secs .

What i want to know is how/where you get this figure from

In the design guide it notes App 3 from Bs7671 time curves for type B 32 amp pg 249

But am i reading the graphs right ie all breakers up to 125 amp would be 0.1secs(t)

Not sure if i'm interpreting the information correctly

Scott

 

[maddfridge]

Thanks guys its not the equaton it self its how you arrrive at the time "t"part

By doing a bit of research it seems t is 0.1 secs .

What i want to know is how/where you get this figure from

In the design guide it notes App 3 from Bs7671 time curves for type B 32 amp pg 249

But am i reading the graphs right ie all breakers up to 125 amp would be 0.1secs(t)

Not sure if i'm interpreting the information correctly

Scott

hi there

The actual disconnection time of 0.1 second is prettey academic as any cuurent higher than the 0.4 sec time so disconnection at this speed is faster so along as tyhe withstand current is not breached it is ok

so as soon as it reaches say 280 from above 32 amp will go out at 160 amps so the higher current will make it go quicker in theory lol

cheers

 

[trebor]

Thanks guys its not the equaton it self its how you arrrive at the time "t"part

By doing a bit of research it seems t is 0.1 secs .

What i want to know is how/where you get this figure from

In the design guide it notes App 3 from Bs7671 time curves for type B 32 amp pg 249

But am i reading the graphs right ie all breakers up to 125 amp would be 0.1secs(t)

Not sure if i'm interpreting the information correctly

Scott

Any fault current in excess of 5 x In of any type b cb will result in instaneous operation so t will equal 0.01 for the adabatic equation
likewise 10 x In for C type and 20 x In for D type

 

[Keny-boy]

Hiya guy I was wound if you can help, been to a house where there is PME, it has the old re wireable fuses, but has a 80 amp 30mA RCD, before the DB. The main earths are 6mm to the gas and main earth on water earth. I know Im bond the water. But was asking weather I need to up grade the earths to the adiabatic equation???

Am I right in doing this??

230/.35=657

657*657*.4=172659

Squared=415

415/115=3.61

is that correct???

so 6mm is fine???

 

[Keny-boy]

I'm mean there is no water bond.

 

[JUD]

Hiya guy I was wound if you can help, been to a house where there is PME, it has the old re wireable fuses, but has a 80 amp 30mA RCD, before the DB. The main earths are 6mm to the gas and main earth on water earth. I know Im bond the water. But was asking weather I need to up grade the earths to the adiabatic equation???

Am I right in doing this??

230/.35=657

657*657*.4=172659

Squared=415

415/115=3.61

is that correct???

so 6mm is fine???

Almost.

First thing you need is a measured Ze. 0.35 on a TN-C-S is the maximum it should be. In reality it may be lower than this which would give you a higher fault current.

You then need to know the BS No. of the main fuse and it's rating.

The k value for insulated protective conductors ≤ 300mm² not incorporated in a cable is actually 143, not 115. (Table 54.2 page 129 BRB)

Example

Ze = 0.1
Fuse = 80A BS1361
Fault current = 230 / 0.1 = 2300A (2.3kA)

An 80A BS1361 fuse will go in 0.1s at 1100A (1.1kA) so we use this as our time in the equation. (Time/Current characteristics page 244 BRB)

√(2300² x 0.1) = 727
727 / 143 = 5.1mm²

 

[Amp David]

Hiya guy I was wound if you can help, been to a house where there is PME, it has the old re wireable fuses, but has a 80 amp 30mA RCD, before the DB. The main earths are 6mm to the gas and main earth on water earth. I know Im bond the water. But was asking weather I need to up grade the earths to the adiabatic equation???

Come again

 

[Keny-boy]

Lol sorry.

I been to a house and theses and old rewirable fuses, It dose have a 80amp 30mA rcd, I will have to double cheek the main fuse rating. But what I was asking was in the house 6mm gas bond and 6mm main earth. I will run in an water bond. But is the 6mm ok or will I need to up grade to 10mm, adiabatic equation?? Would give me the answer

 

[Amp David]

No, the regs state the size of you main equipotential bonds.

The adiabatic equation check the size of the CPCs of circuits and main earth conductors.