Could anyone please help me with this adiabatic equation?

[bpmservices]

Hi, and thank you to anyone who's taking the time to read my question. i'm struggling with the outcome of an adiabatic equation i'm doing and would be incredibly grateful of any guidance or help anyone could offer. its for a design project and i'd like to know if i'm doing something wrong somewhere. here are the details of the circuit etc...

garage board Zdb = 0.22 ohms
ring final circuit wired in 2.5mm²/1.5mm² twin and cpc
circuit length = 30 meters
protective device B32 61009-1 RCBO
2.5/1.5 T+E milliohms per meter = 19.51
temp correct factor = 1.2
K=115

my calculations are as follows; (figures rounded up to two decimal places)
r1 = 7.41 x 30 x 1.2 ÷ 1000 = 0.27 ohms
r2= 12.10 x 30 x 1.2 ÷ 1000 = 0.44 ohms

(R1+R2) for ring final circuit = r1+r2÷4 = 0.27+0.44÷4 = 0.18 ohms (R1+R2)= 0.18 ohms
Zs= Zdb+(R1+R2) 0.22+0.18=0.4 Zs = 0.4 ohms
Ipefc= Uo x Cmax ÷ Zs = 230x1.1÷0.4 = 632.5 amps Ipefc= 632.5 amps

heres where i think im going wrong...
using log graph in bs7671 fig 3A4 disconnection time for BSEN61009-1 B32 with fault current of 632.5amps = 0.1 sec

632.5 x 632.5 = 400,056.25
400,056.25 x 0.1 = 40,005.625
square root of 40,005.625 = 200.014
200.014 ÷ 115 = 1.73 mm²

So S is equal to 1.73 mm² which is the CSA of my CPC which means that my simple 30 meter ring final circuit wired in 2.5/1.5 twin and cpc has now got to be changed to 2.5mm² singles which will then comply with 543.1.4 and table 54.7 AND the adiabatic equation.... OR..... because my fault current is in excess of those causing instantaneous operation of protective device, i.e less than 0.1 second, do i need to start comparing I2t and K2S2 energy let through data etc? i'm sure im going wrong somewhere and would be incredibly grateful if anyone could shed some light or point me off in the right direction with this.

thank you again to anyone that has taken the time to read this.

 

[DomB78]

Hi, and thank you to anyone who's taking the time to read my question. i'm struggling with the outcome of an adiabatic equation i'm doing and would be incredibly grateful of any guidance or help anyone could offer. its for a design project and i'd like to know if i'm doing something wrong somewhere. here are the details of the circuit etc...

garage board Zdb = 0.22 ohms
ring final circuit wired in 2.5mm²/1.5mm² twin and cpc
circuit length = 30 meters
protective device B32 61009-1 RCBO
2.5/1.5 T+E milliohms per meter = 19.51
temp correct factor = 1.2
K=115

my calculations are as follows; (figures rounded up to two decimal places)
r1 = 7.41 x 30 x 1.2 ÷ 1000 = 0.27 ohms
r2= 12.10 x 30 x 1.2 ÷ 1000 = 0.44 ohms

(R1+R2) for ring final circuit = r1+r2÷4 = 0.27+0.44÷4 = 0.18 ohms (R1+R2)= 0.18 ohms
Zs= Zdb+(R1+R2) 0.22+0.18=0.4 Zs = 0.4 ohms
Ipefc= Uo x Cmax ÷ Zs = 230x1.1÷0.4 = 632.5 amps Ipefc= 632.5 amps

heres where i think im going wrong...
using log graph in bs7671 fig 3A4 disconnection time for BSEN61009-1 B32 with fault current of 632.5amps = 0.1 sec

632.5 x 632.5 = 400,056.25
400,056.25 x 0.1 = 40,005.625
square root of 40,005.625 = 200.014
200.014 ÷ 115 = 1.73 mm²

So S is equal to 1.73 mm² which is the CSA of my CPC which means that my simple 30 meter ring final circuit wired in 2.5/1.5 twin and cpc has now got to be changed to 2.5mm² singles which will then comply with 543.1.4 and table 54.7 AND the adiabatic equation.... OR..... because my fault current is in excess of those causing instantaneous operation of protective device, i.e less than 0.1 second, do i need to start comparing I2t and K2S2 energy let through data etc? i'm sure im going wrong somewhere and would be incredibly grateful if anyone could shed some light or point me off in the right direction with this.

thank you again to anyone that has taken the time to read this.

Hi

I am by no means a master at this.

But based on your info:

garage board Zdb = 0.22 ohms
ring final circuit wired in 2.5mm²/1.5mm² twin and cpc
circuit length = 30 meters
protective device B32 61009-1 RCBO
2.5/1.5 T+E milliohms per meter = 19.51
temp correct factor = 1.2
K=115

You said your ZS is 0.4 ohms

I did the calculations too

PEFC = Uo / ZS

230/0.4 = 575A

√(575^2x0.1)/115 = 1.5811mm

I got 1.58mm, which is annoying as it is so close but it does not comply. What's the installation method ? Can you not run the circuit in in a 2.5mm hi tuff or SWA (all cores are 2.5mm) this will give you the 2.5mm CPC you need. If it is a garage DB it would not be unreasonable to be installed in Hi Tuff / SWA, just a thought.

Otherwise you could run in 2.5mm singles, is there a reason why you do not want to run in conduit ?

 

[Pretty Mouth]

1: You need to calculate the conductor CSA for a fault occurring at the most onerous point of the circuit. This is the point of the circuit where, should a fault occur there, let through energy (I²t) would be highest.

For MCBs and RCBOs, let through energy (I²t) tends to increase as fault current increases.

Fault current would be highest for a fault occurring right at the start of the circuit, immediately downstream of the RCBO.

So, you need to calculate it for a fault right at the origin of the circuit, not at the furthest point. Use Zdb to calculate the fault current.

2: Using MCB/RCBO data from appendix 3 will return a pessimistically large minimum conductor size for any fault current higher than the minimum needed to operate the magnetic trip. I calculate yours at over 3mm², way oversized. Get I²t data for your chosen RCBO from the manufacturer, and use that.

For example, for a hager B32, for a 1150A fault, I²t= ~5000A²s, giving a conductor CSA of about 0.6mm².

 

[DefyG]

With a fault current of 632A, the characteristics of the circuit breaker would be instantaneous and therefore a disconnection time of 0.01s would be more appropriate.
Using 0.01 would give a minimum conductor size of 0.55sqmm ........ more in line with Pretty Mouths calc using let through energy of 0.6sqmm.

 

[pc1966]

heres where i think im going wrong...
using log graph in bs7671 fig 3A4 disconnection time for BSEN61009-1 B32 with fault current of 632.5amps = 0.1 sec

While you can use the current/time plots for I2t, for the "instant" part of a MCB they are not really usable.

MCB (and some MCCB) are energy-limiting so they are able to disconnect in under a cycle of the supply, in fact often significantly under 10ms at high currents. So the 100ms from the 3A4 is not the MCB's opening time, just the lowest time they chose to plot.

For fuses the IET regs plots also only go down to 0.1s (e.g. Fig 3A3) but many manufacturers will show times going down to well below 1ms for I2t computation, as this is "virtual time" computed from the PFC and resulting I2t. The actual blowing time for the fuse under high currents is longer but as it current-limits the I2t is much less.

As @Pretty Mouth has already covered, you typically get plots of I2t let-through for MCB/MCCB for this sort of computation. But in the majority of cases the standard circuits of the OSG (i.e. combinations of cable size, OCPD type/rating, and max length) is enough to know that you are meeting the safety requirements.

EDIT: Just to add, not all manufacturers provide such useful plots as the Hager example, but the generic standard for energy limiting MCBs is also give in Table B7 of the OSG where for most domestic cases (PFC <= 3kA) you are OK on the standard T&E CPC sizes.