Help with Adiabatic equation

[soulman]

Hi, Could anyone check my calculation please.
Protective device 60A 1361
E.S = TN-C-S
Cable 16mm T & E
CPC 6.0mm
Zdb = 0.43ohms
Multiplier temp 1.20
0.43 x 1.20 = 0.52ohms
230/0.52 = 442A
SR 442 squared x 2secs/115 = 5.43mm

Cheers

 

[Ian1981]

Hi, Could anyone check my calculation please.
Protective device 60A 1361
E.S = TN-C-S
Cable 16mm T & E
CPC 6.0mm
Zdb = 0.43ohms
Multiplier temp 1.20
0.43 x 1.20 = 0.52ohms
230/0.52 = 442A
SR 442 squared x 2secs/115 = 5.43mm

Cheers

Applying the 1.20 correction factor is only used at the design stage when we use the calculated mv/a/m which are given at 20 degrees and we wish to correct to 70 degrees.
Is it zdb or the actual ze of 0.43 ohms?

 

[soulman]

Hi, the supply goes from the intake to a switched fuse, then to the consumer unit. Ze is 0.35. The multiplier is for the rising temperature of the conductor under fault conditions.
Cheers

 

[Ian1981]

Hi, the supply goes from the intake to a switched fuse, then to the consumer unit. Ze is 0.35. The multiplier is for the rising temperature of the conductor under fault conditions.
Cheers

If you have took a measured zs and got 0.43 for the sub main circuit then use that value.
Remember we are testing at site temperatures which is why we apply the 80% rule of thumb to the given max zs value of bs7671 when testing.
There is no need to apply the 1.20 value as that’s why we apply the 80% value.
As long as when it’s measured the value is equal to or less than the the 80% value of the 100% bs7671 value then You are good.

 

[soulman]

When using the adiabatic in cable design calculations, I don't use any temperature correction, but i'm almost certain when calculating if the cpc is large enough, temperature rise under fault conditions must be taken into consideration. But im probably wrong

 

[Ian1981]

When using the adiabatic in cable design calculations, I don't use any temperature correction, but i'm almost certain when calculating if the cpc is large enough, temperature rise under fault conditions must be taken into consideration. But im probably wrong

The basic equation is just I2t square root/k.
K is the value of the conductor at an assumed operating temp such as a single cpc not bunched operating at max temp of 70 degrees, k is given as 143
No correction factors are applied.

 

[Richard Burns]

Fault current would be higher at lower temperatures as the resistance will decrease so the shortest disconnection time would be starting at 20°C, ambient temperature, a longer disconnection time because of increased resistance (at 70°C) may stress a conductor more due to the increased time it has to carry the fault current.
Generally a measured PEFC is what I might use, but the k values given in the tables show the assumed initial temperature, which is 70°C for a conductor incorporated in a cable, as in your case, so the fault current should be that experienced at 70°C.

However with a fault current of 442A I would give a disconnection time of 1.6s rather than 2s for a 60A BS1361.
(With a fault current of (230/0.43)=535A I would apply a disconnection time of 0.65s for a 60A BS1361.)

As the cable is the cpc of a twin and earth the k value is 115 as you have stated.
Using the figures you have used the calculation is mathematically correct.
Using a disconnection time of 1.6s would calculate to a minimum cpc size of 4.86 mm².
Using a disconnection time of 0.65s and current of 535A would calculate to 3.75 mm².

In all cases the cpc is large enough to carry the fault current.

 

[Ian1981]

As for temperature rise if your concerned,you just need to confirm that t=K2S2/I2
The time t , in which a given fault current will raise the live conductors from the highest permissible operating temperature in normal duty to the limiting temperature before damage to the conductors and surrounding insulation occurs

 

[soulman]

Thanks richard, Ian.
I used the 2 secs as that was the closest in the curve charts. is 1.6s the actual disconnection at 442A
So temperature correction is not to be used.
Where have you got the 0.65s from?

Cheers much appreciated

 

[Richard Burns]

Thanks richard, Ian.
I used the 2 secs as that was the closest in the curve charts. is 1.6s the actual disconnection at 442A
So temperature correction is not to be used.
Where have you got the 0.65s from?

Cheers much appreciated

The 1.6s is what I estimated from the graph at 442A.

With a fault current of 535 A as calculated from a Zdb of 0.43 Ω and a design voltage of 230 V, I estimate that the disconnection time on the graph would be 0.65 s.

 

[Leesparkykent]

I calculated it using 63A bs88-3 as didn’t have anything else to hand and come out at 3.89mm using 0.7 disconnection time.

 

[soulman]

Thats excellent, thanks for your help.

Just out of interest what would the adiabatic equation look like on a radial circuit with a 30ma rcd in circuit.

 

[Richard Burns]

If an RCD is being used for fault protection then the adiabatic equation would not apply to an earth fault as the requirement is then (for a TT) that Ra * Idn <50V.

However you could consider that the RCD disconnection time at say 442 A (perhaps using the 5 x idn value as measurement would be dangerous above this) could be used to ensure that the cpc was not damaged. This is just my thoughts not a recommendation.

 

[soulman]

Cheers.

 

[Leesparkykent]

You could use 0.04 as your max disconnection time as required by the standards, if the fault current exceeds 5xId. If the cable had a 1.5mm CPC and the value of K was 115.

115 squared= 13225

1.5 squared= 2.25

13225 x 2.25=29756.25 A2

Zs=0.43 ohms

230V/0.43=535A

535 squared=286225 A2

286225 x 0.04=11499 A2s

Therefore handling the thermal and mechanical stresses of the earth fault current during disconnection.

 

[soulman]

Hi Lee, that doesn’t make sense at all to me. I was told similar to what Richard burns has said. Rcd in circuit, line & neutral loop is all I should be concerned about. An electrician went through the Adiabatic with me with a 30ma Rcd in circuit, which he said was pointless, he used the pefc at the far end of the circuit. Time used was 300ms.
Cheers

 

[telectrix]

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