adiabatic equation a lot of questions.

[ulfr]

Good evening everyone,
I'm doing 04A1 practical assessment at the moment, and I'm confused with adiabatic equation. It seems easy when tutor is around, but when you're trying to sort it out on your own, a lot of questions appears) I would really appreciate if you could help me out.

Here is one question, please delete it if it's inappropriate here.

8)The circuit supplying lighting to the W.C, office, etc (Gr4) will be wired in PVC singles in conduit and trunking to reference method B.The circuit length is 20m, and the trunking contains 3 other circuits. Due to the positioning of the conduit under the light panels in the roof the ambient temperature will be taken as 25[SUP]o[/SUP]C.

a)Using BS7671, design the circuit to supply this equipment. Note all tables you reference.

Procedure:

Ib- design current.

WC = 28W, Store(flourescent) =36W, Office(flourescent) = 58W, PIR = 84W
28*1.8 + 36*1.8 + 58 + 84 =257.2W
Ib = Watts / Volts = 257.2W /230V = 1.12A

In- rating of the protection.
In ≥Ib ( Appendix F 433.1.1) ;
Circuit breaker: 6A 60898 Type B breaker.
In = 6A

It- tabulated current-carrying capacity.
Reference method B trunking,singles, L=20m, 4 total circuits (1 + 3), ambient temp = 25oC;
Ca = 1.03 (Table F1, ambient temperature 25[SUP]o[/SUP]C),
Cg = 0.65 (Table F3. 4 circuits,reference method B)
It ≥In/CaCg; It ≥6/1.03x0.65; It ≥8.96A

Cable size:
From Table F4(i)(Reference method B): 1mm2 = 12A.
12A ≥9.23A
So 1mm2 is sufficient for the circuit.

Voltage drop:
Appendix F
voltage drop= ((mV/A/m)xLbxL)/1000
Table F4(ii)Reference method (6) A&B = 38 mV/A/m for 1mm2 cable, hence:
vd=(38*1.12*20)/1000=0.85V and allowed is 3% (6.9V);

b)Select an appropriate minimum sized CPC, and confirm your selection by calculation. Note all table syou reference.

Liveconductors are 1mm2 according to 543.1.1(The cross-sectional area of every protective conductor, other than a protective bonding conductor, shall be: (i) calculated in accordance with Regulation 543.1.3, or (ii) selected accordance with Regulation 543.1.4) in accordance with Regulation 543.1.4 the minimum CPC should be the same as the live conductor = 1mm[SUP]2[/SUP]

But in accordance with Regulation 543.1.3, calculations should be the following:
Check the CPC by using the adiabatic equation:

Table 4.6 of OSG Column 3 for 1mm[SUP]2[/SUP] gives 1mm[SUP]2[/SUP].
Table B7 of OSG for Type B gives 1mm[SUP]2[/SUP] for fault leve <= 3kA

k=115 (Table 54.3)
Zs=Ze+(R1+R2), Ze = 0.35Ω(TN-C-S max), R1+R2 = 36.20mΩ * 20m = 724mΩ = 0.724Ω (Table I1 for mΩ/m)
Zs=0.35 + 0.724 = 1.074Ω
I=Uo/Zs = 230 / 1.074 = 214.2A
t=0.1s (Fig 3A4)

S=(sqrroot I[SUP]2[/SUP]t)/k ; S=(sqrroot 214.2[SUP]2[/SUP]*0.1)/115=0.59mm[SUP]2[/SUP] therefore 1mm[SUP]2[/SUP] is enough.

Questions are:
1) If I (earth fault current) is 214.2A do we have to change our Circuit breaker: 6A 60898 Type B (30A) to a breaker that handles 214.2A+ ?
2) Our tutor said that the minimum CPC in singles should be 1.5mm[SUP]2[/SUP] do I have to use that even if 1mm[SUP]2[/SUP] is sufficient?
3) I've read on the forum that:
If you are using the tables in Appendix 3, remember to use the current values in the table, rather than the measured value. So what about graph values?
4) Tutor showed another calculation method without Zs, he toot I=circuit breakers max current(30A for TypeB), and t=max TN-C-S time(because in the assessment it's that) = 5s (or it could be that he took the maximum TypeB time which is 5 as well), and S= (sqrroot 30[SUP]2[/SUP]*5)/115 = 0.5833mm[SUP]2
[/SUP]It seems confusing to me that there are 2 ways of calculating the equation.

Much appreciated.

 

[Simonslimline]

Use the time it takes the OCPD to operate with the PFC value you have.

The times can be found in Appendix 3 of BS7671, Time/current graphs.

 

[Leesparkykent]

The adiabatic equation is given in Regulation 543-01-03:

S = the cross-sectional area of the protective conductor
I= the fault current
t =the operating time of the overcurrent device corresponding to the fault current
k = a factor which takes account of the resistivity, temperature coefficient and heat capacity of the conductor material

The disconnection time, t, is found by reference to the appropriate time/current characteristic of the OCPD. This is found in Appendix 3. These time/current curves make use of logarithmic scales, which enable large scales to be used in a relatively small area. The value of k is obtained from The BGB. Fault current (I)= volts/Zs.

Determining the cross-sectional areas of protective conductors can be selected using the adiabatic equation or using Table 54G.

 

[ulfr]

Thanks guys, much appreciated.
Can you please elaborate this:
If I (earth fault current) is 214.2A do we have to change our Circuit breaker: 6A 60898 Type B (30A) to a breaker that handles 214.2A+ ?
Does it means that a simple 6A circuit will need 50A Type B breaker if PFCC is 214A ?!

 

[Leesparkykent]

No, the value of the kA rating determines how much current the OCPD can withstand under fault conditions. The OCPD only has to withstand this for a brief period of time.

 

[ulfr]

No, the value of the kA rating determines how much current the OCPD can withstand under fault conditions. The OCPD only has to withstand this for a brief period of time.

Table B7?

 

[Leesparkykent]

Table B7?

Sorry mate haven't got the book on me. For example the KA rating is normally stamped on the front of the MCB, the most common KA ratings on an MCB is either 6KA or 10KA (KA meaning kiloamps).

 

[ulfr]

Thanks a lot mate, it starting to make sense.
So, for example Type D RCBO 6A - can protect a 6A circuit against overcurrent up to 120A, but against PFC up to 3KA-6KA ?
Can't see the real difference between them(

 

[Richard Burns]

In your example question you mention 38 mV/A/m for volt drop but that is for 3 phase circuits; for single phase it is 44 mV/A/m.
Table 4.6 of the OSG is for supplementary bonding and not cpc.

A 6 A type B circuit breaker to BSEN60898 has specific characteristics defined by the standard:
It can carry 6 A indefinitely without tripping (the current rating of the MCB).
It will trip "instantaneously" at or above a value of fault current between 3 to 5 times its current rating (i.e. between 18 A and 30 A)
One uses the worst case for calculations (and the graphs in appendix 4) so 30 A is the value used as the minimum current to cause instantaneous tripping.
It will be designed to safely handle a very large fault current for a short period of time this is the maximum 6000 A (6 kA) rating.

Regarding your questions
1) The 6A circuit breaker can handle fault currents up to 6kA so 214.2A is not a problem.
2) Singles cables are manufactured only down to 1.5mm² mainly for the mechanical strength required as the cable is manipulated so this would be the minimum physical size available.
3) and 4) Calculation using the adiabatic equation:

There are a variety of ways to look at this / calculate this because you can use different levels of fault current in the calculation. The important item to remember is that the time used in the calculation should match the fault current used in the calculation, i.e they should derive from the same source of information.

From the tables and graphs in Appendix 4.
In your example you have a fault current of 214.2 A.
If you look on the graph at the 214.2 A level you will see that the values is way beyond the line for a 6A breaker so you cannot read off a disconnection time from that graph to use in the adiabatic equation for that fault current.

If you look at the table to the right for a 6 A breaker, at a fault current of 30 A the breaker will disconnect in 0.1 seconds, this matches the line on the graph at 30 A.

Because a circuit breaker disconnects instantly at a certain fault current the 30 A value applies to both 0.4 s and 5 s maximum disconnection times required by table 41.1. This can lead to difficulty because you could theoretically also use that 30 A value for a disconnection time of 5 s (max for TNCS /TNS distribution circuits and so on) as your tutor may have done in your post.

So from the information in figure 3.4 you can use:
I=30 A, t=0.1 s, K=115 and obtain a minimum cross sectional area of 0.08 mm²
This may seem very small but remember this is only to determine if the cable can carry 30 A for 0.1 s.

If you were using a cpc incorporated in a cable then you could use a 1 mm² cpc, this is the minimum size stipulated in table 52.3.
Because you are using singles the minimum manufactured size is 1.5 mm².

For your last post
Type D RCBO 6A - can protect a 6A circuit against overcurrent up to 120A, but against PFC up to 3KA-6KA
A 6 A type D RCBO can carry 6 A indefinitely without tripping, will trip instantaneously at 120 A and can safely break the circuit at a fault current up to a maximum of 3kA to 6kA, above this level the circuit breaker may not be able to contain the energy and could cause danger by arcing, burning or bursting, etc.

 

[ulfr]

Thank you so much for your time. You've helped a lot! Much appreciated.
Can buy you a pint if you're from London ))