Can someone help?

[Phil Griffiths]

so currently doing my level 3 diploma and have this question I'm stuck on can anyone help?

a radial circuit is protected by 20A type B circuit breaker and is wired using 2.5mm2 70C thermoplastic single core live conductors with 1.5mm CPC to a length of 28mm in steel trunking. ambient temp is 20C supply is 230V single phase TNCS system. Ze & prospective fault current is recorded at 0.1 ohms and 2.3KA.

- already worded out R1 and R2 for the circuit which is 0.546.
- NEED to find - value of short circuit at end point of the circuit?
- AND will the circuit breaker disconnect within the required time under fault conditions.?

Cheers

 

[telectrix]

i assume you mean the value of short circuit current\/ if so, ohm's law.V =230V , R = 0.546, so what is I ?

then read off the graph for time/current for the 20A type B breaker in app. 3 of BS7671 and see what the time to break is for that calculated current.

 

[Phil Griffiths]

i assume you mean the value of short circuit current\/ if so, ohm's law.V =230V , R = 0.546, so what is I ?

then read off the graph for time/current for the 20A type B breaker in app. 3 of BS7671 and see what the time to break is for that calculated current.

thanks dude, so heres what i have so far, for a 20A type b breaker from table 3A4 i got 100A. so .. 230 divided by 100x0.546 = 4.212 this sound about right ?? and will this disconnect the circuit breaker within the required time? thanks again

 

[westward10]

You have your "I" calculation wrong.
I=V÷R
Calculate this first then refer Figure 3A4.
You will also need to factor in the value of Ze.
I=V÷(R+Ze)

 

[telectrix]

what westwood10 means is that your calc. for Isc is V/R, where V =230 and R = 0.546 + 0.1..... ( R1+R2 + Ze). i forgot the Ze, but then I'm 70 and entitled to a bit of senile dementia coz i've earned it and nobody's going to take it away. me away, yet.

 

[westward10]

I edited in the Ze

 

[Phil Griffiths]

You have your "I" calculation wrong.
I=V÷R
Calculate this first then refer Figure 3A4.
You will also need to factor in the value of Ze.
I=V÷(R+Ze)

right ok silly mistake, so heres what i have done so far, I = 230 Divided by 0.546 + 0.1 = 421.35. Looking up the information in figure 3A4, for a 20A type B circuit breaker the prospective current is 100A does this mean the RCD will continually trip due to high demand? thanks, really appreciate it!

 

[telectrix]

no. at 100A fault current level, the breaker will trip within 0.4 secs, but you have 421A, so it will trip even quicker. so, basically any fault current above 100A will comply with BS7671. bear in mind that the breaker will only trip once, as once tripped, it will not reset until the fault has cleared. do not confuse fault current ( dead short ) with over current ( current greater than the circuit is designed for,but not a short. ).

 

[westward10]

Still calculating it wrong. The short circuit value of a line/earth fault is dictated by the combination of your 0.546 (which is R1+R2) and Ze. This basically gives you the value of Zs for the circuit. The magnitude of fault current is dictated by this resistance(Zs). In other words if you get a live to earth fault you need sufficient current to flow to disconnect the circuit, this is your "earth fault loop path".
Ignore the 20A device and Figure 3A4 for now and calculate your fault current.
I=V÷(0.546+Ze).

 

[Phil Griffiths]

no. at 100A fault current level, the breaker will trip within 0.4 secs, but you have 421A, so it will trip even quicker. so, basically any fault current above 100A will comply with BS7671. bear in mind that the breaker will only trip once, as once tripped, it will not reset until the fault has cleared. do not confuse fault current ( dead short ) with over current ( current greater than the circuit is designed for,but not a short. ).

thanks very much mate!

 

[Phil Griffiths]

Still calculating it wrong. The short circuit value of a line/earth fault is dictated by the combination of your 0.546 (which is R1+R2) and Ze. This basically gives you the value of Zs for the circuit. The magnitude of fault current is dictated by this resistance(Zs). In other words if you get a live to earth fault you need sufficient current to flow to disconnect the circuit, this is your "earth fault loop path".
Ignore the 20A device and Figure 3A4 for now and calculate your fault current.
I=V÷(0.546+Ze).

right ok, you have lost me now if I'm honest, don't forget I'm still learning haha, i have a fault current at 2.3KA and a Ze of 0.1 ohms, do i apply the 2.3 KA to the equation? sorry for my knowledge

 

[soms]

I am studying for my L3 NVQ Diploma and we were doing the same sort of exercises the other day.

We made reference to BS7671 table 41.3, which gives you the maximum acceptable Zs values for the circuit breakers to operate in an 0.4s disconnection time. When looking at one protective device this saves looking at characteristic curves.

These values are for 70 degree C operation, so you need to apply a multiplier for the R1+R2 calculation to change your 20 deg C values to 70 deg C values. Look at OSG Table I3.

The earth fault current (not short circuit) in Amps for the end of the circuit can be calculated by dividing the nominal voltage to earth (U0) by the calculated Zs.

 

[telectrix]

no.that fault curren of 2.3kA is at the origin i.e. 230/Ze. at the end of your circuit, you have an extra 0.546 oms to add to the 0.1 Ze, so the fault current will be lower, due to the extra resistance. aw wetie said, Ipf = V/ ( Ze + (R1+R2), so it's 230/ (0.1+0.546) = 230/0.646 = 356A.

 

[westward10]

Have you been shown the principles of automatic disconnection of supply or the earth fault loop path.

 

[Toneyz]

Pete where is one of you video's he does a good one on this