An Awful Feeling This

[gap30]

A BS EN 60898 circuit breaker has an lcn of 10 kA and an Ics rating of 6 kA.
State
a) how the circuit breaker is likely to perform if the circuit developed a fault current of 8kA (1 mark)
b) the likely operational condition of the circuit breaker following the fault (1 mark)
c) the action that would need to be taken to ensure the circuit was protected should a similar fault occur in the future. (1 mark)

This is one topic i have not covered in depth, correct me if i am wrong here

Uo/Zs = Ia

 

[La Poste]

This is one topic i have not covered in depth, correct me if i am wrong here
Uo/Zs = Ia

Correct.

 

[IQ Electrical]

A BS EN 60898 circuit breaker has an lcn of 10 kA and an Ics rating of 6 kA.
State
a) how the circuit breaker is likely to perform if the circuit developed a fault current of 8kA (1 mark)
b) the likely operational condition of the circuit breaker following the fault (1 mark)
c) the action that would need to be taken to ensure the circuit was protected should a similar fault occur in the future. (1 mark)

This is one topic i have not covered in depth, correct me if i am wrong here

Uo/Zs = Ia

Ics is the max fault value that a device can clear and be put back into service again.

Icn is the max fault value that a device can clear but the device will not function again.

A) The device will interrupt the fault.

B) The device will not be operational.

C) The protective device should be replaced.

 

[gap30]

Ics is the max fault value that a device can clear and be put back into service again.

Icn is the max fault value that a device can clear but the device will not function again.

A) The device will interrupt the fault.

B) The device will not be operational.

C) The protective device should be replaced.

i have just paid for a copy of the regs online so i will have a bit of reference material, so the circuit breaker would be faulty, thanks

i have been looking in gn3 for information on 100mA rcds, i know they are tested at times half and at times 1 but i cant find any info on tripping times, if i remember correctly its 200 ms for the bs one and is it 300 ms for 61009

 

[tony mc]

Just remember BS EN RCD have 300ms trip time

BS numbered RCD is 200 ms..

Remember the EN as extra time to trip.

You only do the 5x test on 30mA RCD

 

[gap30]

Just remember BS EN RCD have 300ms trip time

BS numbered RCD is 200 ms..

Remember the EN as extra time to trip.

You only do the 5x test on 30mA RCD

thanks

Determine the maximum permitted value of voltage drop for each of the following circuits.

a) A 230V lighting circuit supplied directly from a public low-voltage distribution system. (1 mark)
b) A 400V three-phase motor circuit supplied from a factory transformer. (1 mark)
c) A 230V immersion heater circuit in a domestic installation.

a) 3% 6.9v
b)
c) 5% 11.5v

i dont know the answer to b, i am assuming its 5%, but why the difference between a and c? i would have thought all 230v circuits would be 3%?

there is not a lot of info on vd in gn3, i am going to have a look through amberleafs sticky

 

[La Poste]

Value of resistivity for 1.5mm = 12.10 milliohms per metre
Value of resistivity for 10mm = 1.83 milliohms per metre

Therefore (R1&R2) resistivity in milliohms per metre = 12.10+1.83 = 13.93 milliohms per metre
Length = 28m
Factors = 1.02 & 1.20 (from tables 9B and 9C OSG)

Therefore, R1&R2 = (milliohms per metre x Length x Factor) / 1000

= (13.93 x 28 x 1.02 x 1.20) / 1000 = 0.48 ohms.

Zs = Ze + (R1&R2)

= 0.14 + 0.48 = 0.62 ohms

I have been thinking about this.

Value of resistivity for 1.5mm = 12.10 milliohms per metre
Value of resistivity for 10mm = 1.83 milliohms per metre

Therefore (R1&R2) resistivity in milliohms per metre = 12.10+1.83 = 13.93 milliohms per metre
Length = 28m

I agree, these figures are taken from table A.1 and are the value at 20C
Therefore at 20C our circuit resistance would be
13.93*28/1000 = 0.39 Ohms.

We have designed the circuit at 25C so if we were to measure the R1+R2 value at 25C we would need to add a correction factor because at 25C the resistance would be slightly higher, this correction factor is 1.02.
So the realistic resistance figure at 25C for our circuit would be:
.39 Ohms * 1.02 = 0.398 Ohms

This is what the correction factors for ambient temperature are for, they are to be applied to the tables EG table A.1 to change the values in those tables measured at 20C to whatever temperature the circuit is being measured at in reality.

Now The rule of thumb (the .8 rule) is used to multiply the tabulated figures in BS7671 which are measured at 70C
The 0.8 is used to bring the figures measured at 70C down to a figure that they would be if the circuit were measured at 20C.
Conversely if we want to convert our figure of R1+R2 measured at 20C to a figure measured at 70C we would multiply it by 1.2 this figure would then become our R1 + R2 measured at 70C and we would add Ze to this. We would then be able to compare this figure directly to the tables in BS 7671.

So in this instance we need the figure for the circuit we are designing, the R1 + R2
measured at 20C.
We calculate what this figure is by using the above formula EG:
13.93*28/1000 = 0.39 Ohms.
This is the figure at 20C.
We then multiply the figure at 20C by a factor of 1.2.
0.39 * 1.2 = 0.468 Ohms.
Add this to Ze
Zs = .14 + .468 = 0.608 Ohms.
This is the figure we would use to compare to figures in BS 7671.
We would not use the ambient temperature figure in this calculation.
The only time we would use the ambient temperature correction factor of 1.02 was if we were measuring our R1+R2 at 25C and we needed to compare these measured figures with those in the book which were recorded at 20C
We would take the figures in the book measured at 20C multiply them by a factor of 1.02 and this would give the value of the table figures at 25C.
We would then be able to compare our measured reading at 25C directly to these tabulated figures.
We would not use both the factors of 1.02 and 1.2 together in the same calculation.

This is how it appears to me, shoot me down if I am wrong.

One more thing, I noticed a nice little fact today.
The resistance of copper will increase by 2% for every 5C increase in temperature.
So if in doubt use this way to measure what the resistances would be at different temperatures.
If a cable increases in temperature from 20C to 70C its resistance will increase by 20%

 

[La Poste]

thanks

Determine the maximum permitted value of voltage drop for each of the following circuits.

a) A 230V lighting circuit supplied directly from a public low-voltage distribution system. (1 mark)
b) A 400V three-phase motor circuit supplied from a factory transformer. (1 mark)
c) A 230V immersion heater circuit in a domestic installation.

a) 3% 6.9v
b)
c) 5% 11.5v

i dont know the answer to b, i am assuming its 5%, but why the difference between a and c? i would have thought all 230v circuits would be 3%?

there is not a lot of info on vd in gn3, i am going to have a look through amberleafs sticky

B is also 5% I think, it's not a lighting circuit so it is 5%

I do not know the reason for the difference in the 3% and the 5% I just know it's a fact set in stone so I don't argue with it.

 

[IQ Electrical]

thanks

Determine the maximum permitted value of voltage drop for each of the following circuits.

a) A 230V lighting circuit supplied directly from a public low-voltage distribution system. (1 mark)
b) A 400V three-phase motor circuit supplied from a factory transformer. (1 mark)
c) A 230V immersion heater circuit in a domestic installation.

a) 3% 6.9v
b)
c) 5% 11.5v

i dont know the answer to b, i am assuming its 5%, but why the difference between a and c? i would have thought all 230v circuits would be 3%?

there is not a lot of info on vd in gn3, i am going to have a look through amberleafs sticky

It's 8% from a private supply source.

The reason for the lower level for lighting is that some types of lighting cannot strike up on lower voltages so tolerances are tighter.

 

[brucelee]

Hi

for single phase 3% for lighting 6.9v
5% for power circuits and distribution 11.5v
and for the answer to b) its 20V which is 5% of 400v

no sorry i am wrong its not 20V as its there own supply not sure

 

[tony mc]

thanks

Determine the maximum permitted value of voltage drop for each of the following circuits.

a) A 230V lighting circuit supplied directly from a public low-voltage distribution system. (1 mark)
b) A 400V three-phase motor circuit supplied from a factory transformer. (1 mark)
c) A 230V immersion heater circuit in a domestic installation.

a) 3% 6.9v
b)
c) 5% 11.5v

i dont know the answer to b, i am assuming its 5%, but why the difference between a and c? i would have thought all 230v circuits would be 3%?

there is not a lot of info on vd in gn3, i am going to have a look through amberleafs sticky

Have a look at Appendix 12 in BS 7671 it covers VD only.

 

[La Poste]

Hi

for single phase 3% for lighting 6.9v
5% for power circuits and distribution 11.5v
and for the answer to b) its 20V which is 5% of 400v

no sorry i am wrong its not 20V as its there own supply not sure

I'm guessing it's 5% per phase so it's 11.5V at each phase where the equipment is installed, not measured between phases.
But I'm guessing.
Sorry it's 8% from a private supply source (This is a new one for me) so I am guessing it is 8% per phase = 18.4 V
But it's just a guess so hopefully I'll be put right by someone.

 

[tony mc]

Have a look at the question as it tells you the voltage is 400v three phase.

So its 8% of the 400v if the low voltage installation is supplied from a private low voltage supply.

Hope this helps.

 

[IQ Electrical]

Have a look at the question as it tells you the voltage is 400v three phase.

So its 8% of the 400v if the low voltage installation is supplied from a private low voltage supply.

Hope this helps.

Spot on, the percentage is actually of 400 Volts ie. 32 Volts

 

[gap30]

the loop length for circuit 1 is 50 m long the ze is is 0.11 ohms

determine showing all calculations the r1 + r2 and the expected zs

7.41 x 50/1000 = 0.37 12.10 x 50/1000 = 0.60

zs = ze + (r1 = r2)

0.11 + (0.37 + 0.60) = 1.08

have i done this correctly?

 

[IQ Electrical]

the loop length for circuit 1 is 50 m long the ze is is 0.11 ohms

determine showing all calculations the r1 + r2 and the expected zs

7.41 x 50/1000 = 0.37 12.10 x 50/1000 = 0.60

zs = ze = (r1 = r2)

0.11 + (0.37 + 0.60) = 1.08

have i done this correctly?

I can't confirm the mOhm values but the maths is correct.

You MUST use (R1+R2) when doing these calculations and remember that r1, rn and r2 are reserved for ring continuity figures, this exam is very big on correct terminology.

 

[IQ Electrical]

Hang on, is this a ring final circuit?

 

[gap30]

Hang on, is this a ring final circuit?

yes i just shortened the question the m/ohm/ for 2.5 is 7.41 and 1.5 is 12.10

i am not great at the terminology but i am getting there with the calculations

in some crazy way i am actually finding the long questions easier than the short ones atm

 

[IQ Electrical]

If it's a ring then the R1+R2 value is r1+r2/4

R1+R2=0.37+0.60/4

R1+R2=0.24 Ohms

Zs=(R1+R2)+Ze

Zs=0.35 Ohms

 

[gap30]

Have a look at the question as it tells you the voltage is 400v three phase.

So its 8% of the 400v if the low voltage installation is supplied from a private low voltage supply.

Hope this helps.

just ordered a set of regs from fleabay so they should be with me in the next couple of days - the only problem i found with the regs in the past was navigating them, i always favoured the on site guide for obvious reasons