An Awful Feeling This

[gap30]

that was the sum as a whole doing the lot and dividing by 1000

12.10 + 18.10 = 30.20

Zs = Ze + m/ohms/m x mf x length/1000
Zs = 0.55 + 30.20 x 1.2 x 10 =362.95
362.95/1000 = 0.36295

Zs = Ze + (R1 +R2)
Zs = 0.55 + 0.36 = 0.91
Zs for the circuit is 0.91 ohms

so i have to work out the sum BEFORE i divide it by 1000

i could do with a few more of these questions to get it right

 

[gap30]

) State ↔TWO↔ likely conditions that may have an effect on the value of earth fault loop impedance ( **** )
c) State ↔ ONE ↔ method that can be used to compensate for the conditions in ↔ b ) ( ***

1. conductor temperature
2.ambient temperature

applying correction factors or multipliers

 

[malcolmsanford]

BODMAS look it up,

You do the brackets first which is R1 + R2 so 30.2 x 1.2 x 10 / 1000 which is 0.36 then you do the whole equation Zs = 0.55 + 0.36 and the answer is 0.91.

 

[gap30]

so i already had the R1 = R2 figures i then had to add muliply factor and lenth/1000

 

[gap30]

d) Determine whether the value in ↔ a ) is acceptable if the max ( Zs ) for a 10A BS-EN 60898 circuit breaker is 4.6Ω ( **** )

using the rule of thumb

0.8 x 4.6 = 3.68 so the value is acceptable

 

[malcolmsanford]

Yes they gave you the m/ohm/m for each cable 1.5 which was 12.10 and the 1.00 which was 18.10 so instead of doing
R1 = 12.10 x 1.2 x 10/1000 = 0.14
R2 = 18.10 x 1.2 x 10/1000 = 0.21 add them together and you get 0.35

The normal way to do it is add the 12.10 + 18.10 to make 30.20 and then 30.20 X 1.2 X 10 / 1000 = ​0.36.

 

[malcolmsanford]

d) Determine whether the value in ↔ a ) is acceptable if the max ( Zs ) for a 10A BS-EN 60898 circuit breaker is 4.6Ω ( **** )

using the rule of thumb

0.8 x 4.6 = 3.68 so the value is acceptable

Yep it is

 

[Sparky Ninja]

i could do with a few more of these questions to get it right

I'll make one up..

Your designing a circuit 28m in length.
Ambient temperature is 25 degrees
You are using A 10mm Line conductor and a 1.5mm cpc.
Cables will be installed in surface fixed pvc conduit.
Ze at the origin is 0.14 ohms.

Calculate Zs?

 

[malcolmsanford]

I'll make one up..

Your designing a circuit 28m in length.
Ambient temperature is 25 degrees
You are using A 10mm Line conductor and a 1.5mm cpc.
Cables will be installed in surface fixed pvc conduit.
Ze at the origin is 0.14 ohms.

Calculate Zs?

please sir me sir ............................:smart:

 

[La Poste]

Zs = Ze + ( R1 + R2) so you know your Ze is 0.55

so Zs = 0.55 + ( R1 + R2 ) therefore 0.55 + m/ohm/m x mf x 10mt /1000 your m/ohm/m is in table 9A of the OSG page 166 and also the same table is in your GN 3 so for 1.5/1.0 the m/ohm/m factor is 30.20

so the formula is 30.20 x 1.2 x 10/1000 which is 0. 36

Zs = 0.55 + 0.36 = 0.91

Please tell me if I am reading this right.
To calculate the Zs in the first part of the equation you use a multiplying factor of 1.2 ?
This multiplying factor is because r1+r2 are measured at 20 degrees C ?
This multiplying factor will compensate the fact the r1 + r2 was measured at 20 degrees C and make the result comparable to the tables in BS7671.

d) Determine whether the value in ↔ a ) is acceptable if the max ( Zs ) for a 10A BS-EN 60898 circuit breaker is 4.6Ω ( **** )

using the rule of thumb

0.8 x 4.6 = 3.68 so the value is acceptable

When you compare this figure to the figure stated in BS7671 which is 4.6 Ohms
you use another correction factor of .8
Why are you using two sets of correction factors?
firstly a 1.2 and then secondly a 0.8?

I thought the reason the 1.2 multiplier was used is so the results of the R1 + R2 calculation could be directly compared to the tables in BS 7671 which are measured at 70 C.
If the multiplier of 1.2 is not used then we have to use the multiplier of 0.8 applied to the tables in BS 7671 to adjust the results in the tables down to 20 C, then we can compare these tables to our measured ZS at 20 C.

I thought the 1.2 was to correct the resistance from 20 C up to 70 C.
I thought the 0.8 was to correct the resistance from 70 C down to 20 C.

I do not know why we are using both sets of multipliers, could someone please explain?
Thanks.

 

[La Poste]

Table A.3 Page 105 Guidance note 3:

"Table A.3 gives the multipliers to be applied to the value given in table A.1 for the purpose of calculating the resistance at the maximum operating temperature of the line conductors and /or CPC in order to determine compliance with the earth fault loop impedance of tables 41.2,41.3,41.4 of BS 7671."

The multipliers being 1.2 for 70 C thermoplastic insulation.

So this says to me that if you use the 1.2 multiplier then there is no need to also use the 0.8 multiplier.

You either use one or the other not both.

If you use 0.8 you do not use 1.2
If you use 1.2 there is no need to use 0.8

Any thoughts people?
I have to get this right myself also.

 

[malcolmsanford]

Your correct yes there is no need to use both methods. The 1.2 factor is normally a design criteria used for designing your circuit and so if you used the 1.2 you would not need to use the rule of thumb.

I would use the rule of thumb on an actual measured Zs test. In the 2391 often questions will be one or the other. The question that the OP posted would be more of a design question and the C & G would expect to see you use the factor.

Another question would be, especially in section B, a list of maximum Zs values taken from the tables in BS 7671-2008 and then a list of measured Zs, they would then ask if these results are acceptable and show your calculations, and that is when you use your 0.8 rule of thumb.

 

[La Poste]

One question.
When a 2391 question asks for me to calculate the ZS of a circuit and gives me the Ze and the resistivity of copper plus the circuit lengths much the same as the previous question we are talking about here would it be correct to use the 1.2 multiplier.
Do City and guilds want the 1.2 multiplier used in a general Zs calculations or do they just want the Zs measured at 20 C?

 

[malcolmsanford]

It's been a long time since I did my 2391-001 so I may be wrong, and I'm sure Widdler will correct me if I'am wrong but if they have given you Ze and the resistivity of the cable and most likely the 20c temperature then yes I would calculate it with the temperature factor of 1.2.

 

[Sparky Ninja]

It would depend on if the question is asking for it measured or at operating temperature.If you work out both and label them then you can't go wrong.