CALCULATION of Zs

[GaryMo]

Zs = (Ze+(R1+R2/1000))x L x (9A figure) x (9C figure)
Therefore
Zs = 0.35+((19.51/1000)x 20 x 0.96 x 1.2)

Ze shouldn't be multiplied by the length of cable, only the (R1+R2)/1000 figure. You may know that and it's probably just how you've typed the formula that's the problem.

 

[Spark123]

I'd avoid using R1+R2/1000 in your calculations as it may be confusing, it is only /1000 if it is a milliohm value (for example from 9a) which are usually from the tables.
9b is quoted as figures at 20ºC
9b would be applied to table 9a if you were wanting to use those figures at 10ºC, for example if you wanted to know what the expected MEASURED value at 10ºC for the calculation before you could work it out by
19.51mΩ/1000 x 0.96 x 20m = 0.374592Ω

0.35Ω + 0.374592Ω = 0.724592Ω , so you could expect a 20m 2.5/1.5mm radial with a Ze of 0.35 to measure somewhere in the region of 0.72 ohms when measured at 10ºC. This is comparable with the figures for maximum measured in the OSG or GN3.
They cannot be direcly compared with the tables in chapter 41 of BS7671:2008.

 

[SgtTrojan]

Agreed. It can be confusing for the kids (as well as me atm ) to have all that in.

This leads me intop my next question on the subject. Let's assume your 2.5mm T+E and expected temp at 10*C

1. It will be easier to calculate R1+R2 first so is this correct?

R1+R2 = (table 9a values)/1000 x L x (table 9B value) x (table 9C value)
Therefore
R1+R2 = (19.51/1000) x 20 x 0.96 x 1.2 = 0.44 Ohms

Add that to ya Ze and BOb's ya uncle, Zs is calculated??

FIngers and toes crossed

 

[SgtTrojan]

The more I look at this the more I am thinking you only use either 9B or 9C and not both together.

Applying multiplier 9C "corrects" 9A's values to 70*C so why apply another factor to anticipate the expected value at 10*C

Gahhhh

 

[17thman]

Assuming a site ambient tempareature of 10C


the 0.96(b) corrects 9a values from 20C down 10C and the 1.2(c) corrects from 20C to 70C

Table 9a refers to resistance at 20C, so you would need to adjust by 0.96 to get to what expected at 10C. BS 7671(table41) values are at 70C so you need to adjust up from 20 C to 70C.
So a combination of both adjust it from 10C to 70C

I must say that i initially got confused about verifying that your circuit complied with what you really wanted which was designing a circuit and i got in a bit of a mess. With the risk of giving you another bum steer i think what i have typed will clear it up

 

[SgtTrojan]

Excellent. (My initial title of Verification may have been misleading)

So does it make more sense to only apply 9C at the design stage?

I mean.. why work out a test reading at 10*C by correcting the 9A value down to using 9B ..... only then to push it back up to 70*C using 9C?

It doesn't add up

9B Ambient temperature factor
9C Multipliers to conductors max imum operating temperature

 

[17thman]

i suppose if you designed at 9c then the tester could use 9b if temperature was 10C on site.

 

[SgtTrojan]

Ok..that makes sense. So do we conclude with...

9B is used for verification of MEASURED values but can be used for calculating an expected value of a given temperature and
9C is used to CALCULATE the worst case scenario of the conductor's maximum operating temperature and is not used together (in the same calculation) with 9B?

 

[Spark123]

That is about the size of it, you wouldn't want to use 9B and 9C together as the result is the resistance of the cable at 60ºC

 

[SgtTrojan]

Glorious. Thanks you for your expert knowledge and patience with my level of understanding.

I'm going to put this into my notes. I will remember you both in the bibliography (you may recognise some of this )

Why use multipliers?
Values in 9A are quoted as figures at 20ºC only.

E.g. You wish to install a circuit consisting of 20m of 2.5mm2 twin and earth (1.5mm2 CPC) rated to 70ºC and you measure a Ze of 0.35Ω Table 9A of the OSG 2.5mm2 twin and earth has a resistance value of 19.51mΩ/m

9B values could be applied to table 9A values if you want to calculate 9A values at 10ºC for example. This means you want to know what the expected MEASURED value will be at 10ºC. For the circuit above you could work it out by:
19.51mΩ/1000 x 0.96 x 20m = 0.374592Ω

As stated above the figures in the OSG Table 9A are for a conductor at 20ºC which are not much use when calculating at the design stage as we need to know what the resistance is at the maximum operating temperature. We can use 9B or 9C for this.
E.g. Factor for expected test value at 70oC is 1 + (70 – 20) x 0.004 = 1.2

This means that when the cable is at 70ºC the resistance of the conductors will be 1.2 x greater. So our circuit conductor resistance at 70ºC will be 19.51mΩ/1000 x 1.2 x 20m = 0.46824Ω.Add this to Ze and we now know what the expected total earth fault loop impedance will be at 70ºC. 0.35Ω + 0.46824Ω = 0.81824Ω

As the above value is for a circuit at its max operating temperature it is directly comparable to the figures in the tables in Chapter 41 of BS7671:2008

9B is used for verification of MEASURED values but can be used for calculating an expected value of a given temperature
9C is used to CALCULATE the worst case scenario of the conductor's maximum operating temperature and is not used together (in the same calculation) with 9B

 

[Spark123]

In the final para, you can use 9B to correct the figures in table 9A to 10ºC which when added to Ze are comparable to the maximum measured values of EFLI in appx 2 of the OSG or in GN3.
Similarly, you can use the figures in table 9C to correct the figures in table 9A to the resistance at their maximum operating temperature.
Technically, 70ºC is only the worst case if it is a 70ºC cable.
If the cable is 90ºC then the figure can be higher, the tables for max efli (chapter 41) then need to apply to the cable at 90ºC, unless you choose to limit the cable temp to 70ºC by using the 70ºC current carrying capacities.

I'll take another peep tomorrow when I get a bit more time.