# CALCULATION of Zs

Discussion in 'Electrics UK' started by SgtTrojan, 18 Mar 2009.

1. ### SgtTrojan

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Hello chaps,

Would someone be able to confirm something for me (I am writing course notes for the training centre I work at).

When calculating Zs and we want to be sure that disconnection times are met we use the calculated value Zs=Ze+((R1+R2) x multilpier) and ensure it is below the relevant value listed in the relevant table in Section 41.

When measuring Zs we compare our measured value to a "corrected"** value.
(**corrected by applying 0.8 to the relevant value in the relevant table listed in Section 41)

Not including the other means of verification, is the above correct?

2. ### 17thman

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Your measued or calculated Zs =(Ze+(R1+R2)) value must be at or lower than 0.8 of the value stated tables 41.

So for a type b 32A your Ze value will need to be at or lower than 0.8 x 1.44 ohms

ie 1.15 ohms.

I see the multiplier added in your formula. if your multiplier was 1.25 then you would just need to be on or under the tabulated value

3. ### SgtTrojan

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Is your multiplier (1.25) from 9B or 9C? The multipliers to be applied to the R1 and R2 values in table 9A?

4. ### 17thman

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i was assuming a site ambient temperature of 10 degrees

Ztest =Ztable 41 x 0.96(from 9b)/1.2(from 9c)

so alternatively 1.2/.96 = 1.25 for applying to calculated Zs

10 degrees is more likely as an ambient temperature. in my view.

5. ### SgtTrojan

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I'm slightly confused by that representation there

To calculate Zs for a circuit where conductors are 70*PVC and incorporated and expected ambient temp of 10 degrees is it:

Zs = Ze + ((R1+R2/1000) x 0.96 x 1.20)

6. ### 17thman

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The 0.8 multiplier referred to earlier is a good rule of thumb to see if your Zs is <or= to the BS tabulated requirments.

Zs <or= .8 x Zstable.

or Zs x 1.25 <or=Zstable.

The 0.8 is a rough figure. There are the other factors like correcting for ambient temperature 0.96 for 10*C

Zs =(Ze+(R1 + R2)) then apply your multipliers/divisors accordingly.

My last post was about seeing if your measured Zs complied. Your last post is seeing if your calculated Zs with mulipliers complied. So i think it should be

Zs of (Ze+(R1+R2/1000)) x 1.2 / .96 must be equal to or less than Zstable 41.

as you have added /1000 to the equation hence if you are taking the values for R1 and R2 from 9A alone you will need to incorporate a multiplier "L" for the length of conductor in metres.

7. ### SgtTrojan

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Ahhaaaa
I see where the 1.25 comes from now. It is 1/0.8
Thank you

What I am fishing for is why one of the multipliers is a divisor and not a multiplier

And oops.. yes I forgot to put the length in the formula

8. ### 17thman

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if Zs of (Ze+(R1+R2/1000))x L x 1.2 / .96 must be equal to or less than Zstable 41.

You convert 1.2 to .833 and move it to the other side of the equation.

You get.
Zs of (Ze+(R1+R2/1000)) xL must be equal to or less than (Zstable 41 x.833 x.96)

Measured Zs is what it is
Calculated Zs is what it is without the mutlipliers.

Apply the multipliers to Zstable 41 figure to see if your circuit is compliant.

ie 1.15 = 1.44 x .833 x .96.

or 1.15 x 1.2/.96 = 1.44

9. ### SgtTrojan

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Ok NOw I am really confused

1. To calculate Zs we need to use the multipliers. That must be correct?
2. Is the formula below correct to calculate Zs at the design stage?
Zs = Ze + ((R1+R2/1000 x L x 0.96 x 1.20))

Bold relates to R1+R2 only

10. ### Spark123

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OK, lets start at the beginning!
For the tables for max Zs in BS7671:2008 are the maximum permissable at the maximum operating temperature of the cable.
For example for a 32A B type MCB it is 1.44ohms.
As the conductor will not be at its maximum temperature when measured, for example a 70ºC cable may be at 10ºC meaning there is a 60ºC difference between the maximum and the measured values.
From the simplified temperature coefficient for copper conductors, we can feed this into the equation which gives us a correction factor 1+(0.004 x 60) = 1.24, hence the cable will a 1.24 x lower resistance at its cooler temperature.
If we take the figure we had for the 32A B type MCB before (1.44ohms) and divide this by 1.24 we are left with 1.16ohms which is therefore the maximum permissable measured EFLI @ 10ºC for a type B 32A breaker.

Do you have a copy of guidance note 3? This is explained in the one of the final 2 appendixes in the back iirc, there is another method for correction of temperature explained in there for adjusting the Zs of the conductors owing to temperature, in this case where Ze is known it can be deducted at the start of the equation and added back on at the end.

11. ### SgtTrojan

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I understand that. Thanks I can get a GN 3from somewhere I'm sure.

But I still don't know how to calculate the expected Zs at the design stage.

12. ### 17thman

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it is Zs = (Ze+(R1+R2/1000))x L

the factors are there to take into account the difference in measurement when testing the circuit and it being in operation as table 41 refers to conductors resistance at maximum permitted operating temperature.

13. ### Spark123

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Depends wether or not you are using the max of measured figures.
If you know Ze, you can calculate the expected Zs by the type of cable and the maximum operating temperature of it.
For example if you have a Ze of 0.35&#937; and you install a circuit consisting of 20m of 2.5 twin and earth (1.5mm CPC) rated to 70ºC.
From chapter 9 of the OSG 2.5mm twin and earth has a resistance value of 19.51m&#937;/m

The figures in the OSG are for a conductor at 20ºC which are not much use as we need to know what the resistance is at the maximum operating temperature.

So we then need to apply our old friend to give us a correction factor again;
1+(temp change x 0.004).
1+(50 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 now be 19.51m&#937;/1000 x 1.2 x 20m = 0.46824&#937;
Add this to Ze and we now know what the expected total earth fault loop impedance will be at 70ºC, 0.46824&#937; + 0.35&#937; = 0.81824&#937;

As the figure is for a circuit at its max operating temperature it is directly comparable to the figures in the tables in chapter41 of BS7671:2008.

14. ### 17thman

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I think we are all singing off the same hymn sheet here.

I wrote Zs of (Ze+(R1+R2/1000)) x L x 1.2 / .96 and Gareth queried why one was a divisor and one was a multiple and i did not glorify myself with my 2.50pm post (though i think it awkwardly makes sense.

spark wrote
My 1.2 multiplier was for 70C to 20C
Then the reason i divided by .96 is cos that is the factor for 10C
As Spark eloquently did. i, more awkwardly get the extra 10c by 1+(0.004 x10) = 1.04. So i suppose much easier to represent it like Gareth said as two multipliers

Need to brush up on my presentation skills.

15. ### SgtTrojan

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Wonderful. As you have only mentioned the 9C value what do we do about the Table 9b value? Is that also a multiplier in this calc? If so it will be (for expected temp of 10*C)

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)

Fingers crossed

PS I have one more question about R1 and R2 after this

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