loop impedance, temperature, cable factor....

[Gary Tollison]

Hello
I'm not sure if this is the correct forum area to post this in, apologies if it isn't...

I've been working through some mock questions on-line, and found myself confused by this:

The measured value of loop impedance for a circuit is 0.83ohm.
If the temperature at the time of the test was 20C, and the cable is 70C (factor 1.2) what is the corrected value. Ze = 0.4ohm:
a) 0.43 ohm
b) 0.86 ohm
c) 0.996 ohm
d) 0.916 ohm

I just cant get my head around this!

Any help would be great!

Thanks in advance,

Gary

 

[Ian1981]

yeah, but the loop impedance of a circuit includes Ze.

The external impedance is not subject to the 1.2 correction factor
When calculating R1+R2 using 20 degrees tables during the design you apply the 1.2 correction factor to compensate for 70 degree operating temperature
External ze is not a factor during this

 

[Risteard]

yeah, but by definition, the loop impedance of a circuit includes Ze.

But the temperature correction factor does not apply to the external part of the loop.

 

[Richard Burns]

The generally accepted method of correcting for temperature is to apply it to the whole of the Zs for the circuit, however this does not , as said above provide an accurate assessment as the external supply may not be running at 70°C.
In that case the separation of Ze and R1+R2 may be undertaken and the correction factor applied to the R1+R2 only and the uncorrected Ze applied.
In the case of the question the method of calculation is not clarified.
Also the correction is from ambient (20°C) to 70°C rather than the usual correction of the tabulated value down to 20°C.
Basically you can apply the 1.2 correction factor to the 0.83Ω and get 0.996Ω
or you can apply the correction factor just to the back calculated R1+R2 (which should not be done) of 0.43Ω and add it to the Ze of 0.4Ω and get 0.916Ω.

Because the question specifies Ze you might assume they mean the more accurate correction but since it requires an unapproved calculation it seems odd.

 

[telectrix]

The generally accepted method of correcting for temperature is to apply it to the whole of the Zs for the circuit, however this does not , as said above provide an accurate assessment as the external supply may not be running at 70°C.
In that case the separation of Ze and R1+R2 may be undertaken and the correction factor applied to the R1+R2 only and the uncorrected Ze applied.
In the case of the question the method of calculation is not clarified.
Also the correction is from ambient (20°C) to 70°C rather than the usual correction of the tabulated value down to 20°C.
Basically you can apply the 1.2 correction factor to the 0.83Ω and get 0.996Ω
or you can apply the correction factor just to the back calculated R1+R2 (which should not be done) of 0.43Ω and add it to the Ze of 0.4Ω and get 0.916Ω.

Because the question specifies Ze you might assume they mean the more accurate correction but since it requires an unapproved calculation it seems odd.

that was my interpretation. correction factors are usually applied to the Zs value in BS7671 , not jusrt to the R1+R2 values.

by calcularting i the way the question implies, it would be possible to get a borderline value above the recommended value to below it.

 

[Richard Burns]

that was my interpretation. correction factors are usually applied to the Zs value in BS7671 , not jusrt to the R1+R2 values.

My first approach was to do just that as this is normally fine and is the first thing you would do and would only apply further accuracy if the value was out of range, but because they specify Ze, which is usually immaterial, I do wonder what they are going for.
Especially since they give both possible answers.

 

[westward10]

I suspect the answer lays with what he has been taught in the classroom but I would say that by being given Ze it should be deducted or maybe a red herring.

 

[Ian1981]

They must just want the R1+R2 value applied to the correction factor otherwise what's the point in stating the Ze value?

 

[NDG Elecs]

Yeah I thought because they have the Ze it was there for a reason. The question should have maybe stated that there were no possible parallel earth paths on the cct so you could accurately make an assumption as to the r1r2.

I forgot all about the tricky wording they use at C&G. Shame it's multiple choice, otherwise you could write both possible answers and the reasoning behind them.

 

[sozib2024]

It's 0.43 x 1.2 = 0.516+ 0.4= 0.916 D

where get R1+R2 value

 

[cireland]

Can I ask where the 1.2 multiplication factor comes from?
If you are using it to go from 80% to 100% it should be 1.25, is it not?

 

[PS Electrics]

Can I ask where the 1.2 multiplication factor comes from?
If you are using it to go from 80% to 100% it should be 1.25, is it not?

Change in resistance is 0.004 Ohms per degree

so 50 x 0.004 = 0.2 a 0.2 increase so correction factor is 1.2

 

[cireland]

Change in resistance is 0.004 Ohms per degree

so 50 x 0.004 = 0.2 a 0.2 increase so correction factor is 1.2

Did not know that!!
Cheers.