2394 question

LittleEd · Sep 17, 2014

Hi I am due to start the city and guilds 2394 course next month. I thought that I would have a look at some past questions to see what to expect. I am having trouble with one of the questions and was wondering if anyone could explain the theory to me.

16) The measured value of loop impedance for a circuit is 0.83Ω. 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 = 0.4Ω
a) 0.43Ω
b) 0.86Ω.
c) 0.996Ω.
d) 0.916Ω.

I am using this formula to work out the question ZStest=1/1.20 x {Zs +Ze (F-1)}

I know the answer is 0.916 by cheating and looking at the answers, but i have no idea how they came to that answer. am I using the wrong formula and if i am where can i find the right on?

Thank you in advance

tony mc · Sep 17, 2014

I would of said it was (c) 0.996.

The way I see it is the resistance of the copper changes by 2 % for every 5 degrees so the resistance will rise by 20 % if its temp rises to 70 degrees.

You have your measured loop impedance at 0.83 ohms at 20C and by multiplying by 1.2 (20 % )with give you the corrected value for 70 degrees .

IMO the resistance value at operating temp of 70 degrees would be 0.83 x 1.2 = 0.096 ohms.

Richard Burns · Sep 18, 2014

I would have gone for the 0.996 as well.

The question practically gives you the answer they have given though.

Having a Ze of 0.4 leaves 0.43Ω as the resistance of the cable (sort of, and not how I would consider it)
0.43 *1.2 = 0.516
0.516 + 0.4 = 0.916

The factor given for the cable is 1.2 @ 70°C, if the cable is running at 70°C this does not mean that the supply from the substation is running at that level since the current carrying capacity is much greater on the supply than on a final circuit.
e.g. you have 16A circuit wired in 1.5mm² cable in conduit that is fully loaded then the cable should be close to 70°C, however the 100A supply is at less than 20% of its CCC and so would barely heat up.

However all this said, the general arrangement for temperature correction from 20°C to 70°C would be to multiply the resistance by 1.2.

cprfenom · Sep 18, 2014

I just want to say I really find Richard Burns to be a huge asset. I don't often see negativity from him and he is always very informative in his answers.

Here is a huge Thank you from me Richard.

LittleEd · Sep 18, 2014

thank you for the help. I guess I read the question wrong.

D Skelton · Sep 18, 2014

You're only correcting your R1+R2 values and as such don't take in to account the external earth fault loop impedance.

Zs = 0.83
Ze = 0.40
Zs - Ze = R1+R2 (0.43)

R1+R2 (0.43) x 1.2 = 0.516

0.516 + 0.40 = 0.916

The answer is (d)

Richard Burns · Sep 18, 2014

Thank you cprfenom.

LittleEd, I did forget to mention that the formula you were using is one to determine the maximum value of Zs (20°Ctest) permitted when you have the Ze and the tabulated maximum Zs from BS7671 (effectively converting the tabulated value from 70°C down to 20°C.

The actual formula you should have used (although the data is not really available in the question as it requires you to use the R1+R2 = Zs-Ze which is not accurate at all unless there is no bonding in place)
(rather than using the (divide by) 0.8 rule of thumb) would be:

Zs (70°C) = Ze +F(R1+R2test)

gdr7671 · Sep 20, 2014

i always use this formulea Zs = Ze+ (( (r1+r2)*Ca))
M

Where Ca is the correction for ambient temperature and M is the multiplication factor. In this case !.2

I hope this has helped you and good luck.