Help with revision question please 2392

[Rbevo]

Currently doing 2392 theory and am having a brain melt and need some help to explain how to do the the following:-

A radial socket circuit is supplied by 2.5mm cable and is 20m long. Given that Ze is .43 ohms and resistivity of cable is 7.41mOhms/m at 20degrees, what is the Zs value?

Many thanks in advance .

Rob

 

[Paul.M]

20 meters X 7.41mOhms + 0.43 Ohms (Ze) = the Zs.

 

[telectrix]

assuming the cpc is 2.5mm , you R1+ R2 is (7.41 x 20 x 2)/1000 = 0.296. round up to 0.3.

so Zs = Ze + (R1+R2) = 0.43+0.3 = 0.73ohms

 

[Rbevo]

Thanks for the quick reply. Staring to make sense, however, apparently the answer is supposed to be .58ohms.
R1+ R2 is (7.41 x 20 x 2)/1000 = 0.296. round up to 0.3. Where does the x 2 part in the equation come from?

But 7.48 x 20 (to convert in to correct length)/1000 (to convert mOhms to OHms) =0.1482 +0.3=.578 or .58.
Have I understood this right?

Thanks

 

[Silly Sausage]

Thanks for the quick reply. Staring to make sense, however, apparently the answer is supposed to be .58ohms.
R1+ R2 is (7.41 x 20 x 2)/1000 = 0.296. round up to 0.3. Where does the x 2 part in the equation come from?

But 7.48 x 20 (to convert in to correct length)/1000 (to convert mOhms to OHms) =0.1482 +0.3=.578 or .58.
Have I understood this right?

Thanks

Paul is right.

The resistivity value you are given is for R1 + R2 of the cable, look them up in your On site guide.
So you don't need the x2 above.

 

[Sparky Ninja]

If the answer is apparently 0.58 ohms then 7.41 represent the R1&R2 resistivity.

 

[Paul.M]

Paul is right.The resistivity value you are given is for R1 + R2 of the cable, look them up in your On site guide.So you don't need the x2 above.

"Given that Ze is .43 ohms and resistivity of cable is 7.41mOhms/m at 20degrees, what is the Zs value?"That's how I read the question that 7.41mOhms/m is for the combined r1 and r2 then X by 20 meters. Because there's no mention of the csa size of the cpc I assumed that it was a total Ohm figure and not an individual conductor. Good old C&G making the questions clear as mud.

 

[oxocube]

Currently doing 2392 theory and am having a brain melt and need some help to explain how to do the the following:-

A radial socket circuit is supplied by 2.5mm cable and is 20m long. Given that Ze is .43 ohms and resistivity of cable is 7.41mOhms/m at 20degrees, what is the Zs value?

Many thanks in advance .

Rob

First of all, its a daft question (not your fault), the calc can't be done without knowing the size of the cpc. If they mean the cable length of the live and earth added together is 20 mts and both are 2.5 then the answer is 0.58.

2
10 ×
7.41 ×
1,000 ÷
.43 +
--------------------------
0.5782 =

The question should have said -A radial socket circuit is supplied by a 2.5mm live cable and has a 2.5 earth cable and is 10m long. Given that Ze is .43 ohms and resistivity of cable is 7.41mOhms/m at 20degrees, what is the Zs value?

 

[Silly Sausage]

In the table in the OSG where it doesn't quote values for the protective conductor I can only assume it's referring to something like singles in steel conduit where the csa of the conduit is so vast that the ohmage is negligible (to 2 dps anyway).

 

[Paul.M]

I think he is referring to appendix 9 table 9A.

 

[Rbevo]

The thing is I am trying to do this with our any reference material!

 

[Amp David]

Paul is right.

The resistivity value you are given is for R1 + R2 of the cable, look them up in your On site guide.
So you don't need the x2 above.

Just forgot the divide by 1000 though

 

[telectrix]

stupid question. i took the res. of each conductor to be 7.41mOhms.