Answers on a postcard please...

[JUD]

They give you the length of the ring in the question -- 54m from origin, all the way around the ring and back to origin.

If you were to link line and cpc of one of the legs and take a continuity reading between line and cpc of the other leg you would get (19.51 ÷ 1000) x 54 = 1.05Ω. This is your r1 + r2.

When you join the legs back together as it would be in service, as you rightly said, you have halved the length and doubled the csa.

Half the length, half the resistance -- ½ of 1.05 = 0.52

Double the csa, half the resistance -- ½ of 0.52 = 0.26

so R1 + R2 = (r1 + r2) ÷ 4 = 0.26Ω

 

[JamesBrownLive]

Ok, I think I am begining to understand, but it is the physics that still don't seem to make sense....


Zs is a measurement of the resistance of the complete earth fault path. i.e from the fault, back to the cu and beyond (Ze).
When a fault occurs on a ring circuit, surely the fault takes the path of least resistance?

So for arguements sake if the fault is 30m in on a 50m ring, the fault would travel back down the part of the ring that is 20m away from the CU?

Does that make sense, and is that correct?

 

[JUD]

Ok, I think I am begining to understand, but it is the physics that still don't seem to make sense....


Zs is a measurement of the resistance of the complete earth fault path. i.e from the fault, back to the cu and beyond (Ze).
When a fault occurs on a ring circuit, surely the fault takes the path of least resistance?

So for arguements sake if the fault is 30m in on a 50m ring, the fault would travel back down the part of the ring that is 20m away from the CU?

Does that make sense, and is that correct?

No. The 2 line conductors and 2 cpcs are connected together at the CU, so in effect they are the same conductor. Therefore the fault current would travel both ways around the ring until it gets back to the CU.

 

[JUD]

Think about it this way.

Take your 50m ring. Let's say you have socket outlets at 5m, 10m and 15m on one leg of the ring.

Plug a 13A load into each outlet -- 13 x 3 = 39A.

If the current took the path of least resistance, in this case 15m of 2.5mm² cable with a 27A current-carrying-capacity, it would overload the cable.

**EDIT** And take out the MCB.....eventually :banghead:

 

[JamesBrownLive]

No. The 2 line conductors and 2 cpcs are connected together at the CU, so in effect they are the same conductor. Therefore the fault current would travel both ways around the ring until it gets back to the CU.

Ok, thanks for that.

It was obviously my belief in the 'Path of least resistance' that was clouding my understanding.

Out of curiosity....
If the circuit was a radial one, then the Zs would simply be a calculation of the resistance of the length of the cable from the point of the fault back to the CU. So if we use the original question and say the fault occured at 54m, then the resistance would be 54 X 19.51 /1000 = 1.05 Ohms??

Cheers

 

[JUD]

Ok, thanks for that.

It was obviously my belief in the 'Path of least resistance' that was clouding my understanding.

Out of curiosity....
If the circuit was a radial one, then the Zs would simply be a calculation of the resistance of the length of the cable from the point of the fault back to the CU. So if we use the original question and say the fault occured at 54m, then the resistance would be 54 X 19.51 /1000 = 1.05 Ohms??

Cheers

Correct. Though Zs reading is taken at the last point in a radial circuit or in the case of a ring circuit the highest reading (incl. spurs).