2391 Question

[parm]

Hi Guys, going through some past papers and my brain won't work this one out any help appreciatted,

Single phase power circuit with a full load current of 25A circuit is installed using 6.0mm2 copper conductors with a circuit length of 35m. (give Resistance of 6.0mm2 copper is 3.08megohm/metre)

Q/ The line and nuetral combined resistance for the circuit is 0.26ohm, determine whether the circuit meets voltage drop requ. if the voltage drop to local db is 4.5V?

Help me please!!

 

[telectrix]

?????? resistance megohms?????

 

[parm]

sorry to much revision i think the correction is microhm per metre still happy to be corrected and chances are I'm wrong!!

 

[telectrix]

i,d have thought milli not micro. small "m" is milli

 

[Guest123]

It is milliohm for sure.

So with a VD of 4.5V to this DB, this leaves a permissable additional VD of 7V the max being 5% of 230V = 11.5V.

Using the values you have given above, with the resistance per meter being for both L&N conductors the formula is - L x Ib x R / 1000 = 35M x 25A x 0.26ohms / 1000 = 0.23V.

So total VD = 4.5V + 0.23V = 4.73V. The circuit complies.

 

[parm]

thanks Lenny, I can sleep now!!

 

[Sintra]

Is the formula not 2 x L x Ib x R / 1000

Where R = 3.06 mOhms / M

This gives me 2 x 35 x 25 x 3.06 / 1000 = 5.355 V + 4.5V = 9.85V

Although the values in the question don't quite tally up

ie circuit length L + N = 2 x 35 M = 70 M

Given R1 + R2 = 0.26 ohms

Therefore mOhms/M should be 0.26 / 70 x 1000 = 3.714 mOhms/M which in turn gives a figure of 7.43 mV/A/M.

Think the tabulated value for 6mm is 7.3 mV/A/M so this is slightly over.

 

[Guest123]

Why would it be 2 x length mate??? as he states in the question that the line & neutral combined resistance is 0.26ohms which should equate to the milliohm/meter value.

As I said I used the values he gave which as you say dont quite tally up but I didn't have any literature to hand to check.

 

[Sintra]

The total circuit length is 35M from source to appliance so the total cable length of L + N is 70M or 2L when using the mOhms/M figure of 3.08.

Using the figure of 0.26 ohms overall resistance the formula is

Vd = Ib x R

Vd = 25 x 0.26

Vd = 6.5V

This is probably the way that the examiner wanted it answered as the original Vd was 4.5V + the additional Vd of 6.5V equals 11V which is just below the max.

Using the tabulated value for a 6mm of 7.3 mV/A/M you get

7.3 x 25 x 35 /1000 = 6.39V

I just wish C&G would research their questions correctly and have all values corresponding to each other so it could be worked out each way and get the same result.

 

[SimpleSimon]

Ohms law Vd= IR
Vd= 25 x 0.26
Vd= 6.5V
6.5V + 4.5V= 11V. Just complies
They try and throw you off the scent by adding circuit length in there when it's not really needed as resistance has been calculated for you.

 

[SimpleSimon]

Even still. If you were to work out the resistance
3.08 x 70/1000 = 0.2156 Still not the figure given!!

 

[Sintra]

Exactly the point I have made.


Sent from my iPhone using Tapatalk

 

[pushrod]

The question gives a bit more info than is needed for this part - maybe other parts of the question refer to the theoretical value. I would use the given figure of 0.26 Ω as simple simon said (now what does that remind me of?), it's not really surprisising that a calculated figure is is slightly lower. Circuit just passes at 11V.