Solution to 2391 Q26 (june 10 paper)

[GStueyXR]

here is my solution... and a definitive one too

cable is 80m @ 1.87 mΩ....
Resistance of line conductor= (80 x 1.87) / 1000 =0.15 Ω

Load is 45A. Expressed as a resistance 230/45 = 5.1 Ω

Resistance of Neutral conductor= (80 x 1.87) / 1000 =0.15 Ω

Total circuit resistance of 5.4 Ω


42.6A would flow in this circuit

So we have 3 resistors in parralel.

--------| 0.15Ω |-----| 5.1 Ω |-----| 0.15Ω|-------


Vdrop across Line conductor would be ----- 0.15 x 42.6 = 6.4V

Vdrop across load would be ------------------- 5.1 x 42.6 = 217.2V

Vdrop across neutral conductor would be --- 0.15 x 42.6 = 6.4V


So, Vrop in circuit conductors would be - 12.8 V

 

[acat]

Pity you didnt post that on the 9th of June good answer all the same.


Chris

 

[GStueyXR]

Would you agree with these calcs ???? If so, I got it wrong on the night... I got 13.5 Vdrop....

Dont know how I got this wrong... It is afterall very basic stuff

 

[CRAGS]

You will still possibly get marks if your calculations and methods are correct

 

[Guest123]

Was there not any mention of temperature in this question???

 

[pushrod]

here is my solution... and a definitive one too

cable is 80m @ 1.87 mΩ....
Resistance of line conductor= (80 x 1.87) / 1000 =0.15 Ω

Load is 45A. Expressed as a resistance 230/45 = 5.1 Ω

Resistance of Neutral conductor= (80 x 1.87) / 1000 =0.15 Ω

Total circuit resistance of 5.4 Ω


42.6A would flow in this circuit

So we have 3 resistors in parralel.

--------| 0.15Ω |-----| 5.1 Ω |-----| 0.15Ω|-------


Vdrop across Line conductor would be ----- 0.15 x 42.6 = 6.4V

Vdrop across load would be ------------------- 5.1 x 42.6 = 217.2V

Vdrop across neutral conductor would be --- 0.15 x 42.6 = 6.4V


So, Vrop in circuit conductors would be - 12.8 V

depends very much on the wording of the actual question - what is the value 1.87m ohms? is it a an (R1+Rn)/metre for a twin core cable or for just a single core cable. Usually it is for a twin core cable in which case it is just

Vd = Ib x L x m ohms/m
1000

= 45 x 80 x 1.87
1000

= 6.73V
and you don't worry about the neutral.
Also as Lenny said if temperature is mentioned it could just be that there are correction factors to applied to the resistance value

 

[GStueyXR]

and you don't worry about the neutral.

Wrong in my opinion.

If there was no PD in the neutral, then current cannot flow. since current is eqaul at all point in a circuit there must be a potential between the termination at load, and the center of the supply transformer.

Every circuit is basically 3 resistors in series.... The line conductor resistance, the load, and the neutral conductor resistance.

 

[pushrod]

Wrong in my opinion.

If there was no PD in the neutral, then current cannot flow. since current is eqaul at all point in a circuit there must be a potential between the termination at load, and the center of the supply transformer.

Every circuit is basically 3 resistors in series.... The line conductor resistance, the load, and the neutral conductor resistance.

I think you are thinking about it too deeply.

As i said it all depends on what that 1.87 figure represents - did it say for a single or a twin core cable? If it is the latter then in all likelyhood you forget the neutral.

don't know if you saw my reply to an earlier post of yours but mV/A/m is resistance per metre.

 

[CRAGS]

I totally messed this question up the only chance of a mark I have on this one is for putting 5%

 

[GStueyXR]

I think you are thinking about it too deeply.
If it is the latter then in all likelyhood you forget the neutral.

If you read the question (just posted) it will clear some things up...

 

[brucelee]

I think you are thinking about it too deeply.

As i said it all depends on what that 1.87 figure represents - did it say for a single or a twin core cable? If it is the latter then in all likelyhood you forget the neutral.

don't know if you saw my reply to an earlier post of yours but mV/A/m is resistance per metre.

26
Durig the course of the inspection one of the single phase radial circuits supplying an item of fixed equipment is to be checked for voltage drop compliance. the circuit has the following charicteristics
. load current =45A
. circuit length =80m
. live conductors =10mm

a) using the info in fig 1 1.83 for 10mm determine wether the circuit meets the requirments for voltage drop.
show all calculations
b) state
i) a suitable observation to be recorded on the periodic report. if the voltage drop fails to comply
ii) one action which could be taken to rectify the situation in b)i above

 

[pushrod]

If you read the question (just posted) it will clear some things up...

cheers
I see it says" line conductors" so i would multiply length by 2 to take account of neutral in this case.

 

[brucelee]

cheers
I see it says" line conductors" so i would multiply length by 2 to take account of neutral in this case.

yeah mate i only knew how to do it by mv/A/M X IB X L divided by 1000
but as there was no mv table something was niggling at me saying it was something to do with V=I X R
but i proceeded to work out the resistance by M0hm/M X L divided by 1000 ie 1.83 x 2 x 80 divided by 1000
= 292.8 divided by 1000 = 0.2982OHm = 0.3 ohm then from there not sure if i put 0.2982 X 45 =13.419 = 13.5v 0r 0.3 x 45 =13.5v
then went onto mess the next bit as it asked for a observation and i put something like its not acceptable instead of saying volt drop to high to comply with 5% ie 11.5v
and for last bit of question where it asked what you could do i put increase csa of conductor
head had gone completly by then just hope i did enough earlier on

 

[Sintra]

here is my solution... and a definitive one too

cable is 80m @ 1.87 mΩ....
Resistance of line conductor= (80 x 1.87) / 1000 =0.15 Ω

Load is 45A. Expressed as a resistance 230/45 = 5.1 Ω

Resistance of Neutral conductor= (80 x 1.87) / 1000 =0.15 Ω

Total circuit resistance of 5.4 Ω


42.6A would flow in this circuit

So we have 3 resistors in parralel.

--------| 0.15Ω |-----| 5.1 Ω |-----| 0.15Ω|-------


Vdrop across Line conductor would be ----- 0.15 x 42.6 = 6.4V

Vdrop across load would be ------------------- 5.1 x 42.6 = 217.2V

Vdrop across neutral conductor would be --- 0.15 x 42.6 = 6.4V


So, Vrop in circuit conductors would be - 12.8 V

Formula to use is

2 x L x Ib x R /1000 = VD

2 x 80 x 45 x 1.87 /1000 = 13.464

 

[GStueyXR]

Formula to use is

2 x L x Ib x R /1000 = VD

2 x 80 x 45 x 1.87 /1000 = 13.464

Hope you are right mate.... I did this through intuition, after a 5 minute think.....

But now I doubt myself as you can see in the OP