Tonights 2391

[GStueyXR]

Pathetic..... C&G need shooting. It's almost childish.

 

[brucegenerator]

Thought it was not too bad except the last 15 point volt drop question which stumped me.

 

[GStueyXR]

what was your answwer to 26 ???

I came to 13.5V drop

 

[brucegenerator]

Just drew a complete blank hopefully did enough on the other questions.Hoping to get the 90 marks elsewhere.

 

[pushrod]

Pathetic..... C&G need shooting. It's almost childish.

easy or hard or?? just finished my last 2330 exam today and hoping to do the 2391 in september.

 

[sams]

my answer was 6.7v
length x ib x mvam/1000
80x45x1.87/1000=6.7v

 

[GStueyXR]

But 1.87 was the resistance per metre and not the mV/A/M drop


heres how I worked it...

Resistance of Line conductor = 1.87 X 80
Resistance of Neutral conductor = 1.87 X 80

Total resistance of 299.2 milliohms
299.2 / 1000 = 0.3 ohms

45 X 0.3 =13.5 volts (ohms law)

 

[GStueyXR]

And I think I was right.... If you include the 45A load as a resistive load of 5.1 ohms add it in series to the resistive load of the cables... do the calculations and it works out to exactly 230V.

so volt drop over cable = 13.5V leaving 216.5V over a resistive load of 5.1 and all the figures seem to work

 

[brucelee]

hi mate i didnt have a clue
reisistance of conductor x 2 then multiplied by 45 then by 80 divided by 1000 and ended up with 13.176 or 13.2 v so it looks like i completly trashed that one i only know its mv/A/m x IB x length divided by 1000 didnt have a clue how to turn it into millivolts so looks not good?

 

[GStueyXR]

hi mate i didnt have a clue
reisistance of conductor x 2 then multiplied by 45 then by 80 divided by 1000 and ended up with 13.176 or 13.2 v so it looks like i completly trashed that one i only know its mv/A/m x IB x length divided by 1000 didnt have a clue how to turn it into millivolts so looks not good?

God knows mate.... Lookin in BS7671 it says the vdrop on 10mm is 4.4 mV/A/m

that works out to about 16V

 

[brucelee]

exactly thats the way i was shown how to do it last year when i did a private training course cost me lots but passed everything and it all made sense where this exam was tosh what type of conductors where they i have even forgot that soon as i got to that question my head completley went and i still had about a hour left

 

[Sintra]

But 1.87 was the resistance per metre and not the mV/A/M drop


heres how I worked it...

Resistance of Line conductor = 1.87 X 80
Resistance of Neutral conductor = 1.87 X 80

Total resistance of 299.2 milliohms
299.2 / 1000 = 0.3 ohms

45 X 0.3 =13.5 volts (ohms law)

2 x l x ib x r / 1000

2 x 80 x 45 x 1.87 / 1000 = 13.46

Think you got that one spot on

 

[GStueyXR]

2 x l x ib x r / 1000

2 x 80 x 45 x 1.87 / 1000 = 13.46

Think you got that one spot on

Where did you get that formula from sintra? I had to work it out from scraps they provided and a bit of intuition....

 

[Sintra]

Learnt that one many moons ago as an apprentice.

How long did it take you to work it out?

 

[GStueyXR]

Learnt that one many moons ago as an apprentice.

How long did it take you to work it out?

to be honest i only had about 5 mins at the question......

But the only formula i can find for Vdrop is L X Ib X mV/A/m / 1000