Solution to 2391 Q26 (june 10 paper)

[trebor]

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

this is the correct answer from a tutor

the question qouted 10mm at 1.83milliohms per metre at 20deg C

the volt drop for a single phase circuit appears in both line and neutral conductors, the resistance should be corrected to 70deg c for thermplastic max conductor temperature, if you do this calc iyou will find it corresponds with BS7671 table 4d1b at 4.4milliohms per metre so the answer should have been written like this

1.83 x 2 x 1.20 x 80/1000 = 0.351 ohms

45 x 0.351 = 15.79V

observation Vd exceeds permitted 11.5V

possible correction increase csa of circuit

 

[GStueyXR]

I always wonder where the 1.2 figure comes from.

 

[JUD]

I always wonder where the 1.2 figure comes from.

The resistance of copper and aluminium conductors increases/decreases from its 20°C value by 0.4% for every °C.

The difference between 20°C and 70°C is an increase of 50°C which works out at 20%.....(0.4 ÷ 100) x 50 = 0.2[20%]

To add 20% to something you multiply it by 1.2.....SIMPLES

this is the correct answer from a tutor

the question qouted 10mm at 1.83milliohms per metre at 20deg C

the volt drop for a single phase circuit appears in both line and neutral conductors, the resistance should be corrected to 70deg c for thermplastic max conductor temperature, if you do this calc iyou will find it corresponds with BS7671 table 4d1b at 4.4milliohms per metre so the answer should have been written like this

1.83 x 2 x 1.20 x 80/1000 = 0.351 ohms

45 x 0.351 = 15.79V

observation Vd exceeds permitted 11.5V

possible correction increase csa of circuit

So trebor's calculation could be written like this:

1.83 x 2 x (1 + (0.4 ÷ 100) x 50) x 80 ÷ 1000 = 0.351Ω

45 x 0.351 = 15.79V

 

[GStueyXR]

The resistance of copper and aluminium conductors increases/decreases from its 20°C value by 0.4% for every °C.

The difference between 20°C and 70°C is an increase of 50°C which works out at 20%.....(0.4 ÷ 100) x 50 = 0.2[20%]

To add 20% to something you multiply it by 1.2.....SIMPLES

Sorry I should have elaborated a little...

What I dont understand is if we take the Max figures from the BRB we have to multiply it by 80%.

It seems they think the reverse calculation of this is to multiply it by 1.2. But it aint, the reverse calc is to multiply it by 1.25.

So if the table is for conductors at 70deg the the calc to find the resistance at 20 is for instance 7.87 X 0.8 = 6.296 . Now if you correct the other way and multiply by 1.2 youl get 7.552. Which of course is wrong.

Maybe I am missing something but to me the recipricol of 5/4 is 4/5.

 

[JUD]

Sorry I should have elaborated a little...

What I dont understand is if we take the Max figures from the BRB we have to multiply it by 80%.

It seems they think the reverse calculation of this is to multiply it by 1.2. But it aint, the reverse calc is to multiply it by 1.25.

The 80% is only a "rule of thumb". The more accurate figure would be 83%....1 ÷ 1.2 = 0.8333333......

The 1.2 is worked out by the resistance-temperature coefficient of 0.004 [0.4%] per °C at 20°C (see note 3 under table 9C page 168 OSG).

 

[GStueyXR]

The 80% is only a "rule of thumb". The more accurate figure would be 83%....1 ÷ 1.2 = 0.8333333......

The 1.2 is worked out by the resistance-temperature coefficient of 0.004 [0.4%] per °C at 20°C (see note 3 under table 9C page 168 OSG).

That's what i was missing then

 

[JUD]

Sorry I should have elaborated a little...

What I dont understand is if we take the Max figures from the BRB we have to multiply it by 80%.

It seems they think the reverse calculation of this is to multiply it by 1.2. But it aint, the reverse calc is to multiply it by 1.25.

So if the table is for conductors at 70deg the the calc to find the resistance at 20 is for instance 7.87 X 0.8 = 6.296 . Now if you correct the other way and multiply by 1.2 youl get 7.552. Which of course is wrong.

Maybe I am missing something but to me the recipricol of 5/4 is 4/5.

The reverse calculation of multiplying something by something is to divide the answer by the same number.

7.87 x 0.8 = 6.296
6.296 ÷ 0.8 = 7.87

 

[GStueyXR]

As I said, that's what I was getting at. It seems the rule of thumb tends to treat sparks as buckteethed rednecks that cant punch out a simple calc.

 

[brucelee]

this is the correct answer from a tutor

the question qouted 10mm at 1.83milliohms per metre at 20deg C

the volt drop for a single phase circuit appears in both line and neutral conductors, the resistance should be corrected to 70deg c for thermplastic max conductor temperature, if you do this calc iyou will find it corresponds with BS7671 table 4d1b at 4.4milliohms per metre so the answer should have been written like this

1.83 x 2 x 1.20 x 80/1000 = 0.351 ohms

45 x 0.351 = 15.79V

observation Vd exceeds permitted 11.5V

possible correction increase csa of circuit

i to did not include the multiplying factor for operating temp think the only bit i will have got right for this is 1 mark for the last part of the question
just hope we did enough eh mate with the earlier questions

 

[Dan oh!]

Did that question, got 6.73v. was that right or wrong??

 

[Danmc069]

hi could u help me with this
a motor runs at 700rpm at 50hz what motor is suitable for this
a.2pole
b.4pole
c.8pole
d.12pole

 

[Dan oh!]

If I remember right I put 8 pole, but it was pure guess work sorry!

 

[pushrod]

hi could u help me with this
a motor runs at 700rpm at 50hz what motor is suitable for this
a.2pole
b.4pole
c.8pole
d.12pole

It is answer C) which is 4 pairs

Because synchronous speed = frequency ÷ Pairs of poles

so pairs of poles = frequency ÷ by speed

(since Hz is per second change speed to rps which is 700÷60 = 11.7

then pairs of poles = 50 ÷ 11.7 = 4.2 pairs which mean the closest number of actual poles is 8

working it back 8 poles means 4 pairs. then speed = frequency ÷ pairs of poles =50 ÷ 4 = 12.5rps

to change to rpm x60 = 750 rpm.

the reason you don't get exactly 700 is that 750 is the synchronous speed and actual speed is affected

by a slip factor, usually around 4 or 5%.

 

[TPES]

Pushrod.. I think your D.Watters in disguise! lol