OP
thorneellis
eek things arnt looking good for me. i just waffled loads of figures. i found that one of the harder questions.
Discuss Solution to 2391 Q26 (june 10 paper) in the Periodic Inspection Reporting & Certification area at ElectriciansForums.net
get a life mate f u got it wrong move on the exam was 3 weeks ago
get a life mate f u got it wrong move on the exam was 3 weeks ago
If you read the question (just posted) it will clear some things up...
I always wonder where the 1.2 figure comes from.
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
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.
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).
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.
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
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
Reply to Solution to 2391 Q26 (june 10 paper) in the Periodic Inspection Reporting & Certification area at ElectriciansForums.net
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