2391 exam question. pls give an answer

[patrao]

A single phase power circuit supplying an item of equipment,having a full load current of 25 A,is to be checked to confirm voltage drop compliance.The circuit is installed using 6.0 mm2 cooper conductors,with a circuit length of 35m.
a) Calculate the maximum permissible voltage drop for this circuit.

b) The line and neutral combined resistance for the circuit is 0.26 Ohms . Determine whether the circuit meets the voltage drop requirements if the voltage drop to the local distribution board 4.5 V.

c)State one other method which could be used to confirm voltage drop.

 

[malcolmsanford]

A/ The equation is mv/m x Ib X length / 1000

What is the method of erection as in the 2391-10 or 20 they will give you this as the mv/m differes between method


B/ Reading the question means ( I think ) U= I x R where U is the volt drop, R is the resistance and I is the load so in your circuit

25 x 0.26 = 6.5 volts so therfore you 4.5 volts are acceptable


C/ U = I x R

Fingers crossed

 

[telectrix]

L and N combined is 0.26. should we not use res.of L only to determine volt drop, i.e.0.13?

 

[IQ Electrical]

A) Is asking for the maximum permissable voltage drop which is 5% for circuits other than lighting so equals 11.5 Volts

The other method for me, would be measurement.

 

[IQ Electrical]

L and N combined is 0.26. should we not use res.of L only to determine volt drop, i.e.0.13?

No, it is calculated on both conductors.

 

[telectrix]

ywah, i saw that and thought stupid question how do you calculate permissible volt drop when it's set in stone as 5% ( 3% if lighting). trick question or badly worded.

 

[malcolmsanford]

L and N combined is 0.26. should we not use res.of L only to determine volt drop, i.e.0.13?

No mate you use both

Sorry didn't see the IQ post pesky Arsenal supporters.

 

[malcolmsanford]

ywah, i saw that and thought stupid question how do you calculate permissible volt drop when it's set in stone as 5% ( 3% if lighting). trick question or badly worded.

Fooled me as well Tel as I saw calculated and thought it wanted a result rather than a value.

 

[IQ Electrical]

No mate you use both

Sorry didn't see the IQ post pesky Arsenal supporters.

Lol, so you take your calculated volt drop for the sub circuit, add it to the volt drop given at the DB and if the total is less than 11.5 Volts, the circuit meets volt drop requirements.

Oh Arsenal we love you....

 

[malcolmsanford]

I liked it much better when it was either 2391 and then the 2400 but now it is 2391-10 and 2391-20 and so when I see what I think is design question I'm off on design route bull in a china shop fashion lol .............................

Cheers mate and Barcelona tonight no problems

 

[telectrix]

of course, silly me. the table values of mV/A/m are dependent on the length of the circuit, there and back, if you like. but the actual volt drop is at the point of utilisation.

 

[IQ Electrical]

of course, silly me. the table values of mV/A/m are dependent on the length of the circuit, there and back, if you like. but the actual volt drop is at the point of utilisation.

And that caught out many taking the June 2010 2391-10 exam, the conductor resistances were given, not the tabulated values from the reg's.

 

[telectrix]

there you have it. 64 and still learning.

 

[morph]

does not meet ?
its 4.5 v at db + the circuit volt drop 7.8 v = 12.3
0.26 x1.2x25 = 7.8v

 

[morph]

1.2 because its the multiplier