Calculating design current

[atto2007]

Hi people,

currently studying to become electrician and have got stuck on one of my tma questions.

Q. A 16mm squared, two core armoured, 70 degrees C PVC insulated cable (copper conductors), is clipped direct to a surface, and forms the two wire radial distributor shown in the diagram below. The distributor is fed at end A 230 volts and the cable lengths are as follows

A to B -45m
B to C -35m
C to D -20m

a. find the current in each section of the diagram
b. use IET Wiring Regulations to calculate the overall voltage drop

45m 35m 20m
A-------B-------C-------D
20A 10A 15A

Can someone explain how to calculate the current without either the Power or resisitance?

Ib = power/Uo I=V/R

just need some help, think im missing something obvious.

Thanks

 

[Guest55]

No way of calculating current with those figures , need a power ( kW ) for the load at the end.
something missing there pal.

 

[Guest123]

The current in each section is already given....A-B = 20A, B-C = 10A, C-D = 15A.

Total power = 20+10+15 =45A.

 

[Guest55]

Ah didnt see your current figures.
oops lol.

 

[ezzzekiel]

We no longer use formulas for calculations, it's all got to be guessed now.

Think they slipped up with question a. Find current as its already given below the question.
vd would be vd. Mva x ib x L / 1000

 

[atto2007]

thanks guys for your replys,

quite possible they have slipped up. however i still cant seem to work out the design current (Ib) in order to calculate the voltage drop.

 

[wallyanker]

Design current = Add the current in each section = 45A

Volt drop formula = Mva x ib x L / 1000

Add all the lengths together and work out total vd or work out each section seperately.

 

[Knobhead]

You have to work out each section separately, the voltage and current varies along the length.

 

[sedgy34]

its like being back behind the table and snoring my head off :snore:

 

[Guitarist]

its like being back behind the table and snoring my head off :snore:

Afternoon lectures following a trip to the pub nearly always had that result...

 

[Vole]

It sounds like I'm doing the same course and got to the same question. With the overall voltage drop do you calculate the voltage drop for each section then add the results, or do you take the Ib as 45A, L is 100m and calculate with those figures. I'm trying it both ways and getting 2 different answers so wonder which is correct

 

[pencilpusher]

It sounds like I'm doing the same course and got to the same question. With the overall voltage drop do you calculate the voltage drop for each section then add the results, or do you take the Ib as 45A, L is 100m and calculate with those figures. I'm trying it both ways and getting 2 different answers so wonder which is correct

Hmm, interesting. That gets the old grey matter going!

 

[steveberry11]

You need to Carry out 3 seperate calculations
1st
A-B @ 45A @ 45M
2nd
B-C @ 25A @ 35M
3rd
C-D @ 15A @ 20M

Use the volt drop figures for the cable used I think this is 16mm for the whole circuit.

The 3 totals should be added for your answer.

Good luck

 

[ackbarthestar]

Just to throw in confusion to chaos ......
You could still run your Garden lights from the final voltage supplied by a rural 240V+ supply.

Its pretty straight forward really. There are 2 ways of doing this calculation

Method 1

1/ Find the mV/A/m volt drop for the conductor size (use the regs)

2/ Calculate the total current (as shown above) for each leg at each distance

3/ Add the voltage drops at each leg

4/ Take the total voltage drop from the supply

Method 2

1/ Find the m.ohm/m for the conductor size (use OSG)

2/ Multiply by 2

3/ As 2 above (method 1)

4/ Divide the m Ohms/m by 1000 to convert to Ohms/m

5/ Multiply the length by 4/ above (method 2). This gives total ohms for each length

6/ Multiply by an ambient temperature correction, given in the next table (OSG)

7/ Multiply the current flow in each length by the ohms in each leg.

8/ As 3 and 4 in method 1