How to work out min cross sectional area of live conductors?

[dyel]

Hey need some help with some homework the question is

'Determine for one of the high bay lighting circuits within the workshop, the minimum cross sectional area of live conductors for current carrying capacity and voltage drop.'

So if I have a lighting circuit with 5 high bay 150W luminaires, where do I go from here?

 

[Darkwood]

Hi dyel, welcome to the forum, you are down in your profile as in training, can you expand on this, if you are in an active electrical course then we have a dedicated area for Trainee's and you would be best off applying for entry to that for this level of question.

If you are doing this electrical course work then refer to your on-site guide BS7671 or the full regulation book BS7671, its is clearly drafted out in there on how to calculate both cable size and volts drop.

 

[dyel]

Why do I have to apply for the training forum? Can't a mod move this there
Anyway is page is 151 and 154 what you're talking about

 

[dyel]

Are you talking about page 149/150etc and 154 on OSG? So it's 4mm2 and thats it? No calculations?

 

[Darkwood]

Why do I have to apply for the training forum? Can't a mod move this there
Anyway is page is 151 and 154 what you're talking about

You do not have to apply, it was a recommendation regarding that you state your a Trainee and the level of question you asked.

I mis-read you initial question so my apologies and see its an homework question so my recommendation to join the Trainee section still applies, the Trainee section has selected Mentors who can help with these types of question and give you guidance, we do not do homework questions on the main forum as it doesn't help with your training to be given the answers without trying to show your calcs yourself, like I said, if you try to show us the calcs then we can guide you but will stop short of just giving trainee's direct answers to questions.

Ill will post a link for an application to the Trainee section then you can choose to join or not, if you decide to apply we need to ensure you are who you say you are and cross check the course you are enlisted on as we have to ensure applicants are not trying to access that section under false pretences if you follow the links you will be requested for personal details and course details although to note only staff and yourself will have access to the info you provide.

http://www.electriciansforums.co.uk...ment-how-apply-access-trainee-only-forum.html

 

[Pete999]

Oh er no need for that sort of response, Darkwood was only offering some advice, flak jacket on, foxhole secured.

 

[Simonslimline]

Are you talking about page 149/150etc and 154 on OSG? So it's 4mm2 and thats it? No calculations?

There is not enough info in your original post to work a suitable cable size out yet.

Length of circuit etc.

 

[Darkwood]

Are you talking about page 149/150etc and 154 on OSG? So it's 4mm2 and thats it? No calculations?

You need to find all the info required to do your calcs here, your tutor will need to see you understand how to arrive at any answer, to determine a cable CSA you need to know installation methods, length of run, design current etc etc and other things that may be applicable like ambient Temp, grouping factor, type of protective device, all have a bearing on which cable is chosen, so in a nutshell, no!, you cannot just put 4mm2.

If your been asked to calculate the answer to this question then you should have been taught how to do so, so refer to you college notes first or use the BS7671 to look up the calculations needed, page 145 onwards of the Yellow on site guide (appendix F).

 

[dyel]

You need to find all the info required to do your calcs here, your tutor will need to see you understand how to arrive at any answer, to determine a cable CSA you need to know installation methods, length of run, design current etc etc and other things that may be applicable like ambient Temp, grouping factor, type of protective device, all have a bearing on which cable is chosen, so in a nutshell, no!, you cannot just put 4mm2.

If your been asked to calculate the answer to this question then you should have been taught how to do so, so refer to you college notes first or use the BS7671 to look up the calculations needed, page 145 onwards of the Yellow on site guide (appendix F).

Thanks for replying, the only notes I have on this specific thing are (mV/A/m)xIbxIl divided by 1000
other notes are
Ib=5.87A
L=19.8
In=6amp breaker
Iz=1.5mm2

 

[Richard Burns]

So your first formula is for voltage drop
mV/A/m are the volt drop figures given in appendix 4 of BS7671 for one amp of current and one metre of cable.
Ib is your design current i.e. the current you expect as a maximum on the circuit.
Il does not appear to be correct and should probably be L i.e the length of the circuit.

Because you have a volt drop figure in mV and you will want to know the volts dropped then you divide by 1000 to change from mV to V.

BS7671 Appendix 4, section 6 tells you about volt drop.

BS7671 Appendix 4, section 5 tells you about selecting the correct size of cable for current carrying capacity.

 

[dyel]

So your first formula is for voltage drop
mV/A/m are the volt drop figures given in appendix 4 of BS7671 for one amp of current and one metre of cable.
Ib is your design current i.e. the current you expect as a maximum on the circuit.
Il does not appear to be correct and should probably be L i.e the length of the circuit.

Because you have a volt drop figure in mV and you will want to know the volts dropped then you divide by 1000 to change from mV to V.

BS7671 Appendix 4, section 6 tells you about volt drop.

BS7671 Appendix 4, section 5 tells you about selecting the correct size of cable for current carrying capacity.

It's single phase ac, reference method B. So I go to the table on page 149, conductor cross sectional area is 1.5mm2 so the current carrying capacity is 17.5A I don't think this is right
What about page 155?
Btw yes I meant L

 

[Robson689]

It's single phase ac, reference method B. So I go to the table on page 149, conductor cross sectional area is 1.5mm2 so the current carrying capacity is 17.5A I don't think this is right
What about page 155?
Btw yes I meant L

Why don't you think it's right?

 

[Richard Burns]

Page 155 is ccc for twin and earth cable.

Start from the beginning:
You have 5 off 150W light fittings, presumably as they are hi bay they are discharge lamps, so from appendix A of the OSG you multiply by 1.8 to account for power losses in the control gear and power factor.

I = P/V *1.8 = 5*150 * 1.8 / 230 = 5.87A

So Ib is 5.87A

As a starting point you have said you want to use 1.5mm² cable, from your last post this is single core cables in conduit or trunking.
The voltage drop tabulated on page 151 in OSG for 1.5mm² cable is 29 mV/A/m

You have also said the length of circuit is 19.8 (presumably metres)

Using volt drop = mV/A/m * Ib * L / 1000 = 29 * 5.87 *19.8 /1000 = 3.37 V

Check if this complies using page 146 of the OSG.

Then if this is too high choose a bigger cable, if it is well within then choose a smaller cable and check again.

Once you have you initial cable size start calculating for current carrying capacity, always ensuring you are complying with BS7671 524.1 minimum csa.

In is the next size up of circuit breaker above the design current Ib, that is 6A.

Using Iz > In > Ib the current carrying capacity of the cable must be greater than 6A.

Then you need to consider the installation methods and how they may affect the current carrying capacity of the cable and ensure that once all factors have been taken into account (page 145 in OSG) that the adjusted CCC is above 6A.

Once you have done this and the current carrying capacity is OK then you have your minimum cable size.

 

[dyel]

Deleted

 

[Spoon]

Deleted

Just read post #13. It's very informative. If there is anything you don't understand then just ask.

 

[Simonslimline]

Page 155 is ccc for twin and earth cable.

Start from the beginning:
You have 5 off 150W light fittings, presumably as they are hi bay they are discharge lamps, so from appendix A of the OSG you multiply by 1.8 to account for power losses in the control gear and power factor.

I = P/V *1.8 = 5*150 * 1.8 / 230 = 5.87A

So Ib is 5.87A

As a starting point you have said you want to use 1.5mm² cable, from your last post this is single core cables in conduit or trunking.
The voltage drop tabulated on page 151 in OSG for 1.5mm² cable is 29 mV/A/m

You have also said the length of circuit is 19.8 (presumably metres)

Using volt drop = mV/A/m * Ib * L / 1000 = 29 * 5.87 *19.8 /1000 = 3.37 V

Check if this complies using page 146 of the OSG.

Then if this is too high choose a bigger cable, if it is well within then choose a smaller cable and check again.

Once you have you initial cable size start calculating for current carrying capacity, always ensuring you are complying with BS7671 524.1 minimum csa.

In is the next size up of circuit breaker above the design current Ib, that is 6A.

Using Iz > In > Ib the current carrying capacity of the cable must be greater than 6A.

Then you need to consider the installation methods and how they may affect the current carrying capacity of the cable and ensure that once all factors have been taken into account (page 145 in OSG) that the adjusted CCC is above 6A.

Once you have done this and the current carrying capacity is OK then you have your minimum cable size.

A good detailed explanation as usual RB. Have you ever considered teaching the subject?

 

[Simonslimline]

Also not forgetting that as this is a lighting circuit, the the maximum Voltage drop permitted is 3% (6.9 volts)

 

[Bobster]

Also not forgetting that as this is a lighting circuit, the the maximum Voltage drop permitted is 3% (6.9 volts)

Just to throw a spanner in.

If you had a site with a known higher voltage, say 255V L-N. Would you use 3% of that (7.65V) or stick to 3% of 230V (6.9V)?

 

[Simonslimline]

Just to throw a spanner in.

If you had a site with a known higher voltage, say 255V L-N. Would you use 3% of that (7.65V) or stick to 3% of 230V (6.9V)?

I see where you coming from and it makes much more sense to use a percentage of the voltage that you actually have, as the DNO Tx will be tapped @ 250/433V.

But when it comes to college work they will expect students to be using 230V.

 

[hightower]

I asked at college recently if when working out Ib (design current) do we use the nominal voltage (which will be a constant of 230v in most instances) or do we use the MI chosen voltage. As an example I've seen appliances rated at 3kW/240v according to the packaging. So is the design current 3000/240=12.5A or is it 3000/230=13A?

The answer the lecturer gave was you always work out Ib at the nominal voltage, ignoring what the manufacturer instructions said. I was a bit suspect of this answer, so Gods.....

EDIT: Bit more info as to why I'm asking. It might not make much difference in most cases, but take the example of a 9.5kW shower. If you work out Ib at 240v you can safely select In as a 40A breaker. However, work out Ib at 230v and you have 41.3A, which would require a 50A breaker and in turn would likely affect the size of the cable you have to install too.