Discuss Flourescent start up current in the Australia area at ElectriciansForums.net

C

Colin Integrale

Chaps...

Our lights in the garage keep tripping every now and again...

We have 16 6' 70W fittings all on the same circuit, I found the landlords and saw they're on a B10...

I'm wondering if they should be on a Type C.
 
Well total circuit amps is 8.7A (assuming 6ft singles) so a 10 is definately needed.

If the existing B type works then I can see no real reason to change it. If there are tripping problems however then Zs permitting swap it for a C type.
 
Are they electronically or inductively ballasted? If they're inductive then with efficiency and PFC taken into consideration they would be pretty close for a 10Amp MCB irrespective of its curve.
 
It's all in 20mm metal pipe, not sure on the ballasts and they're all singles... There might actually be more than 16 but I'm not sure if the bin cupboards are on the same circuit or not.

To be fair, it's not really my problem.. All included in the ground rent but I was just wondering...
 
Florescent lighting has to have the current x 1.8 to find the correct figure to allow for power factor, unless it has a power factor capacitor fitted (which I think a lot of modern ones do)
 
Cold mornings don't help either.

The only real and best way to sort that out is split into a minimum of 2 circuits. As you say it's not your property and therefore not your call. I would personally tell the landlord what is happening and ask him if there is anything he could do about it. After that and he asnwers no then it's just a matter of how much it get's on your nerves
 
This is a situation where your college learning always lets you down as its an area not explained deep enough which i think should be, we can safely say that a single 70w flourescent has an inrush to be taken into consideration but been on its own on a 6amp or 10amp mcb its too small to be of concern, the problem arises when you have multiples of fitting with all their accumilative inrush currents as you basic load calcs to mcb size simply no longer are effective, consider 14 X 58watt single fittings which will work out at approx 3.5amps.... so by your basic teachings you would fit a 6amp mcb type b or c but then applying the 1.8 factor pushes us to 6.3amps thus a 10amp is needed after working this out the last step would be to reference technical data on the mcb with manufacturer (merlin gerin in this example) the maximum no' of 58 watt fitting on a 6amp mcb is noted as 13 so clearly a 6amp mcb is too small, using the inductive factor of 1.8 should normally get you the correct size mcb but occasionally it dosnt, unfortunately it dosnt go higher than 58watts in the tech notes so i cant give you a definitive answer but my rule of thumb is double the load instead of applying 1.8factor and then see what mcb you need and in your case a 10amp should be ok so i suspect a problematic ballast or a type c mcb may cure it, when designing circuits with accumulative inductive loads you need to reference the mcb limits for your lighting before you decide how many per circuit, the above example is for 58w flourescents only and results vary with different lamp types and also with mcb brand so always refer to tech' tables or call them if designing such circuit or it can be an expensive error.
The 1.8 factor is usually good for standard flourescents and working out mcb's but when you drop onto metal hallides and high pressure mercury vapour then the 1.8 factor developes cracks thus refering to manufacturing data is a must, and this applies to all inductive loads like motors, transformers etc
 
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Florescent lighting has to have the current x 1.8 to find the correct figure to allow for power factor, unless it has a power factor capacitor fitted (which I think a lot of modern ones do)
You apply 1.8 factor for inductive loads which are fitted power factor correction capacitors if the fittings dont have them then a different formula needs to be used to work out the no of fittings permitted.... (( mcb rating x 0.8) (u x 0.6)) divided by (p x 1.25)

Luckily we dont come across this type of fitting anymore.
 
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You apply 1.8 factor for inductive loads which are fitted power factor correction capacitors if the fittings dont have them then a different formula needs to be used to work out loading.... (( mcb rating x 0.8) (u x 0.6)) divided by (p x 1.25)

Luckily we dont come across this type of fitting anymore.

But I did when I did my 2400 many years ago as part of my project, bloomin nightmare
 

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