Fluorescent Light Fitting Load Calculations

[tomtom27]

hello every one i was wondering if you can help.

im a maintnenance engineer and only usualy diagnose and rectify electrical faults on production machinary, very rarely instalation so im a bit rusty with calculations and probably way out ith these.

we are instaling some 5ft fluorscent lights at work (i say some its about 100), running trunking, cables the lot. i need to work out the load of the fittings.

this is whats on the fitting.

Lamp           In 230v/50hz       PF

T5 2x49w     0.452                  0.98

so i would say 500ma per fitting, i even rigged one up on the bench with a meter and yes it was drawing .45amps.

how would i work this out on a formular to check this is correct?

i used this, 

VAxPF/V so volt amps = 230x.452= 104

so 104x0.98/230= 0.443, is this formular correct??

my boss said when calculating the load of fluro fittings you use 1.8 as a figue for somthing and he said thats were people go wrong so now im a bit lost, do i use this instead

104x1.8/230? = amps.

like i said i might be way out with this anyway but i hope somone with more knowledge on this type of thing can help.

many thanks to all

tom.

 

[Technician]

Reading that note it would mean that if that factor was not applied,

the actual current demand would be under-estimated by nearly 50%.

The total demand from the figures would suggest then, 17.6kVA.

(If that is 100 fittings with 2 tubes per fitting)

 

[kerching]

1.8 is the fiddle factor used for discharge lighting

You are using T5 HF units ata a guess?

Best option is to ring the manufacturers and ask what the running current is per fitting. As there should be F little losses

On HF T5s........but I may well be wrong, see wife for details

 

[Technician]

+1 OSG 1.8 is the factor for all lighting and as said get manufacturer data and it could be a lot less especially with todays modern fittings and energy efficiency

So would this mean that it need not be applied to an LED

array, the likes of which (2 foot by 2 foot) might be a direct

replacement for the 4 by 2 foot tube luminaire that is common

in suspended ceilings?

 

[kerching]

So would this mean that it need not be applied to an LED

array, the likes of which (2 foot by 2 foot) might be a direct

replacement for the 4 by 2 foot tube luminaire that is common

in suspended ceilings?

Again I would get info from the manufacturers. I found this out recently when I err sort of err got it err a bit err wrong with some LED panel fittings. I managed to get it to be the manufacturers info that was wrong and they covered the cost of the remedial works!...just saying

 

[Steve3948]

Again I would get info from the manufacturers. I found this out recently when I err sort of err got it err a bit err wrong with some LED panel fittings. I managed to get it to be the manufacturers info that was wrong and they covered the cost of the remedial works!...just saying

Any chance of expanding on that  :innocent

 

[Evans Electric]

1.8 is the worst case scenario , without applying the PF your results will be over engineered.

 

[Technician]

Again I would get info from the manufacturers. I found this out recently when I err sort of err got it err a bit err wrong with some LED panel fittings. I managed to get it to be the manufacturers info that was wrong and they covered the cost of the remedial works!...just saying

OH.......Dear.

Caveat Elector.