sodium lighting

[hodaire]

how would i work out the design current for 9 x 600 watt sodium lights at 230v?

 

[telectrix]

rough guide. take the wattage total, work out the current from that, then double it.

 

[Electricalserv]

how would i work out the design current for 9 x 600 watt sodium lights at 230v?

iS IT NOT JUST USUALLY CALCS WITH DIVERSITY?

 

[telectrix]

i assume all 9 would be on at the same time, so no diversity allowed 9 x 600= 5400 watts or 23.5A double up 47A.

 

[hodaire]

cheers, had roughly estimated about 1000 watts allowing for discharge lighting, couple of blokes im working with have reckoned that is only for working the startup current.

design current is about 42 amps?

 

[hodaire]

all be on the same time so no diversity

 

[telectrix]

seems a lot. is the 600 the actual lamp wattage or the wattage of the fitting including control gear?

 

[Electricalserv]

i assume all 9 would be on at the same time, so no diversity allowed 9 x 600= 5400 watts or 23.5A double up 47A.

Why do you double it?

 

[hodaire]

600 watt lamps

 

[telectrix]

was thinking of 1.8 correction, but unsure now if the 600 already allows for that.

 

[hodaire]

surely a 600 watt lamp uses 600 watts, then need to add for ballast

 

[hodaire]

would it be acceptable to wire a few up and use a clamp meter to see what the fittings actually draw and use this figure?

 

[telectrix]

if it's going to be a tight calculation re. cable sizing, i would enquire of the manufacturer the actual current drawn. you don't want to install 10mm when 6mm is adequate.

 

[telectrix]

would it be acceptable to wire a few up and use a clamp meter to see what the fittings actually draw and use this figure?

good idea, was about to suggest that. give it a generous safety factor though.

 

[james king]

10mm for light jesus

 

[Dunc]

Design wise your loading would be 9x600=4500W so 19.57A at 230V. Then you need to remember it's 1.8 times the running current on start up so you need to size your breaker properly to allow for this. Im assuming it's similar in some respects to designing a circuit for a motor, if the motor runs at 5A you wouldn't allow for a cable that is capable of handling 6x that just for start up so the same would apply for the lighting. Think youll need atleast a couple of circuits for that lot and maybe a contactor to bring it all on with a type C breaker.

 

[telectrix]

5400 watts 23.5A. assuming SWA cable. i would use 6mm on 40A type C. need to check voltr drop though . depends on the length of the run. about 38meters is your limit

 

[hodaire]

So the 1.8 figure for discharge lighting is for start up only

Will be running at 600 watts when warmed up?

 

[Dunc]

So the 1.8 figure for discharge lighting is for start up only

Will be running at 600 watts when warmed up?

Yep, and i've just noticed that my maths is sh**e! 9x600 is 5400 not 4500, must have been a typo, honest.....

 

[telectrix]

Yep, and i've just noticed that my maths is sh**e! 9x600 is 5400 not 4500, must have been a typo, honest.....

LOL. it's only us old buggers can still remember chanting out times tables parrot fashion, but 50 years on i can usually beat a calculator on simple sums.

 

[davelerave]

the old 1.8 multiplier isn't for 'startup'

-it's for losses ,power factor ,harmonics

http://www.tlc-direct.co.uk/Book/6.2.1.htm

switches and breakers are rated for arcing and startup


with newer lighting i think you may not need 1.8 for design current and volt-drop- if they have electronic gear and high pf

prob best to find out if possible

 

[hodaire]

Thats how i always understood it, think i will be wiring a few up and stickimg a clamp on.

Quite a big difference in results if some use the 1.8 as start up and some for full running load

 

[davelerave]

they don't

if that was the case you'd be using 7*FLA for design current for motors instead of 100-125%

 

[hodaire]

so the total current is either 19.57A or 47 A or somewhere inbetween

 

[parko]

Hodaire,

You'll measure the RMS Amps with a clamp but you won't measure the Harmonic current. The 1.8 is as Dave said and mainly used to compensate for Harmonics.

Electronic ballasts can modify the supply AC waveform producing Harmonic distortion. The Harmonic currents produced as a result of this distortion, particularly the 3rd harmonic, add to the supply - hence the 1.8 factor.

Your current calc should be 42A.

As a rule I always apply the 1.8 factor unless the manufacturer of the fitting says otherwise.