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Topic Title: Yard Lighting installation Topic Summary: Calculation Of Design Current (Ib) Created On: 07 November 2010 10:56 AM Status: Read Only 
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07 November 2010 10:56 AM


Morning all
I have been asked to install 4 of 400w metal halide light assys in an outside yard. without manufacturers data would the following calculation be reasonable and acceptable to calculate the design load using 0.8 as a pf. i remember back from college that this was the factor to use then but is it still ok as a rule of thumb at present. 400w x 4 = 1600w 1600w / 230 x0.8= Ib Could this same calculation also be used for high pressure sodium and fluorescent lighting assy,s many thanks Regards Martin 



07 November 2010 12:05 PM


In the absence of manufacturers data the factor for discharge lighting is 1.8
4 x 400 x 1.8 / 230 = Ib regards  "Take nothing but a picture, leave nothing but footprints!"  "Oh! The drama of it all."  "You can throw all the philosophy you like at the problem, but at the end of the day it's just basic electrical theory!"  



07 November 2010 12:19 PM


The 1.8 factor that RnR has mentioned is to account for control gear losses which result in additional current,and the possible introduction of harmonics from the control gear.
 Regards Dave(not Cockburn) 



07 November 2010 02:32 PM


Many thanks to you both.
just out of curiosity when do you use the following P / V x 0.8 ( based on 230v supply) 



07 November 2010 02:50 PM


You cant just use it on its own, you have to use it in conjuction with data about power losses, harmonic currents and p.f.
Lets take some live data from a manufacturer; 400W lamp 20% control gear losses (in real terms this is more like 25% or 100W) Harmonic current of 0.598A Power factor of 0.9 If we do the calc using the recommended 1.8 factor all we need to do is, 400 x 1.8 / 230 = an Ib of 3A. Now using the manufacturers data above, 400 + 100 (losses) / 230 x 0.9 (p.f.) = 2.4A 2.4A + 0.589A (harmonic current) = an Ib of 2.98A How close is that then Note that the 1.8 factor is only good for a power factor down to 0.65/0.7 after that then you need to use manufacturers data. regards  "Take nothing but a picture, leave nothing but footprints!"  "Oh! The drama of it all."  "You can throw all the philosophy you like at the problem, but at the end of the day it's just basic electrical theory!"  Edited: 07 November 2010 at 03:32 PM by rocknroll 



07 November 2010 05:44 PM


You cant just use it on its own, you have to use it in conjuction with data about power losses, harmonic currents and p.f. Lets take some live data from a manufacturer; 400W lamp 20% control gear losses (in real terms this is more like 25% or 100W) Harmonic current of 0.598A Power factor of 0.9 If we do the calc using the recommended 1.8 factor all we need to do is, 400 x 1.8 / 230 = an Ib of 3A. Now using the manufacturers data above, 400 + 100 (losses) / 230 x 0.9 (p.f.) = 2.4A 2.4A + 0.589A (harmonic current) = an Ib of 2.98A How close is that then Note that the 1.8 factor is only good for a power factor down to 0.65/0.7 after that then you need to use manufacturers data. regards Many thanks RnR for your time on this one. Regards Martin 



07 November 2010 05:44 PM


You cant just use it on its own, you have to use it in conjuction with data about power losses, harmonic currents and p.f. Lets take some live data from a manufacturer; 400W lamp 20% control gear losses (in real terms this is more like 25% or 100W) Harmonic current of 0.598A Power factor of 0.9 If we do the calc using the recommended 1.8 factor all we need to do is, 400 x 1.8 / 230 = an Ib of 3A. Now using the manufacturers data above, 400 + 100 (losses) / 230 x 0.9 (p.f.) = 2.4A 2.4A + 0.589A (harmonic current) = an Ib of 2.98A How close is that then Note that the 1.8 factor is only good for a power factor down to 0.65/0.7 after that then you need to use manufacturers data. regards Many thanks RnR for your time on this one. Regards Martin 


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