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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: Post and Reply |
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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 |
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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!" ------------------------- |
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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) |
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Many thanks to you both.
just out of curiosity when do you use the following P / V x 0.8 ( based on 230v supply) |
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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 |
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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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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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