Calculating Illuminance of a panel lightby GearlooseHC Tags: flourescent tubes, illuminance, lux, reflector 

#1
Mar1313, 04:50 PM

P: 2

Hello,
I would like to build a panel light with four flourescent tubes. The tubes have a diameter of 26 mm and are 1200 mm long. They have a distance of 42 mm one to another. The tubes are mounted on a board, that is covered with aluminium foil as a reflector. Each tube produces a luminous flux of 2340 Lumen. I would like to calculate the illuminance in Lux in one meter perpendicular distance of the panel light. How can I do that ? The result does not have to be perfect – a result within 10 – 20 % of the actual value would be satisfactory. 



#2
Mar1313, 07:29 PM

P: 2

An attempt that I made, under the assumption that the light photons spread out evenly from the flouro tubes in all directions:
1. Direct light from the tubes There is the area directly above the board a. 0.32 x 1.3 = 0.416 m² then there is the surface area of the four eighthspheres in the corners, each with a radius of 1 m: b. 4 x ( 0.5 x pi x 1²) = 2 x pi = 6.28 m² then the surface of the four quarter cylinders alongside the length and width of the board: c. 2 x ( 0.5pi x 1 x 1.3 ) + 2 x (0.5pi x 1 x 0.32) = pi x 1.3 + pi x 0.32 = 5.09 m² The sum of this surfaces is 11.19 m² Only half of the photons radiates directly of the tubes  the other half are reflected by the aluminum foil. Thus: ( 2340 Lumen x 4) :2 : 11.19m² = 4680 Lumen : 11.19m² = 418.2 Lux That's the direct component , 1m above the board. 2. The reflected light That are all the photons that bounce off the aluminum foil. I assume a reflection rate of 85 %. Again all is evenly distributed in all directions. Then the reflection component of illuminance is 0.85 x 418.2 Lux = 355.5 Lux Thus the sum of the illuminance of the direct and reflected light one meter above the board is 418.2 Lux + 355.5 Lux = 773.67 Lux Do my assumptions and calculations make sense ? Regards, Werner 


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