|Mar13-13, 04:50 PM||#1|
Calculating Illuminance of a panel light
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.
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|Mar13-13, 07:29 PM||#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 eighth-spheres 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 ?
|flourescent tubes, illuminance, lux, reflector|
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