Register to reply

Calculating Illuminance of a panel light

by GearlooseHC
Tags: flourescent tubes, illuminance, lux, reflector
Share this thread:
Mar13-13, 04:50 PM
P: 2

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.
Phys.Org News Partner Physics news on
Physicists discuss quantum pigeonhole principle
First in-situ images of void collapse in explosives
The first supercomputer simulations of 'spin?orbit' forces between neutrons and protons in an atomic nucleus
Mar13-13, 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 m

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 1) = 2 x pi = 6.28 m

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 m

The sum of this surfaces is 11.19 m

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 : 11.19m = 4680 Lumen : 11.19m = 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

Register to reply

Related Discussions
Finding Spectral Illuminance from total Illuminance Classical Physics 0
Luminance and illuminance of multiple point light sources General Physics 5
Light detection panel for laser - help! Electrical Engineering 3
Calculating water flow over a solar panel Engineering Systems & Design 9
Calculating pressure coeff. in triangular panel method Mechanical Engineering 6