Artificial lighting. Dr Torwong Chenvidyakarn

Transcription

1 Artificial lighting Dr Torwong Chenvidyakarn

2 Aims To understand basic concepts of light measurement. To be able to tackle simple design problems relating to artificial lighting.

3 Format Measurement of light Basic lighting design

4 How do we measure light?

5 How do we measure light? Point source

6 How do we measure light? Luminous intensity, I (candela, cd) Point source The power of a light source given out in a particular direction

7 How do we measure light? Luminous flux, F (lumen, lm) The rate of flow of the light energy

8 How do we measure light? ω Area R 2 One lumen I R = Luminous flux emitted within a solid angle of one steradian (strerad) by a point source of one candela So, the total luminous flux is given by F = Iω lumen cd Total solid angle (sterad)

9 How do we measure light? ω F A lux The illuminance, E (lux), on that surface is given by E = F A lm m 2 1 lux = 1 lm / m 2

10 How do we measure light? ω d A I F

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12 Lighting design : localised lighting Surface perpendicular to the light source E = F A = Iω = A I A d A 2 (for small ω) E A = d I 2 cd lux m 2 small ω I d A

13 Lighting design : localised lighting For a surface not perpendicular to the light source (or an inclined light source)... I θ θ I B Y θ I θ θ X The effective area is increased by the ratio XB/XY, so that that the illuminance falls by the factor XY/XB. B d θ d E B = I d θ 2 θ XY XB = I θ cosθ = d 2 θ I θ cosθ d cosθ 2 E B = I θ cos d 2 3 θ

14 From a single lamp...

15 ...to the lighting of the whole room From a single lamp...

16 How many lamps? How should they be arranged and controlled?

17 How many lamps? The lumen method Total illuminance (lux) Area at working plane height (m 2 ) N Number of lamp fittings = F Luminous flux of each lamp (lm) E UF A LLF Light loss factor (deterioration, dirt) Utilisation factor (luminaire, room surfaces, room dimensions (RI) )

18 How many lamps? % 40 60% DLOR Downward light output ratio 60 90% = Downward light output of luminaire Light output of its lamp(s) 10 40% 0 10%

19 How many lamps? Room length (m) RI = H L W ( L W ) m + Room width (m) Vertical distance from lamp to working plane (m)

20 How many lamps? Ensure that lamps are well distributed and the light blends together well. No dark patches. Layout S For fluorescent tubes with diffusing luminaires Smax = 1. 5H m Working plane H m For filament lamps in direct luminaires Smax = 1. 0H m

21 Summary We have discussed The measurement of light: Luminous intensity, I (cd) Luminous flux, F (lm) Illuminance, E (lux) Basic artificial lighting design Localised lighting Distributed lighting Lumen method Want to know more? Lighting by DC Pritchard (1999 or newer), Longman

22 Exercise 1. An ancient manuscript is kept in a horizontal position in a display cabinet in a museum. The illuminance level required to allow the manuscript to be read without being damaged is 100 lux. For visual effect, the manuscript is to be lit at an angle of 45º by a small lamp. How far should the lamp be located away from the manuscript, if it emits light of intensity 100 cd in the direction of the manuscript (i.e. 45º). Illustrate your answer with a diagram. 2. An office measuring 40 x 12 x 3.75 m in height requires an illuminance of 500 lux on the desks set 75 cm above the floor. The 65 W tubular fluorescent lamps chosen have a luminous efficacy of 80 lm/w. They are to be mounted on the ceiling in direct lighting luminaries which have a DLOR of 50%. The room reflectances are 0.5 for the ceiling and 0.3 for the walls. The office is located in a dirty industrial area. a) Use the lumen method to calculate the number of lamps required. b) Suggest a suitable layout for the lamp fittings. c) Suggest an energy-efficient manual control system.

23 Answers Lamp 100 = 100 x (cos45º) 3 / H 2 H = 0.60 m 45º I θ =100cd H 45º 100 lux H Manuscript

24 Answers N = F E A UF LLF RI = H L W ( L W ) m + S = 1. 5H max fluo S = 1. 0H max incand m m E = 500 lux A = 40 x 12 = 480 m 2 F = 80 lm/w x 65 W = 5200 lm/ lamp RI = (40 x 12) / [3x (40+12)] 3 table, r ceiling = 0.5, r wall = 0.3 UF = 0.46 LLF table 0.7 So, N = (500 x 480) / (5200 x 0.46 x 0.7) = lamps Proposed a layout of 9 rows of 16 luminaires each this gives spacing of 12/9 = 1.3 m across the width, and 40/16 = 2.5 m along the length Checking: S max = 1.5 x 3 m = 4.5 m > 2.5 m > 1.3 m so the proposed layout is fine. To separate the switches of the inner area lamps from those along the perimeter for PSALI and energy efficiency.