Physics of Light. Light: electromagnetic radiation that can produce a visual sensation.

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Lighting systems

Physics of Light Light: electromagnetic radiation that can produce a visual sensation. Speed of propagation (c) λ Wave frequency (ƒ) (300 000 km/s) ١

hv λ Less than 1% of the Electromagnetic Spectrum is perceived by the human eye (380 780 nm) Visible Spectrum.

Human Eye The retina (where the image is projected in the eye) comprises detector cells, which convert the energy of the light into nerve impulses making possible human vision. These light sensitive cellsareoftwo types, rods and cones ٣

Light Measurement Units LuminousFlux (Ф): Defined as the flow of visible energy, it is measured in lumens (lm). A light source that equally irradiates 1candela of luminous intensity in all directions will produce a luminous flux of 1 lumen persteradian. ٤

Light Measurement Units The steradian (sr) is the standard international unit of solid angular measure (ω).there are 4π,or approximately 1.5664, steradians in a complete sphere. A steradian is defined as conical in shape, asshown below: Where: P represents the center of the sphere. The solid (conical) angle q, represents one steradian, The Area A of the subtended portion of the sphere is equal to r, where r is the radius of the sphere. 1 steradian Area (A) / r ٥

Luminous Intensity (I): Light Measurement Units φ I Lm / ω Itis defined as the force that generates the lightwesee. Measured in candelas (cd) or lumens per steradian (lm/sr), the luminous intensity is the luminous flux emitted from a point source per unit solid angle. Sr An ordinary wax candle produces a luminous intensity of one candela on all directions. 1 Candela 1 Candle Power (cp) Polar curve ٦

Light Measurement Units Illuminance (Illumination) (E): Illumination is the density of luminous flux on a surface. This parameter shows how "bright" the surface point appears to the human eye. E φ A lux One lumen of luminous flux that uniformly incidents over one square meter [or square foot], produces as illuminance one lux [or 1 foot candle (fc)]. ٧

Laws of Illumination Inverse Square Law : The inverse square law tells is that the illumination is inversely proportional to the square of the distance between the point source and the surface, i.e.: E E 1 I D 1 I D ٨

Laws of Illumination Lambert cosine law: Effective illumination is proportional to the cosine of the angle of incidence of the light on the surface (angle between the direction of the light and the perpendicular to the surface) Illumination at Similarly E E C D I h I h E B : cos cos 3 3 I h θ θ 3 I cosθ d 1 cos 3 1 θ 1 ٩

Outdoor Lighting Design Point by Point Method: -Amethod to calculate the horizontal illumination on aworkplane withinaspace, as well as itsuniformity. -Onlythe direct illumination from the luminaires (lamp fittings) to theworkplane istaken intoaccount. -The illumination at each point is determined by applying the inversesquare law and thecosinelaw to theintensitydataforeach luminaireunder thegiven angle. The mounting height of the lighting luminaire above the ground, the horizontal spacing of one pole to the next, and the cutoff angle of the luminaire are all important issues in outdoor lighting design, just as much as is the choice of the luminaire, the lamp type, and wattage. ١٠

Example 1: Four lamps each 00 cd, suspended at a height of 5m. The distance between them are 10m. Find the illumination level at a point at the center between the second and third lamps. Due I cosθ1 Q EL 1 h 1 15 Qθ1 tan 71.5 5 00 3 EL 1 cos θ1 0.5 lux 5 Due to second lamp L : E E L T to first lamp L 00 3 cos θ 5 ( E + E ) L1 L 1 :.83 lux 6.16 lux ١١

Indoor Lighting scheme Total flux method, or lumen method: This method is used in areas where a uniform light intensity of the work area if the light fittings (luminaires) are to be mounted in a regular pattern. Complete uniformity is impossible in practice, but an acceptable standard is for the minimum to be at least 70% of the maximum illumination level. What we have to do is solve an equation of the type: Where: E is the illumination level required at the work surface, A is the total area of the plane where the work is done. is the flux of light received on the working surface. It is not the flux emitted by the lamps. the flux needs to be installed. ١٢

Total flux method, or lumen method...1 1-The first step is to determine how much flux is to be received by multiply the illumination required by the surface area. The table below is a guidance for recommended light level in different work spaces Activity Illumination (lux, lumen/m) Public areas with dark surroundings 0-50 Homes, Theaters, Archives 150 Easy Office Work, Classes 50 Normal Office Work, PC Work, Study Library, Groceries, Show Rooms, Laboratories Normal Drawing Work, Detailed Mechanical Workshops, Operation Theatres Detailed Drawing Work, Very Detailed Mechanical Works Performance of visual tasks of low contrast and very small size for prolonged periods of time Performance of very prolonged and exacting visual tasks Performance of very special visual tasks of extremely low contrast and small size 500 1,000 1500-000 000-5000 5000-10000 10000-0000 ١٣

Total flux method, or lumen method... -The received flux is related to the installed flux by a simple formula: Where: MF Maintenance factor (Light Loss Factor LLF) UF Utilization factor The maintenance factor gives an estimate of how lighting conditions will deteriorate through use. Some factors are: Dust and dirt inside luminaire surfaces. Aging of light bulbs emitting less light. Cleaning of room surfaces, e.g. ceiling. Without detailed knowledge of a maintenance plan, MF0.80. ١٤

Total flux method, or lumen method...3 This utilization factor is somewhat more complicated to determine than the maintenance factor. The utilization factor will depend on the following: 1-The reflectance of the room surfaces: Surfaces (walls, ceiling) colors are light or dark. Reflectances are available from manufacturers of paints and furniture finishing. One typically recommends in offices: Ceiling 70-90% Wall 50-70% Floor 0-50% ١٥

Total flux method, or lumen method...4 - The geometric proportion of the room: Calculating Room Index factor K: k H L W ( L W ) m + The mounting height, Hm is the vertical distance from the work place to the luminaire. ١٦

Total flux method, or lumen method...5 -There exist data sheets of UF for rooms of different shapes and luminaires of a common type. -Catalogues of luminaires often provide data sheets of UF for rooms. Getting number of lamps : Qφ φ ins N rec EA UF EA UF MF φ lamp MF EA UF MF φ ins ١٧

Total flux method, or lumen method...6 Getting uniform illumination: The space to height ratio (S/Hm):The spacing to height ratio (S/Hm) is the ratio of the centre-to-centre distance between adjacent luminaires, to their height above the working plane. (S/Hm must be around 1 and<1.5 ) ١٨

Example : A production area in a factory measures 60 metres x 4 metres. Find the number of lamps required if each lamp has a Lighting Design Lumen output of 18,000 lumens. The illumination required for the factory area is 00 lux. Utilization factor 0.4 Lamp Maintenance Factor 0.75 The mounting height (H m ) 4 metres. Q N N φ lamp EA UF MF 00lux 60m 4m 18000lm 0.4 0.75 N 53.33 N 54lamps ١٩

Example : Dividing 54 lamps to 16 lamp per row 3 rows: So: as : S 1 / H m 60 m /16 4 4 m lamps m / 3lamps 0.9375 and : S / H m 1.5 4m So increasing number of lamps to 56, dividing them to 14 lamp per row 4 rows : as : and S : 1 / H S m / H m 60 m /14 lamps 4m 4 m / 4lamps 4m 4.8 m 4m 6m 4m 1.5 1.0714 ٢٠

Example : For more even spacing the layout should be re-considered. If 5 rows of 11 lamps were used then the spacing would be: Spacing between rows 4 / 5 4.8 metres S1/Hm1. Spacing between lamps 60/ 11 5.45 metres S1/Hm1.36 ٢١

Example 3: A factory measuring 50m x 10m has a lighting scheme consisting of 4 rows of 5 lighting fittings each housing (65-Watt fluorescent lamps, 4400 lm). -Find the installed flux in total. -What is the installed flux per m of floor area. Installed flux (lm) N x no. lamps/fitting x Ф 4 x 5 x x 4400lm 880,000 lumens The floor area 50 x 10 500 m. Installed flux per m 880,000 / 500 1760 lm/m. ٢٢

Example 4: A room measures 15m x 7m x 3.6m high and the design illumination is 00 lux on the working plane (0.85 metres above the floor). The Utilisation factor is 0.5 and the Maintenance factor is 0.8. If the flux output of each fitting is 70 lumens, calculate; -The number of fittings required. -The fittings layout. -If the spacing/mounting height ratio is < 1 determine whether the current design is acceptable. Q N N φ lamp 19.3 EA UF MF N 0lamps 00lux 15m 3.6m 70lm 0.5 0.8 ٢٣

Example 4: The mounting height (H m ) is taken as the distance form the ceiling to the working plane. H m 3.6-0.85 H m.75 metres If 3 rows of 7 fittings are considered then the spacing is; fittings 15m / 7 fittings.33 m as : S1 / H m 0.847.75 m.75 m 7m / 3 fittings.14 m and : S / H m 0.778.75 m.75 m ٢٤

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