1 lighting learning outcome from this topic, the student is able to – recognize the definitions...

1

Lighting

• Learning outcome from this topic, the student is able to

– recognize the definitions and meanings for terms used in lighting– calculate lighting levels by Lumen and Point-by-point Methods– design the layout of luminaires for an application

2

Colour Temperature

• Colour temperature, or correlated colour temperature, is the colour gradation of the light compared with the light emitted by an intensely heated iron bar of which the temperature is known. There are four categories

– 2500 – 30000 K - Warm– 3000 – 40000 K - Neutral-white– 4000 – 49000 K - Cool-white– 50000 K and above - Daylight and cool-daylight

3

Colour rendering

• The effect of a light source on the colour appearance of objects compared with their colour appearance under a reference light source

• Incandescent lamps and daylight are assumed to have colour rendering of 100%, or colour rendering index (CRI) or Ra 100

4

Colour rendering

• CRI between Ra 90 and 100 – Excellent colour rendering properties – Applications

• mainly at places where correct colour appraisal is a critical task

• CRI between Ra 80 and 90 – Good colour rendering properties – Applications

• in areas where critical colour appraisal is not the primary consideration, yet where good rendition of colours is still desirable

• CRI lower than Ra 80– Moderate to poor colour rendering properties – Applications

• in areas where the quality of the colour rendering is not important

5

Luminous Flux and Efficacy • Luminous flux

– The light emitted by a source, or received by a surface– It is a luminous quantity taking into account the actual power radiating and sensitivity of human eye to different colours – The SI unit of luminous flux is lumen (lm), symbol

• Efficacy – The ratio of lumen output to the electrical power input to lamp– The unit is lm/W

inputpower electrical

output lumen lamp=efficacy Lamp

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Solid Angle (symbol: ω) Plane angle can be drawn on a flat surface

and can be referred to as a planar angle, measured in radians or degrees

A solid angle can only be drawn to suggest a solid and is measured in steradians (sr)

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Solid Angle (symbol: ω)• The angle subtended by an area at a point and is e

qual to the quotient of the projected area (A) on a sphere, centred on the point, by the square of the radius (r) of the sphere, expressed in steradian (sr)

ω = Area / (radius)2 = A / r2

• Solid angle of any sphere subtended at centre with radius r

= surface area of sphere / (radius)2

= 4Πr2 / r2

= 4Π

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Luminous intensity

• Luminous intensity (symbol: I)– This is the illuminating power of the light source to radiate luminous flux in a particular direction– The SI unit is the candela (cd)– See illustration on next slide

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Luminous intensity

– For point source with uniform intensity,

I = F / ω

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Luminous Intensity

• The relation between radiating flux (F) and uniform intensity (I) in a cone is given by

F (lm) = I (cd) x (sr)

Light radiates in three dimensions

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Illuminance

• Illuminance (or lighting level or illumination level) (symbol: E)

– The incident luminous flux (F) per unit of area of surface– The unit is lux (lx)– Expressed as : E = F / A

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Luminance

• Luminance (symbol: L)– The intensity of the light emitted in a given direction per projected area of a luminous or reflecting surface– With unit in candela per square metre

– Expressed as: L = I / A┴ (where A┴ is the luminous area seen by observer)

13

Maintenance & Utilisation factor

• Maintenance factor MF– A factor used in illumination calculation to allow for the reduction of light output from a source of fitting due to lamp lumen depreciation, lamp not being replaced right after burning out, dust and foreign matter being deposited on lamp, etc

•Utilisation factor UF– A factor used in planning lighting schemes, which allows for loss of light by absorption in reflectors, ceilings, walls, floor, etc

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Utilisation factor

•Utilisation factor UF– The ratio of

sourcelight thefromoutput lumens

plane workingon the received lumens

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Uniformity - In the design of a lighting installation, due consideration should be paid to the uniformity of the resulting illuminance of the working place

– It indicates how uniform the lighting level is and has a value between 0 and 1– A value of 1 indicates a 100% uniform working surface– Uniformity is usually defined as:

– For acceptable uniformity, it should not be less than 0.8

uniformityminimum illuminance on the working plane

average illuminance on the working plane

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Lumen Method

• Illumination level calculation – Interior lighting design (the Lumen method)

• In an interior space where it is enclosed by ceiling, walls and floor, the light received by a working plane due to the lighting installation consists of the following components:

– light incident directly from the light sources onto the working plane– light reflected onto the plane via the wall, ceiling and floor

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Lumen Method• The average illuminance produced by a lighting installation in a rectangular room can be determined by the following formula

– where E = average illuminance over the working plane N = number of luminaires in the room n = number of lamps in each luminaire F = bare lamp flux output per lamp MF = maintenance factor UF = utilisation factor of the room at the working plane A = area of the working plane

– In order to use the formula, the room index and the room reflectances must be known in order that the utilisation factor UF can be determined

EN n F MF UF

A

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Room Index• Room Index

– The room index K is a measure of the proportions of the room. For rectangular rooms, the room index is defined as:

where L = length of the roomW = width of the roomHm = height of luminaires above working plane

– If the room is re-entrant (e.g. L-shape) then it should be divided into two or more non re-entrant sections which can be treated separately

)( WLH

WLK

m

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Spacing Height Ratio

• Spacing Height Ratio (SHR)– The ratio is defined as

• Distance between the source point of two luminaries / height of the luminaries above a working plane• Axial and Transverse SHRs

– The correct spacing of luminaires is important since large spaces between the fittings may result in a fall-off of illuminance at the working plane midway between adjacent fittings– The illuminance between the luminaires must not be allowed to fall below 70% of the value directly below the fitting– For most installations a spacing to mounting height ratio of 1:1 to 2:1 above the working surface is usually considered adequate and the working surface is normally taken as 0.8m above the floor level for an office or classroom

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Layout of luminaires

• Layout of luminaires

To maintain an even distribution of illuminance from the luminaires, those adjacent to the walls of the room should be fixed at half the spacing distance. This is because a point in the middle of the room receives luminous flux from two adjacent luminaires, whilst a point close to the wall is illuminated mainly from only one luminaire

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Worked Example

It is proposed to illuminate an electronic workshop, 9m (L) x 8m (W) x3m (H) to an average illuminance of 500 lux at the bench level (0.8m above floor level). The specification calls for luminaires having two 1200mm 36W natural tubes each having an output of 3200 lumens. Use the data of Thorn Popular Pack TLL 297 and assume reflectances of the ceiling, wall and floor are 0.7, 0.5 and 0.2 respectively.

– Determine the number of luminaires required for this installation when the MF is assumed to be 0.7– Sketch the layout of the ceiling plan.– Determine the spacing height ratio. Comment on this value.

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Worked Example

•

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Worked Example • Apply the following equation

E = average illuminance = 500 luxN = number of luminaires in the room = ?n = number of lamps in each luminaire = 2

F = bare lamp flux per lamp = 3200 lm MF = maintenance factor = 0.7

RI = (8x9)/[(3-0.8)(8+9)] = 1.92UF = utilisation factor from the data sheet = 0.54 (By

interpolation) A = area of the working plane = 8 x 9 = 72 m2

From the equation, we get N = 14.83

EN n F MF UF

A

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Worked Example

• One of the methods is to install the luminaires in parallel with the length of the wall, i.e. 9 m

– If we take 16 luminaires, then we will have four columns by four rows of luminaires– The luminaires near the edge of the wall is half the distance from the others.– The layout is shown overleaf with 16 luminaires

– The axial SHR = 2/(3-0.8) = 0.9– The transverse SHR = 2.25 / (3-0.8) = 1.02– Since they are both less than SHR NOM (= 1.75) – The room is adequately lit with acceptable uniformity.

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Worked Example

2 m

2.25 m

0.525 m

1 m

9 m

8 m

1.2 m

1 m

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Point-by-point Method -Rays of light falling upon a small surface from some distance (d) will illuminate that small surface with an illuminance-The illumination of a small surface from a point source follows the inverse square law

E90 = I / d2 and E = I cos α / d2

where:E90 = Illuminance of the surface ┴ to the light ray (in lx)

E = Horizontal illuminance (in lx)

I = Luminous intensity (in candela)

d = Distance (in m)

α = Angle of incidence

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Point-by-point Method

• Given a light source of negligibly small size compared with the distance of the working plane, the horizontal illuminance at a point P is given by the following expressions:

where is the angle of incident from the light source and h is the vertical distance from the light source to the plan of incident

2

3)(cos

h

IE

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Point-by-point Method

Horizontal illuminance at Point P = E

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Point by Point Method

• For a situation where there are more than one light source (all located at the same height), the formula can be extended to

EI

hi i (cos ) 3

2

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Polar Diagram

• The characteristic of the light distribution of a fitting can be found in the 'polar luminous intensity diagram‘

– In the diagram, the luminous intensities at different angles of the transverse and axial planes of the light fitting is represented

• In planning lighting installations, it is important to know how a fitting type influences the light distribution of a source• With such data as the polar luminous intensity diagram, illumination level can be calculated using point-by-point method

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Polar Diagram

• Plots of intensity distribution are shown on a flat diagram (polar co‑ordinate graph) and this refers to the pattern of intensity distribution in a specific plane

– A: Axial plan – B: Transverse plan

• The two variable dimensions are intensity (cd) and direction (degrees)•Angle at downward vertical being designated as zero degree

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Polar Diagram

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Worked Example

• The table gives a list of intensities at different vertical angles. Plot the polar curve.

Angle from downward vertical (degree)

0 15 30 45 60 75 90

Intensity (cd) 620 570 400 200 50 10 0

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Worked Example

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Worked Example

• A ceiling head light with light output of 2500 lm and is used to illuminate a rectangular desk as shown. The principle axis of projection is aimed at the centre point of the desk. The data sheet of the luminaire is given in the Table. • Calculate the

– Illuminance at X (mid-point of the desk)– Illuminance at Y

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Worked Example

1 mX

Y

2 m

DeskLight

Light

Desk

2.4 m

3 m

XY

Principle axis

Z

O

O

P

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Worked Example

• Applying the equation of

– to get the value of I from the data sheet• calculate (the out of principle axis angle)• read I corresponding to the values of

– are obtained by trigonometry» hypotenuse theorem [z2 = x2 + y2]» cosine rule [a2 = b2 + c2 – 2bc cos ]

2

3)(cos

h

IE

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Worked Example • Calculate the illuminance at point X

– Since point X is in the line of the principle axis, = 0

• From the data sheet I per 1000 lm = 244• I = 244 x 2.5 = 610 cd

• Angle of incidence, = tan-1(3/2.4) = 51.30

• EX = 619 cos3 51.3/2.42 = 25.9 lux

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Worked Example• Calculate the illuminance at point Y

Y is off the line of the principle axis, is the angle included between the planes of OX and OY

XY2 = OX2 + OY2 –2[OX*OY*cos ]YZ2 + XZ2 = OX2 + (PY2 + OP2) – 2 *OX * √(PY2+OP2) cos YX2+XZ2=OX2+(PZ2+YZ2+OP2)-2*OX*√(PZ2+YZ2+OP2) cos (0.52+1) = (2.42+32) + (2.42+42+0.52) – 2*√ (2.42+32) *√ (2.42+42+0.52) cos 1.25 = 14.76 + 22.01 – 2 ( 3.84 * 4.69 cos )cos = 0.986 = 9.730

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Worked Example

• Calculate the illuminance at point Y– Y is off the line of the principle axis, is the angle subtended between the vertical plane through the light and OY

• = cos-1 Height/OY = cos-1 2.4/4.69 =590

• The value of I per 1000 lm at = 9.50 is 248.6 cd (by interpolation)• The intensity at point Y= 248.6 * 2.5 = 621.5 cd• The illuminance EY = 621.5 * cos 3 59o / 2.42 = 14.7 lux