Construction Studies Resources

Construction Studies Resources
Light in Buildings By Des Kelly

HOME Light intensity Understanding how to measure the brightness of light can be difficult to grasp. The human eye is more sensitive/receptive to some colours (frequencies) than others We perceive some colours to be brighter than they actually are. The human eye is more sensitive to the colours in the middle of the spectrum of human visibility (yellows and greens) than those at the limit of the spectrum of human visibility (reds and blues). Hi-viz vests and signage! relative intensities of the spectrum

HOME Light intensity The human eye has a varying sensitivity to "brightness." Walk from a dark room into the sunlight. Walk from a brightly lit room into the sunlight. Therefore the human eye may not be a very scientific means of determining “Brightness” levels. When describing how much light is given off by a particular source it is expressed in terms of “Luminous Intensity.” Luminous intensity is measured as a unit of power and is essentially a measure of the power given off from a luminous source. The unit used is the candela (cd) which measures the wattage emitted within a specific angle. Equivilent to approx one candle power of light.

Measuring the intensity of Light
Construction Studies Resources HOME Measuring the intensity of Light Because light is emitted in all directions the solid angle at which the light is emitted is measured. The measure of the solid angle is called the Steradian Take a sphere and inscribe a cone with the tip at the centre of the sphere and its base along the surface of the sphere. The solid angle subtended when the area of the surface is equal to the square of one of the legs is called a steradian. 1 Steradian = radius of the sphere squared (R2) A complete sphere contains 4∏steradians Or as defined by Wikipedia: “For a general sphere of radius r, any portion of its surface with area A = r2 subtends one Steradian”

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Luminous intensity (l)
Construction Studies Resources HOME Luminous intensity (l) A measure of the power/ability of a light source to emit light in a particular direction. As mentioned, the candela (cd) is the measure of luminous intensity that is radiated outward from this angle. However, the candela only measures the luminous intensity of the source. The candela does not consider what happens to the light after it has left this source. For measuring light as it travels through space we use the term lumen. Luminious Flux (F) – See next slide. A lumen is calculated by taking the candela measure of the source and multiplying it by solid angle (steradian) through which it is being emitted. To measure the luminous intensity of a light source on an object we divide the number of lumens by the area over which the light is shining. This unit is called lux. Lux is how we measure the luminous intensity of a light source on any surface area.

HOME Luminious Flux (F) The rate of flow of light energy. Measured in Lumen (lm) 1 Lumen is the luminous flux emitted within one steradian by a point source of 1 Candela. F = I x 4∏ where: F = Luminous flux (lm) I Intensity of source (cd)

HOME Illuminance The amount/density of illunination reaching a surface. Measured in Lux 1Lux= 1 Lumen/metre2 E= F/A where E= Illuminance on surface (Lux) F= Luminous flux ( lumen) A= Area (m2)

HOME Luminous Intensity (Candela) Power of light source Luminous Flux (Lumen) Flow of energy illuminance (Lux) Illumination on surface

HOME Worked example A bulb emits a total luminous flux of 1500 lm in all directions. Calculate the luminous intensity of the source: F = 1500lm W = 4∏ I Using formula I = F / w I = 1500 / 4∏ I = 119 Candela

Inverse square law of illumination
Construction Studies Resources HOME Inverse square law of illumination Naturally the farther light shines the more it spreads out. this is seen by an increase in the area effected by the luminous flux. Because the area that the flux reaches is increased the amount of flux per surface area decreases. The relationship between the two is called the inverse square law. E = I / d2 E.g. as the distance doubles the illuminance is essentially quartered.

Worked example A lamp has a luminous intensity of 1200cd and acts as a point source. Calculate the illuminance produced on surfaces at the following positions at 2m from the lamp At 6m from the lamp Using the formula E = I / d2 a) for a distance of 2m E = 1200 / 22 E = 300lx b) for a distance of 6m E = 1200 / 62 E = 33.33lx

HOME Worked example A small lamp has a luminous intensity of 80 cd. Calculate the illuminance produced on surfaces at the following positions at 0.8m from the lamp at 3.2m from the lamp Using the formula E = I / d2 a) for a distance of 0.8m E = 80 / 0.8*0.8 E = 125lx b) for a distance of 3.2m E = 80 / 3.2*3.2 E = 7.81lx

Light falling at an angle
Construction Studies Resources HOME Light falling at an angle When the luminous flux of a light source falls at an angle the area that it hits will be larger than if it fell directly onto the surface To account for this we use the formula E = (I/d2)*CosѲ where d is the distance of the edge from the point source and Ѳ is the angle that the light falls at from the normal Ѳ d

HOME A lamp acts as a point source with a mean spherical intensity of 1500cd. It is fixed 2m above a circuliar table with a radius of 1.5m. Calculate the illuminance provided at the edge of the table I = 1500cd E= ? Using Pythag Theorem H2 = A2 + O2 H2 = 22 = 1.52 H = 2.5 Cos Ѳ = Adjacent / Hypotenuse Cos Ѳ = 2 / 2.5 Using illumination formulia E = (I/d2)*CosѲ E = (1500 / 6.25) * 0.8 = 192 Illuminance at edge of table = 192 lx H a o

Designing for natural light
Construction Studies Resources HOME Designing for natural light Mirrors on walls and light colours reflect the light about the room South facing windows to catch the available sun Design rooms according to light availablity e.g bedroom on east side of building ot recieve morning light Skylights Clerestry windows Light shelfs Light Tubes

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HOME Selecting Windows When it comes to selecting windows a number of design factors must be considered. The number of panes The spacing of the planes Low-emittance coating (low-E) Gas fill Window frame design Spacer design

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HOME Window frames Window frames can be made from many materials: Timber Aluminum PVC Fiberglass

HOME FR1 Timber FR2 PVC FR3 Aluminium FR4 Fiberglass

HOME Window Rating (BFRC) Indicates: U-Value for window G-value for window Air leakage Other useful information not indicated on the BFRC is the windows Visible Light Transmittance (VLT)

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Open spaces outside a window
Construction Studies Resources HOME Open spaces outside a window The wall must be beyond the 3.6 meter space and under the 30° line as shown No part of any structure, wall, dwelling, plant, etc. can be inside the restricted zone.

HOME 30⁰ 3.6m

HOME 3.6m 3.6m

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HOME Light Calculations Light is measured in units called lux. The size of a window required to provide the appropriate daylight factor for a room can be calculated as follows using the degree of efficiency method: FORMULA: Lux inside = lux outside x window factor x efficiency coefficient x window area/floor area where, lux inside = required level of light inside lux outside = standard overcast sky (ie 5000 lx as discussed previously) window factor = the reduction in incident light due to the fact that the window is in the vertical plane. This reduction is typically assumed to be in the region of 50% and so, is given a constant value of 0.5 for calculations. efficiency coefficient = the reduction in incident light due to factors relating to the glazing (eg transmittance, cleanliness, reflection). For calculations, this is given a constant value of 40% or 0.4. window area = length x height (of window) floor area = length x width (of room)

HOME Window factor Because most windows are orientated vertically, the maximum window area is not orientated at 90 degrees to the sunlight. Hence the window area effectively appears less than it really is and less light can enter. The proportion of light that is lost in this manner is called the window factor (f). It is generally taken to be 50% or .5 A standard overcast sky, EA, is 5000 lux. Therefore because of the window factor only 2500 lux is available to shine in the window.

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HOME Energy co-efficient Because of dirt, imperfections and the nature of glass only approx. 40% (.4) of the available light ends up passing through it. This is called the energy co efficient The symbol for the energy co efficient is n.

HOME Window Area The area of glazing / windows in the room, determines the amount of light available to light the room Because of this we find the total window area The symbol for window area is As the luminance is to be spread around the room it shines into the size of the room will matter as to the overall average luminance To account for this we calculate the total floor area in the room The symbol for floor area is Floor area

Window Area Calculation Worked example #1
Construction Studies Resources HOME Window Area Calculation Worked example #1 A room measuring 3m by 4m, with an unobstructed view, requires an illumination of 150 lux. Determine, using the degree of efficiency method, the approximate area of the glazing required. 150 = 5000 x 0.5 x 0.4 x Ff/12 150 = 1000 x Ff/12 150 x 12 = 1000 x Ff 150 x 12 / 1000 = Ff 1.8m2 = (Window Area) Ff

Construction Studies Resources HOME Window Area Calculation Worked example #2 A room measuring 3m X 4m with an unobstructed view requires an average illumination of 100 Lux. Determine by degree of efficiency method the size of window required ot provide this level of illuminance. Illumination required inside = 100 lux Ea Illumination outside = 5000 lux Fb Floor size = 4 x 3 = 12m2 f Window factor = .5 n Efficiency coefficient = .4 Using formila 100 = 5000 X .5 X .4 X (Ff /12) 100 = 1000 X (Ff /12) 100/ 1000 = Ff /12 .1 X 12 = Ff Ff = 1.2metres 2 LC 1985

Construction Studies Resources HOME Window Area Calculation Worked example #3 The average illumination in a room is to be increased from 90 to 150 lux by enlarging an existing window. Determine by degree of efficiency method the size of the window required if the room is 3800mm long x 4800mm wide. Illumination inside = 150 lux Ea Illumination outside = 5000 lux Fb Floor size = 3.8m x 4.8m = 18.24m2 f Window factor = .5 n Efficiency coefficient = .4 Using formula 150 = 5000 X .5 X .4 X (Ff /18.24) 150 = 1000 X (Ff /18.24) 150/ 1000 = Ff /18.24 .15 X = Ff Ff = metres 2 LC 1987

Construction Studies Resources HOME Window Area Calculation Worked example #4 An architects office requires an average luminance of 750 lux on the working plane. If the drawing office is metres long and 7.2 metres wide, calculate the area of window required ot illuminate the room to this level. Assume the following: An unobscured sky A standard sky of 5000 lux 60% of incident light is reflected or absorbed by the window or dirt on the window Illumination inside = 750 lux Ea Illumination outside = 5000 lux Fb Floor size = 28.8m x 7.2m = m2 Combined window factor and efficiency coefficient = .6 750 = 5000 X .6 X (Ff /207.36) 750 = 3000 X (Ff /207.36) 750/ 3000 = Ff /207.36 .75 X = Ff Ff = metres 2 LC 1993

HOME Question 1 A small lamp emits a total luminous flux of 1257 lm in all directions. Calculate the luminous intensity of this light source. FORMULA: F = I x 4∏ where: F = Luminous flux (lm) I Intensity of source (cd) Example taken from Environmental Science in Buildings page 142

HOME Question 2A A point source of light has an intensity of 410 cd and radiates in all directions uniformly. Calculate the quantity of flux flowing into a hemisphere. FORMULA: I = F / w where: F = Luminous flux (lm) I = Intensity of source (cd) w = solid flux angle /steradian Example taken from Environmental Science in Buildings page 142

HOME Question 2B For the previous example, calculate the average illuminance produced on the inside surface of this hemisphere if its radius was 1.5m. FORMULA: E = I / d where: E = Illumination (lx) I = Intensity of source (cd) d = distance (m) Example taken from Environmental Science in Buildings page 142

HOME Question 3 (A+B) A small lamp has a mean luminous intensity of 80cd. Calculate the maximum direct illuminance the lamp produces on a surface under the following conditions: (A) At a distance of .8m from the lamp. (B) At a distance of 3.2m from the lamp.

HOME Question 4 (A+B) A street lamp has a uniform intensity of 1200cd. It is positioned 7m above the centre line of a road which is 8m wide. (A) Calculate the illuminance on the road surface directly below the lamp. (B) Calculate the illuminance at the edge of the roadway.

HOME Question 5 A uniform point source of light emits a total flux of 2500lm. It is suspended 800mm above the centre of a square table with sides 600mm. Calculate the minimum and maximum illuminances produced on the table.

HOME A uniform point source of light emits a total flux of 2500 lm. It is suspended 800mm above the centre of a square table of sides 600mm in length Calculate the minimum and maximum illuminance provided on the table Flux = 2500 lm E= ? Remember I = F / W I = 2500 / 4 ∏ = Maximum Using illumination formulia E = I/d2 E = / .82 E = lx Minimum Using Triangle laws H2 = A2 + O2 H2 = H2 = H = Cos Ѳ = Adjacent / Hypotenuse Cos Ѳ = .8 / .906 Using illumination formulia E = (I/d2)*CosѲ E = ( / .820) * = Illuminance at edge of table = lx 800 905.63 424.26 848.52 600 600

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