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The Concept of Daylighting Lighting buildings prior to the late 1800s relied chiefly on daylight from windows. Lighting buildings prior to the late 1800s.

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Presentation on theme: "The Concept of Daylighting Lighting buildings prior to the late 1800s relied chiefly on daylight from windows. Lighting buildings prior to the late 1800s."— Presentation transcript:

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2 The Concept of Daylighting Lighting buildings prior to the late 1800s relied chiefly on daylight from windows. Lighting buildings prior to the late 1800s relied chiefly on daylight from windows. Today in America, non-residential buildings consume nearly 60% of electrical energy. Why in America and not the rest of the world? Today in America, non-residential buildings consume nearly 60% of electrical energy. Why in America and not the rest of the world? –Electric lighting was developed in America –America soon learned it had plenty of energy –Modern building design evolved in America –Energy consumption played a major role in improving the comfort of users of buildings. –Until the 1970s. We realized a source of free energy.

3 The realization of need for conserving energy brought a new awareness of daylight as free energy – something that Europe and the rest of the world knew all along. They did not have the abundance of petroleum energy that came and went in America. The realization of need for conserving energy brought a new awareness of daylight as free energy – something that Europe and the rest of the world knew all along. They did not have the abundance of petroleum energy that came and went in America. Daylighting is a different concept: Daylighting is a different concept: –A process of designing from outside in –A careful consideration of light without glare, and energy without heat radiation –Daylight is healthy –Daylight is free

4 A concept developed in Europe considered daylight to be primary, artificial to be secondary A concept developed in Europe considered daylight to be primary, artificial to be secondary Since most of the workday was during the light of day, why not utilize daylight for the work tasks at hand? Since most of the workday was during the light of day, why not utilize daylight for the work tasks at hand? Utilize daylight to its maximum and supplement that with artificial light only as needed. Utilize daylight to its maximum and supplement that with artificial light only as needed. The P.S.A.L.I Concept -- Permanent Supplementary Artificial Lighting Inside. The P.S.A.L.I Concept -- Permanent Supplementary Artificial Lighting Inside.

5 Three Theorems of PSALI : Three Theorems of PSALI : –1 Large variation of light during workday does not adversely affect visual performance. Lightness Constance Lightness Constance Variations in daylight occur slowly over a period of time Variations in daylight occur slowly over a period of time Contrast ratios remain constant Contrast ratios remain constant –2 Daylight and artificial light are easily combined –3 Supplemental systems must be coordinated so lighting levels do not change abruptly.

6 Example Problem in Daylighting As with the artificial light solution, consider Room 102 of the Architecture Building: As with the artificial light solution, consider Room 102 of the Architecture Building: –North exposure of windows –3 Units, each 5’ wide x 8’ high –Reflection criteria; 80% ceiling, 50% walls –Room is 32’ x 31’ –Glass is bronze tinted with transmittance of 80% –The average available exterior illumination for Lubbock at 34 degrees north latitude is 10,000 lux on a cloudy, overcast day.

7 Lubbock is located at 34 degrees north latitude. (bottom of chart) Lubbock is located at 34 degrees north latitude. (bottom of chart) Right side of chart is in thousands of lux of available light on a cloudy day. Right side of chart is in thousands of lux of available light on a cloudy day. Intersect lux with latitude and find available light is in a range of 6,000 to 13,000 lux. Intersect lux with latitude and find available light is in a range of 6,000 to 13,000 lux. Use 10,000 lux for the example problem Use 10,000 lux for the example problem

8 The example problem will follow the guidelines of the C.I.E. method of daylight design. A European method known as “Commission Internationale de l’Eclairage” The example problem will follow the guidelines of the C.I.E. method of daylight design. A European method known as “Commission Internationale de l’Eclairage” Originated in Europe, recognized worldwide. Originated in Europe, recognized worldwide. The method assumes the worst condition for daylight, which is a cloudy, overcast condition, calculating light levels available. Any improvement in the condition will mean additional daylight. The method assumes the worst condition for daylight, which is a cloudy, overcast condition, calculating light levels available. Any improvement in the condition will mean additional daylight.

9 Sequence of Problem: Sequence of Problem: 1 Find the available minimum illumination at a distance of 20’ from the window. 1 Find the available minimum illumination at a distance of 20’ from the window. 2 Find the distance from the windows where the daylight illumination is twice the minimum 2 Find the distance from the windows where the daylight illumination is twice the minimum 3 Find the distance from the windows where the daylight illumination is four times minimum 3 Find the distance from the windows where the daylight illumination is four times minimum 4 Determine the percentage of the daylight workday when the calculated values are available. 4 Determine the percentage of the daylight workday when the calculated values are available.

10 C.I.E. method C.I.E. method considers min considers min daylight to be daylight to be 2’ from wall 2’ from wall opposite the opposite the windows windows ROOM 102 ARCHITECTURE BUILDING ROOM 102 ARCHITECTURE BUILDING

11 A Determine the Basic Daylight Factor A Determine the Basic Daylight Factor First, on the window wall, find the total WIDTH of the wall, then the AGGREGATE width of the windows. Calculate the percentage width of windows, –Width of windows divided by total width of wall... –The total window width equals 15’. –the total window wall width is 32’ So, percentage of windows is 15/32 of window wall =.47, or 47%. Then on the DAYLIGHT FACTOR chart, find the point between the curved, dashed lines that would be 47.

12 Second, find the depth of the room in terms of multiples of window heights: The depth of the room measured from the window wall is 31 feet. The CIE method calculates the minimum daylight available a distance of two feet from the back wall. The depth of the room equals 31’ – 2’ = 29’. But in terms of multiples of window height (8’) = 29/8 = In other words, the point where daylight will be the minimum amount is window heights from the window wall. Enter the DAYLIGHT FACTOR chart to find basic Daylight Factor.

13 Third, at the bottom of the DAYLIGHT FACTOR chart, left to right, find Then follow vertically upward until you reach the percentage of 47. Then follow horizontally left to the border of the chart and read the figure, approximately 1.1. Remember that this number is a percent, which represents the percentage of available daylight on the exterior that will reach the point 29’ into the room opposite the windows. but you must consider the clarity of the window glass.

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15 The Basic daylight factor is found to be 1.1, but since the number is a percentage, mathematically the number is Since light passes through glass in a building, corrections must be made in the daylight factor to compensate for the amount of light the glass diminishes, plus an allowance for how much dirt might collect on the outside of the glass to reduce the amount of light even more. Four, find the chart for TRANSMISSIBILITY of glass:

16 This is a Correction Factor because the glass is tinted, which means it does not allow as much light through as it would if it were clear. This is a Correction Factor because the glass is tinted, which means it does not allow as much light through as it would if it were clear. So the correction factor for a transmissibility of 80% equals 0.95

17 Apply this correction factor to the basic daylight factor: Apply this correction factor to the basic daylight factor: Corrected DF (t) = 1.1 x.95 = Five, Correction Factor for dirt accumulation. If the building is located in an area where the surrounding elements may cause excessive dirt particles to collect on the glass, in which case would further diffuse the amount of daylight that penetrates the glass.

18 Note that the Architecture Building is located in a clean outer suburban area. (no industry or farming) The glass is installed 90 degrees to horizontal, so dirt does not rest on the surface. Note that the Architecture Building is located in a clean outer suburban area. (no industry or farming) The glass is installed 90 degrees to horizontal, so dirt does not rest on the surface. Correction factor from the chart is 0.90

19 Apply the dirt accumulation correction factor to the DF corrected for glass transmissibility: Apply the dirt accumulation correction factor to the DF corrected for glass transmissibility: Corrected DF = x 0.90 = Corrected DF = x 0.90 = Remember this number is a percentage, so mathematically it is written, Remember this number is a percentage, so mathematically it is written, The result is the Daylight Factor to be used in the C.I.E. formula for minimum interior illumination: The result is the Daylight Factor to be used in the C.I.E. formula for minimum interior illumination: So the minimum amount of daylight equals, Corrected Daylight Factor x Available Exterior Illumination. So the minimum amount of daylight equals, Corrected Daylight Factor x Available Exterior Illumination.

20 From the formula, the available interior illumination at a point 29’ from the windows equals x 10,000 = lux. Remember that 10,000 lux is the available amount of outside daylight. Six: Convert the available light from lux, to footcandles, since we use English notation. A lux is one lumen on one square meter, and a footcandle is one lumen on one square foot. So look at the relationship of a square foot and a square meter.

21 A square meter is 3.28’ x 3.28’, which equals square feet. So, if each area has only one lumen, that makes a footcandle times brighter than a lux; or a lux equal.092 footcandle. Since a footcandle is times as bright as a lux, the figure reverts to / = 8.74 footcandles is the amount of daylight available at a point 29’ from the window wall.

22 Logic says that light from the windows is brightest at the window, and diminishes in intensity to the minimum amount at the back of the room. The C.I.E. method establishes three levels of light within the room: The C.I.E. method establishes three levels of light within the room: –The minimum amount, which is 2’ from the back wall –An amount at a point where the DF is twice that of minimum –An amount at a point where the DF is four times minimum

23 Two charts are used to find the distances from the windows where DF is twice minimum, and where DF is four times minimum: Consequently, the amount of available daylight is the same multiple: Two charts are used to find the distances from the windows where DF is twice minimum, and where DF is four times minimum: Consequently, the amount of available daylight is the same multiple:

24 Seven; In the chart 19.18(a) find at the bottom the point that Seven; In the chart 19.18(a) find at the bottom the point that is window heights From that point From that point follow vertically until it intersects with zero degree obstructions. Then move to the Then move to the left horizontally to find the distance of 1.87 window heights. The distance is equal The distance is equal to 1.87 x 8’ = 14.9 feet

25 At a point 14.9 feet from the window wall, the amount of daylight available is equal to twice the minimum amount at 29’ from the window wall. Eight; Next, calculate from the chart 19.18(b), the distance from the window wall where the available amount of light will equal four times the minimum amount at 29’ from the window wall.

26 In the chart find at In the chart find at bottom the point of window heights. From that point follow vertically until it From that point follow vertically until it intersects with zero degree obstructions. Then move to the left horizontally to find Then move to the left horizontally to find the distance of 1.14 window heights. The distance is equal The distance is equal to 1.14 x 8’ = 9.12 feet

27 From the calculations, the distances from window to: From the calculations, the distances from window to: 2 x minimum light = 14.9’, which equals 8.74 x 2 = FC 2 x minimum light = 14.9’, which equals 8.74 x 2 = FC 4 x minimum light = 9.12’, which equals 8.72 x 4 = FC 4 x minimum light = 9.12’, which equals 8.72 x 4 = FC 14.9 ’ 9.12’

28 Realize that none of the 3 figures calculated are sufficient daylight to serve the room if 70 footcandles are desired. However, each of the levels of light is a percentage of that required, which, if daylighting is primary, then less artificial light, and therefore energy, is needed for the task. –At minimum, daylight = 10/70 = 14% –At 2x min, daylight = 17.48/70 = 25% –At 4x min, daylight = 34.96/70 = 50% –The average daylight in the room = FC = 30%, so a saving of 30% of energy can be realized if daylight is utilized. so a saving of 30% of energy can be realized if daylight is utilized.

29 PRACTICE PROBLEM PRACTICE PROBLEM A room that is 40’ x 32’ has a windows 6’ tall on one 40’ side. Total width of the windows is 32 feet. The total correction factor (transmissibility and dirt accumulation) is Available outside light = 10,000 lux. Find: 1 Basic daylight factor 2 Corrected daylight factor 3 Minimum illumination from daylight 4 Distance where daylight is 2 x minimum 5 Distance where daylight is 4 x minimum

30 SOLUTION SOLUTION Room depth is 32’ – 2’ = 30’. Room depth in terms of window heights = 30 / 6 = 5.0 Percentage of window wall is 32’ / 40’ = 80%. 1Basic Daylight Factor from chart = 1.0 2Corrected daylight factor =.855 x 1.0 =.855, but since it is a percentage, then mathematically written is Minimum illumination = x 10,000 = 85 lux, which = 85/10.76 = 7.9 footcandles.

31 4Distance where daylight is twice minimum; go to 5 window heights on the chart and read at 0 obstruction, So 2.6 x window height = 2.6 x 6’ = 15.6 feet. Amount of light = 7.9 x 2 = 15.8 footcandles. 5Distance where daylight is 4 x minimum; go to 5 window heights on the chart and read at 0 obstruction, So 1.37 x window height = 1.37 x 6’ = 8.22’. Amount of light = 7.9 x 4 = 31.6 footcandles.

32 THE FINAL consideration for the P.S.A.L.I. technique of daylighting is to find how much of the workday can be expected to produce the calculated amount of daylight. THE FINAL consideration for the P.S.A.L.I. technique of daylighting is to find how much of the workday can be expected to produce the calculated amount of daylight. The chart labeled, Figure maintains external illumination as a function of latitude for a given percentage of the normal working day – assuming the workday period is approximately centered within the daylight hours. This concept approximates the reasoning behind “daylight savings time.”

33 The chart at right The chart at right shows the per- centage of the workday, for which levels of illumination will be available. The latitude of The latitude of Lubbock is shown on a vertical dashed line. Find 10,000 lux on vertical column on Find 10,000 lux on vertical column on right, follow to left to intersect with latitude line. Available Available illumination in Lubbock will be approx 88% of the day.

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