Day Light And Artificial Lighting
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- Words: 481
- Pages: 6
By AJAY PATEL NITIN LADHA RICHARD TALBOT KHUSHBU DESAI
Contents Page
Introduction……………………………………………………………………….1 Equipment………………………………………………………………………...1 Part a Procedure……………………………………………………………………… ….1 Method…………………………………………………………………………….1-2 Results……………………………………………………………………………..2 Part b Results Conclusion…………………………………………………………………………3
Day light and artificial lighting Introduction This experiment was carried out to observe and understand the principle behind day and artificial lighting in a room. The exploitation of daylight is recognised as an effective means to reduce the artificial lighting requirements of non-domestic buildings. Equipment Light Level Meter (Lux Meter) ~ HAGNER EC1 Surveyor’s measuring tape Procedure 1. We Drew a plan and section of the room. From the centre of the window. 2. We calculated the sky angle (θ). 3. Measurements of the room were taken (length, width and height). We then measured the window area. From this we found out the area of the room and window. 4. The light transmission for the windows was calculated and then we measured the Lux level outside the window and the Lux level inside. We calculated the average daylight factor from the following formula below, which is shown in our table of results. _ Ag θT Average Daylight Factor: D = A 1− R2 5. We measured the lux level from the centre of the room. 6. We compared this level to the outside level and we calculated the Average Daylight level percentage for the room centre by using this formula (D = Luxin/Luxout x 100)
(
)
Method Room ID and Date
Outside sky conditio n
Window Orientat ion
Sky angle Ө
Room M421 27TH Feb 2008
Sunny no clouds
west
52
Total area with windo w 10 * 5.5 = 55 m²
Window area
Window Transmi ssion
Average Daylight Factor
Daylight level at room centre
3.8 * 1.8 = 6.84 m²
0.569
4.9062
0.78
Results for day light (Grid) 021 010
014
011
008
018
019
023
017
012
042
037
033
025
015
085
075
061
028
016
078
077
057
027
021
073
070
055
036
023
048
054
051
029
018
052
058
047
030
017
017
025
028
019
013
Results _
Average Daylight Factor: D =
Ag θT
(
)
A 1− R2 = 6.84 * 52 * (0.569) 55 (1-0.5²) = 202.382/41.25 = 4.9062
Lux level in the middle of the room = 0.078 Lux level outside = 420 Lux level inside = 239 (D = Luxin/Luxout x 100) = 078/420 * 100 = 18.6%
PART B Results for artificial lux level (Grid) 023
012
015
012
011
021
022
027
024
017
048
042
038
033
024
089
073
054
043
032
084
080
054
036
031
075
064
057
042
033
051
055
049
033
022
050
055
050
037
025
022
032
034
020
019
Conclusion To conclude, this experiment has proven that daylight and artificial light is very important in buildings because sufficient lights need to be adequately supplied in buildings for people to feel comfortable in.
Contents Page
Introduction……………………………………………………………………….1 Equipment………………………………………………………………………...1 Part a Procedure……………………………………………………………………… ….1 Method…………………………………………………………………………….1-2 Results……………………………………………………………………………..2 Part b Results Conclusion…………………………………………………………………………3
Day light and artificial lighting Introduction This experiment was carried out to observe and understand the principle behind day and artificial lighting in a room. The exploitation of daylight is recognised as an effective means to reduce the artificial lighting requirements of non-domestic buildings. Equipment Light Level Meter (Lux Meter) ~ HAGNER EC1 Surveyor’s measuring tape Procedure 1. We Drew a plan and section of the room. From the centre of the window. 2. We calculated the sky angle (θ). 3. Measurements of the room were taken (length, width and height). We then measured the window area. From this we found out the area of the room and window. 4. The light transmission for the windows was calculated and then we measured the Lux level outside the window and the Lux level inside. We calculated the average daylight factor from the following formula below, which is shown in our table of results. _ Ag θT Average Daylight Factor: D = A 1− R2 5. We measured the lux level from the centre of the room. 6. We compared this level to the outside level and we calculated the Average Daylight level percentage for the room centre by using this formula (D = Luxin/Luxout x 100)
(
)
Method Room ID and Date
Outside sky conditio n
Window Orientat ion
Sky angle Ө
Room M421 27TH Feb 2008
Sunny no clouds
west
52
Total area with windo w 10 * 5.5 = 55 m²
Window area
Window Transmi ssion
Average Daylight Factor
Daylight level at room centre
3.8 * 1.8 = 6.84 m²
0.569
4.9062
0.78
Results for day light (Grid) 021 010
014
011
008
018
019
023
017
012
042
037
033
025
015
085
075
061
028
016
078
077
057
027
021
073
070
055
036
023
048
054
051
029
018
052
058
047
030
017
017
025
028
019
013
Results _
Average Daylight Factor: D =
Ag θT
(
)
A 1− R2 = 6.84 * 52 * (0.569) 55 (1-0.5²) = 202.382/41.25 = 4.9062
Lux level in the middle of the room = 0.078 Lux level outside = 420 Lux level inside = 239 (D = Luxin/Luxout x 100) = 078/420 * 100 = 18.6%
PART B Results for artificial lux level (Grid) 023
012
015
012
011
021
022
027
024
017
048
042
038
033
024
089
073
054
043
032
084
080
054
036
031
075
064
057
042
033
051
055
049
033
022
050
055
050
037
025
022
032
034
020
019
Conclusion To conclude, this experiment has proven that daylight and artificial light is very important in buildings because sufficient lights need to be adequately supplied in buildings for people to feel comfortable in.
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