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10/19/2014 An example of calculating the number of indoor
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electrical-engineering-portal.com
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iguparmar
An example of calculating the number of indoor lighting
fixtures
An examp le of cal culati ng the number of i ndoor l ighti ng f
ixtures (on photo: Col umbus, OH Ohio History Center reading room;
by
army.arch via Flickr)
Input information
These are the input data for the following calculation:
1. An office area has length: 20 meter; width: 10 meter; height:
3 meter.
2. The ceiling to desk height is 2 meters.
3. The area is to be illuminated to a general level of 250
luxusing twin lamp 32 watt CFL luminaires with a
SHR of 1.25.
4. Each lamp has an initial output (Efficiency) of 85 lumen per
watt.
http://electrical-engineering-portal.com/http://electrical-engineering-portal.com/brightness-lamp-life-and-efficiency-of-led-fixtureshttp://electrical-engineering-portal.com/an-example-of-calculating-the-number-of-indoor-lighting-fixtureshttp://electrical-engineering-portal.com/
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5. The lamps Maintenance factor (MF)is 0.63 ,Utilization Factor
is 0.69 and space height ratio (SHR) is
1.25.
Calculation in 8 steps
1. Total wattage of fixtures:
Total wattage of fixtures = Number of lamps x each lamps
watt.
Total wattage of fixtures = 2 32 = 64 Watt.
2. Lumen per fixtures
Lumen per fixtures = Lumen efficiency (Lumen per Watt) x each
fixtures watt
Lumen per fixtures = 85 x 64 = 5440 Lumen
3. Number of fixtures
Required number of fixtures = Required Lux x Room area / MF x UF
x Lumen per fixture
Required number of fixtures = (250 x 20 x 10) / (0.63 0.69
5440)
We will need 21 fixtures
4. Minimum spacing between each fixture
The ceiling to desk height is 2 meters and space height ratio is
1.25, so:
Maximum spacing between fixtures = 2 1.25 = 2.25 meter.
5. Number of required rows of fixtures along with width of
room
Number of rows required = Width of room / Max. spacing = 10 /
2.25
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Number of rows required is there fore = 4.
6. Number of fixtures required in each row
Number of fixtures required in each row = Total Fixtures /
Number of rows = 21 / 4
There fore, we have 5 fixtures in each row.
7. Axial spacing between each fixture:
Axial spacing between fixtures = Length of room / Number of
fixtures in each row
and that would be: 20 / 5 = 4 Mete r
8. Transverse spacing between each fixture:
Transverse spacing between fixtures = Width of room / Number of
fixtures in row
and that would be: 10 / 4 = 2.5 Meter.
5 things to conclude
So far, we have been calculated the following:
Number of rows with lighting fixtures = 4
Number of lighting fixtures in each row = 5
Axial spacing between fixtures = 4.0 meter
Transverse spacing between fixtures = 2.5 meter
Required total number of fixtures = 21
About Author //
Jignesh Parmar
iguparmar- Jignesh Parmar has completed his B.E(Electrical) from
Gujarat University. He is member of
Institution of Engineers (MIE),India. Membership No:M-1473586.He
has more than 12 years experience in
Transmission -Distribution-Electrical Energy theft
detection-Electrical Maintenance-Electrical Projects
http://www.electricalnotes.wordpress.com/
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lighting fixtures | EEP
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Calculated room for the number of lighting fixtures
(Planning-Designing-Technical Review-coordination -Execution).
He is Presently associate with one of the
leading business group as a Assistant Manager at
Ahmedabad,India. He has published numbers of Technica
Articles in "Electrical Mirror", "Electrical India", "Lighting
India", "Industrial Electrix"(Australian Power
Publications) Magazines. He is Freelancer Programmer of Advance
Excel and design useful Excel base
Electrical Programs as per IS, NEC, IEC,IEEE codes. He is
Technical Blogger and Familiar with English, Hindi
Gujarati, French languages. He wants to Share his experience
& Knowledge and help technical enthusiasts to
find suitable solutions and updating themselves on various
Engineering Topics.
