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Electrical Notes 

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91

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60‘C’

 

Lumen Method 

 The quantity of light reaching a certainsurface is usually the main consieration inesigning a lighting system!

 This quantity of light is s"eci#e

$y illuminance measure in lux, an as thisle%el %aries across the working plane& ana%erage #gure is use!

 

CIBSE Lighting Guides give values of illuminance that

are suitable for various areas.

The section - Lighting Levels in these notes also

gives illuminance values.

 

 The lumen metho is use to etermine thenum$er of lam"s that shoul $e installe fora gi%en area or room!

 

Calculating for the Lumen Method

 

 The metho is a commonly use technique of lighting esign& 'hich is %ali& if the light#ttings (luminaires) are to $e mounteo%erhea in a regular "attern!

 

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 The luminous (u) out"ut (lumens) of eachlam" nees to $e *no'n as 'ell as etails of the luminaires an the room surfaces!

 

+sually the illuminance is alreay s"eci#ee!g! o,ce -00 lu)& *itchen 300 lu)& theesigner chooses suita$le luminaires anthen 'ishes to *no' ho' many are require!

 

 The number of lamps is gi%en $y theformula.

 

'here&N / num$er of lam"s require!E / illuminance le%el require lu) / area at 'or*ing "lane height m2 / a%erage luminous (u) from each

lam" lm+/ utilisation factor& an allo'ance for thelight istri$ution of the luminaire  an the room surfaces!4/ maintenance factor& an allo'ance

for reuce light out"ut $ecause ofeterioration an irt!

 

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Example 1

 

"rouction area in a factory measures 60metres ) 25 metres!

in the num$er of lam"s require if eachlam" has a ighting 7esign umen 7out"ut of 18,000 lumens.

 The illumination require for the factory areais 00 lux.

+tilisation factor / 0!5

am" 4aintenance actor / 0!8-

 

N / 200 lu) ) 60m ) 25m 1:&000 lumens ) 0!5 ) 0!8-

N / -3!33

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N / !" lamps.

 

#pacing

 

 The aim of a goo lighting esign is toa""roach uniformit$ in illumination o%er the'or*ing "lane!

Com"lete uniformity is im"ossi$le in "ractice&$ut an acce"ta$le stanar is forthe minimum to $e at least 80; of  the maximum illumination le%el.

 

 This means& for e)am"le& that for a room 'ithan illumination le%el of -00 lu)& if this is ta*enas the minimum le%el& then the ma)imumle%el in another "art of the room 'ill $e nohigher than 815 lu) as sho'n $elo'!

 

-00 0!8 / 815 lu)

 

7ata in manufacturer<s catalogues gi%es thema)imum ratio $et'een the spacing centreto centre of the #ttings antheir height to lam" centre a$o%e the'or*ing "lane 0!:- metres a$o%e f!f!l!

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Example

 

+sing ata in the "re%ious e)am"le sho' thelighting esign layout $elo'!

 The s"acing to mounting height ratio is & ' .

 The mounting height Hm / 5 metres!

 The s"acing $et'een lam"s is calculatefrom from ="acingHm ratio of 3 . 2!

>f the mounting height is 5 m then thema)imum s"acing is.

 

3 2 / ="acing 5

 

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="acing / 1!- ) 5 / 6 metres

 

 The num$er of ro's of lam"s is calculate $yi%iing the 'ith of the $uiling 25 m $ythe s"acing.

 

25 6 / 5 ro's of lam"s

 

 This can $e sho'n $elo'! Half the s"acing isuse for the ens of ro's!

 

 The num$er of lam"s in each ro' can $ecalculate $y i%iing the total num$er of lam"s foun in e)am"le 1 $y the num$er of ro's!

 

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 Total lam"s -5 5 / 13!- goes u" tonearest 'hole num$er / 15 lam"s in eachro'!

 

 The longituinal s"acing $et'een lam"s can$e calculate $y i%iing the length of the$uiling $y the num$er of lam"s "er ro'!

  ength of $uiling 60m 15 / 5!2: metres!

 

 There 'ill $e half the s"acing at $othens / 5!2: 2

  /2!15 metres

 This can $e sho'n $elo'!

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 The total array of #ttings can $e seen $elo'!

or more e%en s"acing the layout shoul $e

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re?consiere!

 The s"acing "re%iously 'as 6 m $et'een

ro's an 5!2: m $et'een lam"s!

 

>f - ro's of 11 lam"s 'ere use then thes"acing 'oul $e.

 

="acing $et'een

ro's / 25 - / 5!: metres

="acing $et'eenlam"s / 60 11 / -!5- metres

 

nstalled lux

 

=ometimes it is useful to *no' the totalamount of light or *ux, +hich has to $e "utinto a s"ace!

nstalled *ux (lm) -umber of ttings(-) x -umber of lamps pertting x L./.L. output of each lamp ()

 

Example &

 

factory measuring -0m ) 10m has alighting scheme consisting of 5 ro's of 2-lighting #ttings each housing 2No! 6-?@att(uorescent lam"s!

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a in the installe (u) in total!

$ @hat is the installe (u) "er m2 of (oor area!

 

 The out"ut of the lam"s in the a$o%e

e)am"le may $e foun from catalogues! or a6-?@att (uorescent lam" the ighting 7esignumens 7 is 5500 lm!

 

a

>nstalle (u) lm / N ) no!lam"s#tting )

 / 5 ) 2- ) 2 ) 5500

  / 880,000 lumens

 

$

 The (oorarea / -0 ) 10 / -00 m2!

>nstalle (u) "erm2  / ::0&000 -00

  / 10 lm2m.

 

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Example "

 

room measures 1-m ) 8m ) 3!6m high anthe esign illumination is 200 lu) on the'or*ing "lane 0!:- metres a$o%e the (oor!

 The +tilisation factor is 0!- an the4aintenance factor is 0!:!

>f the 7 out"ut of each #tting is 2820lumens& calculateA

a the num$er of #ttings require!

$ the #ttings layout!

c >f the s"acingmounting height ratio is 1. 1 etermine 'hether the current esign isacce"ta$le!

 

a Num$er of #ttings!

  N / 200 ) 1- ) 8 2820 ) 0!- ) 0!:

  N / 19!3

  N / 0 lamps

 

$ ittings layout

 

or shallo' #ttings& the mounting height Hm

may $e ta*en as the istance form the ceiling

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to the 'or*ing "lane!

 

 Therefore Hm  / 3!6 ? 0!:-

  Hm  / 2!8- metres

>f 3 ro's of 8 #ttings are consiere thenthe s"acing isA

 

c ="acing mounting height!

 

="acing Hm ratio.

2!33 2!8- / 0!:58 Thereforeratio is 0.8! ' 1.0

2!15 2!8- / 0!88: Thereforeratio is 0.8 ' 1.0

 

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Example !

 

room& as sho'n $elo'& has a esignillumination is -00 lu) on the 'or*ing "lane0!:- metres a$o%e the (oor!

 The +tilisation factor is 0!- an the4aintenance factor is 0!:!

>f the 7 out"ut of each #tting is 2820lumens& calculateA

a the num$er of #ttings require!

$ the #ttings layout!

c >f the s"acingmounting height ratio is 1. 1 etermine 'hether the current esign is

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acce"ta$le!

 

a

N / -00 ) 10 ) 12 2820) 0!- ) 0!:

N / --!1-

N / ! lamps!

 

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$

="acing& say : lam"s ) 8 ro's!

="acing along 12 m 'all / 12 : / 1!-0m

="acing along 10 m 'all / 10 8 / 1!53m

 

c

4ountingheight / 3!0 ? 0!:- / 2!1- m

7esire Batio / 1.1

ctualratio / 1!- 2!1- / 0!69 Therefore ratio is 0.3 ' 1.0

ctualratio / 1!53 2!1- / 0!68 Therefore ratio is 0. ' 1.0

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