Lighting Design
Example Solution
Part I
Develop Lighting Scheme(s)
Given Data
Classroom 20
20’ x 30’
30 x 9’
9
E=50 fc
WP= 2’-6” AFF
ρc= 80%
hcc= 0
0.0
0’
ρw= 50%
hrc= 6.5’
ρf= 20%
hfc= 2
2.5
5’
fixture: fluorescent (#47)
maintenance: yearly
replacement: on burnout
voltages
lt
& ballast:
b ll t normall
environment: clean
Confirm fixture data
S: T.15.1 p. 632
Complete #1-7
Part I
1. Project
j
Identification: Office
2. Average Maintained illumination: 50 fc
/
3. Manufacturer: N/A
4. Catalog No.: #47 from Table 15.1, p. 632
5. Type and Color: T-8,
T 8, warm white
6. Lamps/luminaire: 4
7 Lumens per luminaire: 4@2900=11600 Lm
7.
0’
ρc= 80%
ρw= 50%
ρf= 20%
8. Record
dimensional
data
6.5’
2.5’
30’
20’
9. Calculate
Cavity
Ratios
Calculate Cavity Ratios
CR = 5 H x (L+W)/(L x W)
RCR = 5 (6.5) x (30+20)/(30x20) = 2.71
CCR = 5 (0) x (30+20)/(30x20) = 0
FCR = 5 (2.5) x (30+20)/(30x20) = 1.04
10. Calculate
Effective
Ceiling
Reflectance
Effective Ceiling Cavity Reflectance
3. Obtain effective ceiling reflectance:
S: T.15.2 p. 657
11. Calculate
Effective
Floor
Reflectance
Stein: T.15.2 P. 657
Effective Floor Cavity Reflectance
3. Obtain effective floor reflectance:
CU= 0.19 Æ 0.20
S: T.15.2 p. 657
12. Select CU
from
#47 T15.1
T15 1
Coefficient of Utilization (CU)
RCR
2.00
2.71
3.00
CU
0.55
X
0.48
S: T.15.1
S
5 p
p. 63
632
CU= 0.50
13-21
Calculate LLF
Light Loss Factor(LLF)
Facto (LLF)
13-16
All factors not known Æ
0.88
Light Loss Factor(LLF)
Facto (LLF)
17 Room Surface Dirt
17.
(based on 24 month cleaning cycle,
normal maintenance)
Direct
0.92 +/- 5%
Li ht Loss
Light
L
Factor(LLF)
F t (LLF) Calculation
C l l ti
18 Lamp Lumen Depreciation
18.
Fluorescent
Group
0.90
Burnout
0.85
Light Loss Factor(LLF)
Facto (LLF)
19 Burnouts
19.
Burnout
0 95
0.95
Light Loss Factor(LLF)
Facto (LLF)
20 Luminaire Dirt Depreciation (LDD)
20.
Verify
y maintenance category
g y
S: T.15.1 p. 632
Light Loss Factor(LLF)
Facto (LLF)
20 Luminaire Dirt Depreciation (LDD)
20.
S: F.15.34 p. 653
LDD=0.83
Light Loss Factor(LLF)
Facto (LLF)
LLF = [a x b x c x d] x e x f x g x h
LLF = [0.88] x 0.92 x 0.85 x 0.95 x 0.83
LLF = 0.54
22. Calculate
Number of
Luminaires
22
23
Calc late Number
Calculate
N mbe of Luminaires
L minai es
No. of Luminaires =
(E x Area)/(Lamps/luminaire x Lumens/Lamp x CU x LLF)
(50 X 600)/(4 X 2900 x 0.50 x 0.54) = 9.6
luminaires
E ample 1
Example
Goal is 50 fc +/- 10% Æ 45-55 fc
Luminaires
L
i i
8
9
10
11
E (f
(fc))
41.8
47 0
47.0
52.2
57.4
x
ok
k Æ 3 rows off 3
ok Æ 2 rows of 5
x
Verify S/MH for fixture, space geometry
S/MH Ratio
Verify S/MH ratio
MH=9.0-2.5=6.5’
S/MH = 1.7 Æ S ≤ 11.05’
11.05
S: T.15.1 p. 632
Spacing Scheme 1
Spacing:
Try 3 rows of
3 luminaires
S/2
S/2
S/2+2S+S/2=20
Æ S=6.67’ ≤ 11.05 ok
S
S
S
30
S/2+S+S+s/2=30
Æ S=10’ ≤ 11.05 ok
S
S/2
20
S/2
Spacing:
p
g Scheme #2
Try 2 rows of
5 luminaires
S/2
S/2
S
S
S/2+S+S/2=20
Æ S=10’ ≤ 11.05 ok
S
30
S/
S/2+4S+s/2=30
S s/ 30
Æ S=6’ ≤ 11.05 ok
S
S
S/2
20
S/2
Part II
Energy Code Compliance
Energy
Ene g Code Compliance
Scheme 1: 9 fixtures @128 w/fixture
Total Power: 1152 watts
P
Power
D
Density:
it 1152/600
1152/600= 1
1.92
92 w/sf
/ f
1.92 w/sf<2.0 w/sf ok, meets code
Energy
Ene g Code Compliance
Scheme 2: 10 fixtures @128 w/fixture
Total Power: 1280 watts
P
Power
D
Density:
it 1280/600
1280/600= 2
2.13
13 w/sf
/ f
2.13 w/sf>2.0 w/sf X, fails code
Energy
Ene g Code Compliance
With just a light switch, only Scheme #1
meets the requirements
q
of the energy
gy
code.
Part III
Assess DS Control
Assess effect of DS 33-step
p
controller
For a DS with a 33-step controller, the power
adjustment factor (PAF) is 0.20
Scheme #1:
Connected Lighting Power (CLP)= 1.92
Adj t d Li
Adjusted
Lighting
hti Power
P
(ALP) is:
i
ALP=(1--PAF)x CLP= (1ALP=(1
(1-0.2)x 1.92= 1.54 w/sf
Assess effect of DS 33-step
p
controller
Scheme #2:
Connected Lighting
g
g Power (CLP)=
(
) 2.13
Adjusted Lighting Power (ALP) is:
ALP=(1--PAF)x CLP= (1ALP=(1
(1-0.2)x 2.13= 1.71 w/sf
Scheme #2 now also meets energy
gy code
Energy
Ene g Code Compliance
Both Scheme #1 and #2 pass the energy
code with the addition of the daylight
y g
control system.
Part IV
Annual Operating Cost
Part
Pa t IV
IV: Ann
Annual
al Operating
Ope ating Cost
8:00 AMAM-5:00 PM x 52 wks/yr x 5 days/wk
is equivalent to 2340 hrs/yrs .
Part
Pa t IV
IV: Ann
Annual
al Operating
Ope ating Cost
W/O Daylight Control
Scheme #1 power consumption is:
2340 hrs/yr x1152 w x1 kwh/1000wkwh/1000w-h x $0.081/kwh
$218.35/yr..
$218.35/yr
Scheme #2 power consumption is:
2340 hrs/yr x1280 w x1 kwh/1000w
kwh/1000w--h x $0.081/kwh
$242.61/yr..
$242.61/yr
Part
Pa t IV
IV: Ann
Annual
al Operating
Ope ating Cost
With Daylight Control
Scheme #1 power consumption is:
2340 hrs/yr x1.54 w/sf x 600 sf x1 kwh/1000w
kwh/1000w--h x $0.081/kwh
$174.76/yr..
$174.76/yr
Scheme #2 power consumption is:
2340 hrs/yr x 1.71 w/sf x 600 sf x1 kwh/1000w
kwh/1000w--h x $0.081/kwh
$194.47/yr.
$194.47/yr.
Part V
Economic Analysis and
Recommendation
Pa t V
Part
V: Economic Analysis
Anal sis
Simple Payback ≤ 3 years is acceptable
Scheme #1
W/O DS
With DS
Added
First Cost
--------100.00
Annual
Cost
218.35
174.76
Annual
Savings
---------43.59
Payback
-------2.3
Payback= $100.00 / $43.59/yr = 2.3 years
Scheme #2
W/O DS
With DS
Added
First Cost
--------100.00
Annual
Cost
242.16
194.47
Annual
Savings
---------48.14
Payback
-------2.1
Payback= $100.00 / $48.14/yr = 2.1 years
Economic Analysis
Anal sis
Recommendation:
Adding daylight control to either scheme has a
payback of ≤ 3 years and therefore is acceptable.
Scheme #2 has higher operating costs and only
meets the energy code with the daylight control
i t ll d
installed.
Therefore use Scheme #1 with daylight control