Technology in ArchitectureTechnology in Architecture
Lecture 4Lecture 4Lighting Design ExampleLighting Design Example
Lecture 4Lecture 4Lighting Design ExampleLighting Design Example
Example 1Example 1
Room Layout CalculationRoom Layout CalculationExample 1Example 1
Room Layout CalculationRoom Layout Calculation
Example 1Example 1Classroom 20’ x 27’ x 12’ E=50 fcWP= 2’-6” AFFρc= 80% hcc= 0.0’
ρw= 50% hrc= 9.5’
ρf= 20% hfc= 2.5’
fixture: fluorescent (#38)maintenance: yearlyreplacement: on burnout voltages & ballast: normalenvironment: medium clean
Example 1Example 1
Confirm fixture data
S: T.15.1 p. 641
Example 1Example 1
Complete #1-6
Example 1Example 1
7. Determine lumens per luminaire
Obtain lamp lumens from manufacturer’s data (or see Stein: Chapter 12)
S: T. 12.5 p. 546
Lumen Flux Lumen Flux MethodMethod
8. Record dimensional data
20’
27’
0’
9.5’
2.5’
ρc= 80%ρw= 50%ρf= 20%
Coefficient of Coefficient of Utilization Utilization Factor(CU) Factor(CU) CalculationCalculation9. Calculate Cavity Ratios
Example 1: Cavity RatiosExample 1: Cavity Ratios
CR = 5 H x (L+W)/(L x W)
RCR = 5 Hrc x (L+W)/(LxW) = 4.1
CCR = 5 Hcc x (L+W)/(LxW) = 0
FCR = 5 Hfc x (L+W)/(LxW) = 1.1
Coefficient of Coefficient of Utilization Utilization Factor(CU) Factor(CU) CalculationCalculation10. Calculate EffectiveCeiling Reflectance
3. Obtain effective ceiling reflectance:
Example 1: Coefficient of Utilization (CU)Example 1: Coefficient of Utilization (CU)
S: T.15.2 p. 667
Example 1Example 1
11. Calculate EffectiveFloor Reflectance
Stein: T.15.2 P. 666
3. Obtain effective ceiling reflectance:
Example 1: Coefficient of Utilization (CU)Example 1: Coefficient of Utilization (CU)
S: T.15.2 p. 667
CU= 0.19 0.20
Example 1Example 1
12. Select CU from mfr’s data or see
CU=0.32
Example 1: Coefficient of Utilization (CU)Example 1: Coefficient of Utilization (CU)
RCR CU 4.0 0.394.1 X5.0 0.35
CU= 0.386
S: T.15.1 p. 641
Example 1Example 1
13-21 Calculate LLF
Example 1: Light Loss Example 1: Light Loss Factor(LLF)Factor(LLF)
13-16
All factors not known 0.88
Example 1: Light Loss Example 1: Light Loss Factor(LLF)Factor(LLF)
17. Room Surface Dirt(based on 24 month cleaning
cycle, normal maintenance)
Direct 0.92 +/- 5%
Light Loss Factor(LLF) Light Loss Factor(LLF) CalculationCalculation
18. Lamp Lumen Depreciation
Group BurnoutFluorescent 0.900.85
Example 1: Light Loss Example 1: Light Loss Factor(LLF)Factor(LLF)
19. Burnouts
Burnout 0.95
Example 1: Light Loss Example 1: Light Loss Factor(LLF)Factor(LLF)
20. Luminaire Dirt Depreciation (LDD)
Verify maintenance category
S: T.15.1 p. 641
Example 1: Light Loss Example 1: Light Loss Factor(LLF)Factor(LLF)
20. Luminaire Dirt Depreciation (LDD)
S: F.15.34 p. 663
LDD=0.80
Example 1: Light Loss Example 1: Light Loss Factor(LLF)Factor(LLF)
LLF = [a x b x c x d] x e x f x g x hLLF = [0.88] x 0.92 x 0.85 x 0.95 x 0.80LLF = 0.52
Example 1Example 1
22. Calculate Number of Luminaires
22
23
Example1: Calculate Number Example1: Calculate Number of Luminairesof Luminaires
No. of Luminaires =
(E x Area)/(Lamps/luminaire x Lumens/Lamp x CU x LLF)
(50 X 540)/(4 X 2950 x 0.386 x 0.52) = 11.4 luminaires
Example 1Example 1
Goal is 50 fc +/- 10% 45-55 fc
Luminaires E (fc) 10 43.9 x11 48.2 ok 2 rows of 4, 1 row of 3
12 52.6 ok 3 rows of 413 57.0 x
Verify S/MH for fixture, space geometry
Example 1: S/MH RatioExample 1: S/MH RatioVerify S/MH ratioMH=12.0-2.5=9.5’ S/MH = 1.0 S ≤ 9.5’
S: T.15.1 p. 641
Try 3 rows of 4 luminaires
S/2+3S+S/2=20 S=5’
S/MH=5/9.5 ≤ 1.0 ok
S/2+S+S+s/2=27 S=9’S/MH=9/9.5 ≤ 1.0 ok
Example 1: SpacingExample 1: Spacing
S/2 S S S S/2
20
27
S/2
S
S
S/2
Try 4 rows of 3 luminaires
S/2+2S+S/2=20 S=6.67’
S/MH=6.67/9.5 ≤ 1.0 ok
S/2+3S+s/2=27 S=6.75’S/MH=6.75/9.5 ≤ 1.0 ok
Example 1: SpacingExample 1: Spacing
S/2 S S S/2
20
27
S/2
S
S
S
S/2
Example 2Example 2
Economic AnalysisEconomic AnalysisExample 2Example 2
Economic AnalysisEconomic Analysis
Example 2: Economic Example 2: Economic AnalysisAnalysis
Operation: 8AM-5PM, M-F, 52 wks/yr
9 x 5 x 52 = 2,340 hrs/yr
Operating Energy: 128 watts/luminaire
Lighting Control: Daylighting sensor with 3- step controller
Example 2: Economic Example 2: Economic AnalysisAnalysis
Connected Lighting Power (CLP):
CLP=12 x 128= 1,536 watts (2.8 w/sf)
Adjusted Lighting Power (ALP):
ALP=(1-PAF) x CLP
Example 2: Economic Example 2: Economic AnalysisAnalysis
Power Adjustment
Control Factor (PAF)Daylight Sensor (DS), 0.30 continuous dimmingDS, multiple-step dimming 0.20DS, On/Off 0.10Occupancy Sensor (OS) 0.30OS, DS, continuous dimming 0.40OS, DS, multiple-step dimming 0.35OS, DS, On/Off 0.35
Source: ASHRAE 90.1-1989
Example 2: Economic Example 2: Economic AnalysisAnalysis
Adjusted Lighting Power (ALP):
ALP=(1-PAF) x CLPALP=(1-0.20) x 1536ALP= 1229 watts (2.3 w/sf)
Example 2: Economic Example 2: Economic AnalysisAnalysis
Energy = 1,229 watts x 2,340 hrs/yr =2,876 kwh/year
Electric Rate: $0.081/kwh
Annual Energy Cost = 2,876 kwh/yr x $0.081/kwh
= $232.94/yr
Example 2: Economic Example 2: Economic AnalysisAnalysis
An alternate control system consisting of a daylighting sensor, with continuing dimming and an occupancy sensor can be substituted for an additional $150.
Using the simple payback analysis method, determine if switching to this control system is economically attractive.
Example 2: Economic Example 2: Economic AnalysisAnalysis
Power Adjustment
Control Factor (PAF)Daylight Sensor (DS), 0.30 continuous dimmingDS, multiple-step dimming 0.20DS, On/Off 0.10Occupancy Sensor (OS) 0.30OS, DS, continuous dimming 0.40OS, DS, multiple-step dimming 0.35OS, DS, On/Off 0.35
Source: ASHRAE 90.1-1989
Example 2: Economic Example 2: Economic AnalysisAnalysis
Adjusted Lighting Power (ALP):
ALP=(1-PAF) x CLPALP=(1-0.40) x 1536ALP= 922 watts (1.7 w/sf)
Example 2: Economic Example 2: Economic AnalysisAnalysis
Energy = 922 watts x 2,340 hrs/yr = 2,157 kwh/year
Annual Energy Cost = 2,157 kwh/yr x $0.081/kwh = $174.72/yr
Annual Savings = 232.94 – 174.72= $58.22/year
Simple Payback = Additional Cost/Annual Savings= 150.00/58.22= 2.6 years < 3 years Economically attractive
Example 3Example 3
Point Source CalculationPoint Source CalculationExample 3Example 3
Point Source CalculationPoint Source Calculation
Example 3Example 3
Spot Lighting – lamp straight down
S: F.15.48 p. 677
S: F.15.49 p. 677
Example 3Example 3
Spot Lighting – lamp pointed at object
S: F.15.49 p. 677
Cp at 90 = 9600
Horizontal illumination=
9900(0.643)3 = 25.5 fc 102
Vertical illumination= 9900(0.766)3 = 30.3 fc 122
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