Technology in Architecture Lecture 4 Lighting Design Example Lecture 4 Lighting Design Example.

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:


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


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


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


19. Burnouts

Burnout 0.95


20. Luminaire Dirt Depreciation (LDD)

Verify maintenance category

S: T.15.1 p. 641


20. Luminaire Dirt Depreciation (LDD)

S: F.15.34 p. 663

LDD=0.80


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


Connected Lighting Power (CLP):

CLP=12 x 128= 1,536 watts (2.8 w/sf)

Adjusted Lighting Power (ALP):

ALP=(1-PAF) x CLP


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



ALP=(1-PAF) x CLPALP=(1-0.20) x 1536ALP= 1229 watts (2.3 w/sf)


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


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.


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



ALP=(1-PAF) x CLPALP=(1-0.40) x 1536ALP= 922 watts (1.7 w/sf)


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

Technology in Architecture Lecture 4 Lighting Design Example Lecture 4 Lighting Design Example.

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