1. Technology in Architecture
Lecture 4
Lighting Design Example

2. Room Layout Calculation
Example 1
Room Layout Calculation

3. Example 1 Classroom 20’ x 27’ x 12’ E=50 fc WP= 2’-6” AFF
ρc= 80% hcc= 0.0’
ρw= 50% hrc= 9.5’
ρf= 20% hfc= 2.5’
fixture: fluorescent (#38)
maintenance: yearly
replacement: on burnout
voltages & ballast: normal
environment: medium clean

4. Example 1
Confirm fixture data
M: T.16.1 p. 709

5. Example 1
Complete #1-6
M: F p. 733

6. Example 1 7. Determine lumens per luminaire
Obtain lamp lumens from manufacturer’s data (or see Stein: Chapter 12)
M: T p. 653

7. Lumen Flux Method 8. Record dimensional data 0’ ρc= 80% ρw= 50% 27’
9.5’
20’
2.5’
8. Record dimensional data
M: F p. 733

8. Coefficient of Utilization Factor(CU) Calculation
9. Calculate Cavity Ratios
M: F p. 733

9. Example 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

10. Coefficient of Utilization Factor(CU) Calculation
10. Calculate Effective Ceiling Reflectance
M: F p. 733

11. Example 1: Coefficient of Utilization (CU)
M: T.16.2 p. 734
3. Obtain effective ceiling reflectance:

12. Example 1 11. Calculate Effective Floor Reflectance M: T.16.2 p. 734
M: F p. 733

13. Example 1: Coefficient of Utilization (CU)
3. Obtain effective ceiling reflectance:
ρfc=  0.20
no CU adjustment needed
M: T.16.2 p. 734

14. Example 1
12. Select CU from mfr’s data
M: F p. 733

15. Example 1: Coefficient of Utilization (CU)
RCR CU X
M: T.16.1 p. 709
CU= 0.386

16. Example 1
Calculate LLF
M: F p. 733

17. Example 1: Light Loss Factor(LLF)
13-16
All factors not known  0.88

18. Example 1: Light Loss Factor(LLF)
17. Room Surface Dirt
(based on 24 month cleaning cycle, normal maintenance)
Direct /- 5%

19. Light Loss Factor(LLF) Calculation
18. Lamp Lumen Depreciation
Group Burnout
Fluorescent

20. Example 1: Light Loss Factor(LLF)
19. Burnouts
Burnout 0.95

21. Example 1: Light Loss Factor(LLF)
20. Luminaire Dirt Depreciation (LDD)
Verify maintenance category
M: T.16.1 p. 709

22. Example 1: Light Loss Factor(LLF)
20. Luminaire Dirt Depreciation (LDD)
LDD=0.80
M: F p. 730

23. Example 1: Light Loss Factor(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 0.80
LLF = 0.52

24. Example 1
22. Calculate Number of Luminaires
M: F p. 733

25. Example1: Calculate Number of Luminaires
No. of Luminaires =
(E x Area)/(Lamps/luminaire x Lumens/Lamp x CU x LLF)
(50 X 540)/(4 X 2950 x x 0.52) = 11.4 luminaires

26. Example 1 Goal is 50 fc +/- 10%  45-55 fc Luminaires E (fc) 10 43.9 x
ok  2 rows of 4, 1 row of 3
ok  3 rows of 4 x
Verify S/MH for fixture, space geometry

27. Example 1: S/MH Ratio Verify S/MH ratio
MH= =9.5’ S/MH = 1.0  S ≤ 9.5’
M: T.16.1 p. 709

28. Example 1: Spacing 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
S/2 S
S/ S S S S/2 27 20

29. Example 1: Spacing 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
S/2 S
S/ S S S/2 27 20

30. Example 2
Economic Analysis

31. Example 2: Economic Analysis
Operation: 8AM-5PM, M-F, 52 wks/yr x 5 x 52 = 2,340 hrs/yr
Operating Energy: 128 watts/luminaire
Lighting Control: Daylighting sensor with step controller

32. Example 2: Economic Analysis
Connected Lighting Power (CLP):
CLP=12 x 128= 1,536 watts (2.8 w/sf)
Adjusted Lighting Power (ALP):
ALP=(1-PAF) x CLP

33. Example 2: Economic Analysis
Power
Adjustment
Control Factor (PAF)
Daylight Sensor (DS), continuous dimming
DS, multiple-step dimming 0.20
DS, On/Off 0.10
Occupancy Sensor (OS) 0.30
OS, DS, continuous dimming 0.40
OS, DS, multiple-step dimming 0.35
OS, DS, On/Off 0.35
Source: ASHRAE

34. Example 2: Economic Analysis
Adjusted Lighting Power (ALP):
ALP=(1-PAF) x CLP
ALP=(1-0.20) x 1536
ALP= 1229 watts (2.3 w/sf)

35. Example 2: Economic Analysis
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

36. Example 2: Economic Analysis
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.

37. Example 2: Economic Analysis
Power
Adjustment
Control Factor (PAF)
Daylight Sensor (DS), continuous dimming
DS, multiple-step dimming 0.20
DS, On/Off 0.10
Occupancy Sensor (OS) 0.30
OS, DS, continuous dimming 0.40
OS, DS, multiple-step dimming 0.35
OS, DS, On/Off 0.35
Source: ASHRAE

38. Example 2: Economic Analysis
Adjusted Lighting Power (ALP):
ALP=(1-PAF) x CLP
ALP=(1-0.40) x 1536
ALP= 922 watts (1.7 w/sf)

39. Example 2: Economic Analysis
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 = – = $58.22/year
Simple Payback = Additional Cost/Annual Savings
= /58.22
= 2.6 years < 3 years
Economically attractive

40. Point Source Calculation
Example 3
Point Source Calculation

41. Example 3 Spot Lighting – lamp straight down M: F.16.47 p. 744

42. Example 3 Spot Lighting – lamp pointed at object M: F.16.47 p. 744
Cp at 90 = 9600
Horizontal illumination=
9900(0.643)3 = 25.5 fc
Vertical illumination=
9900(0.766)3 = 30.3 fc