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## Technology in Architecture

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### Technology in Architecture Lecture 4 Lighting Design Example * * * * * * * * * * * * * * * * * * * * * * Example 2: Economic Analysis Adjusted Lighting Power (ALP ... – PowerPoint PPT presentation

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Title: Technology in Architecture

1
Technology in Architecture
• Lecture 4
• Lighting Design Example

2
• Example 1
• Room Layout Calculation

3
Example 1
• Classroom 20 x 27 x 12 E50 fc
• WP 2-6 AFF
• ?c 80 hcc 0.0
• ?w 50 hrc 9.5
• ?f 20 hfc 2.5
• fixture fluorescent (38)
• maintenance yearly
• replacement on burnout
• voltages ballast normal
• environment medium clean

4
Example 1
• Confirm fixture data

S T.15.1 p. 641
5
Example 1
• Complete 1-6

6
Example 1
• 7. Determine lumens per luminaire
• Obtain lamp lumens from manufacturers data (or
see Stein Chapter 12)

S T. 12.5 p. 546
7
Lumen Flux Method
0
27
?c 80 ?w 50 ?f 20
9.5
20
2.5
• 8. Record dimensional data

8
Coefficient of Utilization Factor(CU)
Calculation
• 9. Calculate Cavity Ratios

9
Example 1 Cavity Ratios
• CR 5 H x (LW)/(L x W)
• RCR 5 Hrc x (LW)/(LxW) 4.1
• CCR 5 Hcc x (LW)/(LxW) 0
• FCR 5 Hfc x (LW)/(LxW) 1.1

10
Coefficient of Utilization Factor(CU)
Calculation
• 10. Calculate Effective Ceiling Reflectance

11
Example 1 Coefficient of Utilization (CU)
• 3. Obtain effective ceiling reflectance
• S T.15.2 p. 667

12
Example 1
• 11. Calculate Effective Floor
Reflectance Stein T.15.2 P. 666

13
Example 1 Coefficient of Utilization (CU)
• 3. Obtain effective ceiling reflectance

CU 0.19 ? 0.20
S T.15.2 p. 667
14
Example 1
• 12. Select CU from mfrs data or see

15
Example 1 Coefficient of Utilization (CU)
• CU0.32

S T.15.1 p. 641
RCR CU 4.0 0.39 4.1
X 5.0 0.35
CU 0.386
16
Example 1
• 13-21 Calculate LLF

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 0.92 /- 5

19
Light Loss Factor(LLF) Calculation
• 18. Lamp Lumen Depreciation
• Group Burnout
• Fluorescent 0.90 0.85

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

S T.15.1 p. 641
22
Example 1 Light Loss Factor(LLF)
• 20. Luminaire Dirt Depreciation (LDD)

LDD0.80
S F.15.34 p. 663
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

22 23
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 0.386 x 0.52) 11.4
luminaires

26
Example 1
• Goal is 50 fc /- 10 ? 45-55 fc
• Luminaires E (fc)
• 10 43.9 x
• 11 48.2 ok ? 2 rows of 4, 1 row of 3
• 12 52.6 ok ? 3 rows of 4
• 13 57.0 x
• Verify S/MH for fixture, space geometry

27
Example 1 S/MH Ratio
• Verify S/MH ratio
• MH12.0-2.59.5 S/MH 1.0 ? S 9.5

S T.15.1 p. 641
28
Example 1 Spacing
• Try 3 rows of
• 4 luminaires
• S/23SS/220
• ? S5
• S/MH5/9.5 1.0 ok
• S/2SSs/227
• ? S9
• S/MH9/9.5 1.0 ok

S/2 S S S/2
S/2 S S
S S/2
27
20
29
Example 1 Spacing
S/2 S S S S/2
• Try 4 rows of
• 3 luminaires
• S/22SS/220
• ? S6.67
• S/MH6.67/9.5 1.0 ok
• S/23Ss/227
• ? S6.75
• S/MH6.75/9.5 1.0 ok

S/2 S S
S/2
27
20
30
• Example 2
• Economic Analysis

31
Example 2 Economic Analysis
• 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

32
Example 2 Economic Analysis
• Connected Lighting Power (CLP)
• CLP12 x 128 1,536 watts (2.8 w/sf)
• Adjusted Lighting Power (ALP)
• ALP(1-PAF) x CLP

33
Example 2 Economic Analysis
• Power
• Control Factor (PAF)
• Daylight Sensor (DS), 0.30 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 90.1-1989

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
• Using the simple payback analysis method,
determine if switching to this control system is
economically attractive.

37
Example 2 Economic Analysis
• Power
• Control Factor (PAF)
• Daylight Sensor (DS), 0.30 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 90.1-1989

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 232.94 174.72 58.22/year
• Simple Payback Additional Cost/Annual Savings
• 150.00/58.22
• 2.6 years lt 3 years
• Economically attractive

40
• Example 3
• Point Source Calculation

41
Example 3
S F.15.49 p. 677
• Spot Lighting lamp straight down

S F.15.48 p. 677
42
Example 3
S F.15.49 p. 677
• Spot Lighting lamp pointed at object

Cp at 90 9600 Horizontal illumination
9900(0.643)3 25.5 fc
102 Vertical illumination
9900(0.766)3 30.3 fc 122
43
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