Presentation is loading. Please wait.

Presentation is loading. Please wait.

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

Similar presentations


Presentation on theme: "Technology in Architecture Lecture 4 Lighting Design Example Lecture 4 Lighting Design Example."— Presentation transcript:

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

2 Example 1 Room Layout Calculation Example 1 Room Layout Calculation

3 Example 1 Classroom 20 x 27 x 12E=50 fc WP= 2-6 AFF ρ c = 80%h cc = 0.0 ρ w = 50%h rc = 9.5 ρ f = 20%h fc = 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 p. 546

7 Lumen Flux Method 8. Record dimensional data ρ c = 80% ρ w = 50% ρ f = 20%

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

9 Example 1: Cavity Ratios CR = 5 H x (L+W)/(L x W) RCR = 5 H rc x (L+W)/(LxW) = 4.1 CCR = 5 H cc x (L+W)/(LxW) = 0 FCR = 5 H fc x (L+W)/(LxW) = 1.1

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

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

12 Example Calculate Effective Floor Reflectance Stein: T.15.2 P. 666

13 3. Obtain effective ceiling reflectance: Example 1: Coefficient of Utilization (CU) S: T.15.2 p. 667 CU=

14 Example Select CU from mfrs data or see

15 CU=0.32 Example 1: Coefficient of Utilization (CU) RCRCU X CU= S: T.15.1 p. 641

16 Example Calculate LLF

17 Example 1: Light Loss Factor(LLF) 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) Direct0.92 +/- 5%

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

20 Example 1: Light Loss Factor(LLF) 19. Burnouts Burnout0.95

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) S: F p. 663 LDD=0.80

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 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 x 0.52) = 11.4 luminaires

26 Example 1 Goal is 50 fc +/- 10% fc Luminaires E (fc) x ok 2 rows of 4, 1 row of ok 3 rows of 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 S: T.15.1 p. 641

28 Try 3 rows of 4 luminaires S/2+3S+S/2=20 S=5 S/MH=5/ ok S/2+S+S+s/2=27 S=9 S/MH=9/ ok Example 1: Spacing S/2 S S S S/ S/2 S S/2

29 Try 4 rows of 3 luminaires S/2+2S+S/2=20 S=6.67 S/MH=6.67/ ok S/2+3S+s/2=27 S=6.75 S/MH=6.75/ ok Example 1: Spacing S/2 S S S/ S/2 S S/2

30 Example 2 Economic Analysis 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): 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 ControlFactor (PAF) Daylight Sensor (DS),0.30 continuous dimming DS, multiple-step dimming0.20 DS, On/Off0.10 Occupancy Sensor (OS)0.30 OS, DS, continuous dimming0.40 OS, DS, multiple-step dimming0.35 OS, DS, On/Off0.35 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 ControlFactor (PAF) Daylight Sensor (DS),0.30 continuous dimming DS, multiple-step dimming0.20 DS, On/Off0.10 Occupancy Sensor (OS)0.30 OS, DS, continuous dimming0.40 OS, DS, multiple-step dimming0.35 OS, DS, On/Off0.35 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 Example 3 Point Source Calculation Example 3 Point Source Calculation

41 Example 3 Spot Lighting – lamp straight down S: F p. 677 S: F p. 677

42 Example 3 Spot Lighting – lamp pointed at object S: F p. 677 Cp at 90 = 9600 Horizontal illumination= 9900(0.643) 3 = 25.5 fc 10 2 Vertical illumination= 9900(0.766) 3 = 30.3 fc 12 2

43


Download ppt "Technology in Architecture Lecture 4 Lighting Design Example Lecture 4 Lighting Design Example."

Similar presentations


Ads by Google