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Environmental Controls II/IIG

Environmental Controls II/IIG. Lecture 4 Lighting Design Example. Example 1 Room Layout Calculation. Example 1. Classroom 20’ x 27’ x 12’ E=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

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Environmental Controls II/IIG

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  1. Environmental Controls II/IIG Lecture 4 Lighting Design Example

  2. 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 S: T.15.1 p. 631

  5. Example 1 Complete #1-6

  6. Example 1 7. Determine lumens per luminaire Obtain lamp lumens from manufacturer’s data (or see Stein: Chapter 12) S: T. 12.5 p. 538

  7. Lumen Flux Method 0’ 27’ ρc= 80% ρw= 50% ρf= 20% 8. Record dimensional data 9.5’ 20’ 2.5’

  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 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 EffectiveCeiling Reflectance

  11. Example 1: Coefficient of Utilization (CU) 3. Obtain effective ceiling reflectance: S: T.15.2 p. 657

  12. Example 1 11. Calculate EffectiveFloor ReflectanceStein: T.15.2 P. 657

  13. Example 1: Coefficient of Utilization (CU) 3. Obtain effective ceiling reflectance: CU= 0.19  0.20 S: T.15.2 p. 657

  14. Example 1 12. Select CU from mfr’s data or see

  15. Example 1: Coefficient of Utilization (CU) CU=0.32 S: T.15.1 p. 631 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. 631

  22. Example 1: Light Loss Factor(LLF) 20. Luminaire Dirt Depreciation (LDD) LDD=0.80 S: F.15.34 p. 653

  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 MH=12.0-2.5=9.5’ S/MH = 1.0  S ≤ 9.5’ S: T.15.1 p. 631

  28. Example 1: Spacing S/2 S S S/2 Try 3 rows of 4 luminaires S/2+3S+S/2=20  S=5’ S/MH=5/9.5 ≤ 1.0ok S/2+S+S+s/2=27  S=9’ S/MH=9/9.5 ≤ 1.0 ok 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/2+2S+S/2=20  S=6.67’ S/MH=6.67/9.5 ≤ 1.0ok S/2+3S+s/2=27  S=6.75’ S/MH=6.75/9.5 ≤ 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): 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), 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 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), 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 < 3 years Economically attractive

  40. Example 3 Point Source Calculation

  41. S: F.20.58 p.1266 Example 3 Spot Lighting – lamp straight down S: F.15.48 p. 667

  42. S: F.20.58 p.1266 Example 3 Spot Lighting – lamp pointed at object Cp at 90 = 9600 Horizontal illumination= 9900(0.643)3 = 25.5 fc102 Vertical illumination= 9900(0.766)3 = 30.3 fc122 S: F.15.48 p.667

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