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1) This classroom has _______. Too much lighting Not enough lighting About the right amount of lighting. 2) For which of the following conditions would you increase the weighting factor by +1 in a lighting design?. A room used exclusively by high school students

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1 this classroom has
1) This classroom has _______.
  • Too much lighting
  • Not enough lighting
  • About the right amount of lighting
slide2
2) For which of the following conditions would you increase the weighting factor by +1 in a lighting design?
  • A room used exclusively by high school students
  • A circular room with mirrors for walls
  • A room used exclusively by government employees (i.e. speed/accuracy of task not important)
  • A room used for cutting dark stone
objective
Objective
  • Use room geometry to calculate coefficient of utilization (CU)
now what
Now what?
  • We now know how much light we need.
  • How do we get it?
    • Zonal cavity method
      • Calculate CU
      • How much light makes it from the fixture to the work surface of interest
    • Graphical methods (similar to stress strain)
    • Ray tracing
      • Computationally intensive
illumination calculation
Illumination Calculation
  • Iws = N × LPL × LOF × CU / A
  • N = number of fixtures
  • LPL = rated lumens per fixture
  • LOF = lamp operating factor
    • Ballast, voltage, temperature, position (HID)
  • CU = coefficient of utilization
    • Fraction of light that meets the work surface
  • A = room area
lamps are not the only thing
Lamps are Not the only thing
  • Fixtures (luminaires)
  • Lamp type and number
  • Power requirements
  • Ballast
  • Application requirements
  • Mounting
  • Fixture control
  • Special features
  • Distribution
slide8
S/MH
  • Fixture height to have even illumination
3 in lighting design the coefficient of utilization
3) In lighting design, the coefficient of utilization __________.
  • Determines the fraction of light fixtures in a room that are actually used.
  • Measures the fraction of emitted light that reaches a working surface.
  • Is lower in a room with light-colored walls than in one with dark walls.
  • Depends on the type of task performed, accuracy required by the task, and on the ages of occupants in a room.
zonal cavity method
Zonal Cavity Method
  • Purpose is to get CU “fixture efficiency”
  • What parameters do you need?
figure 16 1
Figure 16-1

Ref: Tao and Janis (2001)

calculate cavity ratios
Calculate Cavity Ratios
  • CR = 2.5 × PAR × h
    • PAR = perimeter to area ratio = P/A
    • PAR = 2 × (L+ W)/(L × W)
    • h = height of cavity
    • What about CR for non-rectangular rooms?
      • CR = 5 × (L+ W)/(L × W)× h
reflectance
Reflectance
  • Experience
    • White ceiling, Rc = 70 – 80 % = ρc
    • White walls, Rw = 60 - 70 % = ρw
    • Medium to light colored walls, Rw = 50 % =ρw
    • Dark wood paneling, Rw = 25 % = ρw
    • Floor, Rf = 10-30 % = ρf
  • Convert to effective reflectances (ρcc, ρw, ρfc)
    • Tables in Tao and Janis (pg 92-93, 102-107) or from manufacturer
calculation procedure
Calculation Procedure
  • Goal is to get CU (how much light from the fixture gets to the work surface)
  • Data collection
    • Room geometry
    • Surface reflectances
    • Fixture tables
  • Preliminary calculations
    • CR for room, floor, and ceiling
calculations continued
Calculations (continued)
  • Table 16.8
    • ρcc and ρfc (assume ρfc = 20% if no other information given)
  • Table 16.9
    • CU Multiplier if ρfc ≠ 20%
  • Fixture table
    • CU based on ρcc , Rw,RCR
  • Use CU by multiplier from step 4.
example
Example
  • Classroom (30 × 30 × 9)
  • White ceiling, blackboards on 2 sides, light floor
  • Students working on desks
  • Fluorescent fixtures at ceiling level
  • Use standard tables
data so far
Data So Far
  • PAR = 2 × (L+ W)/(L × W) = 120ft/900ft2
  • CCR = 2.5 × PAR × hc = 0
  • RCR = 2.5 × PAR × hr = 2.17
  • FCR = 2.5 × PAR × hf = 0.83
  • ρcc = Rc = 70% (b/c CCR = 0)
  • ρrc = Rw = 30%
  • ρfc = 20% (assumption)
variations
Variations
  • Fixture 2 (pg 92), 1 ft from ceiling
  • Actual fixture, original height
  • Original fixture, 30% reflective floor
fixture 2
Fixture 2
  • PAR = 2 × (L+ W)/(L × W) = 120ft/900ft2
  • CCR = 2.5 × PAR × hc = 0.33
  • RCR = 2.5 × PAR × hr = 1.83
  • FCR = 2.5 × PAR × hf = 0.83
  • ρcc = 64% (Table 16-8)
  • ρrc = Rw = 30%
  • ρfc = 20% (assumption, could use Table 16-8)
actual fixture
Actual Fixture
  • PAR = 2 × (L+ W)/(L × W) = 120ft/900ft2
  • CCR = 2.5 × PAR × hc = 0
  • RCR = 2.5 × PAR × hr = 2.17
  • FCR = 2.5 × PAR × hf = 0.83
  • ρcc = Rc = 70% (b/c CCR = 0)
  • ρrc = Rw = 30%
  • ρfc = 20% (assumption)
more reflective floor
More Reflective Floor
  • PAR = 2 × (L+ W)/(L × W) = 120ft/900ft2
  • CCR = 2.5 × PAR × hc = 0
  • RCR = 2.5 × PAR × hr = 2.17
  • FCR = 2.5 × PAR × hf = 0.83
  • ρcc = Rc = 70% (b/c CCR = 0)
  • ρrc = Rw = 30%
  • ρfc = 30% (given, could use Table 16-8 Tao and Janis)
slide23

4) If a building owner hires Persephone to determine the amount of lighting in an existing building, Persephone would need to know which parameters?

  • Type of activity performed, age of occupants, speed needed to perform activities in the building
  • Shape of the rooms, distance from light fixtures to work surfaces, reflectance of surfaces, types of light fixtures in the building
  • Color rendering index, evenness of lighting, thermal properties of lighting in the building
slide24

5) If a developer hires Francisco to determine the required lighting levels for a new building, Francisco would need to know which parameters?

  • Type of activity performed, age of occupants, speed needed to perform activities in the building
  • Shape of the rooms, distance from light fixtures to work surfaces, reflectance of surfaces, types of light fixtures in the building
  • Color rendering index, evenness of lighting, thermal properties of lighting in the building
illumination calculation1
Illumination Calculation
  • Iws = N × LPL × LOF × CU / A
  • N = number of fixtures
  • LPL = rated lamp lumens per fixture
  • LOF = lamp operating factor
    • Ballast, voltage, temperature, position (HID)
  • CU = coefficient of utilization
    • Fraction of light that meets the work surface
  • N = Iws× A / (LPL × LOF × CU)
distribution
Distribution
  • Direct 90 – 100 % downward
  • Semi-direct 60-90% down, rest upward
  • Direct-indirect/general diffuse
  • Semi-indirect
  • Indirect
summary
Summary
  • Calculate number of fixtures need for a specific space
    • Calculate CU
  • Tuesday
    • Accent lighting
    • Daylighting
    • Lighting quality
  • Thursday
    • Review
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