Technology in architecture
Download
1 / 43

Technology in Architecture - PowerPoint PPT Presentation


  • 105 Views
  • Uploaded on

Technology in Architecture. 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

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about ' Technology in Architecture' - hazel


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
Technology in architecture
Technology in Architecture

Lecture 4

Lighting Design Example


Example 1

Room Layout Calculation


Example 1
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


Example 11
Example 1

Confirm fixture data

S: T.15.1 p. 641


Example 12
Example 1

Complete #1-6


Example 13
Example 1

7. Determine lumens per luminaire

Obtain lamp lumens from manufacturer’s data (or see Stein: Chapter 12)

S: T. 12.5 p. 546


Lumen flux method
Lumen Flux Method

0’

27’

ρc= 80%

ρw= 50%

ρf= 20%

8. Record dimensional data

9.5’

20’

2.5’


Coefficient of utilization factor cu calculation
Coefficient of Utilization Factor(CU) Calculation

9. Calculate Cavity Ratios


Example 1 cavity ratios
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


Coefficient of utilization factor cu calculation1
Coefficient of Utilization Factor(CU) Calculation

10. Calculate EffectiveCeiling Reflectance


Example 1 coefficient of utilization cu
Example 1: Coefficient of Utilization (CU)

3. Obtain effective ceiling reflectance:

S: T.15.2 p. 667


Example 14
Example 1

11. Calculate EffectiveFloor ReflectanceStein: T.15.2 P. 666


Example 1 coefficient of utilization cu1
Example 1: Coefficient of Utilization (CU)

3. Obtain effective ceiling reflectance:

CU= 0.19  0.20

S: T.15.2 p. 667


Example 15
Example 1

12. Select CU from mfr’s data or see


Example 1 coefficient of utilization cu2
Example 1: Coefficient of Utilization (CU)

CU=0.32

S: T.15.1 p. 641

RCR CU

4.0 0.39

4.1 X

5.0 0.35

CU= 0.386


Example 16
Example 1

13-21 Calculate LLF


Example 1 light loss factor llf
Example 1: Light Loss Factor(LLF)

13-16

All factors not known 0.88


Example 1 light loss factor llf1
Example 1: Light Loss Factor(LLF)

17. Room Surface Dirt

(based on 24 month cleaning cycle, normal maintenance)

Direct 0.92 +/- 5%


Light loss factor llf calculation
Light Loss Factor(LLF) Calculation

18. Lamp Lumen Depreciation

Group Burnout

Fluorescent 0.90 0.85


Example 1 light loss factor llf2
Example 1: Light Loss Factor(LLF)

19. Burnouts

Burnout 0.95


Example 1 light loss factor llf3
Example 1: Light Loss Factor(LLF)

20. Luminaire Dirt Depreciation (LDD)

Verify maintenance category

S: T.15.1 p. 641


Example 1 light loss factor llf4
Example 1: Light Loss Factor(LLF)

20. Luminaire Dirt Depreciation (LDD)

LDD=0.80

S: F.15.34 p. 663


Example 1 light loss factor llf5
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


Example 17
Example 1

22. Calculate Number of Luminaires

22

23


Example1 calculate number of luminaires
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


Example 18
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


Example 1 s mh ratio
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. 641


Example 1 spacing
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


Example 1 spacing1
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


Example 2

Economic Analysis


Example 2 economic analysis
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


Example 2 economic analysis1
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


Example 2 economic analysis2
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


Example 2 economic analysis3
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)


Example 2 economic analysis4
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


Example 2 economic analysis5
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.


Example 2 economic analysis6
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


Example 2 economic analysis7
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)


Example 2 economic analysis8
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


Example 3

Point Source Calculation


Example 3

S: F.15.49 p. 677

Example 3

Spot Lighting – lamp straight down

S: F.15.48 p. 677


Example 31

S: F.15.49 p. 677

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


ad