Environmental Controls Lecture 19 Predicting Interior Illumination Scale Modeling Toplighting Method Sidelighting Method Scale Modeling Studio Project Daylighting Study 1
Graduate Studio Project Daylighting Study Graduate Studio Project Daylighting Study Shriner s Children s Hospital of Utah 2
Alice Sheets Marriott Center for Dance University of Utah Student Union Toplighting Method 3
Toplighting Method 1. Determine net to gross area ratio (AR) of toplight (typically this is 0.60-0.90). Toplighting Method 2. Determine light well efficiency (LE) Calculate Well Index WI= H x (L + W) 2 x (L x W) H=2, W=3, L=4 WI = 2 x (3+4)/(2 x (3x4)) WI= 14/24= 0.583 H L W Toplighting Method 2. Determine light well efficiency (LE) WI=0.583 DL6 F.16-3 p. 201 4
Toplighting Method 3. Determine transmission of Glazing (TG) Note: This is VLT not SC or SHGF Toplighting Method 4. Determine transmission effectiveness (TE) of louvers or controls at ceiling plane Type TE No controls 1.00 Prismatic Lenses 0.90 Translucent white louvers 0.70 White translucent diffuser 0.50 Toplighting Method 5. Calculate toplight transmittance (TT) TT= AR x LE x TG x TE 5
Toplighting Method 6. Calculate room ratio (RR) RR= W x L. H c x (W + L) H c = Room Height Workplane Height W = Room Width L = Room Length Toplighting Method 6. Calculate room ratio (RR) H c = 12-3 = 9 W = 20 L = 30 RR= (W x L)/ (H c x (W + L)) RR= (20 x 30)/ (9 x (20 + 30)) RR=600/450= 1.33 Toplighting Method 7. Look up room coefficient of utilization (CU) RR=1.33 0.75 DL6 F.16-4 p. 201 6
Toplighting Method 8. Select a light loss factor (LLF) DL6 F.16-2 p. 200 See also: S T14.5 p. 602 Toplighting Method 9. Calculate room area (RA) Toplighting Method 10. Determine horizontal exterior illuminance (E kg + E ug, or ESS, EH). DL4 p. 36; see also DL4 p. 37 7
Toplighting Method 11. Determine net area of toplight (AN) Error in method as printed in DL6 11. Determine gross area of toplight (AG) Toplighting Method 12. Determine illuminance due to toplight (E top ) E top = (CU x LLF x ESS x AN)/RA Error in method as printed in DL6 E top =(CU x LLF x ESS x AG x TT)/RA 8
Given conditions Solar Altitude: 40 Sky: Overcast Room Height: 10 Room Width: 20 Room Length: 20 Work Plane: 2-6 10 20 20 Given conditions Ceiling Reflectance: 0.75 Wall Reflectance: 0.50 Floor Reflectance: 0.20 10 20 20 Given conditions Skylight: 3 x 5 x 5 (H x L x W) Control: Translucent White Diffuser Glazing Transmission: 0.80 Well Reflectance: 0.60 Net Area: 22.5 sf Factory Conditions 3 5 5 10 20 20 9
1. AR= 22.5/25= 0.90 10 20 20 2. Determine light well efficiency (LE) Well Index (WI) WI= H x (L + W) 2 x (L x W) WI= 3 x (5+5)/2 x (5 x 5) WI=30/50= 0.6 10 20 3 5 5 20 2. Determine light well efficiency (LE) WI=0.6 Reflectance=0.60 LE= 0.58 DL6 F.16-3 p. 201 10
3. Determine transmission of Glazing (TG) Given: TG=0.80 4. Determine the transmission effectiveness (TE) of louvers or controls at ceiling plane Type TE No controls 1.00 Prismatic Lenses 0.90 Translucent white louvers 0.70 White translucent diffuser 0.50 TE=0.50 5. Calculate toplight transmittance (TT) TT= AR x LE x TG x TE TT= 0.90 x 0.58 x 0.80 x 0.50 TT= 0.209 20.9% 11
6. Calculate room ratio (RR) RR= W x L. H c x (W + L) RR= (20 x 20)/(7.5 x (20 +20) RR= 400/300= 1.33 H c = Room Height Workplane Height W = Room Width L = Room Length 7. Look up room coefficient of utilization (CU) Ceiling Reflectance= 0.75 Wall Reflectance= 0.50 CU=0.75 RR=1.33 0.75 DL6 F.16-4 p. 201 8. Select a light loss factor (LLF) LLF=0.34 DL6 F.16-2 p. 200 12
9. Calculate room area (RA) RA=20 x 20 = 400 sf 10. Determine horizontal exterior illuminance (E kg ) DL4 p. 36; see also DL4 p. 37 Altitude= 40 E kg =1500 fc DL4 p. 36; see also DL4 p. 37 13
11. Determine gross area of toplight (AG) AG= 5 x 5= 25 sf 12. Determine illuminance due to toplight (E top ) E top =(CU x LLF x ESS x AG x TT)/RA E top =(0.75 x 0.34 x 1500 x 25 x 0.209)/400 E top = 5.0 fc What are the implications? Possible changes: Revise Control to Prismatic Lens: TE: 0.50 0.90 E top = 9.0 fc Revise Well reflectance to lighter color: LE: 0.58 0.73 E top = 6.3 fc Do both: E top = 11.3 fc Other changes: larger lightwell, multiple lightwells, change glazing, clean windows 14
Sidelighting Method Subscripts w=window k=sky g=ground u=sun Parameters R g T g 15
Illumination (E) E kw E kg E uw E ug E kuw E kug Illumination from sky on window Illumination from sky on ground Illumination from sun on window Illumination from sun on ground Illumination from sky and sun on window Illumination from sky and sun on ground Illumination (E) E gw E kwp E gwp R g A g T g Illumination from ground on window Illumination from sky (and sun) on work plane Illumination from ground on work plane Reflectance of ground surface Window area of transmittance Transmittance of glass Coefficients of Utilization (C,K) C os and K os C cs and K cs C us and K us C sv and K sv C ug and K ug C gv and K gv overcast sky clear sky uniform sky sky with venetian blind uniform ground ground with venetian blind 16
Daylighting Components Illumination on window E kuw = E kw + E uw Illumination on ground E kug = E kg + E ug Daylighting Components Illumination from sky (and sun) on work plane E kwp = E kuw x A g x T g x C x K Daylighting Components Illumination from ground on window E gw = E kug x R g x 0.5 Illumination from ground on work plane E gwp = E gw x A g x T g x C x K 17
Total Daylighting Components Illumination from ground on window E TOT = E kwp + E gwp Daylighting Availability DL4 p. 36 Available Skylight 1. Sky conditions 2. Orientation (V, H) 3. Season 4. Solar Altitude 5. Solar Azimuth 18
Available Sunlight 1. Clear Sky only 2. Orientation (V, H) 3. Solar Altitude 4. Solar Azimuth DL4 p. 37 1. Available Daylight Solar Altitude 36 Sky Condition Overcast Window Orientation N/A DL4 p. 19 19
DL4 p. 36 Example 1. E kw = 500 fc E kg = 1250 fc E kw = 500 fc E kg = 1250 fc For overcast sky: E uw =0, E ug =0 Illumination on window E kuw = E kw + E uw = 500 + 0 = 500 fc Illumination on ground E kug = E kg + E ug = 1250 + 0 = 1250 fc Determine E gw E gw = E kug x R g x 0.5 E gw = 1250 x 0.06 x 0.5 E gw = 37.5 fc 38 fc 20
2. Room Conditions Length= 30 Width= 20 Ceiling Height= 12 Sill Height= 3 Ceiling Reflectance= 80% Wall Reflectance= 70% Floor Reflectance= 30% Framing Factor(FF)= 20% T g = 20% 3. Window Transmittance Factors Area (A g ) = Length x Height x (1-FF) A g = 30 x (12-3) x (1-0.20) A g = 216 sf DL4 p. 19 4. Coefficients of Utilization Illumination from sky C os MAX K os MAX MID MID Illumination from ground C ug MAX K ug MAX MID MID 21
4. Coefficients of Utilization Illumination from sky C os MAX K os MAX MID MID Illumination from ground C ug MAX K ug MAX MID MID DL4 p.38 4. Coefficients of Utilization Illumination from sky C os MAX 0.0191 K os MAX 0.111 MID 0.0101 MID 0.111 MIN 0.0063 MIN 0.111 Illumination from ground C ug MAX K ug MAX MID MID 4. Coefficients of Utilization Illumination from ground DL4 p.39 22
4. Coefficients of Utilization Illumination from sky C os MAX 0.0191 K os MAX 0.111 MID 0.0101 MID 0.111 MIN 0.0063 MIN 0.111 Illumination from ground C ug MAX 0.0102 K ug MAX 0.111 MID 0.0094 MID 0.111 MIN 0.0082 MIN 0.111 5. Calculations E kwp = E kuw x A g x T g x C os x K os (MAX) = 500 x 216 x 0.20 x 0.0191 x 0.111 = 45.8 fc E gwp = E gw x A g x T g x C ug x K ug (MAX) = 38 x 216 x 0.20 x 0.0102 x 0.111 = 1.9 fc 6. Total Illumination E tot = E kwp + E gwp (MAX) = 45.8 + 1.9 = 47.7 48 fc (MID) = 24.2 + 1.7 = 25.9 26 fc (MIN) = 15.1 + 1.5 = 16.6 17 fc 23
Example 2 Example 2 1. Available Daylight Solar Altitude 23 Sky Condition Clear Window Orientation 90 azimuth from sun DL4 p. 19 DL4 p. 36 Example 1. E kw = 430 fc E kg = 750 fc 24
Example 2 1. E uw = 0 fc E ug = 2250 fc DL4 p. 37 Example 2 E kw = 430 fc E kg = 750 fc E uw =0 fc E ug =2250 fc Illumination on window E kuw = E kw + E uw = 430 + 0 = 430 fc Illumination on ground E kug = E kg + E ug = 750 + 2250 = 3000 fc Example 2 Determine E gw E gw = E kug x R g x 0.5 E gw = 3000 x 0.74 x 0.5 E gw = 1110 fc 25
Example 2 2. Room Conditions Length= 30 Width= 20 Ceiling Height= 12 Sill Height= 3 Ceiling Reflectance= 80% Wall Reflectance= 70% Floor Reflectance= 30% Framing Factor(FF)= 20% T g = 44% Example 2 3. Window Transmittance Factors Area (A g ) = Length x Height x (1-FF) A g = 30 x (12-3) x (1-0.20) A g = 216 sf DL4 p. 20 Example 2 4. Coefficients of Utilization Illumination from sky C cs MAX K cs MAX MID MID Illumination from ground C ug MAX K ug MAX MID MID 26
Example 2 4. Coefficients of Utilization Illumination from sky C os MAX K os MAX MID MID Illumination from ground C ug MAX K ug MAX MID MID DL4 p. 38 Example 2 4. Coefficients of Utilization Illumination from sky C cs MAX 0.0143 K cs MAX 0.111 MID 0.0100 MID 0.111 MIN 0.0079 MIN 0.111 Illumination from ground C ug MAX K ug MAX MID MID 4. Coefficients of Utilization Illumination from ground Example 2 DL4 p. 39 27
Example 2 4. Coefficients of Utilization Illumination from sky C cs MAX 0.0143 K cs MAX 0.111 MID 0.0100 MID 0.111 MIN 0.0079 MIN 0.111 Illumination from ground C ug MAX 0.0102 K ug MAX 0.111 MID 0.0094 MID 0.111 MIN 0.0082 MIN 0.111 5. Calculations Example 2 E kwp = E kuw x A g x T g x C os x K os (MAX) = 430 x 216 x 0.44 x 0.0143 x 0.111 = 64.9 fc E gwp = E gw x A g x T g x C ug x K ug (MAX) = 1110 x 216 x 0.44 x 0.0102 x 0.111 = 119.4 fc 6. Total Illumination Example 2 E tot = E kwp + E gwp (MAX) = 64.9 + 119.4 = 184.3 184 fc (MID) = 45.4 + 110.1 = 155.5 156 fc (MIN) = 35.8 + 96.0 = 131.8 132 fc Note: If Tg=.20 were used these values would be MAX=84 fc MID=71 fc MIN=60 28
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