has an aluminum frame without a thermal break. The glazing is doublewith a 13 mm gap width filled with air and a low-e coating (e = 0.20).Solution: An approximation can be made by multiplying the U-factorfor a site-assembled sloped/overhead glazing product having the sameframe and glazing features by the ratio of the surface area (includingthe curb) of the barrel vault to the rough opening area in the roof thatthe barrel vault fits over. First, determine the surface area (including thecurb) of the barrel vault:
Area of the curved portion of the barrel vault
= (π × diameter/2) × length= (3.14 × 6/2) × 10 = 94.25 m2Area of the two ends of the barrel vault
= 2π(radius2)/2 = πr2= 3.14 × 32 = 28.27 m2Area of the curb
= perimeter × curb height
= (6 + 10 + 6 + 10) × 0.150 = 4.8 m2Total surface area of the barrel vault
= 94.25 + 28.27 + 4.8 = 127.3 m2
Second, determine the rough opening area in the roof that the barrelvault fits over:
= length × width= 6 × 10 = 60 m2Third, determine the ratio of the surface area to the rough opening area:
= 127.3/60 = 2.12Fourth, determine the U-factor from Table 4 of a site-assembledsloped/overhead glazing product having the same frame and glazingfeatures. The U-factor is 4.06 W/(m2·K) (ID = 17, 12th column on thesecond page of Table 4).
Fifth, determine the estimated U-factor of the barrel vault.U= barrel vault
Usloped overhead glazing × surface area/rough opening for the barrel vault= 4.06 × 2.12 = 8.61 W/(m2·K)
VISIBLE TRANSMITTANCE
SOLAR HEAT GAIN AND Fenestration solar heat gain has two components. The quantity ofdirectly transmitted solar radiation is governed by the solar trans-mittance of the glazing system. Multiplying the incident irradianceby the glazing area and its solar transmittance yields the solar heatentering the fenestration directly. Absorbed solar radiation isremoved from the main beam and absorbed in the glazing and fram-ing materials of the window, and some is subsequently conducted tothe interior of the building.
Shading devices alter the effective solar heat gain coefficient forthe product, generally reducing it. Often, the alteration is angle-dependent. To determine the total solar gain accurately in this case,the angular distribution of the incident solar flux must be known.In addition to the possible complication of angular selectivity,many glazing systems (as well as shades) have optical propertiesthat are spectrally selective (i.e., their properties vary across theelectromagnetic spectrum). Ordinary clear float glass possesses thisproperty, but to a modest degree that is seldom of much concern inload calculations. Tinted and coated glass can exhibit strong spec-tral selectivity, a desirable property for certain applications, and thiseffect must be accounted for in solar heat gain determinations.For example, glazing systems intended for cold climates oftenare provided with a coating having low emittance over the longwavelength spectral region, above 3 or 4 µm, the purpose of whichis to reduce radiative heat loss from a warm interior to a cold exte-rior. These coatings are known generically as high-solar-gain low-ecoatings. In recent years, hot-climate versions have been marketed,in which the low-emittance (accompanied by high reflectance overthe same wavelength range) spectral range is extended below 3 µm,ideally to the edge of the visible spectrum in the vicinity of 800 nm
31.13
of wavelength. These coatings generally exhibit reasonably hightransmittance over the visible portion of the spectrum (approxi-mately 380 to 780 nm) while rejecting by reflection the infraredportion of the incident solar radiation at wavelengths greater than800 nm.
Both the directional and spectral distributions of solar, sky, andground-reflected flux must be known for accurate calculations oftotal solar heat gain through a fenestration system.
DETERMINING INCIDENT SOLAR FLUXSolar Radiation
The flux of solar radiation on a surface normal (perpendicular) tothe sun’s rays above the earth’s atmosphere at the mean earth-sundistance of 149.5 × 106 km (Allen 1973) is defined as the solar con-stant Esc. The currently accepted value is 1367 W/m2 (Iqbal 1983).Because the earth’s orbit is slightly elliptical, the extraterrestrialradiant flux Eo varies from a maximum of 1413 W/m2 on January 3,when the earth is closest to the sun (aphelion), to a minimum of1332 W/m2 on July 4, when the earth-sun distance reaches its max-imum (perihelion).
The earth’s orbital velocity also varies throughout the year, soapparent solar time, as determined by a solar time sundial, variessomewhat from the mean time kept by a clock running at a uni-form rate. This variation, called the equation of time, is given inTable 7. The conversion between local standard time and solar timeinvolves two steps: the equation of time is added to the local stan-dard time, and then a longitude correction is added. This longitudecorrection is four minutes of time per degree difference betweenthe local (site) longitude and the longitude of the local standardmeridian for that time zone. Standard meridians are found every15° from 0° at Greenwich, U.K. (Greenwich Meridian). In theUnited States and Canada, these values are 60° for Atlantic Stan-dard Time, 75° for Eastern Standard Time, 90° for Central StandardTime, 105° for Mountain Standard Time, 120° for Pacific StandardTime, 135° for Alaska Standard Time, and 150° for Hawaii-Aleutian Standard Time.
Equation (10) relates apparent solar time (AST) to local standardtime (LST) as follows:
AST = LST + ET/60 + (LSM – LON)/15
(10)
where
AST=apparent solar time, decimal hoursLST=local solar time, decimal hoursET=equation of time, decimal minutes
LSM=local standard time meridian, decimal ° of arcLON =
local longitude, decimal ° of arc
Because the earth’s equatorial plane is tilted at an angle of 23.45°to the orbital plane, the solar declination δ (the angle between theearth-sun line and the equatorial plane) varies through out the year,as shown in Figure 6, Table 7, and Equation (11). This variationcauses the changing seasons with their unequal periods of daylightand darkness. The following equation can be used to estimate thedeclination from the day of year η, but it is more accurate to look upthe actual declination in an astronomical or nautical almanac for theactual year and date in question.
δ = 23.45 sin {[360(284 + η)]/365}
(11)
The spectral distribution of solar radiation beyond the earth’satmosphere (Figure 7) resembles the radiant energy emitted by ablackbody at about 6000 K. The peak solar spectral irradiance of2130 W/(m2·µm) is reached at 0.451 µm (451 nm) in the green por-tion of the visible spectrum.
In passing through the earth’s atmosphere, the sun’s radiation isreflected, scattered, and absorbed by dust, gas molecules, ozone,water vapor, and water droplets (fog and clouds). The extent of this
31.14
Fig. 6Motion of Earth around Sun
Terrestrial and Extraterrestrial Solar Spectral Irra-Fig. 7Terrestrial and Extraterrestrial Solar
Spectral Irradiances
Table 7Extraterrestrial Solar Irradiance and Related Data
Equation Declina-ABCof Time, tion, (Dimensionless
Io, W/m2
min.degreesW/m2Ratios)Jan1416–11.2–20.012020.1410.103Feb1401–13.9–10.811870.1420.104Mar1381–7.50.011640.1490.109Apr13561.111.611300.1640.120May13363.320.011060.1770.130June1336–1.423.4510920.1850.137July1336–6.220.610930.1860.138Aug1338–2.412.311070.1820.134Sep13597.50.011360.1650.121Oct138015.4–10.511660.1520.111Nov140513.8–19.811900.1440.106Dec
1417
1.6
–23.45
1204
0.141
0.103
Note: Data are for 21st day of each month during the base year of 1964.
alteration at any given time is determined by atmospheric composi-tion and length of the atmospheric path traversed by the sun’s rays.This length is expressed in terms of the air mass m, which is the ratioof the mass of atmosphere in the actual earth/sun path to the massthat would exist if the sun were directly overhead at sea level (m =1.0). For most purposes, the air mass at any time equals the cosecantof the solar altitude multiplied by the ratio of the existing barometricpressure to standard pressure. Beyond the atmosphere, m = 0.
2005 ASHRAE Handbook—Fundamentals (SI)
trum with Direct Beam Spectra Through AtmospheresFig. 8Comparison of Standard Air Mass m = 1.5 Solar Spectrum with Direct Beam Spectra Through Atmospheres
Characteristic of southwest in winter (SWWINT) and southeastern U.S. in summer (SESUMM) for two solar altitude angles (McCluney 1996)
Table 8
Portions of Total Solar Spectral Irradiance Contained
in Portions of Visible Spectrum
Wavelength, nm
PercentPercentStartEndIrradianceIlluminance37077054.4100.038076052.2100.039075050.299.940074047.499.941073044.999.842072041.999.843071039.599.844070036.799.845069035.399.5460
680
31.1
99.1
Note: The integrated total irradiance = 950 W/m2 and illuminance = 100 klx.
Most ultraviolet solar radiation is absorbed by the ozone in theupper atmosphere, and part of the radiation in the short-wave por-tion of the spectrum is scattered by air molecules, imparting the bluecolor to the sky. Water vapor in the lower atmosphere causes thecharacteristic absorption bands observed in the solar spectrum at sealevel (Figure 8). For a solar altitude β of 41.8° (air mass m = 1.5), thetotal solar direct beam flux on a clear day at sea level can be dividedinto spectral regions as follows. Less than 3% of the total is in theultraviolet, 47% is in the visible region, and the remaining 50% is inthe infrared (ASTM Standard E891). The maximum spectral irradi-ance occurs at 0.61 µm, and little solar energy (less than 5% of thespectrum) exists at wavelengths beyond 2.1 µm.
It is interesting to see what fraction of the total solar irradiancelies in the visible part of the spectrum. Because the limits of the vis-ible portion vary from observer to observer (and because the eye isnot very sensitive to radiation at the spectral limits of vision), thefractions of total irradiance and illuminance found between differ-ent spectral limits at the edge of the visible portion of the spectrumcan be calculated. The results are shown in Table 8 for the ASTM airmass m = 1.5 terrestrial spectrum shown in Figure 8.
The solar spectral distribution shown in Figure 7 for m = 0 is theWorld Radiation Center’s 1985 standard extraterrestrial spectrumfor a solar constant of 1367 W/m2 (Wehrli 1985). The one for m =1.5 in Figure 8 is from ASTM Standard E891. This latter takes noaccount of monthly variations in irradiance caused by changes in the
Fenestration
Comparison of Direct and Diffuse Solar Spectra forFig. 9Comparison of Direct and Diffuse Solar Spectra for
Low Solar Altitude Angleearth/sun distance and by variations in the atmosphere’s constituentparticulates and gases.
When variations in atmospheric constituents and air mass areconsidered, the solar spectral distribution is seen to vary, as illus-trated in Figure 9, for two different atmospheric conditions and fortwo solar altitude angles, and in Figure 10 for both direct and diffuseradiation components and a low sun angle. It is clear that the spec-tral distribution for low-sun-angle beam radiation is significantlyshifted toward longer wavelengths. This shift can be seen visually asa reddening of the sun near to the horizon. Clear-sky diffuse radia-tion is generally shifted toward the blue end of the spectrum.
Upon passage through the atmosphere, extraterrestrial solar radi-ation is reduced in magnitude through absorption by atmosphericgases and particulates. The strength of this absorption varies withwavelength, and the terrestrial solar spectrum exhibits definite“dips” in regions of strong absorption, called absorption bands.The most prominent atmospheric gases contributing to this effectare listed below:
•Ozone. Strongest absorption in the ultraviolet, some in the visi-ble. Concentration variable.
•H2O. Strongest absorption in near and far IR. Highly variable.•CO2. Strongest absorption in near and far IR. Slightly variable.•O2, CH4, N2O, CFCs. Strongest absorption mostly in the IR.Concentration almost constant.
•NOluted areas.
2. Strongest absorption in the visible. Highly variable in pol-The effect of aerosols and other particulates on terrestrial solarradiation can be significant. Diffuse sky radiation is solar beam radi-ation that has been multiply scattered out of the direct beam anddownward through the atmosphere to the earth’s surface. This scat-tering is produced by 30 different atmospheric molecules (of whichthe above are the most significant optically) and by larger particlesof different types, including aerosols of water, dust, smoke, and par-ticulates of other kinds.
More information on atmospheric optics can be found in Chapter44 of the Optical Society of America’s Handbook of Optics (Bass1995) and in Iqbal (1983).
Glazing systems exhibiting strong spectral selectivity (strongchanges in their optical properties over the solar spectrum) willselectively pass more or less radiation in different parts of the spec-trum. This effect can cause substantial changes in the solar heat gaincoefficient of the glazing system when the shape of the solar spec-trum shifts appreciably. This, in turn, can cause errors in solar heatgain predictions when the actual solar radiation on a fenestrationsystem has a spectrum that is different from the standard spectrumused to determine the solar heat gain coefficient of that system(McCluney 1996). These errors are typically 5 to 10% but can besubstantially greater in special cases.
31.15
Some short-wavelength radiation scattered by air molecules,dust, and other particulates in the atmosphere reaches the earth inthe form of diffuse sky radiation Ed. Because this diffuse radiationcomes from all parts of the sky, its irradiance is difficult to predictand varies as moisture and particulate content and sun anglechange throughout any given day. For completely overcast condi-tions, the diffuse component accounts for all solar radiant heatgain of fenestrations.
The total short-wavelength irradiance Et reaching a terrestrialsurface is the sum of the direct solar radiation Eradiation EED, the diffuse skysurfaces. The irradiance on the fenestration aperture of the directd, and the solar radiation r reflected from surroundingbeam component EED is the product of the direct normal irradiationDN and the cosine of the angle of incidence θ between the incom-ing solar rays and a line normal (perpendicular) to the surface:
Et=EDNcosθ+Ed+Er
(12)
A method for computing all the factors on the right side of Equa-tion (12) is presented in the sections on Direct Normal Irradianceand Diffuse and Ground-Reflected Radiation. Gueymard (1987,1993), Perez et al. (1986), and Solar Energy (1988), give moredetailed models, which separate the diffuse sky radiation into dif-ferent components. Gueymard (1995) provides a comprehensivespectrally based model for calculating the spectral and broadbandtotals of all three terms in Equation (12), for cloudless sky condi-tions. The Gueymard model allows user input of the concentrationsof a variety of atmospheric constituents, including particulates.The importance of the diffuse component is illustrated in Figure9, which shows that, at low sun angles, the diffuse component con-tains more radiant flux than the direct beam component, even on aclear day, and that the spectral distributions of the two componentsare quite different. Although the total irradiances are relatively mod-est for both of these components, they are not insignificant forannual energy performance calculations.
Vertical windows receive considerable quantities of diffuse skyradiation over the course of a year. The diffuse component is animportant part of solar radiant heat gain.
Determining Solar Angle
The sun’s position in the sky is conveniently expressed in termsof the solar altitude β above the horizontal and the solar azimuth φmeasured from the south (Figure 10). These angles, in turn, dependon the local latitude L; the solar declination δ, which is a function ofthe date [Table 7 or Equation (11)]; and the apparent solar time,expressed as the hour angle H, where
H = 15(AST – 12)
(13)
Equations (14) and (15) relate β and φ to the three angles justmentioned:
sinβ=cosLcosδcosH+sinLsinδ
(14)cosφ=sin--------β-cos---sin-----β---L-cos---–-----sin-δ
L
---------(15)
Figure 10 shows the solar position angles and incident angles forhorizontal and vertical surfaces. Line OQ leads to the sun, thenorth-south line is NOS, and the east-west line is EOW. Line OV isperpendicular to the horizontal plane in which the solar azimuth φ(angle HOS) and the surface azimuth ψ (angle POS) are located.Angle HOP is the surface solar azimuth γ, defined as
γ = φ – ψ
(16)
The solar azimuth φ is positive for afternoon hours and negative formorning hours. Likewise, surfaces that face west have a positivesurface azimuth ψ; those facing east have a negative surface azimuth(Table 9). If γ is greater than 90° or less than –90°, the surface is in
31.16
Fig. 10
Solar Angles for Vertical and Horizontal Surfaces
Table 9Surface Orientations and Azimuths,
Measured from South
OrientationN
NE
E
SE
SSWWNWSurface azimuth ψ
180°–135°–90°–45°
0
45°
90°
135°
the shade. Table 9 gives values in degrees for the surface azimuth ψ,applicable to the orientations of interest.
The angle of incidence θ for any surface is defined as the anglebetween the incoming solar rays and a line normal to that surface.For the horizontal surface shown in Figure 11, the incident angle θis QOV; for the vertical surface, the incident angle θH is related to βV is QOP.
For any surface, the incident angle θ, γ, and the tiltangle of the surface from the horizontal Σ by
cos θ = cos β cos γ sin Σ + sin β cos Σ
(17)
When the surface is horizontal, Σ = 0° and
cos θH = sin β
(18)For a vertical surface, Σ = 90° and
cos θV = cos β cos γ
(19)Direct Normal Irradiance
At the earth’s surface on a clear day, direct normal irradiation, orsolar irradiance EDN, is represented by
EDN=---------------Aexp(B---⁄---sin--------β---)
(20)
where
A = apparent solar irradiation at air mass m = 0 (Table 7)B = atmospheric extinction coefficient (Table 7)Values of A and B vary during the year because of seasonalchanges in the dust and water vapor content of the atmosphere andbecause of the changing earth-sun distance. Equation (20) does notgive the maximum value of EDN that can occur in each month butyields values that are representative of conditions on cloudless daysfor a relatively dry and clear atmosphere. For very clear atmo-spheres, EDN can be 15% higher than indicated by Equation (20),using values of A and B in Table 7.
For locations where clear, dry skies predominate (e.g., at highelevations) or, conversely, where hazy and humid conditions arefrequent, values found by using Equation (20) and Table 7 shouldbe multiplied by the clearness numbers in Threlkeld and Jordan(1958), reproduced as Figure 5 in Chapter 33 of the 2003 ASHRAE
2005 ASHRAE Handbook—Fundamentals (SI)
Table 10
Solar Reflectances of Foreground Surfaces
Incident Angle
Foreground Surface20°30°40°50°60°70°New concrete0.310.310.320.320.330.34Old concrete
0.220.220.220.230.230.25Bright green grass0.210.220.230.250.280.31Crushed rock
0.200.200.200.200.200.20Bitumen and gravel roof0.140.140.140.140.140.14Bituminous parking lot
0.09
0.09
0.10
0.10
0.11
0.12
Adapted from Threlkeld (1962).
Handbook—HVAC Applications. This broadband model shouldonly be used for determining fenestration solar gain when theglazing system is not strongly spectrally selective and when itsangular selectivity closely matches that of single-pane glass.
Diffuse and Ground-Reflected Radiation
The following equations can be used to generate Eall angles are in degrees. The solar azimuth φ and the surface azi-d and Er, wheremuth ψ are measured in degrees from south; angles to the east ofsouth are negative, and angles to the west of south are positive. Val-ues of A, B, and C are given in Table 7 for the 21st day of eachmonth. Values for other dates can be obtained by interpolation.The ratio Y of sky diffuse radiation on a vertical surface to skydiffuse radiation on a horizontal surface is given byY=0.55+0.437cosθ+0.313cos2
θfor cosθ>–0.2
Y=0.45
for cosθ≤–0.2
(21)
Diffuse irradiance Ed is given byEd = CYEDN
for vertical surfaces
(22)
E=CE1--+-----cos2
--------Σ
dDN------for surfaces other than vertical(23)
Ground-reflected irradiance Er for surfaces at all orientations isgiven by
E1–cosΣ
r= EDN(C+sinβ)ρg----------2
----------(24)
where ρture of ground surfaces. For other surfaces, g is ground reflectivity, often taken to be 0.2 for typical mix-Table 10 lists angle-dependent solar reflectances.
Example 5. Find the solar azimuth and altitude at 3:00 P.M. eastern day-light time on July 21 in Atlanta, GA.
Solution: Eastern daylight time of 3:00 P.M. re-expressed in decimalhours at local standard time, is LST = 14.0. The latitude and longitudeof Atlanta, are 33.65°N and 84.42°W, respectively. The local standardmeridian of the eastern time zone is 75°W. The equation of time (Table7) is –6.2 min. From Equation (10), apparent solar time (AST) = 14.0 –6.2/60 + (75 – 84.42)/15 = 13.2687. From Equation (13), H = 15 ×(13.2687 –12) = 19.03°. Table 7 gives the solar declination δ on July 21as 20.60°.
Thus, by Equation (14),
sin β = cos (33.65°) cos (20.60°) cos (19.03°)
+ sin (33.65°) sin (20.60°) = 0.9316 β = 68.68°
Using Equation (15),
cosφ=sin--------(---68.68----)sin33.65)sin()cos---------(--68.68---------(-----)---cos---------(--33.65-–-------------)--20.6------------=0.5430
φ = 57.11°
Fenestration
Example 6. For the conditions of Example 5, find the incident angle at a
window facing 30° south of west.Solution: From Example 5, φ = 57.10°. By interpolation from Table 9,ψ = 60°. From Equation (16), γ = 57.10° – 60° = –2.90°.
A negative surface solar azimuth γ indicates that the sun is south of thenormal to the surface. Thus, using Equation (19),
cos θ = cos (68.68°) cos (–2.90°) = 0.3631θ = 68.71°
Example 7. Find the direct, diffuse, and ground-reflected components of
the solar irradiation on the window in Example 6.
Solution: From Example 5, sin β = 0.9316, and from Table 7, A =1093W/m2 and B = 0.186. Therefore, from Equation (20),
EDN
=-exp---------(---0.186/0.9316----1093
----------------------------2)
=895.18 W⁄mEθ=895.18×0.3631=325.04 W⁄m
2
DNcosFrom Table 7, C = 0.138. From Equation (21), Y = 0.55 + 0.437 ×
0.3631 + 0.313 × (0.3631)2 = 0.7499. From Equation (22),
E2
d=(0.138)(0.7499)(895.18)=92.64 W⁄m
From Equation (24), assuming a ground reflectivity ρg of 0.2,
Er=(895.18)(0.138+0.9316)(0.2)⁄2=95.75 W⁄m
2
Thermal Infrared Radiation
Any material above a temperature of absolute zero emits elec-tromagnetic radiation. The rate of emission depends upon the tem-perature of the material and can be expressed in a simple equation,called the Stefan-Boltzmann law:
E4
b=σT
(25)
where
Eb=hemispherical total emissive power of black body, W/m2T=temperature, °R
σ=Stefan-Boltzmann constant, 5.67 × 10–8 W/(m2·K4)
The emissivity e of a surface is the ratio of the emission of ther-mal radiant flux from the surface to the flux that would be emittedby a blackbody emitter at the same temperature. Given the temper-ature and emissivity of a surface, the emitted irradiance spectrumcan be computed from
E=eσT
4
where E is the emitted irradiance in units of flux per unit area, e isthe emissivity of the surface, σ is the Stefan-Boltzmann constant,and T is the absolute temperature of the surface.
The maximum value of hemispherical emissivity e for anymaterial is 1.0, in which case the surface emits the theoretical max-imum amount of radiation possible. In this case, the surface is calleda blackbody, and the radiation emitted by the surface is calledblackbody radiation.
Occasionally, a related quantity called normal emissivity is used.The relationship between them is illustrated in Figure 11. The spectraldistribution of blackbody radiation is illustrated in Figure 12 for tem-peratures ranging from 300 K (room temperature) to 20000 K.
OPTICAL PROPERTIES
Solar radiation (including both direct rays from the sun and dif-fuse rays from the sky, clouds, and surrounding objects) incident on
31.17
Fig. 11Illustration of Difference Between Normal and
Hemispherical EmissivityDifferent Source Temperatures12Spectral Distributions of Blackbody Radiation atFig. 12Spectral Distributions of Blackbody Radiation at
Different Source Temperatures
a fenestration system is partly transmitted and partly reflected by theglazings of that system. An additional fraction is absorbed withinthe glazings and/or the coatings on their surfaces. The fraction ofincident flux that is reflected is called the Areflectance R, the fractionabsorbed is called the absorptance , and the fraction transmittedis the A, and reflectance transmittanceR of a glazing layer is unity:
T. The sum of the transmittance T, absorptanceT+R+A=1
(26)
However, this is complicated by the fact that radiation incidenton a surface can have nonconstant distributions over the directionsof incidence and over the wavelength (or frequency) scale. Thus,when measuring one or more of the optical properties, the wave-length distribution and direction of incident (and emerging) radia-tion must be specified.
Angular Dependence
The concept of solid angle is needed to understand angulardependence. A solid angle is defined and enclosed by all rays join-ing a point to a closed curve. For a closed curve on a sphere of radiusR, the solid angle ϖ is the ratio of the projected area A on the sphereto the square of R. A sphere has a solid angle of 4π steradians(4πsr); a hemisphere has a 2π sr solid angle.
Radiation incident on a point in a surface comes to that pointfrom many directions in some solid angle. For a cone of half angleαcone is given by
, the solid angle defined by the circular top and point bottom of thatϖ = 2π (1 – cos α)
(27)
In measuring transmittance or reflectance, a sample is illumi-nated over a specified solid angle. The reflected or transmitted flux
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