Solar Energy Fundamentals - Oregon State...

2500 ---~---------------

2000

[ 11500

~ ~ ~ - 1000

-shy _ 5800 K Blackbody

Solar Energy Fundamentals

i

114

05 10 15 20 2 5 30

Wavelength (um)

[]TI INTRODUCTION

This and the next three chapters will examine solar energy The purposes of this chapshyter are to review radiation heat transfer and to present fundamental solar energy information necessary to understand the applications of the next three chapters

~ RADIATION HEAT TRANSFER REVIEW

An understanding of solar energy must be based on familiarity with the spectral nature of solar radiation from the sun This section reviews the fundamental radiashytion heat transfer information needed for studying solar energy engineering processes

From a solar energy standpoint two areas of interest are the characteristics of energy from the sun and the response of a surface to that energy The starting point is to consider the characteristics of radiation heat transfer Thermal radiation may be viewed as the transport of photons or the propagation of an electromagnetic wave For the propagation of radiation in a medium the frequency and wavelength are related as

cA= shy

lJ (6-1)

where c is the speed of light in the medium A is the wavelength and v is the freshyquency In a vacuum C = 2998 X 108 m sec

The electromagnetic continuum is divided into different regimes as a function of the wavelength The regimes and wavelength ranges are delineated in Table 61

Of interest in solar energy engineering is thermal radiation in the wavelength range 10-1 lt A lt 102 -Lm The visible part of the spectrum is in the range 04 -Lm lt A lt 07 -Lm and is bounded by the ultraviolet and the infrared Parts of the ultraviolet and the infrared regimes are also in the thermal radiation regime

~5 3 0

this chapshyif energy 3pters

~ spectral tal radiashygineering

ristics of ing point tion may ~tic wave ngth are

(6-1)

s the freshy

function able 6l Ivelength Ie range Parts of egime

62 Radiation Heat Transfer Review 115

TABLE 61 The electromagnetic continuum and regimes

Regime Range

Gamma rays A lt 10-4 J-Lm X-rays 10-4 lt A lt 10-2 J-Lm Ultraviolet 10-2 lt A lt 04 fLm

Thermal radiation 10-1 lt A lt 102 flm

Infrared 07 lt A lt 102 J-Lm Microwave A gt 102 --lm

The magnitude of thermal radiation varies with the wavelength and gives rise to the spectral distribution In addition to the spectral nature of thermal radiation a key feature is the directional distribution In order for thhmal radiation to be quanshy

tified both the spectral and directional attributes must be examined

A good starting point is to consider the characteristics of a blackbody

(1) A blackbody absorbs all incident radiation regardless of wavelength and direction

(2) For a prescribed temperature and wavelength no surface can emit more energy than a blackbody

(3) The blackbody is a diffuse (independent of direction) emitter

The spectral emissive power density of a blackbody is given by the Planck disshytribution (Incropera and deWitt 1999)

(6-2)EAb(AT) = 5 [ (C2 ) J

A - exp -- - 1 A -T

where C1 = 3742 -108 W J-Lm4jm2 and C2 = 1439 -104 J-LmK The spectral emissive power density EAb is customarily expressed in the units of W j J-Lm m2 and is interpreted as the emissive power per unit area per wavelength The subscripts A and b in Eq (6-2) signify that the power density is given per wavelength (spectral) and is from a blackbody respectively The total emissive power of a blackbody is the integral of the blackbody spectral emissive power density over all wavelengths or

( )O ( 00 C 4 (6-3)Eb(T) = Jo EAbdA = Jo 5 [ (~2) J dA = (TT

A - exp A _T - 1

where (T = 567 -10-8 Wjm2 K4 the Stefan-Boltzmann constant The T4 funcshy

tional dependence of the blackbody emissive power is one of the strongest functional dependencies in physics and emphasizes the importance of temperature in thermal radiation

116 Chapter 6 Solar Energy Fundamentals

5800 K ------r-shy

1106

ltl ~ 0

0- 110

5 ltl sect

1 104W iii U ltl

1103P CJ

---+

100

1 ~--~~~~~------~--~~--~~--~~~ 01 10 100

Wavelength in microns

Figure 61 Blackbody spectral emissive power density for selected surface temperatures

Additional examination of Eq (6-2) is warranted Figure 61 plotted in log-log coordinates presents a graphical study of the blackbody spectral emissive power denshysity for several different temperatures T = 5800 K is one of the temperatures conshysidered because the spectral distribution of thermal radiation from the sun is close to that of a blackbody at 5800 K The pronounced effect of temperature on the blackshybody spectral emissive power is demonstrated by its range-eight orders of magnishytude from 1 W )Lmm2 to nearly 108 W )Lmm2

Also of interest is the location of the maximum spectral emissive power as a function of surface temperature For example at T = 300 K close to room temperature the maximum occurs near A = 10 )Lm but as the temperature is increased the maximum shifts to shorter wavelengths thus at T = 5800 K the maximum is near 05 )Lm Not surprisingly A = 05 )Lm is the middle of the visible spectrum for which vision on Earth is optishymally adapted The thermal radiation range given in Table 61 is congruent with the results of Figure 61 as the 10-1 lt A lt 102 )Lm range of Figure 61 contains virtushyally all of the emitted energy for these relevant temperatures The total blackbody emissive power is the area under the curve and the blackbody emissive power at T = 5800 K is 140000 times that at T = 300 K


= (jj c Q)

o ~ t5 Q) 0 (f)

Q5 s o n

Wavelength

Figure 62 Blackbody radiation function illustration

For many radiation heat transfer investigations the fraction of energy contained between a wavelength interval is needed The basis for obtaining such fractions is the blackbody radiation function Using the Planck distribution the fraction of thershymal energy in the wavelength range 0 to A is cast as follows

tEAb dA lAEAbdA l ATE )b F(O - = = 5 d(AT)AT) Jo co 0 --- - (6-4)r uT4

0 uTJo E)b dA

Figure 62 illustrates the concept graphically Many heat transfer textbooks contain tabulations of the blackbody radiation

function alternatively Eqs (6-2) and (6-4) can be integrated to give the blackbody radiation function values Moreover Siegel and Howell (2002) have devised a closedshyform series that replicates to within a fraction of a percent the accepted blackbody radiation function values and also provides a convenient expression for use in many problem solutions Their series appears as

15 20 e-n( ( 6 2)

F() = - 2- - + 6- +3 + 3- (6-5)4 3 2

7T n=l n n n n

where = C2 AT Table 62 contains a tabular listing of Eq (6-5) values as a funcshytion of AT Table 62 and Eq (6-5) will be used in the solution of example problems requiring the use of the blackbody radiation function Example 61 illustrates the use of the blackbody radiation function

For a blackbody at 5800 K what is the percentage of energy contained in the visishyble part of the spectrum Contrast this with the percentage of energy contained in the visible spectrum for a blackbody at 3000 K

Two different approaches for obtaining the solution will be presented and discussed (1) the use of Table 62 and (2) the use of Eq (6-5) and Mathcad

)0

atures

in log-log ower denshytures conshyIn is close the blackshyof magnishy

ocation of ature For curs near to shorter rprisingly middotth is op tishy1t with the ains virtushyblackbody ~ power at

Solution

11 8 Chapter 6 Solar Energy Fundamentals

TABLE 62 Blackbody radiation function tabulation

AT (ILmmiddot K) F(O - AT)

200 0

400 1855410 12

600 9263middot10-8

800 1639610-5

1000 00003

1200 00021

1400 00078

1600 00197

1800 00393

2000 00667

2200 01008

2400 01402

2600 0183

2800 02278

3000 02731

3200 0318

3400 03616

3600 04035

3800 04433

4000 04808

4200 05159

4400 05487

4600 05792

4800 06074

5000 06336

AT (JLmmiddotK) F(O shy AT)

5200 06579

5400 06802

5600 07009

5800 072

6000 07377

6200 0754

6400 07691

6600 07831

6800 0796

7000 0808

7200 08191

7400 08294

7600 0839

7800 08479

8000 08562

8200 08639

8400 08711

8600 08778

8800 08841

9000 08899

9200 08954

9400 09006

9600 09054

9800 09099

10000 09141

AT (JLmmiddot K) F(O - AT)

10500 09236

11000 09318

11500 09389

12000 0945

12500 09504

13000 09551

13500 09592

14000 09628

14500 09661

15000 09689

15500 09715

16000 09738

16500 09758

17000 09776

17500 09793

18000 09808

18500 09822

19000 09834

19500 09845

20000 09855

AT (ILm middotK) F(O - AT)

25000 09922

30000 09953

35000 0997

40000 09979

45000 09985

50000 09989

55000 09992

60000 09993

65000 09995

70000 09996

75000 09997

80000 09997

85000 09998

90000 09998

95000 09998

110 5 09998

The visible spectrum spans the wavelength range 04 f-I-in S A S 07 f-I-m Figure 63 shows the range of interest for the 5800 K blackbody F(O - A21) proshyvides the fraction of radiant energy from 0 to A2 and F(O - A1T) provides the fracshytion from 0 to AI The fraction of radiant energy contained between A1 and A2 can be expressed as the difference

(6-6)

The Table 62 approach will be illustrated first For the blackbody at 5800 K

AIT = 04 f-I-m 5800 K = 2320 f-I-mK

A2T = 07 f-I-m 5800 K = 4060 f-I-ID K



AT (pmmiddot K) F(O - AT)

200 0

400 1855410-12

600 926310-8

800 1639610-5

1000 00003

1200 00021

1400 00078

1600 00197

1800 00393

2000 00667

2200 01008

2400 01402

2600 0183

2800 02278

3000 02731

3200 0318

3400 03616

3600 04035

3800 04433

4000 04808

4200 05159

4400 05487

4600 05792

4800 06074

5000 06336

AT (pmK) F(O - AT)

5200 06579

5400 06802

5600 07009

5800 072

6000 07377

6200 0754

6400 07691

6600 07831

6800 0796

7000 0808

7200 08191

7400 08294

7600 0839

7800 08479

8000 08562

8200 08639

8400 08711

8600 08778

8800 08841

9000 08899

9200 08954

9400 09006

9600 09054

9800 09099

10000 09141


10500 09236

11000 09318

11500 09389

12000 0945

12500 09504

13000 09551

13500 09592

14000 09628

14500 09661

15000 09689

15500 09715

16000 09738

16500 09758

17000 09776

17500 09793

18000 09808

18500 09822

19000 09834

19500 09845

20000 09855

AT (pmmiddotK) F(O shy AT)

25000 09922

30000 09953

35000 0997

40000 09979

45000 0 9985

50000 09989

55000 09992

60000 09993

65000 09995

70000 09996

75000 09997

80000 09997

85000 09998

90000 09998

95000 09998

110 5 09998

The visible spectrum spans the wavelength range 04 fLin 5 A 5 07 fLm Figure 63 shows the range of interest for the 5800 K blackbody F(O - A2T) proshyvides the fraction of radiant energy from 0 to 12 and F(O - 11 T) provides the fracshytion from 0 to 11 The fraction of radian t energy contained between 11 and 12 can be expressed as the difference

(6-6)


AiT = 04 fLm 5800 K = 2320 fLmK

A2T = 07 fLm 5800 K = 4060 fLmK

F(O -AT)

09922

09953

0997

09979

09985

0 9989

09992

09993

09995

09996

09997

09997

09998

09998

09998

0 9998

07 fLm ~ T) proshyhe fracshyd Az can

(6-6)

800 K


shyw c Q)

o ~ U Q) D

(f)

(jj 3 o

0

Wavelength

Figure 63 Visible spectrum range for a 5800 K blackbody

Interpolating between the entries in Table 62 yields

F( 2320 fLmK) = 01244

F(4060 fLmK ) = 04913

so the fractional radiant energy contained between Al = 04 fLm and Az = 07 fLill becomes

F(AzT - A1T) = 04913 - 01244 = 03669

Thus for a 5800 K blackbody 37 percent of the radiant energy is contained in the visible part of the spectrum

The same procedure is applied for the 3000 K blackbody except that the temshyperature used is 3000 K not 5800 K The results can be surrunarized as

A1T = 04 fLmmiddot 3000 K = 1200 fLmK

AzT = 07 fLmmiddot 5800 K = 2100 fLm K

F(1200 fLmK ) = 00021

F(2100 fLmK ) = 00838

F (A2T - ArT ) = 00830 - 00021 = 00817

Hence for the 3000 K blackbody only 8 percent of the total radiant energy is in the visible range

Figure 64 presents the solution using Eq (6-5) and Mathcad The same procedure is followed with the only major difference being the use of Eq (6-5) for the blackbody radiation function The use of the Siegel-Howell expression is very convenient as it obviates the need for table lookups and interpolations to solve spectral distribution problems As the remainder of the section will demonstrate this ability Will be important for a number of thermal radiation problems not limshyited to blackbody formulations The results in Figure 64 are more accurate than using Table 62 since no interpolation is required but the differences between the two approaches are so negligible that for most purposes either will suffice


-8 W The Stefan-Boltzmann constant and [Jma = 567middot[0 24 m middotK

C2 = 14391OlmK C1 and C2 for the Planck power spectral

J5 [ 20 e-ni [6 t 3 t denSity expres~~n

F(c)= - - - - + 6middot- + t + 3-2Jj t = - Definition of L4 L n 3 2 n AmiddotT

IT n = [ n n

Blackbody at 5800 K

F(~) = 012392 H1

F (~J = 049154 H2

Fvisible = F(~1-F(~1 Fvisible = 036762 H2 H1

Blackbody at 3000 K

C2 1 - 3H1 = OAmiddotpm middot3000K F- =2 131 15xl0( H1

H2= 07middotpmmiddot3000middotK F(~1= 0083 H2

Fvisible = F(~) - F(~) Fvisible = 008087 H2 H1

Figure 64 Mathcad worksheet for the solution to Example 61

The source of solar energy on the Earth is obviously the sun The sun is located 150 X 1011 m from the Earth and has a diameter of 139 X 109 ill The solar constant defined as the flux of solar energy incident on the surface oriented normal to the rays of the sun has a mean value of 1353 W 1m2 However the solar constant is the flu x incident at the outer edge of the atmosphere For terrestrial applications what is important is the solar flux incident on the surface of the Earth after the suns rays have passed through the atmosphere Figure 65 provides a quantitative assessment of the solar flux The spectral distribution is illustrated for extraterrestrial conditions (incident on the edge of the atmosphere) and on the Earths surface lhe extratershyrestrial solar flux has the approximate spectral distribution of blackbody radiation

Ind flm

spectral

s located constant ) the rays the flux what is uns rays sessment )nditions extratershyradiation


2500~--~----~----~----------~----

2000

~ 1000 U 0

()

500

Solar Irradiation

T t I

I -

I

5800 K Blackbody I

I

I

I-II I

I

~I---CO2 I

0U1L-L-__L-~U-~J=~~~~ o 05 10 15 2 0 25 30

Wavelength (um)

Figure 65 Solar irradiation characteristics (Incropera and DeWitt 1999)

at 5800 K but not the total emissive power ofa blackbody-as a comparison of Figures 61 and 62 will confirm

As the solar flux passes through the atmosphere it is attenuated by the presence of oxygen (02) ozone (03) water vapor (H20) and carbon dioxide (C0 2) As a result the solar flux at the Earths surface is significantly less than the extraterresshytrial solar flux As the solar flux traverses the atmosphere both the magnitude and directional distributions are altered These alterations are due to absorption and scatshytering The lower plot in Figure 65 demonstrates the absorption of radiant energy by gases in the atmosphere In the ultraviolet regime 10- 2 lt A lt 04 f1m ozone is responsible for the absorption and results in virtually no solar flux reaching the surshyface for A lt 03 f1m Part of the concern about depletion of the ozone layer is that without the ozone much more of the thermal radiation in the ultraviolet regime would reach the surface of the Earth with long-term deleterious results In the infrared regime the absorption is primarily due to water vapor although CO2 also plays a role For all wavelengths thermal radiation is also absorbed by the particushylate contents of the atmosphere

In addition to absorption scattering occurs resulting in the redirection of the suns rays Rayleigh scattering attributable to the gas molecules provides for virtually


uniform scattering of thermal radiation The uniformity of Rayleigh scattering means that about one-half of the scattered radiation is redirected away from the Earth A t any point on the Earths surface the scattered radiation is incident from all direcshytions Mie scattering by the particulates in the atmosphere does little to reorient the suns rays Incropera and DeWitt (1999) have provided a cogent summary

That portion of the radiation that has penetrated the atmosphere without havshying been scattered (or absorbed) is in the direction of the zenith angle and is termed the direct radiation The scattered radiation is incident from all directions although its intensity is largest for the directions close to that of the direct radishyation However because the radiation intensity is often assumed to be indeshypendent of the direction the radiation is termed diffuse

The total thermal radiation reaching the surface is the sum of the direct and the diffuse amounts On a clear day the diffuse radiation may account for only 10 pershycent of the total while on a cloudy day it is essentially 100 percent of the total

With the characteristics of thermal radiation from the sun understood considshyeration of radiation heat transfer fundamentals for a surface is appropriate Specific topics include (1) what happens when thermal radiation is incident on a surface and (2) the emission characteristics of a surface Emission will be examined first

The emissive characteristics of a blackbody have been established E missions from a non-blackbody can have directional and spectral characteristics Consider as shown in Figure 66 a hemisphere surrounding an elemental area that is emitting thermal radiationThe hemisphere about the emitting area dA is a convenient mechshyanism for visualizing the geometry of thermal radiation The spectral intensity IAe

of the emitted radiation is defined as the rat~ at which radiant energy is emitted at the wavelength A in the (e cent) direction per unit area of the emitting surface normal to this direction per unit solid angle about this direction and per unit wavelength interval dA about A The normal to the emitting surface area in the (e cent) direction is the area dAn in Figure 66 If the blackbody surface is used as the basis the spec tral directional emissivity CAe (A e cent T) is defined as the ratio of the emitted radia tion intensity in a given direction at a given wavelength divided by the radiation intenshysity of a blackbody or

IAe(AecentT) SA e(AecentT) = I (A T) (6-7)

Ab

Equation (6-7) provides a process for computing the actual emission in refershyence to that of a blackbody if SA8(Ae cent T) is known If the spectral directional intenshysity is integrated over 0 S cent S 271 and 0 s e s 712 the spectral hemispherical emissivity CA(A T) results and is defined as

IA e(A T) EA(AT) SA(AT) = I (A T ) =

Ab E (A T )

Ab (6-8)

where EA(A T) is the spectral hemispherical emissive power and SA (A T) is the spectral hemispherical emissivity Since the angular dependence has been inteshy


lllg means nEarth At aU direcshy 0lt elt 7T2

orien t the r

hout havshygle and is jirections irect radishyI be indeshy

t and the ly 10 pershytotal d considshy= Specific rface and st ltmissions Insider as emitting ~nt mechshynsityle mitted at e normal welength direction e spectral radiation on inten-

(6-7)

in refershynal in tenshyspherical

(6-8)

T) is the ~en inte-

~cent

Figure 66 Hemisphere surrounding an area emitting thermal radiation

grated out the word hemispherical rather than directional is used The transishytion from 1e(A T) to E(A T) is seamless since the emission intensity is based on the projected area and the emission is based on the actual area and they are related as E(A T) = 7T Ie(A T) Details are provided in heat transfer textbooks such as Incropera and DeWitt (1999) If the spectral hemispherical emissive power is inteshygrated over aU wavelengths the total hemispherical emissive power E(T) is recovshyered and appears as

seT) = E(T)

(6-9)

where E(T) is the total hemispherical emissive power and seT) is the total hemishyspherical emissivity generally referred to as simply the emissivity The sequence is from spectral directional to spectral hemispherical to total hemispherical Values for the spectral hemispherical and total hemispherical emissivities for various mateshyrials are available from handbooks and textbooks and on a number of websites (wWVicessucsbmodisemis for example)


A wall at T= 500 K has the spectral emissivity illustrated in Figure 67 D etermine the total emissivity and the emissive power of the surface

l middotr-------------------~-------~---~~---~

+ -

3

- _---7-l--t

- -t----

-T-r-t-- I

09 I 1 1 ~ I~ ~~ - -- A ----7=- --- r r------shy- - --shy

ct 06 Imiddots

tshy(Il 05sect -- + shyU 04 C2 A2

~ 03 Q)

02 + _ --__shy01 I

C l 1--+ - ----middot---+shy

I0

0 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Wavelength microns

Figure 67 Spectral emissivity for Example 62

Solution Once the total emissivity is known Eq (6-9) can be used to calculate the emissive power The emiss ive power of the surface is composed of the individual con tribushytion for each wavelength range associated with each spectral hemispherical emisshysivity or

(6-10)

which can be expressed as

8 = 81F(0 - A1T) + 82[F(0 - 1l2T) - F(O - AJT)] + 83[1 - F(O - 1l2T)] (6-11)

Thus by using the spectral emiSSIve description of the surface behavior the blackbody radiation function can be used to make computations for non-blackbody surfaces

The Mathcad worksheet for this problem is presented in Figure 6S The total emissivity is 06099 and the total emissive power is 2161 W1m2 B y comshyparison if the surface were a blackbody the total emissive power would be 3544 W m2

5

etermine

emissive ontribushyal emisshy

( 6-10)

A2T )] (6-11)

vior the ackbody

68 The By comshyould be


Evaluation of the total emissivity

C2 J -6F - =593352xI0( n1

F(~J = 063363n2

E = 060991

Compute the total emissive power per unit area

Ts = 500middotK 3E = 216137x 10 W 2 m

Blackbody emissive power 4 3WEb= aTs Eb = 354375 x 10 2

m

Figure 68 Mathcad solution for Example 62

Equations (6-7) to (6-9) involve surface emissions of thennal radiation Radiation incident on a surface is called irradiation and is denoted G and the spectral irradiashytion (the irradiation at a given wavelength) is denoted GA Three things happen to thershymal radiation incident on a surface (1) some is absorbed (2) some is reflected and (3) some is transmitted Figure 69 schematically illustrates these processe Consider each of the possibilities in turn Irradiation like emitted radiation has a spectral nature but unlike emitted radiation whose spectral characteristics are determined by

Gincident Greflected

Gabsorbed

Gtransmitted

I Figure 69 Irradiation on a surface


the surface temperature the spectral characteristics of irradiation are determined by the source of the irradiation And just like emitted radiation irradiation has direcshytional attributes Surface properties associated with irradiation are described by three terms (1) absorptivity for the absorbed irradiation (2) reflectivity for the reflected irradiation and (3) transmissivity for the transmitted irradiation The same hierarchy of spectral directional to spectral hemispherical to total hemispherical is used Working definitions are as follows

Absorptivity I Abulliabsorbed(A e4gt )

Spectral directional CiAe( A e 4raquo = --I - (-A- e- -)- (6-12) A l If

GA absorbed (A)Spectral hemispherical (6-13)CiA (A) = GA(A)

100

CiAGA(A)dA Total bemispherical Ci= 00 ( 6-14) 1GAdA

where IA i is the incident irradiation flux and G A is spectral irradiation An analogous sequence is defined for the reflectivity and the transmissivity

Reflectivity I A i reflected ( A e 4gt )

Spectral directional ( 6-15) PA e(Ae4raquo = I (A e A)Al yen

GA reilccted( A)Spectral hemispherical (6-16)

GA(A)

001 PAGA (A) dA Total bemispherical p= (6-17)

( Xi GAdA 10

Transmissivity I Ai transmitted( A e 4gt )

Spectral directional T (A e A) - ------ (6-18)108 I - I (A e A)Al yen

GA transmitted (A)Spectral hemispherical T (A ) - ----- (6-19)A - GA(A)

lOO TAGA(A)dA Total hemispherical T= (6-20)

rmined by has direcshyd by three ~ reflected hierarchy

II is used

(6-12)

(6-13)

( 6-14)

malogous

(6-15)

(6-16)

(6-17)

( 6-18)

( 6-19)

(6-20)


For an opaque surface a surface that does not transmit thermal radiation the transshymissivity terms are all zero

Conservation of energy and the above definitions demand that

1 = et + p + T

l=et+p+T (6-21)

If the spectral absorptivity is equal to the spectral emissivity in Example 62 lOA =

etA and if the irradiation G of 750 W m2 has the spectral characteristics of a blackshybody at 2000 K find the total absorptivity the total reflectivity and the irradiation absorbed if the surface is opaque

Solution The spectral absorptivity is known from Example 62 Since the irradiation has the spectral characteristics of a blackbody at 2000 K the blackbody emission function for a temperature of 2000 K can be used to describe the spectral distribution of the irradiation Since the surface is opaque the transmissivity is zero The Mathcad solution is presented in Figure 610 The solution follows the general approach of

The definition of IJm

C 1 and C 2 for the Planck power spectral density expression

C21S [ 20 e - n 1 ( 6 1 3 eJIF()1 =- -- -+6-+ 1 +3middot- 1 =U Definition of C4L n 3 2 n IT n=l n n

G= 7S0 W 2

Irradiation m

H1 = 15-~mmiddot2000K F(~J = 027312 H1 The radiation function for the

absorptivity must be calculated

F(~) =0985SS H2

using the spectral characteristics of the source of the irradiation

a = OlF( ~T~ J+ OS(F( ~T~J -F( ~~ JJ + 08 1 - F( ~T~JJ a = 039S09

p = I-a p =060491 Reflectivity

Gabsorbed = a G Gabsorbed = 29631455 W 2

m

Irradiation absorbed by surface



Example 62 Several things are of interest in this problem Because of the spectral characteristics of the surface (lI e which is an assumption sometimes made for problems such as this The irradiation is not that of a blackbody but it has the specshytral characteristics of p blackbody at 2000 K

In the problem statement for Example 63 eA = (lIA under what conditions is this a viable assertion When is (lI = e acceptable Most heat transfer textbooks address these issues The usual question to ask is when is eM = (lIA8 This is always true since e M and (liM are properties of the surface and are independent of the emission or irrashydiation The spectral hemispherical emissivity and absorptivity are equal under two conditions (1) when the irradiation is diffuse or (2) when the surface is diffuse For many problems of interest one of these conditions is often approximated Moving up the hierarchy when is (lI = e The total absorptivity is equal to the total emissivshyity under two circumstances (1) when the irradiation corresponds to that of blackshybody or (2) when the surface is gray A gray surface is a surface for which eA = (lIA for all A (in the ranges of interest) Many spectrally selective surfaces including the one in Examples 62 and 63 are not gray surfaces and (lI e

Table 63 presents the emissivity and solar absorptivity for selected surfaces The ratio (lI l e of the absorptivity for solar irradiation and the emissivity is a useful solar engineering parameter as small values reject heat and large values absorb solar energy

The foregoing review of radiation heat transfer is sufficient for the needs of this book The next section will explore the motion of the sun with respect to the Earth

TABLE 63 Solar absorptivity emissivity and their ratio for selected surfaces

Surface Solar Absorptivity a Emissivity pound Ratio ale

Aluminium Evaporated film 009 0G3 300 Hard anodized 003 080 004

White paint 021 096 022 Black paint 097 097 100 Black chrome 095 015 647 Red brick 063 093 068 Snow 028 097 029 Teflon 012 085 014 Vegetation (corn) 076 097 078

spectral nade for the spec-

Ions is this s address true since on or irrashyunder two iffuse For d Moving 11 emissivshyt of blackshyCA = OA for 19 the one

faces The eful solar sorb solar

~ needs of ect to the

es

~atio alE

300 004 022 100 647 068 029 014 078

63 Sun Path Description and Calculation 129

] SUN PATH DESCRIPTION AND CALCULATION

An understanding of the position of the sun relative to the Earth is important for solar engineering studies Perhaps the most effective tool for explaining and undershystanding the position of the sun is the sun path - the path of the sun in relation to a location on the surface of the Earth A number of textbooks reference books and websites contain discussions descriptions tables and illustrations of sun paths for varshyious locations However few describe how to calculate sun paths or contain software elements to aid in their computation The purpose of this section is to provide a proshycedure for calculating and understanding sun path lines

Consider the Earth-sun geometric relationship As illustrated in Figure 611 the Earth rotates at an angle of 2345deg with respect to the ecliptic orbital plane This rotashytion angle is responsible for the seasons (and for much of the difficulty in computshying sun paths) Although the Earth moves around the sun the simplest way to understand the motion of the Earth is to adopt a Ptolemaic view - to consider the Earth as stationary and the sun as being in motion In the Ptolemaic view the suns relation to a point on the surface of the Earth is described by the solar altitude angle 0 and the solar azimuth angle as These angles are shown in Figure 612 The altishytude angle is the angle between a line collinear with the sun and the horizontal plane The azimuth angle is the angle between a due-south line and the projection of the

Ecliptic

Polar

Figure 611 Motion of the Earth about the sun


N

E

S Horizon

Figure 612 Solar azimuth angle and solar altitude angle definitions

site-to-sun line on the horizontal plane The sun path for June 21 for Mississippi State University (MSU) 33455deg north latitude is illustrated in Figure 613 In addition to the relation between the azimuth and altitude angles the solar time is shown in the figure At MSU June 21 has almost 14 hours of daylight and at solar noon the sun is nearly vertical (a = 80deg) The sun rises almost 30deg north of east (as = -120deg) and sets almost 30deg north of west (as = +120deg) An interesting aspect of this sun path line is the rapidity of movement of the sun between 1100 am and 100 pm During this time the sun traverses a total azimuth angle of nearly 120deg-almost half of the total azimuth angle traversal of 240deg The motion of the sun is also symmetrical about solar noon with the difference between morning and afternoon being just the sign of the azimuth angle

However as Goswami et al (2000) point out the altitude and azimuth angles are not fundamental but are functions of the location (the latitude) the time (the solar hour angle) and the solar declination (As an aside longitudes and latitudes for locations in the United States can be found at wwwgeonamesusgsgov and those for locations worldwide at wwwastrocom) The solar declination 85 varies between -2345deg and +2345deg and has the same numerical value as the latitude at which the sun is directly overhead at solar noon on a given day The solar declination can be approximated as

360(284 + 17)]Os = 2345deg sin [ 365 (6-22)

where n is the Julian day The Julian day is the number of the day as measured ippi State sequentially from January 1 (Julian day 1) to December 31 (Julian day 365) j dition to Strictly interpreted the number of the day from January 1 is the ordinal day but in wn in the many solar engineering applications the JUlian day is taken as the ordinal day n the sun The solar declination as a function of the Julian day (or ordinal day) is illustrated in L200) and Figure 614 path line uring this 30 the total 25 lout solar ltgt 20 ltll

gn of the ~ ISb1) ltll

-0 10 th angles 5bn cime (the 0laquo latitudes -59 md those lt -)0 between

5 u -IS Q

ltll -20vhich the -25n can be -30

~ middot 1I- t 1 t-1 I

-1 I

r -fIL-i

- L I

I - - --

1 -[ t -Imiddot r -( - r I r

0 30 60 90 120 150 180 210 240 270 300 330 360 390 Julian Date

(6-22) Figure 614 Declination angle as a function of Julian date


The solar hour angle hs is based on the 24 hours required for the sun to move 360deg around the Earth The motion is 15deg per hour with -180deg S hs S +180deg The solar hour angle at solar noon (the time when the azimuth angle is zero and lh sun altitude angle is the maximum for a given day) is defined as zero with values east of south (morning) being negative and values west of south (afternoon) being p sitive

The solar altitude angle a is related to the declination (os) the latitude (L) and the solar hour (hs) angles as

sin(a) = sin(L) sin(() + cos(L) cos(os) cos(hs ) (6-23)

The solar azimuth angle as is given as

cosCos) sin(hs ) sm(as ) = ( ) (6-24)

cos a

For a given day the generation of a sun path line using Eqs (6-23) and (6-24) is straightforward for a latitude greater than the solar declination Care must be taken for solar azimuth angles greater than + 90deg for the afternoon or less than -900 for the morning Since the principal angle range for the arcsin function is +90deg to - 90deg for azimuth angles greateriless than plusmn90deg logic must be used to ensure that a value greaterless than 90deg is obtained One way to accomplish this is to determine the hou angle hlimit that corresponds to as = plusmn90deg (the sun position due westeast) Then for all hour angles greaterless than plusmnhlimit the azimuth angle must be greate rless than plusmn90deg Goswami et al (2000) provide the following expression for hlimit

tan( os)cos(hlimit) = plusmn () (6-25)

tan L

When the hour angle is greaterless than plusmnhlimit the azimuth angle is evalu ted as plusmn1T - arcsin[ sine as)] Lproperly preserving lasl gt 90deg for hour angles greaterless than plusmnhlimit

The relation between standard time and local time must be incorporated if the sun path times are to be cast as local or clock times The relationship between solar time and local (clock) time is presented in the following expressions

Solartime = ST + 4(SL - LL) (minutes) + E (minutes) (6-26)

ST + 1 hour = DST (6-27)

where Solartime is the solar time ST is the standard time SL is the tandard longitude LL is the local longitude DST is daylight savings time and E is the corshyrection in minutes provided by the equation of time The equation of time is

E = 987middot sin(2middot Bn) - 753 cos(Bn) - 15middot sin( Bn) (6-28)

with

B n

360middot (n - 81 ) 1T =

364180 (6-29)

and n is the Julian day (the day number during the year with January 1 b ing day 1) Table 63 provides the standard longitudes for United States time zones

1 to move 180degThe Ld the sun les east of ~ positive ~ (L) and

(6-23)

(6-24)

i (6-24) is be taken -90deg for

to -90deg at a value ~ the hour 1st) Then eaterless nit

(6-25)

duated as ea terless

Ited if the teen solar

(6-26)

(6-27)

standard s the corshy~ is

(6-28)

(6-29)

ng day 1)

Solution


TABLE 63 Standard longitudes for United States time zones

75deg for Eastern 90deg for Central

105deg for Mountain l20deg for Pacific

The equation of time is plotted as a function of the Julian day in Figure 615 Using Eqs (6-26)-(6-28) or Eqs (6-26)-(6-27) and Figure 615 the solar time can

be related to the standard time for any longitude and Julian day In this contextstan shydard time is the local time or the time indicated by a clock

20 I 15 -- r 1 +- -

) ~ 10l

secth

[- 15 ~ ft__ 1

~

h 02 ti ltl) -5

I - I

tt 0 u -10 ~ J

- l 1 -Jshy-15 I

-20 0 30 60 90 120 150 180 210 240 270 300 330 360 390

Julian Date

Figure 615 The equation of time as a function of Julian day

What i the standard time corresponding to LOO pm solar time on August 20 at Mis issippi State University

Mississippi State University is located in the Central time zone at latitude 33455deg (north) and longitude 88789deg (west) The Julian day or ordinal day for August 20 is 232 Daylight savings time is observed Inserting Eq (6-27) into Eq (6-26) and solving for the standard time yields

DST = Solartime + 1 h - 4(SL - LL) min - E min (6-30)

The equation of time correction can be read from Figure 615 or calculated for day 232 using Eq (6-28) with a result of -2946 minutes Inserting the values into Eq (6-30) the local daylight savings time becomes

DST = 100 pm + 1 h - 4(90 - 88789) min - (-295 min)

= 15810 pm

~5 3 0

this chapshyif energy 3pters

~ spectral tal radiashygineering

ristics of ing point tion may ~tic wave ngth are

(6-1)

s the freshy

function able 6l Ivelength Ie range Parts of egime


TABLE 61 The electromagnetic continuum and regimes

Regime Range

Gamma rays A lt 10-4 J-Lm X-rays 10-4 lt A lt 10-2 J-Lm Ultraviolet 10-2 lt A lt 04 fLm

Thermal radiation 10-1 lt A lt 102 flm

Infrared 07 lt A lt 102 J-Lm Microwave A gt 102 --lm

The magnitude of thermal radiation varies with the wavelength and gives rise to the spectral distribution In addition to the spectral nature of thermal radiation a key feature is the directional distribution In order for thhmal radiation to be quanshy

tified both the spectral and directional attributes must be examined

A good starting point is to consider the characteristics of a blackbody

(1) A blackbody absorbs all incident radiation regardless of wavelength and direction

(2) For a prescribed temperature and wavelength no surface can emit more energy than a blackbody

(3) The blackbody is a diffuse (independent of direction) emitter

The spectral emissive power density of a blackbody is given by the Planck disshytribution (Incropera and deWitt 1999)

(6-2)EAb(AT) = 5 [ (C2 ) J

A - exp -- - 1 A -T

where C1 = 3742 -108 W J-Lm4jm2 and C2 = 1439 -104 J-LmK The spectral emissive power density EAb is customarily expressed in the units of W j J-Lm m2 and is interpreted as the emissive power per unit area per wavelength The subscripts A and b in Eq (6-2) signify that the power density is given per wavelength (spectral) and is from a blackbody respectively The total emissive power of a blackbody is the integral of the blackbody spectral emissive power density over all wavelengths or

( )O ( 00 C 4 (6-3)Eb(T) = Jo EAbdA = Jo 5 [ (~2) J dA = (TT

A - exp A _T - 1

where (T = 567 -10-8 Wjm2 K4 the Stefan-Boltzmann constant The T4 funcshy

tional dependence of the blackbody emissive power is one of the strongest functional dependencies in physics and emphasizes the importance of temperature in thermal radiation


5800 K ------r-shy

1106

ltl ~ 0

0- 110

5 ltl sect

1 104W iii U ltl

1103P CJ

---+

100

1 ~--~~~~~------~--~~--~~--~~~ 01 10 100






= (jj c Q)

o ~ t5 Q) 0 (f)

Q5 s o n

Wavelength




0 uTJo E)b dA



15 20 e-n( ( 6 2)

F() = - 2- - + 6- +3 + 3- (6-5)4 3 2

7T n=l n n n n




)0

atures



Solution




200 0

400 1855410 12

600 9263middot10-8

800 1639610-5

1000 00003

1200 00021

1400 00078

1600 00197

1800 00393

2000 00667

2200 01008

2400 01402

2600 0183

2800 02278

3000 02731

3200 0318

3400 03616

3600 04035

3800 04433

4000 04808

4200 05159

4400 05487

4600 05792

4800 06074

5000 06336


5200 06579

5400 06802

5600 07009

5800 072

6000 07377

6200 0754

6400 07691

6600 07831

6800 0796

7000 0808

7200 08191

7400 08294

7600 0839

7800 08479

8000 08562

8200 08639

8400 08711

8600 08778

8800 08841

9000 08899

9200 08954

9400 09006

9600 09054

9800 09099

10000 09141


10500 09236

11000 09318

11500 09389

12000 0945

12500 09504

13000 09551

13500 09592

14000 09628

14500 09661

15000 09689

15500 09715

16000 09738

16500 09758

17000 09776

17500 09793

18000 09808

18500 09822

19000 09834

19500 09845

20000 09855


25000 09922

30000 09953

35000 0997

40000 09979

45000 09985

50000 09989

55000 09992

60000 09993

65000 09995

70000 09996

75000 09997

80000 09997

85000 09998

90000 09998

95000 09998

110 5 09998


(6-6)


AIT = 04 f-I-m 5800 K = 2320 f-I-mK

A2T = 07 f-I-m 5800 K = 4060 f-I-ID K




200 0

400 1855410-12

600 926310-8

800 1639610-5

1000 00003

1200 00021

1400 00078

1600 00197

1800 00393

2000 00667

2200 01008

2400 01402

2600 0183

2800 02278

3000 02731

3200 0318

3400 03616

3600 04035

3800 04433

4000 04808

4200 05159

4400 05487

4600 05792

4800 06074

5000 06336

AT (pmK) F(O - AT)

5200 06579

5400 06802

5600 07009

5800 072

6000 07377

6200 0754

6400 07691

6600 07831

6800 0796

7000 0808

7200 08191

7400 08294

7600 0839

7800 08479

8000 08562

8200 08639

8400 08711

8600 08778

8800 08841

9000 08899

9200 08954

9400 09006

9600 09054

9800 09099

10000 09141


10500 09236

11000 09318

11500 09389

12000 0945

12500 09504

13000 09551

13500 09592

14000 09628

14500 09661

15000 09689

15500 09715

16000 09738

16500 09758

17000 09776

17500 09793

18000 09808

18500 09822

19000 09834

19500 09845

20000 09855


25000 09922

30000 09953

35000 0997

40000 09979

45000 0 9985

50000 09989

55000 09992

60000 09993

65000 09995

70000 09996

75000 09997

80000 09997

85000 09998

90000 09998

95000 09998

110 5 09998


(6-6)


AiT = 04 fLm 5800 K = 2320 fLmK

A2T = 07 fLm 5800 K = 4060 fLmK

F(O -AT)

09922

09953

0997

09979

09985

0 9989

09992

09993

09995

09996

09997

09997

09998

09998

09998

0 9998


(6-6)

800 K


shyw c Q)

o ~ U Q) D

(f)

(jj 3 o

0

Wavelength



F( 2320 fLmK) = 01244

F(4060 fLmK ) = 04913


F(AzT - A1T) = 04913 - 01244 = 03669





F(1200 fLmK ) = 00021

F(2100 fLmK ) = 00838

F (A2T - ArT ) = 00830 - 00021 = 00817








IT n = [ n n

Blackbody at 5800 K

F(~) = 012392 H1

F (~J = 049154 H2


Blackbody at 3000 K






Ind flm

spectral



2500~--~----~----~----------~----

2000

~ 1000 U 0

()

500

Solar Irradiation

T t I

I -

I

5800 K Blackbody I

I

I

I-II I

I

~I---CO2 I

0U1L-L-__L-~U-~J=~~~~ o 05 10 15 2 0 25 30

Wavelength (um)













Ab



Ab E (A T )

Ab (6-8)




orien t the r



(6-7)


(6-8)

T) is the ~en inte-

~cent



seT) = E(T)

(6-9)




l middotr-------------------~-------~---~~---~

+ -

3

- _---7-l--t

- -t----

-T-r-t-- I

09 I 1 1 ~ I~ ~~ - -- A ----7=- --- r r------shy- - --shy

ct 06 Imiddots


~ 03 Q)

02 + _ --__shy01 I

C l 1--+ - ----middot---+shy

I0

0 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Wavelength microns



(6-10)





5

etermine


( 6-10)

A2T )] (6-11)

vior the ackbody




C2 J -6F - =593352xI0( n1

F(~J = 063363n2

E = 060991


Ts = 500middotK 3E = 216137x 10 W 2 m


m




Gabsorbed

Gtransmitted







100






GA(A)


( Xi GAdA 10






II is used

(6-12)

(6-13)

( 6-14)

malogous

(6-15)

(6-16)

(6-17)

( 6-18)

( 6-19)

(6-20)




1 = et + p + T

l=et+p+T (6-21)







G= 7S0 W 2

Irradiation m



F(~) =0985SS H2





m
















es

~atio alE

300 004 022 100 647 068 029 014 078





Ecliptic

Polar



N

E

S Horizon




360(284 + 17)]Os = 2345deg sin [ 365 (6-22)




5 u -IS Q



-1 I

r -fIL-i

- L I

I - - --


0 30 60 90 120 150 180 210 240 270 300 330 360 390 Julian Date








cos a



tan L




ST + 1 hour = DST (6-27)



with

B n

360middot (n - 81 ) 1T =

364180 (6-29)



(6-23)

(6-24)



(6-25)



(6-26)

(6-27)


(6-28)

(6-29)

ng day 1)

Solution







20 I 15 -- r 1 +- -

) ~ 10l

secth

[- 15 ~ ft__ 1

~

h 02 ti ltl) -5

I - I

tt 0 u -10 ~ J

- l 1 -Jshy-15 I

-20 0 30 60 90 120 150 180 210 240 270 300 330 360 390

Julian Date






DST = 100 pm + 1 h - 4(90 - 88789) min - (-295 min)

= 15810 pm


5800 K ------r-shy

1106

ltl ~ 0

0- 110

5 ltl sect

1 104W iii U ltl

1103P CJ

---+

100

1 ~--~~~~~------~--~~--~~--~~~ 01 10 100






= (jj c Q)

o ~ t5 Q) 0 (f)

Q5 s o n

Wavelength




0 uTJo E)b dA



15 20 e-n( ( 6 2)

F() = - 2- - + 6- +3 + 3- (6-5)4 3 2

7T n=l n n n n




)0

atures



Solution




200 0

400 1855410 12

600 9263middot10-8

800 1639610-5

1000 00003

1200 00021

1400 00078

1600 00197

1800 00393

2000 00667

2200 01008

2400 01402

2600 0183

2800 02278

3000 02731

3200 0318

3400 03616

3600 04035

3800 04433

4000 04808

4200 05159

4400 05487

4600 05792

4800 06074

5000 06336


5200 06579

5400 06802

5600 07009

5800 072

6000 07377

6200 0754

6400 07691

6600 07831

6800 0796

7000 0808

7200 08191

7400 08294

7600 0839

7800 08479

8000 08562

8200 08639

8400 08711

8600 08778

8800 08841

9000 08899

9200 08954

9400 09006

9600 09054

9800 09099

10000 09141


10500 09236

11000 09318

11500 09389

12000 0945

12500 09504

13000 09551

13500 09592

14000 09628

14500 09661

15000 09689

15500 09715

16000 09738

16500 09758

17000 09776

17500 09793

18000 09808

18500 09822

19000 09834

19500 09845

20000 09855


25000 09922

30000 09953

35000 0997

40000 09979

45000 09985

50000 09989

55000 09992

60000 09993

65000 09995

70000 09996

75000 09997

80000 09997

85000 09998

90000 09998

95000 09998

110 5 09998


(6-6)


AIT = 04 f-I-m 5800 K = 2320 f-I-mK

A2T = 07 f-I-m 5800 K = 4060 f-I-ID K




200 0

400 1855410-12

600 926310-8

800 1639610-5

1000 00003

1200 00021

1400 00078

1600 00197

1800 00393

2000 00667

2200 01008

2400 01402

2600 0183

2800 02278

3000 02731

3200 0318

3400 03616

3600 04035

3800 04433

4000 04808

4200 05159

4400 05487

4600 05792

4800 06074

5000 06336

AT (pmK) F(O - AT)

5200 06579

5400 06802

5600 07009

5800 072

6000 07377

6200 0754

6400 07691

6600 07831

6800 0796

7000 0808

7200 08191

7400 08294

7600 0839

7800 08479

8000 08562

8200 08639

8400 08711

8600 08778

8800 08841

9000 08899

9200 08954

9400 09006

9600 09054

9800 09099

10000 09141


10500 09236

11000 09318

11500 09389

12000 0945

12500 09504

13000 09551

13500 09592

14000 09628

14500 09661

15000 09689

15500 09715

16000 09738

16500 09758

17000 09776

17500 09793

18000 09808

18500 09822

19000 09834

19500 09845

20000 09855


25000 09922

30000 09953

35000 0997

40000 09979

45000 0 9985

50000 09989

55000 09992

60000 09993

65000 09995

70000 09996

75000 09997

80000 09997

85000 09998

90000 09998

95000 09998

110 5 09998


(6-6)


AiT = 04 fLm 5800 K = 2320 fLmK

A2T = 07 fLm 5800 K = 4060 fLmK

F(O -AT)

09922

09953

0997

09979

09985

0 9989

09992

09993

09995

09996

09997

09997

09998

09998

09998

0 9998


(6-6)

800 K


shyw c Q)

o ~ U Q) D

(f)

(jj 3 o

0

Wavelength



F( 2320 fLmK) = 01244

F(4060 fLmK ) = 04913


F(AzT - A1T) = 04913 - 01244 = 03669





F(1200 fLmK ) = 00021

F(2100 fLmK ) = 00838

F (A2T - ArT ) = 00830 - 00021 = 00817








IT n = [ n n

Blackbody at 5800 K

F(~) = 012392 H1

F (~J = 049154 H2


Blackbody at 3000 K






Ind flm

spectral



2500~--~----~----~----------~----

2000

~ 1000 U 0

()

500

Solar Irradiation

T t I

I -

I

5800 K Blackbody I

I

I

I-II I

I

~I---CO2 I

0U1L-L-__L-~U-~J=~~~~ o 05 10 15 2 0 25 30

Wavelength (um)













Ab



Ab E (A T )

Ab (6-8)




orien t the r



(6-7)


(6-8)

T) is the ~en inte-

~cent



seT) = E(T)

(6-9)




l middotr-------------------~-------~---~~---~

+ -

3

- _---7-l--t

- -t----

-T-r-t-- I

09 I 1 1 ~ I~ ~~ - -- A ----7=- --- r r------shy- - --shy

ct 06 Imiddots


~ 03 Q)

02 + _ --__shy01 I

C l 1--+ - ----middot---+shy

I0

0 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Wavelength microns



(6-10)





5

etermine


( 6-10)

A2T )] (6-11)

vior the ackbody




C2 J -6F - =593352xI0( n1

F(~J = 063363n2

E = 060991


Ts = 500middotK 3E = 216137x 10 W 2 m


m




Gabsorbed

Gtransmitted







100






GA(A)


( Xi GAdA 10






II is used

(6-12)

(6-13)

( 6-14)

malogous

(6-15)

(6-16)

(6-17)

( 6-18)

( 6-19)

(6-20)




1 = et + p + T

l=et+p+T (6-21)







G= 7S0 W 2

Irradiation m



F(~) =0985SS H2





m
















es

~atio alE

300 004 022 100 647 068 029 014 078





Ecliptic

Polar



N

E

S Horizon




360(284 + 17)]Os = 2345deg sin [ 365 (6-22)




5 u -IS Q



-1 I

r -fIL-i

- L I

I - - --


0 30 60 90 120 150 180 210 240 270 300 330 360 390 Julian Date








cos a



tan L




ST + 1 hour = DST (6-27)



with

B n

360middot (n - 81 ) 1T =

364180 (6-29)



(6-23)

(6-24)



(6-25)



(6-26)

(6-27)


(6-28)

(6-29)

ng day 1)

Solution







20 I 15 -- r 1 +- -

) ~ 10l

secth

[- 15 ~ ft__ 1

~

h 02 ti ltl) -5

I - I

tt 0 u -10 ~ J

- l 1 -Jshy-15 I

-20 0 30 60 90 120 150 180 210 240 270 300 330 360 390

Julian Date






DST = 100 pm + 1 h - 4(90 - 88789) min - (-295 min)

= 15810 pm


= (jj c Q)

o ~ t5 Q) 0 (f)

Q5 s o n

Wavelength




0 uTJo E)b dA



15 20 e-n( ( 6 2)

F() = - 2- - + 6- +3 + 3- (6-5)4 3 2

7T n=l n n n n




)0

atures



Solution




200 0

400 1855410 12

600 9263middot10-8

800 1639610-5

1000 00003

1200 00021

1400 00078

1600 00197

1800 00393

2000 00667

2200 01008

2400 01402

2600 0183

2800 02278

3000 02731

3200 0318

3400 03616

3600 04035

3800 04433

4000 04808

4200 05159

4400 05487

4600 05792

4800 06074

5000 06336


5200 06579

5400 06802

5600 07009

5800 072

6000 07377

6200 0754

6400 07691

6600 07831

6800 0796

7000 0808

7200 08191

7400 08294

7600 0839

7800 08479

8000 08562

8200 08639

8400 08711

8600 08778

8800 08841

9000 08899

9200 08954

9400 09006

9600 09054

9800 09099

10000 09141


10500 09236

11000 09318

11500 09389

12000 0945

12500 09504

13000 09551

13500 09592

14000 09628

14500 09661

15000 09689

15500 09715

16000 09738

16500 09758

17000 09776

17500 09793

18000 09808

18500 09822

19000 09834

19500 09845

20000 09855


25000 09922

30000 09953

35000 0997

40000 09979

45000 09985

50000 09989

55000 09992

60000 09993

65000 09995

70000 09996

75000 09997

80000 09997

85000 09998

90000 09998

95000 09998

110 5 09998


(6-6)


AIT = 04 f-I-m 5800 K = 2320 f-I-mK

A2T = 07 f-I-m 5800 K = 4060 f-I-ID K




200 0

400 1855410-12

600 926310-8

800 1639610-5

1000 00003

1200 00021

1400 00078

1600 00197

1800 00393

2000 00667

2200 01008

2400 01402

2600 0183

2800 02278

3000 02731

3200 0318

3400 03616

3600 04035

3800 04433

4000 04808

4200 05159

4400 05487

4600 05792

4800 06074

5000 06336

AT (pmK) F(O - AT)

5200 06579

5400 06802

5600 07009

5800 072

6000 07377

6200 0754

6400 07691

6600 07831

6800 0796

7000 0808

7200 08191

7400 08294

7600 0839

7800 08479

8000 08562

8200 08639

8400 08711

8600 08778

8800 08841

9000 08899

9200 08954

9400 09006

9600 09054

9800 09099

10000 09141


10500 09236

11000 09318

11500 09389

12000 0945

12500 09504

13000 09551

13500 09592

14000 09628

14500 09661

15000 09689

15500 09715

16000 09738

16500 09758

17000 09776

17500 09793

18000 09808

18500 09822

19000 09834

19500 09845

20000 09855


25000 09922

30000 09953

35000 0997

40000 09979

45000 0 9985

50000 09989

55000 09992

60000 09993

65000 09995

70000 09996

75000 09997

80000 09997

85000 09998

90000 09998

95000 09998

110 5 09998


(6-6)


AiT = 04 fLm 5800 K = 2320 fLmK

A2T = 07 fLm 5800 K = 4060 fLmK

F(O -AT)

09922

09953

0997

09979

09985

0 9989

09992

09993

09995

09996

09997

09997

09998

09998

09998

0 9998


(6-6)

800 K


shyw c Q)

o ~ U Q) D

(f)

(jj 3 o

0

Wavelength



F( 2320 fLmK) = 01244

F(4060 fLmK ) = 04913


F(AzT - A1T) = 04913 - 01244 = 03669





F(1200 fLmK ) = 00021

F(2100 fLmK ) = 00838

F (A2T - ArT ) = 00830 - 00021 = 00817








IT n = [ n n

Blackbody at 5800 K

F(~) = 012392 H1

F (~J = 049154 H2


Blackbody at 3000 K






Ind flm

spectral



2500~--~----~----~----------~----

2000

~ 1000 U 0

()

500

Solar Irradiation

T t I

I -

I

5800 K Blackbody I

I

I

I-II I

I

~I---CO2 I

0U1L-L-__L-~U-~J=~~~~ o 05 10 15 2 0 25 30

Wavelength (um)













Ab



Ab E (A T )

Ab (6-8)




orien t the r



(6-7)


(6-8)

T) is the ~en inte-

~cent



seT) = E(T)

(6-9)




l middotr-------------------~-------~---~~---~

+ -

3

- _---7-l--t

- -t----

-T-r-t-- I

09 I 1 1 ~ I~ ~~ - -- A ----7=- --- r r------shy- - --shy

ct 06 Imiddots


~ 03 Q)

02 + _ --__shy01 I

C l 1--+ - ----middot---+shy

I0

0 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Wavelength microns



(6-10)





5

etermine


( 6-10)

A2T )] (6-11)

vior the ackbody




C2 J -6F - =593352xI0( n1

F(~J = 063363n2

E = 060991


Ts = 500middotK 3E = 216137x 10 W 2 m


m




Gabsorbed

Gtransmitted







100






GA(A)


( Xi GAdA 10






II is used

(6-12)

(6-13)

( 6-14)

malogous

(6-15)

(6-16)

(6-17)

( 6-18)

( 6-19)

(6-20)




1 = et + p + T

l=et+p+T (6-21)







G= 7S0 W 2

Irradiation m



F(~) =0985SS H2





m
















es

~atio alE

300 004 022 100 647 068 029 014 078





Ecliptic

Polar



N

E

S Horizon




360(284 + 17)]Os = 2345deg sin [ 365 (6-22)




5 u -IS Q



-1 I

r -fIL-i

- L I

I - - --


0 30 60 90 120 150 180 210 240 270 300 330 360 390 Julian Date








cos a



tan L




ST + 1 hour = DST (6-27)



with

B n

360middot (n - 81 ) 1T =

364180 (6-29)



(6-23)

(6-24)



(6-25)



(6-26)

(6-27)


(6-28)

(6-29)

ng day 1)

Solution







20 I 15 -- r 1 +- -

) ~ 10l

secth

[- 15 ~ ft__ 1

~

h 02 ti ltl) -5

I - I

tt 0 u -10 ~ J

- l 1 -Jshy-15 I

-20 0 30 60 90 120 150 180 210 240 270 300 330 360 390

Julian Date






DST = 100 pm + 1 h - 4(90 - 88789) min - (-295 min)

= 15810 pm




200 0

400 1855410 12

600 9263middot10-8

800 1639610-5

1000 00003

1200 00021

1400 00078

1600 00197

1800 00393

2000 00667

2200 01008

2400 01402

2600 0183

2800 02278

3000 02731

3200 0318

3400 03616

3600 04035

3800 04433

4000 04808

4200 05159

4400 05487

4600 05792

4800 06074

5000 06336


5200 06579

5400 06802

5600 07009

5800 072

6000 07377

6200 0754

6400 07691

6600 07831

6800 0796

7000 0808

7200 08191

7400 08294

7600 0839

7800 08479

8000 08562

8200 08639

8400 08711

8600 08778

8800 08841

9000 08899

9200 08954

9400 09006

9600 09054

9800 09099

10000 09141


10500 09236

11000 09318

11500 09389

12000 0945

12500 09504

13000 09551

13500 09592

14000 09628

14500 09661

15000 09689

15500 09715

16000 09738

16500 09758

17000 09776

17500 09793

18000 09808

18500 09822

19000 09834

19500 09845

20000 09855


25000 09922

30000 09953

35000 0997

40000 09979

45000 09985

50000 09989

55000 09992

60000 09993

65000 09995

70000 09996

75000 09997

80000 09997

85000 09998

90000 09998

95000 09998

110 5 09998


(6-6)


AIT = 04 f-I-m 5800 K = 2320 f-I-mK

A2T = 07 f-I-m 5800 K = 4060 f-I-ID K




200 0

400 1855410-12

600 926310-8

800 1639610-5

1000 00003

1200 00021

1400 00078

1600 00197

1800 00393

2000 00667

2200 01008

2400 01402

2600 0183

2800 02278

3000 02731

3200 0318

3400 03616

3600 04035

3800 04433

4000 04808

4200 05159

4400 05487

4600 05792

4800 06074

5000 06336

AT (pmK) F(O - AT)

5200 06579

5400 06802

5600 07009

5800 072

6000 07377

6200 0754

6400 07691

6600 07831

6800 0796

7000 0808

7200 08191

7400 08294

7600 0839

7800 08479

8000 08562

8200 08639

8400 08711

8600 08778

8800 08841

9000 08899

9200 08954

9400 09006

9600 09054

9800 09099

10000 09141


10500 09236

11000 09318

11500 09389

12000 0945

12500 09504

13000 09551

13500 09592

14000 09628

14500 09661

15000 09689

15500 09715

16000 09738

16500 09758

17000 09776

17500 09793

18000 09808

18500 09822

19000 09834

19500 09845

20000 09855


25000 09922

30000 09953

35000 0997

40000 09979

45000 0 9985

50000 09989

55000 09992

60000 09993

65000 09995

70000 09996

75000 09997

80000 09997

85000 09998

90000 09998

95000 09998

110 5 09998


(6-6)


AiT = 04 fLm 5800 K = 2320 fLmK

A2T = 07 fLm 5800 K = 4060 fLmK

F(O -AT)

09922

09953

0997

09979

09985

0 9989

09992

09993

09995

09996

09997

09997

09998

09998

09998

0 9998


(6-6)

800 K


shyw c Q)

o ~ U Q) D

(f)

(jj 3 o

0

Wavelength



F( 2320 fLmK) = 01244

F(4060 fLmK ) = 04913


F(AzT - A1T) = 04913 - 01244 = 03669





F(1200 fLmK ) = 00021

F(2100 fLmK ) = 00838

F (A2T - ArT ) = 00830 - 00021 = 00817








IT n = [ n n

Blackbody at 5800 K

F(~) = 012392 H1

F (~J = 049154 H2


Blackbody at 3000 K






Ind flm

spectral



2500~--~----~----~----------~----

2000

~ 1000 U 0

()

500

Solar Irradiation

T t I

I -

I

5800 K Blackbody I

I

I

I-II I

I

~I---CO2 I

0U1L-L-__L-~U-~J=~~~~ o 05 10 15 2 0 25 30

Wavelength (um)













Ab



Ab E (A T )

Ab (6-8)




orien t the r



(6-7)


(6-8)

T) is the ~en inte-

~cent



seT) = E(T)

(6-9)




l middotr-------------------~-------~---~~---~

+ -

3

- _---7-l--t

- -t----

-T-r-t-- I

09 I 1 1 ~ I~ ~~ - -- A ----7=- --- r r------shy- - --shy

ct 06 Imiddots


~ 03 Q)

02 + _ --__shy01 I

C l 1--+ - ----middot---+shy

I0

0 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Wavelength microns



(6-10)





5

etermine


( 6-10)

A2T )] (6-11)

vior the ackbody




C2 J -6F - =593352xI0( n1

F(~J = 063363n2

E = 060991


Ts = 500middotK 3E = 216137x 10 W 2 m


m




Gabsorbed

Gtransmitted







100






GA(A)


( Xi GAdA 10






II is used

(6-12)

(6-13)

( 6-14)

malogous

(6-15)

(6-16)

(6-17)

( 6-18)

( 6-19)

(6-20)




1 = et + p + T

l=et+p+T (6-21)







G= 7S0 W 2

Irradiation m



F(~) =0985SS H2





m
















es

~atio alE

300 004 022 100 647 068 029 014 078





Ecliptic

Polar



N

E

S Horizon




360(284 + 17)]Os = 2345deg sin [ 365 (6-22)




5 u -IS Q



-1 I

r -fIL-i

- L I

I - - --


0 30 60 90 120 150 180 210 240 270 300 330 360 390 Julian Date








cos a



tan L




ST + 1 hour = DST (6-27)



with

B n

360middot (n - 81 ) 1T =

364180 (6-29)



(6-23)

(6-24)



(6-25)



(6-26)

(6-27)


(6-28)

(6-29)

ng day 1)

Solution







20 I 15 -- r 1 +- -

) ~ 10l

secth

[- 15 ~ ft__ 1

~

h 02 ti ltl) -5

I - I

tt 0 u -10 ~ J

- l 1 -Jshy-15 I

-20 0 30 60 90 120 150 180 210 240 270 300 330 360 390

Julian Date






DST = 100 pm + 1 h - 4(90 - 88789) min - (-295 min)

= 15810 pm




200 0

400 1855410-12

600 926310-8

800 1639610-5

1000 00003

1200 00021

1400 00078

1600 00197

1800 00393

2000 00667

2200 01008

2400 01402

2600 0183

2800 02278

3000 02731

3200 0318

3400 03616

3600 04035

3800 04433

4000 04808

4200 05159

4400 05487

4600 05792

4800 06074

5000 06336

AT (pmK) F(O - AT)

5200 06579

5400 06802

5600 07009

5800 072

6000 07377

6200 0754

6400 07691

6600 07831

6800 0796

7000 0808

7200 08191

7400 08294

7600 0839

7800 08479

8000 08562

8200 08639

8400 08711

8600 08778

8800 08841

9000 08899

9200 08954

9400 09006

9600 09054

9800 09099

10000 09141


10500 09236

11000 09318

11500 09389

12000 0945

12500 09504

13000 09551

13500 09592

14000 09628

14500 09661

15000 09689

15500 09715

16000 09738

16500 09758

17000 09776

17500 09793

18000 09808

18500 09822

19000 09834

19500 09845

20000 09855


25000 09922

30000 09953

35000 0997

40000 09979

45000 0 9985

50000 09989

55000 09992

60000 09993

65000 09995

70000 09996

75000 09997

80000 09997

85000 09998

90000 09998

95000 09998

110 5 09998


(6-6)


AiT = 04 fLm 5800 K = 2320 fLmK

A2T = 07 fLm 5800 K = 4060 fLmK

F(O -AT)

09922

09953

0997

09979

09985

0 9989

09992

09993

09995

09996

09997

09997

09998

09998

09998

0 9998


(6-6)

800 K


shyw c Q)

o ~ U Q) D

(f)

(jj 3 o

0

Wavelength



F( 2320 fLmK) = 01244

F(4060 fLmK ) = 04913


F(AzT - A1T) = 04913 - 01244 = 03669





F(1200 fLmK ) = 00021

F(2100 fLmK ) = 00838

F (A2T - ArT ) = 00830 - 00021 = 00817








IT n = [ n n

Blackbody at 5800 K

F(~) = 012392 H1

F (~J = 049154 H2


Blackbody at 3000 K






Ind flm

spectral



2500~--~----~----~----------~----

2000

~ 1000 U 0

()

500

Solar Irradiation

T t I

I -

I

5800 K Blackbody I

I

I

I-II I

I

~I---CO2 I

0U1L-L-__L-~U-~J=~~~~ o 05 10 15 2 0 25 30

Wavelength (um)













Ab



Ab E (A T )

Ab (6-8)




orien t the r



(6-7)


(6-8)

T) is the ~en inte-

~cent



seT) = E(T)

(6-9)




l middotr-------------------~-------~---~~---~

+ -

3

- _---7-l--t

- -t----

-T-r-t-- I

09 I 1 1 ~ I~ ~~ - -- A ----7=- --- r r------shy- - --shy

ct 06 Imiddots


~ 03 Q)

02 + _ --__shy01 I

C l 1--+ - ----middot---+shy

I0

0 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Wavelength microns



(6-10)





5

etermine


( 6-10)

A2T )] (6-11)

vior the ackbody




C2 J -6F - =593352xI0( n1

F(~J = 063363n2

E = 060991


Ts = 500middotK 3E = 216137x 10 W 2 m


m




Gabsorbed

Gtransmitted







100






GA(A)


( Xi GAdA 10






II is used

(6-12)

(6-13)

( 6-14)

malogous

(6-15)

(6-16)

(6-17)

( 6-18)

( 6-19)

(6-20)




1 = et + p + T

l=et+p+T (6-21)







G= 7S0 W 2

Irradiation m



F(~) =0985SS H2





m
















es

~atio alE

300 004 022 100 647 068 029 014 078





Ecliptic

Polar



N

E

S Horizon




360(284 + 17)]Os = 2345deg sin [ 365 (6-22)




5 u -IS Q



-1 I

r -fIL-i

- L I

I - - --


0 30 60 90 120 150 180 210 240 270 300 330 360 390 Julian Date








cos a



tan L




ST + 1 hour = DST (6-27)



with

B n

360middot (n - 81 ) 1T =

364180 (6-29)



(6-23)

(6-24)



(6-25)



(6-26)

(6-27)


(6-28)

(6-29)

ng day 1)

Solution







20 I 15 -- r 1 +- -

) ~ 10l

secth

[- 15 ~ ft__ 1

~

h 02 ti ltl) -5

I - I

tt 0 u -10 ~ J

- l 1 -Jshy-15 I

-20 0 30 60 90 120 150 180 210 240 270 300 330 360 390

Julian Date






DST = 100 pm + 1 h - 4(90 - 88789) min - (-295 min)

= 15810 pm

F(O -AT)

09922

09953

0997

09979

09985

0 9989

09992

09993

09995

09996

09997

09997

09998

09998

09998

0 9998


(6-6)

800 K


shyw c Q)

o ~ U Q) D

(f)

(jj 3 o

0

Wavelength



F( 2320 fLmK) = 01244

F(4060 fLmK ) = 04913


F(AzT - A1T) = 04913 - 01244 = 03669





F(1200 fLmK ) = 00021

F(2100 fLmK ) = 00838

F (A2T - ArT ) = 00830 - 00021 = 00817








IT n = [ n n

Blackbody at 5800 K

F(~) = 012392 H1

F (~J = 049154 H2


Blackbody at 3000 K






Ind flm

spectral



2500~--~----~----~----------~----

2000

~ 1000 U 0

()

500

Solar Irradiation

T t I

I -

I

5800 K Blackbody I

I

I

I-II I

I

~I---CO2 I

0U1L-L-__L-~U-~J=~~~~ o 05 10 15 2 0 25 30

Wavelength (um)













Ab



Ab E (A T )

Ab (6-8)




orien t the r



(6-7)


(6-8)

T) is the ~en inte-

~cent



seT) = E(T)

(6-9)




l middotr-------------------~-------~---~~---~

+ -

3

- _---7-l--t

- -t----

-T-r-t-- I

09 I 1 1 ~ I~ ~~ - -- A ----7=- --- r r------shy- - --shy

ct 06 Imiddots


~ 03 Q)

02 + _ --__shy01 I

C l 1--+ - ----middot---+shy

I0

0 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Wavelength microns



(6-10)





5

etermine


( 6-10)

A2T )] (6-11)

vior the ackbody




C2 J -6F - =593352xI0( n1

F(~J = 063363n2

E = 060991


Ts = 500middotK 3E = 216137x 10 W 2 m


m




Gabsorbed

Gtransmitted







100






GA(A)


( Xi GAdA 10






II is used

(6-12)

(6-13)

( 6-14)

malogous

(6-15)

(6-16)

(6-17)

( 6-18)

( 6-19)

(6-20)




1 = et + p + T

l=et+p+T (6-21)







G= 7S0 W 2

Irradiation m



F(~) =0985SS H2





m
















es

~atio alE

300 004 022 100 647 068 029 014 078





Ecliptic

Polar



N

E

S Horizon




360(284 + 17)]Os = 2345deg sin [ 365 (6-22)




5 u -IS Q



-1 I

r -fIL-i

- L I

I - - --


0 30 60 90 120 150 180 210 240 270 300 330 360 390 Julian Date








cos a



tan L




ST + 1 hour = DST (6-27)



with

B n

360middot (n - 81 ) 1T =

364180 (6-29)



(6-23)

(6-24)



(6-25)



(6-26)

(6-27)


(6-28)

(6-29)

ng day 1)

Solution







20 I 15 -- r 1 +- -

) ~ 10l

secth

[- 15 ~ ft__ 1

~

h 02 ti ltl) -5

I - I

tt 0 u -10 ~ J

- l 1 -Jshy-15 I

-20 0 30 60 90 120 150 180 210 240 270 300 330 360 390

Julian Date






DST = 100 pm + 1 h - 4(90 - 88789) min - (-295 min)

= 15810 pm






IT n = [ n n

Blackbody at 5800 K

F(~) = 012392 H1

F (~J = 049154 H2


Blackbody at 3000 K






Ind flm

spectral



2500~--~----~----~----------~----

2000

~ 1000 U 0

()

500

Solar Irradiation

T t I

I -

I

5800 K Blackbody I

I

I

I-II I

I

~I---CO2 I

0U1L-L-__L-~U-~J=~~~~ o 05 10 15 2 0 25 30

Wavelength (um)













Ab



Ab E (A T )

Ab (6-8)




orien t the r



(6-7)


(6-8)

T) is the ~en inte-

~cent



seT) = E(T)

(6-9)




l middotr-------------------~-------~---~~---~

+ -

3

- _---7-l--t

- -t----

-T-r-t-- I

09 I 1 1 ~ I~ ~~ - -- A ----7=- --- r r------shy- - --shy

ct 06 Imiddots


~ 03 Q)

02 + _ --__shy01 I

C l 1--+ - ----middot---+shy

I0

0 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Wavelength microns



(6-10)





5

etermine


( 6-10)

A2T )] (6-11)

vior the ackbody




C2 J -6F - =593352xI0( n1

F(~J = 063363n2

E = 060991


Ts = 500middotK 3E = 216137x 10 W 2 m


m




Gabsorbed

Gtransmitted







100






GA(A)


( Xi GAdA 10






II is used

(6-12)

(6-13)

( 6-14)

malogous

(6-15)

(6-16)

(6-17)

( 6-18)

( 6-19)

(6-20)




1 = et + p + T

l=et+p+T (6-21)







G= 7S0 W 2

Irradiation m



F(~) =0985SS H2





m
















es

~atio alE

300 004 022 100 647 068 029 014 078





Ecliptic

Polar



N

E

S Horizon




360(284 + 17)]Os = 2345deg sin [ 365 (6-22)




5 u -IS Q



-1 I

r -fIL-i

- L I

I - - --


0 30 60 90 120 150 180 210 240 270 300 330 360 390 Julian Date








cos a



tan L




ST + 1 hour = DST (6-27)



with

B n

360middot (n - 81 ) 1T =

364180 (6-29)



(6-23)

(6-24)



(6-25)



(6-26)

(6-27)


(6-28)

(6-29)

ng day 1)

Solution







20 I 15 -- r 1 +- -

) ~ 10l

secth

[- 15 ~ ft__ 1

~

h 02 ti ltl) -5

I - I

tt 0 u -10 ~ J

- l 1 -Jshy-15 I

-20 0 30 60 90 120 150 180 210 240 270 300 330 360 390

Julian Date






DST = 100 pm + 1 h - 4(90 - 88789) min - (-295 min)

= 15810 pm

Ind flm

spectral



2500~--~----~----~----------~----

2000

~ 1000 U 0

()

500

Solar Irradiation

T t I

I -

I

5800 K Blackbody I

I

I

I-II I

I

~I---CO2 I

0U1L-L-__L-~U-~J=~~~~ o 05 10 15 2 0 25 30

Wavelength (um)













Ab



Ab E (A T )

Ab (6-8)




orien t the r



(6-7)


(6-8)

T) is the ~en inte-

~cent



seT) = E(T)

(6-9)




l middotr-------------------~-------~---~~---~

+ -

3

- _---7-l--t

- -t----

-T-r-t-- I

09 I 1 1 ~ I~ ~~ - -- A ----7=- --- r r------shy- - --shy

ct 06 Imiddots


~ 03 Q)

02 + _ --__shy01 I

C l 1--+ - ----middot---+shy

I0

0 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Wavelength microns



(6-10)





5

etermine


( 6-10)

A2T )] (6-11)

vior the ackbody




C2 J -6F - =593352xI0( n1

F(~J = 063363n2

E = 060991


Ts = 500middotK 3E = 216137x 10 W 2 m


m




Gabsorbed

Gtransmitted







100






GA(A)


( Xi GAdA 10






II is used

(6-12)

(6-13)

( 6-14)

malogous

(6-15)

(6-16)

(6-17)

( 6-18)

( 6-19)

(6-20)




1 = et + p + T

l=et+p+T (6-21)







G= 7S0 W 2

Irradiation m



F(~) =0985SS H2





m
















es

~atio alE

300 004 022 100 647 068 029 014 078





Ecliptic

Polar



N

E

S Horizon




360(284 + 17)]Os = 2345deg sin [ 365 (6-22)




5 u -IS Q



-1 I

r -fIL-i

- L I

I - - --


0 30 60 90 120 150 180 210 240 270 300 330 360 390 Julian Date








cos a



tan L




ST + 1 hour = DST (6-27)



with

B n

360middot (n - 81 ) 1T =

364180 (6-29)



(6-23)

(6-24)



(6-25)



(6-26)

(6-27)


(6-28)

(6-29)

ng day 1)

Solution







20 I 15 -- r 1 +- -

) ~ 10l

secth

[- 15 ~ ft__ 1

~

h 02 ti ltl) -5

I - I

tt 0 u -10 ~ J

- l 1 -Jshy-15 I

-20 0 30 60 90 120 150 180 210 240 270 300 330 360 390

Julian Date






DST = 100 pm + 1 h - 4(90 - 88789) min - (-295 min)

= 15810 pm









Ab



Ab E (A T )

Ab (6-8)




orien t the r



(6-7)


(6-8)

T) is the ~en inte-

~cent



seT) = E(T)

(6-9)




l middotr-------------------~-------~---~~---~

+ -

3

- _---7-l--t

- -t----

-T-r-t-- I

09 I 1 1 ~ I~ ~~ - -- A ----7=- --- r r------shy- - --shy

ct 06 Imiddots


~ 03 Q)

02 + _ --__shy01 I

C l 1--+ - ----middot---+shy

I0

0 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Wavelength microns



(6-10)





5

etermine


( 6-10)

A2T )] (6-11)

vior the ackbody




C2 J -6F - =593352xI0( n1

F(~J = 063363n2

E = 060991


Ts = 500middotK 3E = 216137x 10 W 2 m


m




Gabsorbed

Gtransmitted







100






GA(A)


( Xi GAdA 10






II is used

(6-12)

(6-13)

( 6-14)

malogous

(6-15)

(6-16)

(6-17)

( 6-18)

( 6-19)

(6-20)




1 = et + p + T

l=et+p+T (6-21)







G= 7S0 W 2

Irradiation m



F(~) =0985SS H2





m
















es

~atio alE

300 004 022 100 647 068 029 014 078





Ecliptic

Polar



N

E

S Horizon




360(284 + 17)]Os = 2345deg sin [ 365 (6-22)




5 u -IS Q



-1 I

r -fIL-i

- L I

I - - --


0 30 60 90 120 150 180 210 240 270 300 330 360 390 Julian Date








cos a



tan L




ST + 1 hour = DST (6-27)



with

B n

360middot (n - 81 ) 1T =

364180 (6-29)



(6-23)

(6-24)



(6-25)



(6-26)

(6-27)


(6-28)

(6-29)

ng day 1)

Solution







20 I 15 -- r 1 +- -

) ~ 10l

secth

[- 15 ~ ft__ 1

~

h 02 ti ltl) -5

I - I

tt 0 u -10 ~ J

- l 1 -Jshy-15 I

-20 0 30 60 90 120 150 180 210 240 270 300 330 360 390

Julian Date






DST = 100 pm + 1 h - 4(90 - 88789) min - (-295 min)

= 15810 pm



orien t the r



(6-7)


(6-8)

T) is the ~en inte-

~cent



seT) = E(T)

(6-9)




l middotr-------------------~-------~---~~---~

+ -

3

- _---7-l--t

- -t----

-T-r-t-- I

09 I 1 1 ~ I~ ~~ - -- A ----7=- --- r r------shy- - --shy

ct 06 Imiddots


~ 03 Q)

02 + _ --__shy01 I

C l 1--+ - ----middot---+shy

I0

0 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Wavelength microns



(6-10)





5

etermine


( 6-10)

A2T )] (6-11)

vior the ackbody




C2 J -6F - =593352xI0( n1

F(~J = 063363n2

E = 060991


Ts = 500middotK 3E = 216137x 10 W 2 m


m




Gabsorbed

Gtransmitted







100






GA(A)


( Xi GAdA 10






II is used

(6-12)

(6-13)

( 6-14)

malogous

(6-15)

(6-16)

(6-17)

( 6-18)

( 6-19)

(6-20)




1 = et + p + T

l=et+p+T (6-21)







G= 7S0 W 2

Irradiation m



F(~) =0985SS H2





m
















es

~atio alE

300 004 022 100 647 068 029 014 078





Ecliptic

Polar



N

E

S Horizon




360(284 + 17)]Os = 2345deg sin [ 365 (6-22)




5 u -IS Q



-1 I

r -fIL-i

- L I

I - - --


0 30 60 90 120 150 180 210 240 270 300 330 360 390 Julian Date








cos a



tan L




ST + 1 hour = DST (6-27)



with

B n

360middot (n - 81 ) 1T =

364180 (6-29)



(6-23)

(6-24)



(6-25)



(6-26)

(6-27)


(6-28)

(6-29)

ng day 1)

Solution







20 I 15 -- r 1 +- -

) ~ 10l

secth

[- 15 ~ ft__ 1

~

h 02 ti ltl) -5

I - I

tt 0 u -10 ~ J

- l 1 -Jshy-15 I

-20 0 30 60 90 120 150 180 210 240 270 300 330 360 390

Julian Date






DST = 100 pm + 1 h - 4(90 - 88789) min - (-295 min)

= 15810 pm



l middotr-------------------~-------~---~~---~

+ -

3

- _---7-l--t

- -t----

-T-r-t-- I

09 I 1 1 ~ I~ ~~ - -- A ----7=- --- r r------shy- - --shy

ct 06 Imiddots


~ 03 Q)

02 + _ --__shy01 I

C l 1--+ - ----middot---+shy

I0

0 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Wavelength microns



(6-10)





5

etermine


( 6-10)

A2T )] (6-11)

vior the ackbody




C2 J -6F - =593352xI0( n1

F(~J = 063363n2

E = 060991


Ts = 500middotK 3E = 216137x 10 W 2 m


m




Gabsorbed

Gtransmitted







100






GA(A)


( Xi GAdA 10






II is used

(6-12)

(6-13)

( 6-14)

malogous

(6-15)

(6-16)

(6-17)

( 6-18)

( 6-19)

(6-20)




1 = et + p + T

l=et+p+T (6-21)







G= 7S0 W 2

Irradiation m



F(~) =0985SS H2





m
















es

~atio alE

300 004 022 100 647 068 029 014 078





Ecliptic

Polar



N

E

S Horizon




360(284 + 17)]Os = 2345deg sin [ 365 (6-22)




5 u -IS Q



-1 I

r -fIL-i

- L I

I - - --


0 30 60 90 120 150 180 210 240 270 300 330 360 390 Julian Date








cos a



tan L




ST + 1 hour = DST (6-27)



with

B n

360middot (n - 81 ) 1T =

364180 (6-29)



(6-23)

(6-24)



(6-25)



(6-26)

(6-27)


(6-28)

(6-29)

ng day 1)

Solution







20 I 15 -- r 1 +- -

) ~ 10l

secth

[- 15 ~ ft__ 1

~

h 02 ti ltl) -5

I - I

tt 0 u -10 ~ J

- l 1 -Jshy-15 I

-20 0 30 60 90 120 150 180 210 240 270 300 330 360 390

Julian Date






DST = 100 pm + 1 h - 4(90 - 88789) min - (-295 min)

= 15810 pm

5

etermine


( 6-10)

A2T )] (6-11)

vior the ackbody




C2 J -6F - =593352xI0( n1

F(~J = 063363n2

E = 060991


Ts = 500middotK 3E = 216137x 10 W 2 m


m




Gabsorbed

Gtransmitted







100






GA(A)


( Xi GAdA 10






II is used

(6-12)

(6-13)

( 6-14)

malogous

(6-15)

(6-16)

(6-17)

( 6-18)

( 6-19)

(6-20)




1 = et + p + T

l=et+p+T (6-21)







G= 7S0 W 2

Irradiation m



F(~) =0985SS H2





m
















es

~atio alE

300 004 022 100 647 068 029 014 078





Ecliptic

Polar



N

E

S Horizon




360(284 + 17)]Os = 2345deg sin [ 365 (6-22)




5 u -IS Q



-1 I

r -fIL-i

- L I

I - - --


0 30 60 90 120 150 180 210 240 270 300 330 360 390 Julian Date








cos a



tan L




ST + 1 hour = DST (6-27)



with

B n

360middot (n - 81 ) 1T =

364180 (6-29)



(6-23)

(6-24)



(6-25)



(6-26)

(6-27)


(6-28)

(6-29)

ng day 1)

Solution







20 I 15 -- r 1 +- -

) ~ 10l

secth

[- 15 ~ ft__ 1

~

h 02 ti ltl) -5

I - I

tt 0 u -10 ~ J

- l 1 -Jshy-15 I

-20 0 30 60 90 120 150 180 210 240 270 300 330 360 390

Julian Date






DST = 100 pm + 1 h - 4(90 - 88789) min - (-295 min)

= 15810 pm






100






GA(A)


( Xi GAdA 10






II is used

(6-12)

(6-13)

( 6-14)

malogous

(6-15)

(6-16)

(6-17)

( 6-18)

( 6-19)

(6-20)




1 = et + p + T

l=et+p+T (6-21)







G= 7S0 W 2

Irradiation m



F(~) =0985SS H2





m
















es

~atio alE

300 004 022 100 647 068 029 014 078





Ecliptic

Polar



N

E

S Horizon




360(284 + 17)]Os = 2345deg sin [ 365 (6-22)




5 u -IS Q



-1 I

r -fIL-i

- L I

I - - --


0 30 60 90 120 150 180 210 240 270 300 330 360 390 Julian Date








cos a



tan L




ST + 1 hour = DST (6-27)



with

B n

360middot (n - 81 ) 1T =

364180 (6-29)



(6-23)

(6-24)



(6-25)



(6-26)

(6-27)


(6-28)

(6-29)

ng day 1)

Solution







20 I 15 -- r 1 +- -

) ~ 10l

secth

[- 15 ~ ft__ 1

~

h 02 ti ltl) -5

I - I

tt 0 u -10 ~ J

- l 1 -Jshy-15 I

-20 0 30 60 90 120 150 180 210 240 270 300 330 360 390

Julian Date






DST = 100 pm + 1 h - 4(90 - 88789) min - (-295 min)

= 15810 pm


II is used

(6-12)

(6-13)

( 6-14)

malogous

(6-15)

(6-16)

(6-17)

( 6-18)

( 6-19)

(6-20)




1 = et + p + T

l=et+p+T (6-21)







G= 7S0 W 2

Irradiation m



F(~) =0985SS H2





m
















es

~atio alE

300 004 022 100 647 068 029 014 078





Ecliptic

Polar



N

E

S Horizon




360(284 + 17)]Os = 2345deg sin [ 365 (6-22)




5 u -IS Q



-1 I

r -fIL-i

- L I

I - - --


0 30 60 90 120 150 180 210 240 270 300 330 360 390 Julian Date








cos a



tan L




ST + 1 hour = DST (6-27)



with

B n

360middot (n - 81 ) 1T =

364180 (6-29)



(6-23)

(6-24)



(6-25)



(6-26)

(6-27)


(6-28)

(6-29)

ng day 1)

Solution







20 I 15 -- r 1 +- -

) ~ 10l

secth

[- 15 ~ ft__ 1

~

h 02 ti ltl) -5

I - I

tt 0 u -10 ~ J

- l 1 -Jshy-15 I

-20 0 30 60 90 120 150 180 210 240 270 300 330 360 390

Julian Date






DST = 100 pm + 1 h - 4(90 - 88789) min - (-295 min)

= 15810 pm














es

~atio alE

300 004 022 100 647 068 029 014 078





Ecliptic

Polar



N

E

S Horizon




360(284 + 17)]Os = 2345deg sin [ 365 (6-22)




5 u -IS Q



-1 I

r -fIL-i

- L I

I - - --


0 30 60 90 120 150 180 210 240 270 300 330 360 390 Julian Date








cos a



tan L




ST + 1 hour = DST (6-27)



with

B n

360middot (n - 81 ) 1T =

364180 (6-29)



(6-23)

(6-24)



(6-25)



(6-26)

(6-27)


(6-28)

(6-29)

ng day 1)

Solution







20 I 15 -- r 1 +- -

) ~ 10l

secth

[- 15 ~ ft__ 1

~

h 02 ti ltl) -5

I - I

tt 0 u -10 ~ J

- l 1 -Jshy-15 I

-20 0 30 60 90 120 150 180 210 240 270 300 330 360 390

Julian Date






DST = 100 pm + 1 h - 4(90 - 88789) min - (-295 min)

= 15810 pm





es

~atio alE

300 004 022 100 647 068 029 014 078





Ecliptic

Polar



N

E

S Horizon




360(284 + 17)]Os = 2345deg sin [ 365 (6-22)




5 u -IS Q



-1 I

r -fIL-i

- L I

I - - --


0 30 60 90 120 150 180 210 240 270 300 330 360 390 Julian Date








cos a



tan L




ST + 1 hour = DST (6-27)



with

B n

360middot (n - 81 ) 1T =

364180 (6-29)



(6-23)

(6-24)



(6-25)



(6-26)

(6-27)


(6-28)

(6-29)

ng day 1)

Solution







20 I 15 -- r 1 +- -

) ~ 10l

secth

[- 15 ~ ft__ 1

~

h 02 ti ltl) -5

I - I

tt 0 u -10 ~ J

- l 1 -Jshy-15 I

-20 0 30 60 90 120 150 180 210 240 270 300 330 360 390

Julian Date






DST = 100 pm + 1 h - 4(90 - 88789) min - (-295 min)

= 15810 pm


N

E

S Horizon




360(284 + 17)]Os = 2345deg sin [ 365 (6-22)




5 u -IS Q



-1 I

r -fIL-i

- L I

I - - --


0 30 60 90 120 150 180 210 240 270 300 330 360 390 Julian Date








cos a



tan L




ST + 1 hour = DST (6-27)



with

B n

360middot (n - 81 ) 1T =

364180 (6-29)



(6-23)

(6-24)



(6-25)



(6-26)

(6-27)


(6-28)

(6-29)

ng day 1)

Solution







20 I 15 -- r 1 +- -

) ~ 10l

secth

[- 15 ~ ft__ 1

~

h 02 ti ltl) -5

I - I

tt 0 u -10 ~ J

- l 1 -Jshy-15 I

-20 0 30 60 90 120 150 180 210 240 270 300 330 360 390

Julian Date






DST = 100 pm + 1 h - 4(90 - 88789) min - (-295 min)

= 15810 pm




5 u -IS Q



-1 I

r -fIL-i

- L I

I - - --


0 30 60 90 120 150 180 210 240 270 300 330 360 390 Julian Date








cos a



tan L




ST + 1 hour = DST (6-27)



with

B n

360middot (n - 81 ) 1T =

364180 (6-29)



(6-23)

(6-24)



(6-25)



(6-26)

(6-27)


(6-28)

(6-29)

ng day 1)

Solution







20 I 15 -- r 1 +- -

) ~ 10l

secth

[- 15 ~ ft__ 1

~

h 02 ti ltl) -5

I - I

tt 0 u -10 ~ J

- l 1 -Jshy-15 I

-20 0 30 60 90 120 150 180 210 240 270 300 330 360 390

Julian Date






DST = 100 pm + 1 h - 4(90 - 88789) min - (-295 min)

= 15810 pm







cos a



tan L




ST + 1 hour = DST (6-27)



with

B n

360middot (n - 81 ) 1T =

364180 (6-29)



(6-23)

(6-24)



(6-25)



(6-26)

(6-27)


(6-28)

(6-29)

ng day 1)

Solution







20 I 15 -- r 1 +- -

) ~ 10l

secth

[- 15 ~ ft__ 1

~

h 02 ti ltl) -5

I - I

tt 0 u -10 ~ J

- l 1 -Jshy-15 I

-20 0 30 60 90 120 150 180 210 240 270 300 330 360 390

Julian Date






DST = 100 pm + 1 h - 4(90 - 88789) min - (-295 min)

= 15810 pm


(6-23)

(6-24)



(6-25)



(6-26)

(6-27)


(6-28)

(6-29)

ng day 1)

Solution







20 I 15 -- r 1 +- -

) ~ 10l

secth

[- 15 ~ ft__ 1

~

h 02 ti ltl) -5

I - I

tt 0 u -10 ~ J

- l 1 -Jshy-15 I

-20 0 30 60 90 120 150 180 210 240 270 300 330 360 390

Julian Date






DST = 100 pm + 1 h - 4(90 - 88789) min - (-295 min)

= 15810 pm

Date post:	05-Jan-2020
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