Solar Energy Analysis HW 1

8/20/2019 Solar Energy Analysis HW 1

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Nicole Marshall MAE 5240 Solar Energy AnalysisHW #1 9 February 2016

15! "alculae he angle o$ inci%ence o$ he bea& solar ra%iaion a 1400 '2(M! solar i&e on February 10 a laiu%e 4))* on sur$aces o$ he $ollo+ingorienaions,

cosθ=sinδsin cosβϕ −sinδcos sinβcosγ ϕ +cosδcos cosβcosωϕ +cosδsin sinβcosγcosωϕ +cosδsinβsinγsinω

n - 41. - 4))*

δ =23.45sin(360 284+n

365 )=−14.90 °

/ - '15*hr!'2 hours a$er noon! - )0*

a! Horional3 - 0* - 0*

cosθ=sin (−14.90 )sin (43.3 ) cos (0 )−sin (−14.90 ) cos (43.3 )sin (0 ) cos (0 )+cos (−14.90 )cos (43.3 ) c

¿ (−0.257 ) (0.686 ) (1)−(−0.257 ) (0.728 ) (0 ) (1 )+(0.966 ) (0.728 ) (1 ) (0.866 )+ (0.966 ) (0.686 ) (0 ) (1) (0.86

¿−0.176−0+0.609+0+0=0.433

θ=cos−1 (0.433 )=64.36 °

b! sloe% o souh a 60*3 - 60* - 0*

cosθ=sin (−14.90 )sin (43.3 ) cos (60 )−sin (−14.90 ) cos (43.3 ) sin (60 ) cos (0 )+cos (−14.90 ) cos (43.3

¿ (−0.257) (0.686 ) (0.5 )−(−0.257 ) (0.728) (0.866 ) (1 )+(0.966 ) (0.728 ) (0.5 ) (0.866 )+(0.966 ) (0.686 ) (0

¿−0.088−(−0.162 )+0.305+0.497+0=0.875

θ=cos−1 (0.875 )=28.90 °

c! Sloe o$ 60* $acing 40* +es o$ souh3 - 60* - 40*

cosθ=sin (−14.90 )sin (43.3 ) cos (60 )−sin (−14.90 ) cos (43.3 ) sin (60 ) cos (40 )+cos (−14.90 )cos (43

¿ (−0.257 ) (0.686 ) (0.5 )−(−0.257 ) (0.728 ) (0.866 )(0.766)+ (0.966 ) (0.728 ) (0.5) (0.866 )+(0.966 ) (0.6

¿−0.088−(−0.124 )+0.305+0.381+0.269=0.990

θ=cos−1 (0.990 )=8.00 °

%! 7erical $acing souh




3 - 90* - 0*

cosθ=sin (−14.90 )sin (43.3 ) cos (90)−sin (−14.90 ) cos (43.3 )sin (90 )cos (0 )+cos (−14.90 )cos (43.3

¿ (−0.257 ) (0.686 ) (0 )−(−0.257 ) (0.728 ) (1 ) (1 )+(0.966 ) (0.728 ) (0 ) (0.866 )+(0.966 ) (0.686 ) (1 ) (1 ) (0.86

¿0− (−0.187 )+0+0.574+0=0.761

θ=cos−1 (0.761 )=40.44 °

e! 7erical $acing +es3 - 90* - 90*

cosθ=sin (−14.90 )sin (43.3 ) cos (90)−sin (−14.90 ) cos (43.3 )sin (90 )cos (90)+cos (−14.90 )cos (43

¿ (−0.257 ) (0.686 ) (0 )−(−0.257 ) (0.728 ) (1 ) (0 )+(0.966 ) (0.728 ) (0 ) (0.866 )+ (0.966 ) (0.686 ) (1 ) (0) (0.86

¿0−0+0+0+0.483=0.483

θ=cos−1 (0.483 )=61.11°

16!a! When i is 2 (M MS8 on February ) in Norh (lae NE ' - 101* W . -

411* N! +ha is he solar i&e:loc - 101*s - 105*

San%ar% i&e - 14,00n - )4

B=(n−1 )(360365)=(34−1 )(360365)=32.55

E=229.2 (0.000075+0.001868 cosB−0.032077 sinB−0.014615cos2B−0.04089sin 2B )

¿229.2 (0.000075+0.001868∗cos (32.55 )−0.032077∗sin (32.55 )−0.014615∗cos (2∗32.55 )−0.0

¿0.017+0.361−3.955−1.411−8.501=−13.489

solar time=standardtime+4

( L

st

− Lloc )+ E=14:00+4 (105−101)+(−13.489 )=14:00+2.5=14:

b! When i is 2 (M MS8 in ;oise <= ' - 116* W . - 4)6* N! onFebruary ) +ha is he solar i&e:

loc - 116*s - 105*

San%ar% i&e - 14,00




n - )4

B=(n−1 )(360365)=(34−1 )(360365)=32.55


¿229.2 (0.000075+0.001868∗cos (32.55 )−0.032077∗sin (32.55 )−0.014615∗cos (2∗32.55 )−0.0

¿0.017+0.361−3.955−1.411−8.501=−13.489

solar time=standardtime+4 ( Lst − Lloc )+ E=14:00+4 (105−116)+(−13.489 )=14:00−44−13.5

c! Wha Easern =ayligh 8i&e correson%s o solar noon on >uly )1 $or(orlan% ME ' - ?05* W . - 4)?*!:

loc -?05*

s - ?5*Solar i&e - 12,00

n - 212

B=(n−1 )(360365)=(212−1)(360365 )=208


¿229.2 (0.000075+0.001868∗cos (208)−0.032077∗sin (208)−0.014615∗cos (2∗208)−0.04089

standard time=solar time−4 ( Lst − Lloc)− E=12:00−4 (75−70.5 )−(−6.55 )=12:00−18+7=12

%! Wha "enral =ayligh 8i&e correson%s o 10,00 AM on >uly )1 $or <ronMounain M< ' - 90* W . - 45@* N!:

loc -90*s - 90*

Solar i&e - 10,00n - 212

B=(n−1 )( 360365 )=(34−1 )( 360365 )=208


¿229.2 (0.000075+0.001868∗cos (208)−0.032077∗sin (208)−0.014615∗cos (2∗208)−0.04089

standard time=solar time−4 ( L st − Lloc )− E=10:00−4 (90−90)−(−6.55 )=10:00−0+7=12:00




1?! =eer&ine he sunse hour angle an% %ay lengh $or Ma%ison an% $orMia&i $or he $ollo+ing %aes,cosωs=−tan tanδ ϕ

ϕ Madison=43.3 °

ϕ Miami=25.8 °

a! >anuary 1n - 1

δ =23.45sin(360 284+n

365 )=−23.01°

Ma%ison,cosωs=−tan (43.3 )∗tan (−23.01 )=0.40

ωs=cos−1 (0.40 )=66.41°

day length=ωs∗2

15° =

(66.41° )∗2

15° =8.85hours

Mia&i,cosωs=−tan (25.8 )∗tan (−23.01 )=0.21

ωs=cos−1 (0.21 )=78.15°

day length=ωs∗2

15°

=(78.15° )∗2

15°

=10.42hours

b! March 22n - @1

δ =23.45sin(360 284+n

365 )0°

Ma%ison,cosωs=−tan (43.3 )∗tan(0)=0

ωs=cos−1 (0 )=90 °

day length=ωs∗2

15° =

(90° )∗2

15° =12hours

Mia&i,




cosωs=−tan (25.8 )∗tan (0 )=0

ωs=cos−1 (0 )=90 °

day length=ωs∗2

15°

=(90° )∗2

15°

=12hours

c! >uly 1n - 1@2

δ =23.45sin(360 284+n

365 )=23.12 °

Ma%ison,cosωs=−tan (43.3 )∗tan (23.12 )=−0.40

ωs=cos−1 (−0.40 )=113.72°

day length=ωs∗2

15° =

(113.72° )∗2

15° =15.16hours

Mia&i,cosωs=−tan (25.8 )∗tan (23.12 )=−0.21

ωs=cos−1 (−0.21 )=101.91 °

day length=ωs∗2

15° =

(101.91° )∗2

15° =13.59hours

%! Mean %ay o$ Februaryn -4?

δ =23.45sin(360 284+n

365 )=−12.95°

Ma%ison,cosωs=−tan (43.3 )∗tan (−12.95 )=0.22

ωs=cos−1 (0.22 )=77.48°

day length=ωs∗2

15 ° =

(77.48° )∗215 °

=10.33hours

Mia&i,cosωs=−tan (25.8 )∗tan (−12.95 )=0.11

ωs=cos−1 (0.11 )=83.62 °




day length=ωs∗2

15° =

(83.62° )∗2

15° =11.15hours

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Solar Energy Analysis HW 1

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