EGEE 437: HWK #2. Brownson. Yaqdaan Alkayyoomi, Mackenzie Ailes, Nicholas Minutillo, Sheel Vora. Group 17. Due on Thursday, Feb.

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1 EGEE 437: HWK #2 Group 17 Due on Thursday, Feb. 18, 2016 Brownson Yaqdaan Alkayyoomi, Mackenzie Ailes, Nicholas Minutillo, Sheel Vora

2 Contents Problem Problem Problem Problem Problem Problem Problem Problem Problem Problem Problem Problem Problem Problem

3 Group 17 EGEE 437 (Brownson): HWK #2 PROBLEM 5.1 Problem 5.1 What is the sky dome in solar energy? The sky dome is all of the components of light interaction contained in the hemisphere enclosing the terrestrial surface of Earth. It is the primary driver of short-wave scattering and it can be both an absorber and a source of longwave irradiation. The semi-spherical sky dome can be projected onto a flat surface for analysis. Page 2 of 21

4 Group 17 EGEE 437 (Brownson): HWK #2 PROBLEM 5.2 Problem 5.2 Describe the difference between a meteorological air mass and the empirical air mass used to calibrate short-term performance of solar technologies (AM1.5). An air mass can be defined in a meteorological sense as a large body of interacting particles that has homogenous physical characteristics in pressure, chemistry and temperature. These properties are established while the air mass was situated a specific source and can dynamically change as it subsequently moves away. On the other hand, solar energy engineers define air mass as AM = 1 cos(θ z ) where z is the zenith angle in degrees, when using it to calibrate the short-term performance of solar technologies. This describes the direct optical length light must take to travel through the Earth s atmosphere. Here, air mass can be used to describe the solar spectrum after the effects of the atmosphere and characterizes how light is attenuated as a result. Page 3 of 21

5 Group 17 EGEE 437 (Brownson): HWK #2 PROBLEM 5.3 Problem 5.3 How does one calculate AM0 (extraterrestrial irradiance)? The equation to calculate AM0 in W/m 2 is provided in the SECS textbook, p. 203 (8.1): G 0 = G sc [ cos ( )] 360n [sin φ sin δ + cos φ cos δ cos ω] 365 where G sc is the solar constant, n is the day, φ is the latitude, δ is the declination and ω is the hour angle. Page 4 of 21

6 Group 17 EGEE 437 (Brownson): HWK #2 PROBLEM 5.4 Problem 5.4 What is the meaning of clear sky in solar energy conversion? included and excluded from a clear sky calculation. Specify what parameters are The clear sky is used as a baseline condition, to which the effects of clouds and aerosols can further decrease the solar budget by. Parameters that are included in the calculation of a clear sky model include: Surface pressure Zenith angle (θ z ) Precipitable water Reduced ozone and NO 2 Solar constant (G 0 ) Ground albedo (ρ g ) Scattering factors of Ångström s exponents of scattering Parameters that are excluded from a clear sky model include the effects of clouds, trees, terrestrial landscape topology and the effects of urban shading. Page 5 of 21

7 Group 17 EGEE 437 (Brownson): HWK #2 PROBLEM 6.1 Problem 6.1 Assign definitions, correct symbols, and sign convention to the following terms. Angle of Incidence. Solar Azimuth. Collector Azimuth. Solar Altitude. Zenith Angle. Declination. Hour Angle. Angle of Incidence: the angle between a ray and the surface vector (θ). Solar Azimuth: the angle defining the location of the Sun in relation to an observer, located between 0 and 360. The angle is clockwise relative to the northernmost point of the horizon (γ s ). Collector Azimuth: the angle defining the location of the Sun in relation to a SECS, located between 0 and 360. The angle is clockwise relative to the northernmost point of the horizon (γ). Solar Altitude: the angle between the horizon and line existing between an observer and the center of the Sun (α s ). Zenith Angle: the complement of α s. The angle between a vector perpendicular to the Earth s surface and ray of the Sun (θ z ). Declination: the observed or apparent angle between the plane of the equator and the plane the Earth follows in orbiting the Sun. Independent of the observer s location on Earth (δ). Hour Angle: the celestial angular value of the Sun with respect to a meridian on Earth. The local hour angle is 0 at solar noon. Prior to noon is -15 /hr and after noon is +15 /hr (ω). Page 6 of 21

8 Group 17 EGEE 437 (Brownson): HWK #2 PROBLEM 6.2 Problem 6.2 Calculate the solar declination (δ) for the spring and fall Equinoxes and the summer and winter Solstices. [Extra] estimate the error between the calculation and the defined declinations for each event Dates Equinox: Solstice: March 20th (spring), September 22nd (fall) June 20th (summer), December 21st (winter) n values n SE = = 79 n F E = = 265 n SS = = 171 n W S = = 355 The values of δ can be found using the equation δ = sin[ (284+n)]. The problem is solved using the following scilab code: //N values n se = 79 n fe = 265 n ss = 171 n ws = 355 delta se = 23.45*(sind((360/365)*(284+n se))) delta fe = 23.45*(sind((360/365)*(284+n fe))) delta ss = 23.45*(sind((360/365)*(284+n ss))) delta ws = 23.45*(sind((360/365)*(284+n ws))) δ SE = δ F E = δ SS = δ W S = Problem 6.2 continued on next page... Page 7 of 21

9 Group 17 EGEE 437 (Brownson): HWK #2 PROBLEM 6.2 Extra We can then compare these values with the known actual values to compute our error: Spring equinox δ = 0 Fall equinox δ = 0 Summer solstice δ = Winter solstice δ = Error = Calculated-Actual E δse = 0.81 E δf E = 0.61 E δss = 0 E δw S = 0 Page 8 of 21

10 Group 17 EGEE 437 (Brownson): HWK #2 PROBLEM 6.3 Problem 6.3 Calculate the sunrise and sunset times (standard time) for four days in Paris: the spring and fall Equinoxes and the summer and winter Solstices, given 49 N latitude (φ) and 2 E longitude (λ). Do not use daylight savings in any of these solutions. The following equations were used to calculate the specified times along with the δ values obtained from Problem 6.3: t sol = ω ss hr ω ss = cos 1 ( tan(φ) tan(δ)) The following scilab code was used to solve the problem: //Variables phi = 49 // Angle of latitude in degrees delta se = delta fe = delta ss = delta ws = // Hour angle omega se = acosd(-tand(phi)*tand(delta se)) omega fe = acosd(-tand(phi)*tand(delta fe)) omega ss = acosd(-tand(phi)*tand(delta ss)) omega ws= acosd(-tand(phi)*tand(delta ws)) // Solar times of Sunrise tsr ss = -omega se*(1/15)+12 tsr fe = -omega fe*(1/15)+12 tsr ss = -omega ss*(1/15)+12 tsr ws = -omega ws*(1/15)+12 // Solar times of sunset tss se = omega spring*(1/15)+12 tss fe = omega fall*(1/15)+12 tss ss = omega summer*(1/15)+12 tss ws = omega winter*(1/15)+12 Problem 6.3 [] continued on next page... Page 9 of 21

11 Group 17 EGEE 437 (Brownson): HWK #2 PROBLEM 6.3 Sunrise Spring equinox: 6:04 am Fall equinox: 6:03 am Summer solstice: 4:00 am Winter solstice: 8:00 am Sunset Spring equinox: 7:56 pm Fall equinox: 6:55 pm Summer solstice: 8:00 pm Winter solstice: 4:00 pm Page 10 of 21

12 Group 17 EGEE 437 (Brownson): HWK #2 PROBLEM 6.4 Problem 6.4 Calculate the day lengths for four days in Paris: the spring and fall Equinoxes and the summer and winter Solstices, given 49 N latitude (φ) and 2 E longitude (λ). Do not use daylight savings in any of these solutions. The following equations were used to calculate the specified day lengths: ω ss = cos 1 (tan(φ) tan(δ)) Length Day = 2 ω ss ( 1h 15 ) Along with the ω values obtained in the last problem (6.3), the following scilab script was used: // Day lengths DL se = 2* -omega se*(1/15) DL fe = 2* -omega fe*(1/15) DL ss = 2* -omega ss*(1/15) DL ws = 2* -omega ws*(1/15) Spring Equinox: hrs. Fall Equinox: hrs. Summer Solstice: hrs. Winter Solstice: 8.00 hrs. Page 11 of 21

13 Group 17 EGEE 437 (Brownson): HWK #2 PROBLEM 6.5 Problem 6.5 Determine the solar altitude and azimuth angles at 14h00 local time (watch time) for Fargo, North Dakota on August 13. Assume daylight savings is in effect (DST). The following equations were used to calculate the specified angles: δ = sin( 360 (284 + n)) 365 ω = 15 h (t sol 12h) B = (360 )( n ) E t = 229.2( ) ( cos B sin B) 229.2( cos 2B sin 2B) t sol = t std (±4(λ std λloc) + E t 60) α s = sin 1 (sin(φ) sin(δ) + cos(φ) cos(δ) cos(ω)) θ z = cos 1 (sin(φ) sin(δ) + cos(φ) cos(δ) cos(ω)) γ s = sin(ω) cos 1 cos(θ z sin(φ) sin(δ)) sin(θ z cos(φ)) Fargo, ND is located at N (φ) and W (λ loc ) 1. Using these coordinates and the following scilab script, the answer was calculated. 1 according to google.com //Variables n = 225 t std = 14 phi = //latitude lambda = //longitude UTC = -6 // in Hours long std = UTC*15 // B constant B=360*((n-1)/365); // E t equation E t=229.2*( )+229.2*( *cosd(b) *sind(b)) *( *cosd(2*B) *sind(2*B)); // Cacluating t sol t lambda=4*(long std - lambda) TC=(E t+t lambda)/60 tsol=(t std+tc-1) // Get the angles now delta=23.45*(sind((360/365)*(284+n))) omega=(tsol-12)*15 Problem 6.5 [] continued on next page... Page 12 of 21

14 Group 17 EGEE 437 (Brownson): HWK #2 PROBLEM 6.5 //alpha s alpha s=asind(sind(phi)*sind(delta)+cosd(phi)*cosd(delta)*cosd(omega)) theta z=acosd(sind(phi)*sind(delta)+cosd(phi)*cosd(delta)*cosd(omega)) //gamma s gamma s=acosd((cosd(theta z)*sind(phi)-sind(delta))/(sind(theta z)*cosd(phi))) Solar altitude (α s ) = Solar azimuth (γ s ) = Page 13 of 21

15 Group 17 EGEE 437 (Brownson): HWK #2 PROBLEM 6.6 Problem 6.6 Calculate three condition for the day of June 9, at 10h30 solar time in Athens, Greece: (a) the solar zenith and azimuth angles, (b) the sunrise and sunset times, and (c) the day length. a The latitude of Athens, Greece is located at N (φ) and the longitude is located at E (λ loc ) 2. The following equations were used to find the specified angles: n = = 160 δ = sin( 360 (284 + n)) 365 ω = 15 h (t sol 12h) θ z = cos 1 (sin(φ) sin(δ) + cos(φ) cos(δ) cos(ω)) γ s = sin(ω) cos 1 cos(θ z sin(φ) sin(δ)) sin(θ z cos(φ)) These equations are also used in the scilab script form problem 6.5. The same code was used for this problem, with the new solar time, n, φ and λ values as inputs. Solar zenith (θ z ) = Solar azimuth (γ s ) = b The following equations are used to find the specified times: ω = cos 1 (tan(φ) tan(δ)) t sol = ω h These equations were used in the following scliab script to compute the answer: 2 According to google.com //Variables phi = n = 160 //Now find the times using declination angle and phi delta=23.45*(sind((360/365)*(284+n))) omega = acosd(-tand(phi)*tand(delta)) tsr = -omega*(1/15)+12 tss = omega*(1/15)+12 Problem 6.6 [b] continued on next page... Page 14 of 21

16 Group 17 EGEE 437 (Brownson): HWK #2 PROBLEM 6.6 Sunrise : 4:43am Sunset : 7:17pm c Using the equation of L day = 2 ω ss ( 1h 15 ), the day length can be found: Day Length : hours Page 15 of 21

17 Group 17 EGEE 437 (Brownson): HWK #2 PROBLEM 6.7 Problem 6.7 Repeated calculations for two days and two times: Match 15 and September 15, each at 10h00 and 14h30 solar time. Solve for the following two conditions given the locale of the borrough of Brooklyn in New York City, NY.: (a) the solar azimuth (γ s ) and altitude (α s ) angles at the appointed times, and (b) the sunrise and sunset times. a The latitude of Brooklyn, NYC, NY is N and the longitude is located at W (λ loc ) 3. Using t sol1 = 10 and t sol2 = 14.5, the following equations were used: n 1 = = 74 n 2 = = 258 δ = sin( 360 (284 + n)) 365 ω = 15 h (t sol 12h) α s = sin 1 (sin(φ) sin(δ) + cos(φ) cos(δ) cos(ω)) θ z = cos 1 (sin(φ) sin(δ) + cos(φ) cos(δ) cos(ω)) γ s = sin(ω) cos 1 cos(θ z sin(φ) sin(δ)) sin(θ z cos(φ)) The new φ and n values were used as input for the scilab script from problem 6.5. Solar altitude (10h00) (α s ) = Solar azimuth (10h00) (γ s ) = Solar altitude (14h30) (α s ) = Solar azimuth (14h30) (γ s ) = Solar altitude (10h00) (α s ) = Solar azimuth (10h00) (γ s ) = Solar altitude (14h30) (α s ) = Solar azimuth (14h30) (γ s ) = March 15 September 15 3 according to google.com Problem 6.7 continued on next page... Page 16 of 21

18 Group 17 EGEE 437 (Brownson): HWK #2 PROBLEM 6.7 b The following equations are used to find the specified times using the scilab script from Problem 6.6 (b): Sunrise : 6:09 am Sunset : 5:50 pm Sunrise : 5:52 am Sunset : 6:08 pm ω = cos 1 (tan(φ) tan(δ)) t sol = ω h March 15 September 15 Page 17 of 21

19 Group 17 EGEE 437 (Brownson): HWK #2 PROBLEM 6.12 Problem 6.12 A flat plate collector in Pittsburgh, PA, is tilted at 34 from horizontal and pointed 5 West of South. Calculate the angle of incidence (θ) on the collector at 10:30am and 2:30pm solar times for both March 15 and September 15. The latitude and longitude of Pittsburgh, PA are N and W 4.Using t sol1 = 10 and t sol2 = 14.5, the following equations were used to calculate the angle of incidence (θ) on the collector: n 1 = = 74 n 2 = = 258 δ = sin( 360 (284 + n)) 365 ω = 15 h (t sol 12h) θ = cos 1 (sin(φ) sin(δ) sin(β) cos(φ) cos(δ) cos(β) cos(γ) + cos(φ) cos(δ) cos(β) cos(γ) + sin(φ) cos(δ) sin(β) cos(γ) cos(ω) + cos(δ) sin(β) sin(ω) sin(γ)) The following scilab script was used to solve this problem. changed for each given date and time. The n and t values were beta= 34 gamma=5 phi= ; //latitude n=74; //change n value here delta=23.45*sind((360/365)*(284+n)); //declination angle t=10.5 //change time value here omega=15*(t-12) costheta = sind(phi) * sind(delta) * cosd(beta) - cosd(phi) * sind(delta) *.. sind(beta)* cosd(gamma) + cosd(phi) * cosd(delta) * cosd(beta) *.. cosd(omega) + sind(phi) * cosd(delta) * sind(beta) * cosd(gamma) *.. cosd(omega) + cosd(delta) * sind(beta) * sind(omega) * sind(gamma); theta = acosd(costheta); 4 according to google.com Problem 6.12 [] continued on next page... Page 18 of 21

20 Group 17 EGEE 437 (Brownson): HWK #2 PROBLEM 6.12 March 15 10:30am (θ) : :30pm (θ) : September 15 10:30am (θ) : :30pm (θ) : Page 19 of 21

21 Group 17 EGEE 437 (Brownson): HWK #2 PROBLEM 7.1 Problem 7.1 Plot an orthographic projection of a southeasterly tall building interfering with a first floor window of interest, seen in Figure 7.9 (problem adapted To plot the projections, the following calculations were used: α X A = tan 1 9 ( ) = α X B = tan 1 ( 9 8 ) = 48 α X C = tan 1 ( 9 80 ) = 45.2 γ X A = tan 1 ( 7 8 ) 25 = 66 γ X B = 0 25 = 25 γ X C = tan 1 ( 4 8 ) 25 = 1.5 Figure 1: Orthographic projection of a southeasterly tall building Page 20 of 21

22 Group 17 EGEE 437 (Brownson): HWK #2 PROBLEM 7.2 Problem 7.2 Plot a polar projection of the same scenario (using the same critical points). Figure 2: Polar projection of a southeasterly tall building Page 21 of 21

Observer-Sun Angles. ), Solar altitude angle (α s. ) and solar azimuth angle (γ s )). θ z. = 90 o α s

Observer-Sun Angles. ), Solar altitude angle (α s. ) and solar azimuth angle (γ s )). θ z. = 90 o α s Observer-Sun Angles Direction of Beam Radiation: The geometric relationships between a plane of any particular orientation relative to the earth at any time and the incoming beam solar radiation can be