The Sun Solar Factoids (I) The sun, a medium-size star in the milky way galaxy, consisting of about 300 billion stars. (Sun Earth-Relationships) A gaseous sphere of radius about 695 500 km (about 109 times of Earth radius) => by far the largest object in the solar system Mass: 1.989 * 10 30 kg (99.8% of total mass of solar system) Our Sun Solar Factoids (II) The Sun Sun consists of 3 parts of hydrogen, one part of helium. Proportion changes over time. Sun s energy output is produced in the core of the sun by nuclear reactions (fusion of four hydrogen (H) atoms into one helium (He) atom). Sun is about 4.5 billion years old. Since its birth it has used up about half of the hydrogen in its core. Sufficient fuel remains for the Sun to continue radiating "peacefully" for another 5 billion years (although its luminosity will approximately double over that period), but eventually it will run out of hydrogen fuel. The Sun Solar Factoids (III) The Sun's energy output is 3.84 * 10 17 Gigawatts: (a typical nuclear power plant produces 1 Gigawatt) The outer 500 km of the sun ( photosphere ) emits most of radiation received on Earth Radiation emitted by the photosphere closely approximates that of a blackbody of 5777K
The Sun Emission of Sun Effective surface temperature of the sun: 5778 K => Emission B s (per m ) at the sun surface (Stefan-Boltzman law): B s = T 4 =5.67 10-8 Wm - K -4 *(5778K) 4 = 6.3*10 7 Wm - => Total emission of Sun E TOT : E TOT =4 " r s B s with r s =6.955 *10 8 m= radius of the sun: 4 * 3.14* (6.955 *10 8 m) *6.3 10 7 Wm - =3.84 10 6 W= 3.84 10 17 GigaW cf. World s energy cosumption: 15 TerraW (1.5 10 13 W) Area on Sun surface required to cover world s energy cosumption: 1.5 10 13 W / B s = 1.5 10 13 W / 6.3 10 7 Wm - =.5 10 5 m =0.5 km. =>if we could harvest energy directly on the sun surface, 0.5 km would be sufficient to cover world s energy demands. Binding energy per nucleon in He core: 1.1*10-1 J Energy generated by one fusion reaction combining 4 H nuclei into one He core: 4 *1.1 10-1 J= 4.4 10-1 J Total energy per second emitted by sun: E TOT =3.84*10 6 W (Js -1 ) Number of fusion reactions per second = E TOT / energy generated per fusion reaction= 3.84 10 6 Js -1 / 4.4 10-1 J = 0.9*10 38 s -1 1 proton mass= 1.67*10-7 kg Solar fusion => per fusion reaction 4*1.67*10-7 kg of H is consumed. Total emission E TOT of Sun: E TOT = 4 " r s * B s Total Emission of Sun (in W) distributed over a sphere (in m ) with radius a, where a= Earth-Sun Distance (semi major axis of Earth s orbit, 149.6 * 10 9 m), determines the Solar irradiance S per m at the Top of the Earth s atmosphere (Solar Constant) at distance a : S=1366Wm - a r s Total amount of H consumed in the Sun per second: = number of fusion reactions * amount of H consumed per reaction = 0.9*10 38 s -1* 4*1.67e -7 kg = 6 *10 11 kg= 600 Mio Tons => Every second 600 Mio Tons of H are transformed to He S = 4 " r s B s / (4 " a ) = (r s /a) B s =(6.955*10 8 m / 149.6*10 9 m) *6.3*10 7 Wm - = 1366 Wm - Current best estimate from measurements: 1361 Wm - 5 Wm - deviation may to difference from ideal black body and measurement uncertainties
More generally, if a planet is at distance rp from the sun, then the solar irradiance Sp (in Wm-) onto the planet is: Intensity of solar irradiance decreases with distance according to Inverse square law. Examples: rs =6.955 *108m Bs=6.3*107 Wm- =>c=3.057* 105W Planet Distance from Sun (109 m) Intensity of solar radiation (Wm-) Venus 108 60 Earth 149.6 1366 Mars 8 558 Sun Earth relationships Earth s orbit around the Sun: Earth's orbit is an ellipse and the sun is located in one of its focal points. Definition Ellipse: The sum of the distances from any point on the ellipse to the two focal points is constant (equal x semi major axis a) => Sun Earth-distance r varies during the course of the year Definitions: Perihelion P: point on the orbit which is closest to the Sun Aphelion A: point on the orbit which is farthest from the Sun Eccentricity e: Amount by which orbit deviates from a perfect circle, where 0 is perfectly circular, and 1.0 is a parabola. Ratio of the distance between the foci of the ellipse to the length of the major axis of the ellipse. semi major axis
Definitions: Solar constant S (1361 Wm - ): Solar irradiance obtained per m on a plane perpendicular to the sunbeam at distance a (semi major axis) from the Sun. Earth-Sun distance varies over the course of a year: Insolation S r at distance r: Distance a (semi major axis) sometimes also called S S r S 1 Astronomical Unit (AU) # 150 Mio km semi major axis a a: semi major Axis: a r is a function of time of the year: r(t) Special cases: Earth in Aphel: r=a+ae=a(1+e) Earth in Perihel: r=a-ae=a(1-e) S r (perihel) S r (aphel) = S (1" e) S (1+ e) = S (1+ e) (1+ e) = (1" e) S (1" e) Current e=0.0167: S r (perihel) S r (aphel) (1+ e) (1+ 0.0167) = = (1" e) (1" 0.0167) = 1.07 7% difference in insolation between Perihel and Aphel max. e in Earth history: 0.06: => 7% difference in insolation between Perihel and Aphel
Insolation G(r) received per m on average on the Earth sphere Mean insolation on Earth G P over an entire orbital period P (annual mean insolation) Integrating: yields: Total energy taken out of solar flux by Earth disk: S R *" Total solar energy per m distributed over Earth sphere S R *" / (4 R *") = S/4= 340 Wm - G P = S 4P P " a % S ( $ ' dt = 0 # r(t) & 4 1) e where e is the eccentricity of the elliptic orbit of Earth around the sun Current conditions: e=0.0167: =>e effect:0.00014*340wm - =0. 033 Wm - Maximum eccentricity over past Million years: e=0.06: =>e effect:0.0018*340wm - =0. 61 Wm - term negligible for annual mean calculations => G P =S/4= 340Wm - annual mean insolation Total solar energy received on earth: 4"r G p =4"(6.37 10 6 m) 340*Wm - =1.74 10 17 W (174 PetaW) (174,000,000,000,000,000 J per second from the sun) Compare: 1 average swiss nuclear power plant generates power on the order of 1 GigaW=10 9 W Solar energy incident on Earth compares to about 1.7 x 10 8 nuclear power plants (170 Mio. nuclear power plants) Compare: World s current energy consumption: 15 TeraW (1.5$10 13 W) 10 000 times smaller than solar energy incident on the planet solar energy received within less than one hour would be sufficient to cover one year of World s current energy consumption
Desertec: Solar Power from the Desert www.desertec.org Within 6h deserts receive more energy from the sun than humankind consumes within a year Planetary albedo A: Fraction of reflected solar radiation with respect to incoming solar radiation Mean annual energy G A absorbed by the planet per m on the sphere: A = 0.3 for Earth Effective Temperature In equilibrium, absorbed shortwave energy G A (over the Earth disk) is balanced by longwave emission (over the Earth sphere) according to the Stefan-Boltzman law with an effective temperature T eff : Effective Temperature Effective temperature: (blackbody) temperature at which the emitted longwave equals the absorbed shortwave radiation. If the temperature of a planet is below the effective temperature it will emit less radiation than it absorbs => planet will warm until it reaches radiative equilibrium and effective temperature. if its temperature is above the effective temperature it will cool toward radiative equilibrium by emitting more radiation than it absorbs. planet distance from sun (10 9 m) albedo (1-albedo) T eff (K) Mercury 58 0.06 0.94 44 Venus 108 0.78 0. 7 Earth 150 0.30 0.70 55 Mars 8 0.17 0.83 16 Jupiter 778 0.45 0.55 105
Effective Temperature of Planets Exercices Temperature (K) Distance from sun Surface insolation I (irradiance) at a specific location and time: Earth centered Cartesian coordinate system (x; y; z) Determination of zenith angle % solar zenith angle at that position and time Solar Zenith angle % function of: Time of the day, expressed in hour angle H Latitude & Calender day (season), expressed as declination ' r function of time on Earth orbit. z-axis points to the North pole x-axis in the equatorial plane with sun in the x-z-plane n s points to local zenith at P points to the Sun % is zenith angle at observer point P
Declination ": angle between the direction to the sun and to equatorial plane Determination of zenith angle Hour Angle H: Angle in the equatorial plane between the meridian of the observer P and the direction to the sun projected onto the equatorial plane. Determination of zenith angle Declination varies over the year from +3 7 (1. June) to -3 7 (1. Dec) Hour angle in radiance 0: solar noon Determination of zenith angle Local Zenith angle: Zenith angle at point P: Angle between local zenith n and the direction to the sun s Determination of zenit angle Unit vector 1
Determination of zenit angle Unit vector Determination of zenit angle Unit vector 1 1 sin ' Determination of zenit angle Unit vector Determination of zenit angle Unit vector 1 sin"# 1 "# cos"cosh H
direction to the local zenith: direction to the sun at the location of an observer The astronomical sunrise and sunset, +-H 0, are given for the mathematical horizon at %="/ cos" = cos H cos# cos$ + sin # sin$ with " = % when H = H 0 With the scalar product we obtain the zenith angle % : Fundamental equation for zenith angle % 0 = cos H 0 cos# cos$ + sin # sin$ cos H 0 = & sin # sin$ = & tan # tan$ cos# cos$ where H 0 is defined only for -1 ( cosh 0 ( 1. For cos H 0 > 1, we have the polar night with no sunrise and for cosh 0 <-1 we have the polar day with no sunset. Daily insolation Daily insolation I d at a given location and date is obtained by integrating Daily insolation for the hour angle from sunrise at -t 0 to sunset at t 0. Declination is kept constant during one day. Horizon at a zenith angle of 90, => integral is evaluated from sunrise at t 0 to sunset at t 0, with The integral can be evaluated analytically, where the hour angle H 0 is measured in radian, where H 0 is the hour angle for sunrise at the mathematical horizon. The integral can be evaluated analytically.
H o H o $ cos" dh = $ (cos H cos% cos& + sin % sin&)dh = # H o # H o Daily insolation Daily insolation Mean daily insolation at TOA (in Wm - ) H = sin H cos% cos& + sin % sin& ' H] 0 # H 0 = sin H 0 cos% cos& + sin % sin& ' H 0 # (sin(#h 0 )cos% cos& + sin % sin& ' (#H 0 )) = sin H 0 cos% cos& + sin % sin& ' H 0 # (#sin(h 0 )cos% cos& # sin % sin& ' H 0 ) = (sin H 0 cos% cos& + sin % sin& ' H 0 ) ( I d = S 86400 ) = S 86400 ) * a, - + r. / H o $ cos" dh # H o * a, - / (sin H + r. 0 cos% cos& + sin % sin& ' H 0 )