Let’s assume that during the Late Glacial Maximum perihelion occurs on the first
ID: 284505 • Letter: L
Question
Let’s assume that during the Late Glacial Maximum perihelion occurs on the first day of summer in the Northern Hemisphere and that Earth’s polar angle is 24.2°. Because of a combination of strong winds (driven by the ice-covered polar regions) and a supply of particulates (loess) generated by glaciers, the atmospheric transmissivity is 0.40. Calculate the irradiance (in Wm–2) at 15°S at Noon on the summer solstice as measured on a horizontal surface.(use the necessary equation(s) from attached sheet) Then, compare your answer above to the contemporary value at the same location and time (Noon, 15°S on the summer solstice), assuming an atmospheric transmissivity of 51% and that aphelion occurs on the summer solstice.
EQUATIONS AND CONSTANTS Inverse Square Law: Eu d E2 d where E irradiance Wm d distance between two bodies Stephan-Boltzmann's Law: For a blackbody: E -OT4 where T temperature (K) Average irradiance for a spherical, rotating planet that is not a blackbody: 4EOT4 where a albedo E emissivity Wien's Law 2897 (uk) MAKE T (K) where NMAX wavelength at which maximum emission of radiation occurs Planck's Law: Cern)-1 where cl and c2 are constants, is wavelength and Tis temperature ce on a horizontal surface (ER: Irradian EH-Eet, siny Eet Cosz where v is the elevation angle (or altitude) of the sun, Zis the sun's zenith angle, is the transmissivity, and E is the solar constant Irradiance of a sphere: W Radiant Flux. [W] Irradiance 4r it [m ml where r is the radius of the emitting body. Speed of light: c wavelength x frequencyExplanation / Answer
Solar irradiance is the power recieved from the sun per unit area in the form of electromagnetic radiation. This irradiance varies throughout the seasons as well as the day depending on position of sun in the sky. The SI unit of irradiance is watt/square meter. Solar irradiance over a period of time is known as the solar insolation. The solar irradiation on the horizontal surface i.e. theta=0 degree can be calculated by the day number and the star constant value
solar irradiation=solar constant(1+0.033*cos360/365*n)
where n= no of days starting from janurary 1st=356
solar constant=1367W/m2
solar irradiation=1367*(1+0.033*cos360/365*356)
=1367(1+0.033*cos(0.98*356)
=1367(1+0.033*cos(0.34)
=1367(1+0.033*0.99)
=1367(1+0.03)
=1367*1.03
=1408.01W/m2
atmospheric transmissivity=0.40
Therefore=0.40/1408
=0.028
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