the total clear sky irradiation of a surface tilted at an angle of 60° from the horizontal located at ME, on July 21 at 2:00 P.M. solar time. The latitude for the given location is 68.02° N. The surface outhwest. Neglect reflected radiation. Use Gae = CGND Fws for diffuse solar radiation. o = 252 Btu/hr-ft", Gae = 26.58 Btu/hr-ft", Total irradiation = 278.58 Btu/hr-ft

Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)
8th Edition
ISBN:9781305387102
Author:Kreith, Frank; Manglik, Raj M.
Publisher:Kreith, Frank; Manglik, Raj M.
Chapter11: Heat Transfer By Radiation
Section: Chapter Questions
Problem 11.66P
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Calculate the total clear sky irradiation of a surface tilted at an angle of 60° from the horizontal located at
Caribou, ME, on July 21 at 2:00 P.M. solar time. The latitude for the given location is 68.02° N. The surface
faces the southwest. Neglect reflected radiation. Use Gae = CGND Fws for diffuse solar radiation.
Answer: G, = 252 Btu/hr-ft', Gae = 26.58 Btu/hr-ft', Total irradiation = 278.58 Btu/hr-ft
%3D
Transcribed Image Text:Calculate the total clear sky irradiation of a surface tilted at an angle of 60° from the horizontal located at Caribou, ME, on July 21 at 2:00 P.M. solar time. The latitude for the given location is 68.02° N. The surface faces the southwest. Neglect reflected radiation. Use Gae = CGND Fws for diffuse solar radiation. Answer: G, = 252 Btu/hr-ft', Gae = 26.58 Btu/hr-ft', Total irradiation = 278.58 Btu/hr-ft %3D
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