A canopy has a vertical (cylindrical) leaf angle distribution and leaf area index of 4. When the solar elevation is 45◦ and Sb above the canopy is 500 W m^−2, calculate, for surfaces at the base of the canopy, (i) the mean direct solar irradiance per unit surface area, (ii) the fractional area of

Question

A canopy has a vertical (cylindrical) leaf angle distribution and leaf area index of 4. When the solar elevation is 45° and [latex]S_{b}[/latex] above the canopy is 500 W [latex]m^{-2}[/latex] , calculate, for surfaces at the base of the canopy, (i) the mean direct solar irradiance per unit surface area, (ii) the fractional area of sunflecks, (iii) the mean irradiance of sunlit leaves.

Accepted Answer

[latex]\mathcal{K} _{s} = 2\left(\cot \beta ight) / \pi = 0.637[/latex]

i. [latex]S_{b} (L) =500\> exp \> (-0.637 imes 4) = 39.1\; W \; m^{-2}[/latex].

ii. Fractional area of sunflecks is [latex]S_{b} (L) / S_{b} (0) = 39.1 / 500 = 0.078[/latex].

iii. Mean irradiance per unit leaf area is [latex]\mathcal{K} _{s} S_{b} (L) = 0.637 imes 39.1 = 24.9\; W\; m^{-2} [/latex].

Question 8.1: A canopy has a vertical (cylindrical) leaf angle distributio...

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