Question 8.5: Consider a sparse forest canopy with the following character...
Consider a sparse forest canopy with the following characteristics:
Leaf area index L = 1.0.
Theoretical attenuation coefficient \mathcal{K}_{b} = 1.0 (assuming black leaves).
Soil reflection coefficient ρ_{s} = 0.15.
Leaf reflection coefficient (PAR) ρ_{p} = 0.10.
Leaf transmission coefficient (PAR) τ_{p} = 0.10.
Calculate:
a. The canopy attenuation coefficient \mathcal{K} for PAR.
b. The canopy reflection and transmission coefficients (ρ_{c} and τ_{c}) for PAR, assuming that second-order terms can be ignored.
c. The absorption coefficient of the canopy (α_{c}) for PAR
d. The ratio of absorption (α_{c}) to interception (1 − τ_{c}) of PAR by the canopy.
a. \alpha _{p} = 1- \tau _{p} -\rho _{p}
= 1 − 0.10 − 0.10
= 0.80,
\mathcal{K} = \alpha ^{0.5}_{p} \mathcal{K}_{b}= (0.80^{0.5} ) \times 1
= 0.89.
b. Neglecting second-order terms,
\rho _{c} = \rho ^{*}_{c} – (\rho ^{*}_{c} -\rho _{s} )\exp (-2 \mathcal{K}L)and \rho ^{*}_{c} = (1 − \alpha ^{0.5}_{p} )/(1 + \alpha ^{0.5}_{p} ) = 0.056
so \rho _{c} = 0.056 − (0.056 − 0.15) exp (−2 × 0.89 × 1)
= 0.072,
\tau _{c} = \exp (−\mathcal{K}L)= exp (−0.89 × 1)
= 0.41.
c. \alpha _{c} = 1-\rho_{c}-\tau_{c} (1-\rho _{s} )
= 1 − 0.072 − 0.41(1 − 0.15)
= 0.58.
d. Interception is defined as (1 – \tau _{c} ), so \alpha _{c} /(1 − \tau _{c} ) = 0.98. Thus 98% of the PAR radiation intercepted by the sparse canopy is absorbed. If the problem had been set up for total solar radiation, the ratio would have been closer to 0.75.