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Q. 8.5

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.

Verified Solution

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.