Interception and depression storage

1.1   Net rainfall

See Figure 6.17 in Dingman (2015).

\begin{equation} R_n=R-E_i \end{equation} where $R_n$ is the net rainfall and $E_i$ is the total interception loss.

\begin{equation} E_i=E_c+E_l \end{equation} where $E_c$ is the canopy interception loss and $E_l$ is the litter interception loss.

\begin{equation} R=R_t+R_s+E_c \end{equation} where $R_t$ is the throughfall and $R_s$ is the stemflow.

Therefore, \begin{equation} R_n=R_t+R_s-E_l \end{equation}

1.2   Determining factors

• Storm characters
  • Intensity
  • Duration
• Wind
• Atmospheric conditions
• Vegetation species, age, and density
• Season of the year

1.3   How much interception?

Usually, 10-20% of the precipitation during the growing season is intercepted and evaporates.

Pronounced under dense closed forest stands as much as 25% of the total annual precipitation.

Average annual interception losses by

• Douglas fir stands in western Oregon and Washington about 24%,
• a 10-year-old loblolly pine plantation in the South approximately 14%,
• Ponderosa pine forests in California about 12%, and
• hardwood stands in the White Mountains of New Hampshire about 13%.

1.4   Interception loss estimation

See Figure 5.1.

\begin{equation} L_i=S+KEt \end{equation} where $L_i$ is the volume of water intercepted (in), $S$ is the interception storage (usually between 0.01 and 0.05 in), $K$ is the ratio of surface area of intercepting leaves to horizontal projection of this area or the Leaf-Area Index (LAI), $E$ is the evaporation rate per hour (in/hr), and $t$ is time (hr).

To account for the rainfall amount $P$ (in), \begin{equation} L_i=S(1-e^{-P/S})+KEt \end{equation}

1.5   Modeling considerations

Interception losses can be significant in annual or long-term models.

However, for heavy rainfalls for short-term models may be negligible.

Also need to consider areal and temporal variability.

1.6   Example 5.1

Using the following equations developed by Horton for interception by ash and oak trees, estimate the interception loss beneath these trees for a storm having a total precipitation of 1.5 in.

For ash trees, \begin{equation} L_i=0.015+0.23P \end{equation}

For oak trees, \begin{equation} L_i=0.03+0.22P \end{equation}

3.1   Initial depression storage

See Figure 5.2 for a simple depression storage abstraction scheme with the assumption that the initial precipitation will fully be stored until the maximum storage capacity of the depressions is reached. Is it always true?

3.2   Depression storage loss estimation

Linsley et al. (1949) approximated the volume of depression storage $V$ as \begin{equation} V=S_d(1-e^{-kP_e}) \end{equation} where $S_d$ is the maximum storage capacity of the depressions, $P_e$ is the rainfall excess, and $k$ is $1/S_d$.

Why is $k$ $1/S_d$? If the rainfall excess is very small ($P_e\approx 0$), its rate will be the same as that of the storage volume (all the rainfall excess can be stored) (i.e., $d\,V/d\,P_e=1$). \begin{equation} \frac{d\,V}{d\,P_e}=kS_de^{-kP_e}=kS_d=1 \end{equation}

3.3   Time distribution of the depression storage

\begin{equation} \frac{\sigma}{i-f}=\frac{i-f-v}{i-f} \label{eq:depression_rate_ratio} \end{equation} where $\sigma$ is the overland flow supply rate, $i$ is the net rainfall intensity, $f$ is the infiltration rate, and $v$ is the depression storage rate (i.e., $v=d\,V/d\,t$).

From Eq. (\ref{eq:depression_rate_ratio}), we can derive \begin{equation} \frac{\sigma}{i-f}=1-e^{-kP_e} \end{equation}

See Figures 5.3–5.5.

3.4   Problem 5.1

Using the precipitation input of Fig. 5.2, estimate the volume of depression storage for a 3-acre paved drainage area. State the volume in cubic feet and cubic meters. Convert it to equivalent depth over the area in inches and centimeters.