Pesticides in both the particulate and dissolved forms are subject to degradation. The amount of pesticide that is removed from the sediment via degradation is:

$pst_{deg,sed} = k_{p,sed} *pst_{rchsed}$ 7:4.2.8

​where $pst_{deg,sed}$ is the amount of pesticide removed from the sediment via degradation (mg pst), $k_{p,sed}$ is the rate constant for degradation or removal of pesticide in the sediment (1/day), and $pst_{rchsed}$ is the amount of pesticide in the sediment (mg pst). The rate constant is related to the sediment half-life:

$k_{p,sed} =\frac{0.693}{t_{1/2,sed}}$ 7:4.2.9

where $k_{p,sed}$ is the rate constant for degradation or removal of pesticide in the sediment (1/day), and $t_{1/2,sed}$ is the sediment half-life for the pesticide (days).

As in the water layer, pesticides in the sediment layer will partition into particulate and dissolved forms. Calculation of the solid-liquid partitioning in the sediment layer requires a suspended solid concentration. The “concentration” of solid particles in the sediment layer is defined as:

$conc^*_{sed}=\frac{M_{sed}}{V_{tot}}$ 7:4.2.1

where $conc^*_{sed}$ is the “concentration” of solid particles in the sediment layer (g/m$^3$), $M_{sed}$ is the mass of solid particles in the sediment layer (g) and $V_{tot}$ is the total volume of the sediment layer (m$^3$).

Mass and volume are also used to define the porosity and density of the sediment layer. In the sediment layer, porosity is the fraction of the total volume in the liquid phase:

$\phi=\frac{V_{wtr}}{V_{tot}}$ 7:4.2.2

where $\phi$ is the porosity, $V_{wtr}$ is the volume of water in the sediment layer (m$^3$) and $V_{tot}$ is the total volume of the sediment layer (m$^3$). The fraction of the volume in the solid phase can then be defined as:

$1-\phi=\frac{V_{sed}}{V_{tot}}$ 7:4.2.3

where $\phi$ is the porosity, $V_{sed}$ is the volume of solids in the sediment layer (m$^3$) and $V_{tot}$ is the total volume of the sediment layer (m$^3$).

The density of sediment particles is defined as:

$\rho_s=\frac{M_{sed}}{V_{sed}}$ 7:4.2.4

where $\rho_s$ is the particle density (g/m$^3$), $M_{sed}$ is the mass of solid particles in the sediment layer (g), and $V_{sed}$ is the volume of solids in the sediment layer (m$^3$).

Solving equation 7:4.2.3 for $V_{tot}$ and equation 7:4.2.4 for $M_{sed}$ and substituting into equation 7:4.2.1 yields:

$conc^*_{sed}=(1-\phi)*\rho_s$ 7:4.2.5

where $conc^*_{sed}$ is the “concentration” of solid particles in the sediment layer (g/m$^3$), $\phi$ is the porosity, and $\rho_s$ is the particle density (g/m$^3$).

Assuming $\phi = 0.5$ and $\rho_s=2.6*10^6$ g/m$^3$, the “concentration” of solid particles in the sediment layer is $1.3*10^6$ g/m$^3$.

The fraction of pesticide in each phase is then calculated:

$F_{d,sed}=\frac{1}{\phi +(1- \phi)*\rho_s *K_d}$ 7:4.2.6

Pesticide in the dissolved phase is available for diffusion. Diffusion transfers pesticide between the water and sediment layers. The direction of movement is controlled by the pesticide concentration. Pesticide will move from areas of high concentration to areas of low concentration. The amount of pesticide that is transferred between the water and sediment by diffusion is:

7:4.2.11

where is the amount of pesticide transferred between the water and sediment by diffusion (mg pst), is the rate of diffusion or mixing velocity (m/day), is the flow depth (m), is the fraction of total sediment pesticide in the dissolved phase, is the amount of pesticide in the sediment (mg pst), is the fraction of total water layer pesticide in the dissolved phase, is the amount of pesticide in the water (mg pst), and is the flow duration (days). If , is transferred from the sediment to the water layer. If , is transferred from the water to the sediment layer.

The diffusive mixing velocity, , can be estimated from the empirically derived formula (Chapra, 1997):

7:4.2.12

where is the rate of diffusion or mixing velocity (m/day), is the sediment porosity, and is the molecular weight of the pesticide compound.

Pesticide in the sediment layer is available for resuspension. The amount of pesticide that is removed from the sediment via resuspension is:

$pst_{rsp,wtr}=\frac{v_r}{depth}*pst_{rchsed}*TT$ 7:4.2.10

where $pst_{rsp,wtr}$ is the amount of pesticide removed via resuspension (mg pst), $v_r$ is the resuspension velocity (m/day), $depth$ is the flow depth (m), $pst_{rchsed}$ is the amount of pesticide in the sediment (mg pst), and $TT$ is the flow travel time (days). Pesticide removed from the sediment layer by resuspension is added to the water layer.

Pesticide in the sediment layer underlying a reach segment is increased through addition of mass by settling and diffusion from the water. The amount of pesticide in the sediment layer is reduced through removal by degradation, resuspension, diffusion into the overlying water, and burial.

Pesticide in the sediment layer may be lost by burial. The amount of pesticide that is removed from the sediment via burial is:

$pst_{bur}=\frac{v_b}{D_{sed}}*pst_{rchsed}$ 7:4.2.13

where $pst_{bur}$ is the amount of pesticide removed via burial (mg pst), $v_b$ is the burial velocity (m/day), $D_{sed}$ is the depth of the active sediment layer (m), and $pst_{rchsed}$ is the amount of pesticide in the sediment (mg pst).

Table 7:4-2: SWAT+ input variables related to pesticide in the sediment.