Pesticide in a well-mixed water body is increased through addition of mass in inflow, resuspension and diffusion from the sediment layer. The amount of pesticide in a well-mixed water body is reduced through removal in outflow, degradation, volatilization, settling and diffusion into the underlying sediment.

SWAT+ incorporates a simple mass balance developed by Chapra (1997) to model the transformation and transport of pesticides in water bodies. The model assumes a well-mixed layer of water overlying a sediment layer. Figure 8:4-1 illustrates the mechanisms affecting the pesticide mass balance in water bodies.

SWAT+ defines four different types of water bodies: ponds, wetlands, reservoirs and depressional/impounded areas (potholes). Pesticide processes are modeled only in reservoirs.

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

$pst_{rsp,wtr}=v_r*SA*\frac{pst_{lksed}}{V_{tot}}$ 8:4.2.10

where $pst_{rsp,wtr}$ is the amount of pesticide removed via resuspension (mg pst), $v_r$ is the resuspension velocity (m/day), $SA$ is the surface area of the water body (m$^2$), $pst_{lksed}$ is the amount of pesticide in the sediment (mg pst), and $V_{tot}$ is the volume of the sediment layer (m$^3$). The volume of the sediment layer is calculated:

$V_{tot}=SA*D_{sed}$ 8:4.2.11

where $V_{tot}$ is the volume of the sediment layer (m$^3$), $SA$ is the surface area of the water body (m$^2$), $D_{sed}$ is the depth of the active sediment layer (m). Pesticide removed from the sediment layer by resuspension is added to the water layer.

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

$pst_{deg,wtr}=k_{p,aq}*pst_{lkwtr}$ 8:4.1.6

where $pst_{deg,wtr}$ is the amount of pesticide removed from the water via degradation (mg pst), $k_{p,aq}$ is the rate constant for degradation or removal of pesticide in the water (1/day), and $pst_{lkwtr}$ is the amount of pesticide in the water at the beginning of the day (mg pst). The rate constant is related to the aqueous half-life:

$k_{p,aq}=\frac{0.693}{t_{1/2,aq}}$ 8:4.1.7

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

Pesticides will partition into particulate and dissolved forms. The fraction of pesticide in each phase is a function of the pesticide’s partition coefficient and the water body’s suspended solid concentration:

$F_ d=\frac{1}{1+K_d*conc_{sed}}$ 8:4.1.1

$F_p=\frac{K_d*conc_{sed}}{1+K_d*conc_{sed}}=1-F_d$ 8:4.1.2

where $F_d$ is the fraction of total pesticide in the dissolved phase, $F_p$ is the fraction of total pesticide in the particulate phase, $K_d$ is the pesticide partition coefficient (m$^3$/g), and $conc_{sed}$ is the concentration of suspended solids in the water (g/m$^3$).

The pesticide partition coefficient can be estimated from the octanol-water partition coefficient (Chapra, 1997):

$K_d=3.085*10^{-8}*K_{ow}$ 8:4.1.3

where $K_d$ is the pesticide partition coefficient (m$^3$/g) and $K_{ow}$ is the pesticide’s octanol-water partition coefficient (mg $m^{-3}_{octanol}$(mg $m^{-3}_{water})^{-1}$). Values for the octanol-water partition coefficient have been published for many chemicals. If a published value cannot be found, it can be estimated from solubility (Chapra, 1997):

$log(K_{ow})=5.00-0.670*log(pst'_{sol})$ 8:4.1.4

where $pst'_{sol}$ is the pesticide solubility ($\mu$moles/L). The solubility in these units is calculated:

$pst'_{sol}=\frac{pst_{sol}}{MW}*10^3$ 8:4.1.5

where $pst'_{sol}$ is the pesticide solubility ($\mu$moles/L), $pst_{sol}$ is the pesticide solubility (mg/L) and $MW$ is the molecular weight (g/mole).

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:

$pst_{deg,sed}=k_{p,sed}* pst_{lksed}$ 8:4.2.8

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 $pst_{lksed}$ is the amount of pesticide in the sediment (mg pst). The rate constant is related to the sediment half-life:

$k_{p,sed}=\frac{0.693}{t_{1/2,sed}}$ 8:4.2.9

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).

Pesticide in the dissolved phase is available for volatilization. The amount of pesticide removed from the water via volatilization is:

$pst_{vol,wtr}=v_v*SA*\frac{F_d*pst_{lkwtr}}{V}$ 8:4.1.8

where $pst_{vol,wtr}$ is the amount of pesticide removed via volatilization (mg pst), $v_v$ is the volatilization mass-transfer coefficient (m/day), $SA$ is the surface area of the water body (m$^2$), $F_d$ is the fraction of total pesticide in the dissolved phase, $pst_{lkwtr}$ is the amount of pesticide in the water (mg pst), and V is the volume of water in the water body(m$^3$ H$_2$O).

The volatilization mass-transfer coefficient can be calculated based on Whitman’s two-film or two-resistance theory (Whitman, 1923; Lewis and Whitman, 1924 as described in Chapra, 1997). While the main body of the gas and liquid phases are assumed to be well-mixed and homogenous, the two-film theory assumes that a substance moving between the two phases encounters maximum resistance in two laminar boundary layers where transfer is a function of molecular diffusion. In this type of system the transfer coefficient or velocity is:

$v_v=K_l*\frac{H_e}{H_e+R*T_K*(K_l/K_g)}$ 8:4.1.9

where $v_v$ is the volatilization mass-transfer coefficient (m/day), $K_l$ is the mass-transfer velocity in the liquid laminar layer (m/day), $K_g$ is the mass-transfer velocity in the gaseous laminar layer (m/day), $H_{e}$ is Henry’s constant (atm m$^3$ mole$^{-1}$), $R$ is the universal gas constant (8.206 $*$ 10$^{-5}$ atm m$^3$ (K mole)$^{-1}$), and $T_K$ is the temperature ($K$).

For lakes, the transfer coefficients are estimated using a stagnant film approach:

$K_l=\frac{D_l}{z_l}$ $K_g=\frac{D_g}{z_g}$ 8:4.1.10

where $K_l$ is the mass-transfer velocity in the liquid laminar layer (m/day), $K_g$ is the mass-transfer velocity in the gaseous laminar layer (m/day), $D_l$ is the liquid molecular diffusion coefficient (m$^2$/day), $D_g$ is the gas molecular diffusion coefficient (m$^2$/day), $z_l$ is the thickness of the liquid film (m), and $z_g$ is the thickness of the gas film (m).

Alternatively, the transfer coefficients can be estimated with the equations:

$K_l=K_{l,O_2}*(\frac{32}{MW})^{0.25}$ 8:4.1.11

$K_g =168*\mu_w*(\frac{18}{MW})^{0.25}$ 8:4.1.12

where $K_l$ is the mass-transfer velocity in the liquid laminar layer (m/day), $K_g$ is the mass-transfer velocity in the gaseous laminar layer (m/day), $K_{l,O_2}$ is the oxygen transfer coefficient (m/day), $MW$ is the molecular weight of the compound, and $\mu_w$ is the wind speed (m/s). Chapra (1997) lists several different equations that can be used to calculate $K_{l,O_2}$.

The processes described above can be combined into mass balance equations for the well-mixed water body and the well-mixed sediment layer:

$\Delta pst_{lkwtr}=pst_{in}- (pst_{sol,o}+ p st_{sorb,o})-pst_{deg,wtr} -pst_{vol,wtr}- pst_{stl,wtr}+pst_{rsp,wt r}\pm pst_{dif}$

8:4.3.1

$\Delta pst_{lksed}=pst_{deg,sed}+pst_{stl,wtr}-pst_{rsp,wtr }-pst_{bur}\pm pst_{dif}$ 8:4.3.2

where $\Delta pst_{lkwtr}$ is the change in pesticide mass in the water layer (mg pst), $\Delta pst_{lksed}$ is the change in pesticide mass in the sediment layer (mg pst), $pst_{in}$ is the pesticide added to the water body via inflow (mg pst), $pst_{sol,o}$ is the amount of dissolved pesticide removed via outflow (mg pst), $p st_{sorb,o}$ is the amount of particulate pesticide removed via outflow (mg pst), $pst_{deg,wtr}$ is the amount of pesticide removed from the water via degradation (mg pst), $pst_{vol,wtr}$ is the amount of pesticide removed via volatilization (mg pst), $pst_{stl,wtr}$ is the amount of pesticide removed from the water due to settling (mg pst), $pst_{rsp,wt r}$ is the amount of pesticide removed via resuspension (mg pst), $pst_{dif}$ is the amount of pesticide transferred between the water and sediment by diffusion (mg pst), $pst_{deg,sed}$ is the amount of pesticide removed from the sediment via degradation (mg pst), $pst_{bur}$ is the amount of pesticide removed via burial (mg pst)

Pesticide in the sediment layer underlying a water body 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.

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:

8:4.2.1

where is the “concentration” of solid particles in the sediment layer (g/m), is the mass of solid particles in the sediment layer (g) and is the total volume of the sediment layer (m).

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:

8:4.2.2

where is the porosity, is the volume of water in the sediment layer (m) and is the total volume of the sediment layer (m). The fraction of the volume in the solid phase can then be defined as:

8:4.2.3

where is the porosity, is the volume of solids in the sediment layer (m) and is the total volume of the sediment layer (m).

The density of sediment particles is defined as:

8:4.2.4

where is the particle density (g/m), is the mass of solid particles in the sediment layer (g), and is the volume of solids in the sediment layer (m).

Solving equation 8:4.2.3 for and equation 8:4.2.4 for and substituting into equation 8:4.2.1 yields:

8:4.2.5

where is the “concentration” of solid particles in the sediment layer (g/m), is the porosity, and is the particle density (g/m).

Typical values of porosity and particle density for fine-grained sediments are = 0.8-0.95 and = 2.4-2.7 *10 g/m (Chapra, 1997). Assuming = 0.8 and = 2.6*10 g/m, the “concentration” of solid particles in the sediment layer is 5.210 g/m.

The fraction of pesticide in each phase is then calculated:

8:4.2.6

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:

8:4.2.12

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 surface area of the water body (m), is the fraction of total sediment pesticide in the dissolved phase, is the amount of pesticide in the sediment (mg pst), is the volume of the sediment layer (m), is the fraction of total water layer pesticide in the dissolved phase, is the amount of pesticide in the water (mg pst), and V is the volume of water in the water body (m HO). 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):

8:4.2.13

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 particulate phase may be removed from the water layer by settling. Settling transfers pesticide from the water to the sediment layer. The amount of pesticide that is removed from the water via settling is:

8:4.1.13

where is the amount of pesticide removed from the water due to settling (mg pst), is the settling velocity (m/day), is the surface area of the water body (m), is the fraction of total pesticide in the particulate phase, is the amount of pesticide in the water (mg pst), and . is the volume of water in the water body (m HO).

Pesticide is removed from the water body in outflow. The amount of dissolved and particulate pesticide removed from the water body in outflow is:

8:4.1.14

8:4.1.15

where is the amount of dissolved pesticide removed via outflow (mg pst), is the amount of particulate pesticide removed via outflow (mg pst), is the rate of outflow from the water body (m HO/day), is the fraction of total pesticide in the dissolved phase, is the fraction of total pesticide in the particulate phase, is the amount of pesticide in the water (mg pst), and is the volume of water in the water body (m HO).

Table 8:4-1: SWAT+ input variables that pesticide partitioning.

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

8:4.2.14

where is the amount of pesticide removed via burial (mg pst), is the burial velocity (m/day), is the surface area of the water body (m), is the amount of pesticide in the sediment (mg pst), and is the volume of the sediment layer (m).

Table 8:4-2: SWAT+ input variables related to pesticide in the sediment.