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).

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:

$pst_{stl,wtr}=v_s*SA*\frac{F_p*pst_{lkwtr}}{V}$ 8:4.1.13

where $pst_{stl,wtr}$ is the amount of pesticide removed from the water due to settling (mg pst), $v_s$ is the settling velocity (m/day), $SA$ is the surface area of the water body (m$^2$), $F_p$ is the fraction of total pesticide in the particulate 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).

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 $K_{l,O_2}$.

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.

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

$pst_{sol,o}=Q*\frac{F_d*pst_{lkwtr}}{V}$ 8:4.1.14

$pst_{sorb,o}=Q*\frac{F_p*pst_{lkwtr}}{V}$ 8:4.1.15

where $pst_{sol,o}$ is the amount of dissolved pesticide removed via outflow (mg pst), $pst_{sorb,o}$ is the amount of particulate pesticide removed via outflow (mg pst), $Q$ is the rate of outflow from the water body (m$^3$ H$_2$O/day), $F_d$ is the fraction of total pesticide in the dissolved phase, $F_p$ is the fraction of total pesticide in the particulate 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).

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

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).