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:

$pst_{deg,wtr}= k_{p,aq}*pst_{rchwtr}*TT$ 7:4.1.6

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), $pst_{rchwtr}$ is the amount of pesticide in the water at the beginning of the day (mg pst), and $TT$ is the flow travel time (days). The rate constant is related to the aqueous half-life:

$k_{p,aq}=\frac{0.693}{t_{1/2,aq}}$ 7:4.1.7

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 reach segment’s suspended solid concentration:

$F_d=\frac{1}{1+K_d*conc_{sed}}$ 7:4.1.1

$F_p=\frac{K_d*conc_{sed}}{1+ K_d*conc_{sed}}=1-F_d$ 7:4.1.2

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):

$K_d=3.085*10^{-8}*K_{ow}$ 7:4.1.3

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):

$log(K_{ow})=5.00-0.670*log(pst'_{sol})$ 7:4.1.4

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

$pst'_{sol}=\frac{pst_{sol}}{MW}*10^3$ 7:4.1.5

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).

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

$pst_{sol,o}=Q*\frac{F_d*pst_{rchwtr}}{V}$ 7:4.1.13

$pst_{sorb,o}=Q*\frac{F_p*pst_{rchwtr}}{V}$ 7:4.1.14

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 reach segment (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, $pst_{rchwtr}$ is the amount of pesticide in the water (mg pst), and $V$ is the volume of water in the reach segment (m$^3$ H$_2$O).

Table 7:4-1: SWAT+ input variables that pesticide partitioning.

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

7:4.1.8

where is the amount of pesticide removed via volatilization (mg pst), is the volatilization mass-transfer coefficient (m/day), is the flow depth (m), is the fraction of total pesticide in the dissolved phase, is the amount of pesticide in the water (mg pst), and is the flow travel time (days).

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:

7:4.1.9

where is the volatilization mass-transfer coefficient (m/day), is the mass-transfer velocity in the liquid laminar layer (m/day), is the mass-transfer velocity in the gaseous laminar layer (m/day), is Henry’s constant (atm m mole), is the universal gas constant ( atm m (K mole)), and is the temperature (K).

For rivers where liquid flow is turbulent, the transfer coefficients are estimated using the surface renewal theory (Higbie, 1935; Danckwerts, 1951; as described by Chapra, 1997). The surface renewal model visualizes the system as consisting of parcels of water that are brought to the surface for a period of time. The fluid elements are assumed to reach and leave the air/water interface randomly, i.e. the exposure of the fluid elements to air is described by a statistical distribution. The transfer velocities for the liquid and gaseous phases are calculated:

7:4.1.10

where is the mass-transfer velocity in the liquid laminar layer (m/day), is the mass-transfer velocity in the gaseous laminar layer (m/day), is the liquid molecular diffusion coefficient (m/day), is the gas molecular diffusion coefficient (m/day), is the liquid surface renewal rate (1/day), and is the gaseous surface renewal rate (1/day).

O’Connor and Dobbins (1958) defined the surface renewal rate as the ratio of the average stream velocity to depth.

7:4.1.11

where is the liquid surface renewal rate (1/day), is the average stream velocity (m/s) and is the depth of flow (m).

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

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:

$pst_{stl,wtr}=\frac{v_s}{depth}*F_p*pst_{rchwtr}*TT$ 7:4.1.12

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), $depth$ is the flow depth (m), $F_p$ is the fraction of total pesticide in the particulate phase, $pst_{rchwtr}$ is the amount of pesticide in the water (mg pst), and $TT$ is the flow travel time (days).