Plant growth may be reduced due to extreme temperatures, and insufficient water, nitrogen or phosphorus. The amount of stress for each of these four parameters is calculated on a daily basis using the equations summarized in the following sections.

Temperature stress is a function of the daily average air temperature and the optimal temperature for plant growth. Near the optimal temperature the plant will not experience temperature stress. However as the air temperature diverges from the optimal the plant will begin to experience stress. The equations used to determine temperature stress are:

$tstrs=1$ when $\overline T_{av} \le T_{base}$ 5:3.1.2

$tstrs=1-exp[\frac{-0.1054*(T_{opt}-\overline T_{av})^2}{(\overline T_{av}-T_{base})^2}]$ when $T_{base}<\overline T_{av} \le T_{opt}$ 5:3.1.3

$tstrs=1-exp[\frac{-0.1054*(T_{opt}-\overline T_{av})^2}{(2*T_{opt}-\overline T_{av}-T_{base})^2}]$ when $T_{opt}<\overline T_{av}\le 2*T_{opt}-T_{base}$ 5:3.1.4

$tstrs=1$ when 5:3.1.5

where is the temperature stress for a given day expressed as a fraction of optimal plant growth,is the mean air temperature for day (°C), is the plant’s base or minimum temperature for growth (°C), and is the plant’s optimal temperature for growth (°C). Figure 5:3-1 illustrates the impact of mean daily air temperature on plant growth for a plant with a base temperature of 0°C and an optimal temperature of 15°C.

Water stress is 0.0 under optimal water conditions and approaches 1.0 as the soil water conditions vary from the optimal. Water stress is simulated by comparing actual and potential plant transpiration:

$wstrs=1-\frac{E_{t,act}}{E_t}=1-\frac{w_{actualup}}{E_t}$ 5:3.1.1

where $wstrs$ is the water stress for a given day, $E_t$ is the maximum plant transpiration on a given day (mm H$_2$O), $E_{t,act}$ is the actual amount of transpiration on a given day (mm H$_2$O) and $w_{actualup}$ is the total plant water uptake for the day (mm H$_2$O). The calculation of maximum transpiration is reviewed in Chapter 2:2 and the determination of actual plant water uptake/transpiration is reviewed in Chapter 5:2.

As with nitrogen, phosphorus stress is quantified by comparing actual and optimal plant phosphorus levels. Phosphorus stress varies non-linearly between 0.0 at optimal phosphorus content and 1.0 when the phosphorus content of the plant is 50% or less of the optimal value. Phosphorus stress is computed with the equation:

$pstrs=1-\frac{\phi_p}{\phi_p +exp[3.535-0.02597*\phi_p]}$ 5:3.1.8

where $pstrs$ is the phosphorus stress for a given day, and $\phi_p$ is a scaling factor for phosphorus stress. The scaling factor is calculated:

$\phi_p=200*(\frac{bio_P}{bio_{P,opt}}-0.5)$ 5:3.1.9

where $bio_{P,opt}$ is the optimal mass of phosphorus stored in plant material for the current growth stage (kg N/ha) and $bio_P$ is the actual mass of phosphorus stored in plant material (kg N/ha).

Table 5:3-1: SWAT+ input variables that pertain to stress on plant growth.

Nitrogen stress is calculated only for non-legumes. SWAT+ never allows legumes to experience nitrogen stress.

Nitrogen stress is quantified by comparing actual and optimal plant nitrogen levels. Nitrogen stress varies non-linearly between 0.0 at optimal nitrogen content and 1.0 when the nitrogen content of the plant is 50% or less of the optimal value. Nitrogen stress is computed with the equation:

$nstrs=1-\frac{\phi _n}{\phi_n +exp[3.535-0.02597*\phi _n]}$ 5:3.1.6

where $nstrs$ is the nitrogen stress for a given day, and $\phi_n$ is a scaling factor for nitrogen stress. The scaling factor is calculated:

$\phi_n=200*(\frac{bio_N}{bio_{N,opt}}-0.5)$ 5:3.1.7

where is the optimal mass of nitrogen stored in plant material for the current growth stage (kg N/ha) and is the actual mass of nitrogen stored in plant material (kg N/ha).