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4.6.3: Wet Soil Evaporation

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    44379
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    The increased rate of evaporation due to a wet soil surface is influenced by the amount of canopy development, the energy available to evaporate water, and the hydraulic properties of the soil. The factor (Kw) can be used to predict the amount of wet soil evaporation. The total amount of evaporation from a wet soil should be less than the amount of water received by rain or irrigation. The method of Wright (1982) has been adapted to account for wet soil evaporation:

    \(K_w = F_w(K_{cmax} - K_{co})f_t\)

    \(K_{cmax} = 1.2 + K_{cf}\)

    \(f_t = 1-\sqrt{\dfrac{t}{t_d}}\) (4.14)

    where: Kw = wet soil evaporation factor,

    Kcmax = maximum crop coefficient for wet soil evaporation,

    Kcf = crop coefficient adjustment factor (Table 4.4),

    Fw = fraction of surface wetted,

    t = time since last wetting of soil surface (d),

    td = duration of wet soil evaporation (d), and

    ft = wet soil decay function.

    Table 4.5. Fraction of the soil surface wetted for various types of irrigation systems
    Wetting Method Fw
    Rain 1.0
    Sprinkler irrigation:
    Above Canopy sprinklers 1.0
    in-Canopy sprinklers 0.75
    LEPA systems (alternate furrows wetted) 0.5
    Borders and basin irrigation 1.0
    Furrow irrigation:
    Large applications depth 1.0
    Small application depth 0.5
    Alternate furrows irrigated 0.5
    Surface trickle or drip irrigation 0.25
    subsurface drip irrigation:
    Large applications 0.1
    Normal applications 0.0
    Table 4.6. Duration of wet soil evaporation (td) for selected soil textures and values of the wet soil decay function (ft) for time since wetting (t).
    Time Since Wetting, (t), days Soil Texture Clay Soil Texture Clay Loam Soil Texture Silt Loam Soil Texture Sandy Loam Soil Texture Loamy Sand Soil Texture Sand
    10 days of wet soil evaporation (td) 7 days of wet soil evaporation (td) 5 days of wet soil evaporation (td) 4 days of wet soil evaporation (td) 3 days of wet soil evaporation (td) 2 days of wet soil evaporation (td)
    0 ft=1.00 ft=1.00 ft=1.00 ft=1.00 ft=1.00 ft=1.00
    1 ft=0.68 ft=0.62 ft=0.55 ft=0.50 ft=0.42 ft=0.29
    2 ft=0.55 ft=0.47 ft=0.37 ft=0.29 ft=0.18 ft=0.00
    3 ft=0.45 ft=0.35 ft=0.23 ft=0.13 ft=0.00  
    4 ft=0.37 ft=0.24 ft=0.11 ft=0.00    
    5 ft=0.29 ft=0.15 ft=0.00      
    6 ft=0.23 ft=0.07        
    7 ft=0.16 ft=0.00        
    8 ft=0.11          
    9 ft=0.05          
    10 ft=0.00          

    The fraction of the surface wetted depends on the type of irrigation system (Table 4.5). The duration of wet soil evaporation depends on the type of soil. Sandy soils dry quicker than fine-textured soils. Representative values of the drying duration are given in Table 4.6. Local observations can also be used to determine values for the drying duration.

    Example \(\PageIndex{1}\)

    You manage two adjacent fields with the same soil properties and crop conditions but different soil water status. Compute the evapotranspiration rate for each field for the following information.

    Given: Soil water was measured in each field with the following results:

    Field A: Soil water content is 0.22 Field B: Soil water content is 0.12

    Volumetric water content at field capacity and permanent wilting are 0.25 and 0.10.

    The crop root zone is 4 ft deep in both fields.

    The reference ET rate is 0.30 in/d and the basal crop coefficient is 1.1 at this growth stage.

    The crop grown at the sites has a critical soil water threshold (frc) of 0.35.

    The soil has not been wetted during the last week.

    Solution

    Wet soil evaporation is insignificant since there was no irrigation or rain recently, so Kw = 0.

    1. Compute fr for each field:

    \(f_r = \dfrac{\theta_v - \theta_{wp}}{\theta_{fc} - \theta_{wp}}\) (Eq. 2.10)

    \(\text{Field A: }f_r = \left(\dfrac{0.22-0.10}{0.25-0.10}\right) = 0.80 \text{ Field B: }f_r = \left(\dfrac{0.12-0.10}{0.25-0.10}\right)= 0.13\)

    2. Compute the ET for crop in field A:

    From Eq. 4.13 since \(f_r \geq f_c\) (i.e., 0.80 ≥ 0.35), then, \(K_s = 1.0\)

    and the ET rate is \(ET = K_s K_c ET_o = 1.0 \times 1.1 \times 0.3 = 0.33\) in/d

    3. Compute the ET for field B:

    In Field B, \(f_r < f_c\) (0.13 < 0.35), so \(K_s = \frac{f_r}{f_c} = \frac{0.13}{0.35} = 0.37\) (Eq. 4.13)

    and \(ET = K_s K_c ET_o = 0.37 \times 1.1 \times 0.3 = 0.12\) in/d


    4.6.3: Wet Soil Evaporation is shared under a CC BY-NC-ND 4.0 license and was authored, remixed, and/or curated by LibreTexts.