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Lecture 12 Precipitation Interception (1) Interception Processes General Comments Controls on Interception Interception in Woodlands Interception in Grasslands.

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Presentation on theme: "Lecture 12 Precipitation Interception (1) Interception Processes General Comments Controls on Interception Interception in Woodlands Interception in Grasslands."— Presentation transcript:

1 Lecture 12 Precipitation Interception (1) Interception Processes General Comments Controls on Interception Interception in Woodlands Interception in Grasslands Interception by Crops Measurement of Relative Humidity

2 Interception Water abstracted from gross precipitation by leaves and stems of a vegetation canopy and temporarily stored in its surfaces. Interception Loss Intercepted water lost by evaporation to the atmosphere before reaching the soil surface

3 General Comments Accounts for much of the variability in evaporation and transpiration between plant species or associations Precipitation is usually intercepted by: Tree canopy Grass Shrubs Litter Moss Built structures Interception capacity is usually considered to be a fixed amount for a given site: Canopy Shrubs Urban During filling and once storage is full, water passes through the canopy and reaches the soil as: Throughfall (TF) Stemflow (SF)

4 Net Rainfall 1. 2. 3.

5 Terms to Remember 1.Interception loss: Part of the rainfall intercepted by a plant canopy is evaporated back into the atmosphere and takes no part in the land-bound portion of the hydrological cycle 2.Throughfall: raindrops and snowflakes that fall through gaps in the plant canopy and water which drips from leaves, twigs and stems 3.Stemflow: Water run down the main stem or trunk from twigs and branches to the ground 4.Gross rainfall: rainfall on top of plant canopies 5.Net rainfall: The sum of throughfall and stemflow 6.Negative interception: Water intercepted from fogs and mists that contributes to stemflow

6 Controls on the amount of interception 1. Vegetation form/structure Shape Branch/leaf orientation Broad vs. needle leaves Number of leaves/stems Surface texture Flexibility/turgidity/stability 2. Vegetation growth pattern/physiology Seasonal growth Deciduous habit Total biomass Form/structure Age Growth rate Density of stand Leaf Area Index (LAI) 3. Meteorological Conditions Precipitation intensity and duration --Heavy and long duration precipitation will quickly exceed crown capacity leading to greater TF and SF --Conifers intercept more because they coincide with gentler rain or snow --Often possible to relate/predict losses from total P  Phase of precipitation Snow/sleet/rain/hail Wind speed and turbulence Energy balance Albedo related to vegetation type

7 Additional Points to Note These botanical and meteorological factors generally apply to non- botanical surfaces as well (e.g., urban surfaces) Strong dependence on meteorological factors allows interception, TF, or SF to be estimated from empirical relations Originally believed that interception losses were balanced by reduced transpiration losses. This is now believed to be incorrect Interception is not an alternate loss, rather an additional one

8 Interception loss during precipitation event Interception losses are greatest early during a precipitation event Losses decrease when interception storage is filled Interception ratio: (Interception loss) / (total precipitation)

9 Interception Loss from Woodlands Generally: deciduous crown closure > conifer crown closure However, conifer stands tend to exhibit higher interception losses because of higher leaf area density Conifer interception losses: ~25-35% Decid. Interception losses: ~10-30% Potential reasoning: needle shapes and distributions relative to broadleaves Spatially variable: density of trees (spacing)

10 Interception losses from grasses/shrubs LAI of mature, homogeneous grass cover is generally much smaller than that of forests Higher aerodynamic resistance than tall vegetation; thus, less interception loss Grazed or cut grasslands exhibit greatly reduced storage Interception losses vary ~13-26%

11 Interception losses from agricultural crops Usually evenly spaced plants Highly dependent on stage of development Depending on LAI

12 Interception of snow Difficult to measure and highly variable spatially due to wind redistribution Idea: snow accumulation on canopy decreases aerodynamic resistance (smooth) Thus, evaporation rates should be lower than for wet canopy Snow often melts, slides, slips, or is blown off of vegetation Studies indicate only ~15% of intercepted snow sublimates or evaporates Snow-stored water can be much greater than water storage – potential for more evaporation is there but energy requirements are not always met

13 Fog and clouds Deposition of fine water droplets to vegetated surfaces (e.g., mist, fog, clouds) Too fine to precipitate and would not be collected by rain gauges “ Negative interception ” Kittredge (1948) More common in mountainous regions and coastal areas Can be a significant addition of moisture to local vegetation Different process than dew, which is temperature controlled condensation of water vapor

14 Instrument for measuring air humidity

15 Relative humidity: Ratio of the actual amount of moisture in the atmosphere to the amount of moisture the atmosphere can hold Therefore, a relative humidity of 100% means the air can hold no more water (rain or dew is likely) Relative humidity of 0% indicates there is no moisture in the atmosphere. es wb =Saturation vapor pressure at T wb (kPa) es db =Saturation vapor pressure at T db (kPa) e d =Vapor pressure (kPa) Elv =Elevation above sea level (m) P =Air pressure (kPa) T wb =Wet bulb temperature (°C) T db =Dry bulb temperature (°C)

16 Procedure for Calculating Relative Humidity 1. Approximate the air pressure, P in kPa (kiloPascals). If you don't know your elevation, use P = 101.325 kPa. P = 101.325exp(-0.0001184  Elv) 2. Calculate a conversion factor, A. A = 0.00066(1.0 + 0.00115  T wb ) 3. Calculate the saturated vapor pressure at T wb. es wb = exp[(16.78  T wb – 116.9) / (T wb + 237.3)] 4. Calculate the vapor pressure, or the partial pressure of water vapor, e d in kPa. e d = es wb – AP(T db – T wb ) 5. Calculate the saturated vapor pressure at T db. es db = exp [(16.78  T db – 116.9) / (T db + 237.3)] 6. Finally, calculate the relative humidity, RH, in percent. RH = 100  (e d / es db )

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