Answers Problem set 5

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					  Problem Set # 5 EVAPORATION
1 Soil weighing lysimeters are used to measure actual evaporative losses from soil.
  The lysimeter may range in size from 20 cm to 150 cm deep and in mass from several kg
  to several Mg. Changes in moisture content is registered by changes in weight.
  A medium sized rectangular one has been set up with the following mass and dimensions:
              D                    60 cm; depth of lysimeter
              A                  0.25 m2; surface area exposed
              pb                1200 kg/m3; soil bulk density
  The lysimeter weighed on Day 1 and then again 7 days later (Day 8):
  (the weights here are only of the soil and water and does not include the metal container)
              Mt1                 234 kg; total mass of lysimeter on Day 1
              Mt8                 227 kg; total mass of lysimeter on Day 8

  1a) Find the following moisture contents

              property Day 1        Day 8                    Vt                0.15 m3
                  w            0.30        0.26              Ms                 180 kg
                  q            0.36        0.31              Mw1                 54 kg
               mm H2O           216         188              Mw2                 47 kg
              where; w is mass moisture content (kg/kg), q is volumetric moisture content (m3/m3)
              and mm H2O is the depth of water in the lysimeter

  1b) if all the moisture lost was due only to evaporation, then how many mm of water
      evaporated; how many MJ of energy was expended for the lysimeter; and
      how many MJ of energy was expended per square meter on a daily basis?

              Emm                 28 mm, water loss to evaporation
              Qely             17.15 MJ energy used in evaporation for just lysimeter
              Qe                 9.8 MJ/m2/d

2 The Penman method is often the preferred method for determining evaporation. The
  spreadsheet model included here has the equation from your text entered into the
  cells. Instructions for use are on the spreadsheet.
  2a) Find daily potential evaporation (QE and PE) for the following conditions
               T                    20 C; air temperature
               RH                  0.5 relative humidity (DO NOT USE %!)
               Qs                    8 MJ/m2; total incoming solar radiation
               U                     5 km/hr; average daily windspeed
               a                  0.25 albedo for a full crop surface

  2b) Keeping all the conditions the same as above except for that specified
      report the effect of the change on PE for the following. Use a table to present your results.
               T                    30 C; hot day but all the other parameters do not change
               RH                  0.1 dry day, but all the other parameters do not change
               Qs                   25 MJ/m2; bright long sunny day, etc.
               U                    20 km/hr; windy day, etc.
               a                   0.1 albedo for a bare soil surface

  2c) Explain why a change in each of the above conditions affect PE.
 Example: Reducing the albedo means less solar radiation is reflected and thus more
 is available for evaporation of water.

                       2a            T=30          RH=0.1    Qs=25     U=20      a=0.1
QE         MJ/m2/d             9.4          11.3        13.9      17.2      19.3       10.1
PE         mm/d                3.8           4.6         5.7       7.0       7.9        4.1

T increase; an increase in T partily increases energy for evaporation but has a greater effect upon
            the amount of vapour the air can hold. At higher temperatures air can hold more water
            vapour and therefore evaporation is greater
RH decrease; lower the RH the greater the vapour deficit in the air and the higher the evaporation
Qs higher; greater the solar radiation the higher the energy for evaporation
U higher; greater the windspeed the faster the recently evaporated water vapour is removed
            from above the evaporating surface and the greater the evaporation rate
a lower; lower the albedo the lower the amount of reflected energy and the greater the
            amount of energy available for evaporation.
           Penman method for calculation of potential evaporation
This method uses input values of daily temperature, relative humidity, windspeed, albedo, and incoming
solar radiation to calculate daily potential evaporation. These input values are common meteorological
measurements. The model used here is simplified; some of the basic assumptions are: it assumes that
net longwave radiation is zero; that the solar radiation absorbed by the ground (ground heat) is 10%
of net solar and that atmospheric pressure is 101 kPa.
The air temperature and relative humidity are used to calculate psychrometric values and constants.

        Blue cells represent input cells
        Yellow cells contain output for that of evaporation. Do not alter the contents of these cells

INPUT           T          20      C; average air temperature during the day
                U           5      km/hr; average daily wind speed (gets converted to m/s later on)
               RH          0.5     average daily relative humdity
               Qs           8      MJ/m2; total solar radiation received for the day
                a         0.25     surface albedo (reflectivity coefficient)

                         D (Qn - Qg) + g 6.43(1+U)(es-ea)
             Qe =

CALCULATIONS (For your interest only; equations are embedded in cells)
          Qr         2       MJ/m2; reflected solar radiation
          Qn         6       MJ/m2; net radiation
          Qg        0.6      MJ/m2; amount of net radiation lost to ground
          es       2.338 kPa; saturated vapour pressure
          ea       1.169 kPa; actual vapour pressure = RH es
          D        0.145 kPa/C; gradient of saturated vapour pressure
           g       0.067 kPa/C; psychrometric constant; 10C 0.0661, 20C=0.0667, 30C=0.0674

OUTPUT (equations embedded in cells)
            Qe        9.38    MJ/m2; energy expended on daily evaporation
            PE        3.83    mm depth of water evaporated