# Answers Problem set 5 by liwenting

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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.

DIRECTIONS
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
a         0.25     surface albedo (reflectivity coefficient)

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

CALCULATIONS (For your interest only; equations are embedded in cells)
Qr         2       MJ/m2; reflected solar radiation