Evapotranspiration Analysis Project

This project involves estimating potential evaporation and reference crop evapotranspiration using the Penman Monteith equation. It includes calculations based on climatic data, soil properties, and the impact of weather conditions on evapotranspiration rates. The project also requires plotting daily actual evapotranspiration and total soil moisture in a specified soil profile, providing a comprehensive understanding of water dynamics in environmental engineering.

SOLUTIONS TO THE QUESTIONS
1. Question 1 The following information is available for a project site:
• Latitude = 35°S
• Longitude = 152°E
• Elevation = 0 m (at sea level)
• Average Albedo = 0.2
Jun-Aug Sep-Nov Dec-Feb Mar-May
Air Temperature (°C) 17 21 24 20
Relative Humidity (%) 65 70 70 60
Wind Speed (m/s) 2.2 2.4 2.3 2.1
Average sunshine hours 8 8.4 8.7 7.6
Estimate the following using the above information:
(a) Potential Evaporation for an open water body using the Penman Monteith equation
This is given by:
PE= (D/D+Y)xQet+ (Y/D+Y)xEat
Where Qet= Qs(1-r)-Q1
Eat= 0.3(1+0.5U)(Ea-Ed)
D= slope of the saturation vapor pressure
Y= hygrometric constant
Q1= long wave radiation
r = coefficient relating to vegetation cover (r=0.25)
Ta= air temperature
n/N= ratio of actual to possible sunshine hours of bright sunshine
ῥ = density of water (kg/m3)
u= wind velocity
ea= saturation vapor pressure
ed= actual vapor pressure
Y= psychometric constant
Qet= Qs(1-r)-Q
Q1= 0.95{(8.64x107)/ ῥλ} x ꓶ x(273.16+Ta)4x(0.53+0.065(Ed-1)0.5x(0.1+0.9))
The saturation vapor pressure is determined as follows:

Ea1= 610.78Exp{T/(T+238.3)x17.2694}
= 610.78Exp(17/(17+238.3)x17.2694)= 613.14Pa
Ea2= 610.78Exp{21/(21+238.3)x17.2694}= 613.65Pa
Ea3= 610.78Exp{24/(24+238.3)x17.2694}= 614.02Pa
Ea4= 610.78Exp{20/(20+238.3)}x17.2694= 613.52Pa
The long and short wavelengths are given as : Qs= 0.2mm and Ql= 0.4mm
The psychometric constant Y= 0.665x10-3P
Now, since this is at sea level, Y= 0.665x10-3x 101.3x10-3= 67.3645
The actual vapor pressure Ed can be obtained from the following relationship:
RH= P/Ps= Ed/Ea
Ed1= RHxEa1= 0.65x613.14= 398.54Pa
Ed2= 0.7x613.65= 429.56Pa
Ed3= 0.7x614.02= 429.81Pa
Ed4= 0.6x613.52=368.11Pa
Now, Eat values are determined from the equation:
Eat1=0.3(1+0.5U)(613.14-398.54)
Eat1 = 0.3x2.1x214.6=135.198
Eat2= 0.3(1+0.5x2.4)(613.65-429.56)= 121.49
Eat3= 0.3(1+0.5x2.3)(614.02-429.81)= 118.82
Eat4= 0.3(1+0.5x2.1)(614.52-368.11)= 151.54
Qet= 0.2(1-0.25)-0.4= 0.25
The slope of the saturation vapor pressure is given by:
D= 4098{0.6108Exp[17.27T/T+237.3]}/(T+237.3)2
Tmean= (17+21+24+20)/4= 20.5
D= 4098[0.6108Exp(17.27x20.5)/20.5+237.3]/ (20.5+237.3)2
D= 4098x0.6108x3.948/(66 460.84)= 0.1487
Hence the actual potential evaporation is determined using equation below:

PE1= (D/D+Y)Qet+ (Y/D+Y)Eat
= (0.1487/0.1487+67.36)x0.25+ (0.1487/0.1487+67.36)x 135.198= 5.5x10-4+2.203x10-
3x135.198= 0.2984
PE2= = (0.1487/0.1487+67.36)x0.25+ (0.1487/0.1487+67.36)x 121.49= 5.5x10-4+2.203x10-
3x121.49= 0.2682
PE3= = (0.1487/0.1487+67.36)x0.25+ (0.1487/0.1487+67.36)x 118.82= 5.5x10-4+2.203x10-
3x118.82= 0.2623
PE4= = (0.1487/0.1487+67.36)x0.25+ (0.1487/0.1487+67.36)x 113.29= 5.5x10-4+2.203x10-
3x113.29= 0.2501
(b) The reference crop evapotranspiration
ETo= 0.408D(Rn-G)+ Y900/(T+273)U(Es-Ea)/ D+ Y(1+0.34U)
Rn= Net radiation at the crop surface
G=0 (neglected)
Es= saturation vapor pressure
Ea= actual vapor pressure
Net radiation, Rn = Rns-Rnl= (1-D)Rs= (1-0.1487)Rs
But Rs= Krs(Tmax-Tmin)0.5xRa
And Ra= 24x60/(π)Gsec= 24x60/πx 0.0820=37.59
Rs= 0.18(24-17)0.5x 37.59= 0.18x2.646x37.59= 17.90
Rn= (1-0.1487)x 17.90= 15.238
Now, ETo1= 0.408x0.1487x15.238+
67.36x900x2.2/(20.5+273)x(613.14-398.54)/(0.1487+67.36(1+0.34x2.2)= 507.36
ET2= 0.9245+206.56x2.4(613.14-398.54)/ 0.1487+67.36(1+0.34x2.4)= 865.19
ET3= 0.9245+206.56x2.3x(613.65-429.56)/(0.1487+67.36(1+0.34x2.3)= 727.72
ET4= 0.9245+206.56x2.1x(613.52-368.11)/(0.1487+67.36(1+0.34x2.1)) = 920.85
(c) Potential evapotranspiration for a farm growing wheat that was planted in September
PEm= 16Nm[10Tm/I] where I= (20.5/5)(1/5)= 1.326
Nm= 8.4/8.175= 1.0275

PEm= 16 x1.0275x[10x20.5]/1.326= 16x1.0275x154.6= 2541.63
Hence the graph of the estimated evapotranspiration rates is given:
June-Aug Sept-Nov Dec-Feb Mar-May
0
100
200
300
400
500
600
700
800
900
1000
Eto
As illustrated the evapotranspiration rate increased from 500 to about 900 before reducing to
about 700 then rising to slightly passed 900mm. This is mainly attributed to the fluctuating
weather conditions throughout the year so that during cloudy days, lower rates would be
registered while higher rates would be registered during sunshine and windy conditions.
2. Given a fully developed and growing pasture well rooted (evenly) to 0.5m at the location
described in Question 1,
(i) compute and plot the daily actual evapotranspiration and total soil moisture in the
top 0.5m soil profile between October 20 and November 12.
The following soil properties may be used:
θw (soil moisture content at wilting point) = 0.10
θf (soil moisture content at field capacity) = 0.23
θs (soil moisture content at saturation) = 0.42
Assume the soil was at field capacity on October 20 and 40 mm rain occurs on November
1, and another 50 mm on November 8.
(ii) Plot the daily soil moisture percentage for the top 0.5 m root zone and the daily
actual evapotranspiration between October 20 and November 12. Note that any
water above saturation would result in overland flow from the soil. Soil water
content between saturation and field capacity would decrease to field capacity due
to infiltration. So, if there were no evapotranspiration, the soil moisture should

remain indefinitely at field capacity. For calculation purposes assume that the
water comes down to the field capacity at a very fast rate.
Now, the actual evapotranspiration is given by:
Et1= PE(h-hwp)/(hfc-hwp)
Where h= amount of soil moisture =0.1
hfc= amount of soil moisture corresponding to field capacity =0.23
hwp= amount of soil moisture corresponding to wilting=0.42
Et1= 0.2984x(0.1-0.42)/(0.23-0.42)= 0.2984x1.684= 0.503
Et2= 0.2682x1.684=0.452
Et3= 0.2623x1.684= 0.442
Et4= 0.2501x1.684= 0.421
The total soil moisture = h+hfc+hwp= 0.1+0.23+0.42= 0.75
Hence the plot of the grap for the actual evapotranspiration rates:
June-Aug Sept-Nov Dec-Feb Mar-May
0.38
0.4
0.42
0.44
0.46
0.48
0.5
0.52
Eta