PRT4301:
Modelling and computer
simulations in agriculture
Dr. Christopher Teh (Room C202)
chris@agri.upm.edu.my
Tel: 8946 6976
1
Pre-amble


Objective
 understand how mathematics is applied in
agriculture, in particular in crop growth
 understand how computer models are built
and used
2+1 credits
 lab is completely in the computer lab
2

Exams
 20% Test 1
 20% Test 2
 10% Lab work
 50% Final
 Always in essay format (never in multiple
choice @ objective format)
3

Reading materials
 Kropff, M.J. and H.H. van Laar. 1993.
Modelling crop-weed interactions. CAB
International (in association with International
Rice Research Institute), Oxon, UK.
 Goudriaan, J. and H.H. van Laar. 1994.
Modeling potential crop growth processes. A
textbook with exercise. Current issues in
production ecology. Netherlands, Kluwer
Academic.
4

Reading materials
 Monteith, J.L. 1975. Principles of
environmental physics. Edward Arnold,
London.
 Campbell, G.S. and J.M. Norman. 1998. An
introduction to environmental biophysics. 2nd
Edition. Springer-Verlag, New York.
5

Reading materials
 Teh, C. 2006. Introduction to mathematical
modeling of crop growth: how the equations
are derived and assembled into a computer
model. BrownWalker Press, Florida, US.
6
Part 1:
What is mathematical
modelling?
7



Definition of a model
 simplified representation of real systems
Types of models
 pictorial, conceptual/verbal, physical,
mathematical
Definition of a mathematical model
 represents a real system in a mathematical
form (one or more equations)
8

Uses of mathematical models
 help us to understand, predict and control a
system
 identify areas of deficient knowledge
 less experimentation by trial-and-error
 answer various “what if?” scenarios
 add value to experiments
 may replace experiments (in rare cases)
 encourage collaboration among researchers
from various disciplines
9

Characteristics of models
 incomplete description of real systems
 models built from assumptions
 model simplicity vs. model accuracy
 no one best model for all circumstances
 not about computers or ICT
10

Modelling methodology: an example
How to determine the number of leaves
in this tree (or any in trees)?
Most accurate method: manually count
each and every leaf
Problem: tedious and time-consuming
Alternative: N = nl x nb
where N is no. of leaves; nl is average
no. of leaves per branch; and nb is no.
of branches
nl = 153, nb = 99, so N = 15,147 leaves
11

Let’s develop our own leaf count model:
Maximum number of small boxes
that could fit into the large box is
the volume of the large box (Vlarge)
divided by the volume of the small
box (Vsmall)
So max. no. of small boxes that can
fit into the large box is:
N small 
Vl arg e
Vsmall
12

So applied to canopy:
So no. of leaves in canopy (N) is:
N  Vc Vl
Vc = vol. of canopy
Vl = average vol. of one leaf
4 H

W
L
Vc  
 HWL
2
2
2
3
6
   
The geometrical shape of an
ellipsoid is used to represent
the canopy volume of a tree
and the volume a single leaf
  2  2 
4
Vl   l
2
3
l
l


6
l3
Vc  6 HWL HWL
N

 3
3

Vl
l
6l
13



H = 3.5m; W = L = 2.5m; l = 0.14m,
so N is 7,185 leaves
 vs. 15,147 leaves
 model error of 47% (large!)
Why the error?
 check our assumptions
14

Revised model:
Vol. of canopy occupied by leaves is:
Vo  Vc  Ve
   
     
4
W
L 
  H
2
2
2
3
4
 f H H 2 fW W 2 f L L 2
3


6
HWL 1  f H fW f L 
The canopy of a tree is represented by two ellipsoids, where the inner
ellipsoid (or shell) is devoid of any leaves, and the outer shell marks the
canopy edge. The space between the inner and outer shell is where the
leaves are located in the canopy.
15

So number of leaves (N) is now:
N  Vo Vl
Vo = vol. of canopy occupied by leaves
Vl = average vol. of one leaf
But, let’s account for leaf density (how closely packed are the leaves):
Vl 
1  3
l
l 6
where 1/pl is the fraction of a full ellipsoid volume occupied by a single leaf;
the larger the pl , the more closely packed the leaves are together.
16

Finally, we have:

V
N o  6
Vl
HWL 1  f H fW f L 
1  3
 l
l 6

l  HWL 1  f H fW f L 
l3
Measuring, we get: fH = fW = fL = 0.5; and pl = 2, so N is 13,951 leaves.
Error of only 8% (acceptable).
17


Our model advantages:
 no manual leaf counting
 faster and less tedious
Our model disadvantages:
 accurate determination of pl difficult
 assumes uniform distribution of leaves
 assumes ellipsoidal volumetric canopy and
leaves
18
REVIEW OF
MODELLING
METHODOLOGY
19

Types of mathematical models
 Mechanistic (process-based) and Empirical
 Static and Dynamic
 Discrete and continuous
 Deterministic and stochastic
20




Static models
 no time factor
Dynamic models
 has time factor
Discrete models
 time is an integer (1, 2, 3, …)
Continuous models
 time is a real value (1.1, 2.5, 3.0, …)
21


Deterministic models
 no element of randomness
Stochastic models
 has elements of randomness (probabilities)
22


Mechanistic models
 process-based models
 describes and explains the processes
 more difficult to build
Empirical models
 correlative or statistical models
 describes but does not explain the
processes
 easier to build
23

Properties of empirical models
 1) easier to build; curve-fitting exercise
y
y
y
y = A sin (2 x/P)
A
0
P
y = b 0 + b 1x
y = b0 + b1 log(x)
x
x
linear
y
y = b0 + b1x + b2x2
x
logarithmic
sine
y
y = b 0 e b1x
x
quadratic
x
exponential
24
 mechanistic
models difficult to build because
we need to know which and how the factors
interact with one another to produce the
system process
25
 2)
empirical models cannot imply causality
(cause-and-effect)
describes how variable are related but does not
explain why

Relationship
between
the
population of Oldenburg city,
Germany with the number of
bird storks in 1930-36
3
y, population (x 10 )
80
70
60
y = 139.1x + 38183.7
R2 = 0.9
50
100
150
200
250
Other examples:
- sale of ice cream and reservoir level
- electricity bill and weather
300
x, number of storks
26
 3)
empirical models are highly environmentspecific
can only be used in conditions from which they
were derived only
 its use more limited than mechanistic models
(applicable over wider range of environments or
conditions)
 but when used in their environment, empirical
models are often very accurate, more so than
mechanistic models

27
Crop production levels


Four levels of production:
 Level 1: potential growth, limited only by solar
radiation
 Level 2: additionally limited by water
 Level 3: additionally limited by nitrogen
 Level 4: additionally limited by other nutrients
Helps us to focus on developing our models
28
Overview of the Gg (Generic crop growth) model
29
30
Part 2:
Meteorology
31
Importance



Agriculture strongly affected by meteorology
(weather)
Water (rain)
 photosynthesis, respiration, nutrient supply
RH
 40-80% desirable; too high, high disease and
pests
 low crop yield because high vegetative growth
32

Solar radiation
 major energy supplier; photosynthesis, water
uptake
 length of day: flowering


soybean (short day); groundnut (long day)
Air temperature
 every plant has an optimum temp. range
 regulates chemical reactions in
photosynthesis, flowering, germination,
transpiration, respiration
33
Solar position
Solar position with respect to the
observer
 = azimuth (-ve, before solar noon)
 = elevation (or solar height)
  asin sin   sin   cos   cos   cos 
 sin   sin   sin  

cos


cos



  acos 
 = site latitude
 = solar declination
 = hour angle
34
Location of observer on the Earth’s
surface. Np and Sp are the North
Pole and South Pole, respectively.
Solar position with respect to the Earth. Np
and Sp are the North Pole and South Pole,
respectively.
35
Solar declination  varies
depending on the Earth’s position
in orbit around the Sun. Np is the
North Pole.
sun declination (deg.)
30
Cyclical change in solar declination with
the day of year
20
10
  0.4093cos  2  td  10  365
0
-10
td = day of year (1=Jan 1, …, 365=Dec 31)
-20
-30
0
50 100 150 200 250 300 350
day of year
36
Hour angle, :
Earth rotates 360 every 24 hours, so every 1 hour, Earth rotates 15


12
 th  12 
th = local solar time
 is –ve before solar noon, +ve after solar noon
Local solar time vs. local time
local time is determined by
- Standard Meridian
- political boundaries (West Malaysia & Singapore is actually GMT +7)
37

Therefore,
solar elevation
(from horizontal)
  asin sin   sin   cos   cos   cos 
 sin   sin   sin  

 cos   cos 

solar azimuth
(from South)
  acos 
solar inclination
(from vertical)
  acos sin   sin   cos   cos   cos 
solar azimuth
(from North in an
eastward direction)
   
38
Daylength and sunrise and sunset time

Sunset time
 sin   sin  
tss  12  acos  


 cos   cos  
12

Sunrise time

Daylength
 number of hours between sunrise and sunset
 sin   sin  
tsr  24  tss  12  acos  


 cos   cos  
DL  2  tss  12  
12
 sin   sin  
acos  


cos


cos



24
39
Solar radiation
Longer the wavelength, lesser the energy:
PAR (photosynthetically active radiation, 400-700 nm, same as visible light)
UV too high energy
NIR too low energy
40

Terminology:
 radiant flux = amount of energy emitted or
received per unit time (W or J s-1)
 radiant flux density = radiant flux per unit area
(W m-2)
 irradiance = radiant flux density received
(incident)
 radiant emittance = radiant flux density
emitted (transmitted)
41
Daily irradiance
s 

I t , d  I et , d  b0  b1

DL 

s = sunshine hours (no. of hours when solar irradiance >120 W m-2)
DL = daylength
Iet,d = extra-terrestrial solar irradiance (W m-2)
b0 and b1 = empirical coefficients
The Angstrom coefficients b0 and b1 used for calculating daily
solar radiation for different climate zones (Frere and Popov, 1979)
Climate zones
b0
b1
Cold or temperate
0.18
0.55
Dry tropical
0.25
0.45
Humid tropical
0.29
0.42
42
Hourly ET irradiance:
I et  1370   0  sin 
 0  1  0.033cos  2  td  10  / 365
Daily ET irradiance:
tss
I et , d  3600 1370   0  sin  dth
tsr
12 
tss
 sin  dt
tsr
h

12 
12

12

ac os   a / b 

a  b cos  2  th  12  / 24  dth
ac os   a / b 
2
 24  a  acos   a b   b 1   a b  
 

where
a = sinsin
b = coscos
43
Hourly irradiance
It  A cos  2 th / 24  B
-2
solar irradiance (W m )
1000
800
A  b  and B  a 
600
where  
400
 I t , d 86400
a  acos  a b   b 1   a b 
2
200
0
0 2 4 6 8 10 12 14 16 18 20 22 24
a = sinsin
b = coscos
local solar hour
Typical diurnal trend for solar irradiance
44
Net radiation
net = incoming - outgoing
Rn  1  p  It  RnL
s 

 b  1  b 
RnL  RnL
DL 

  RLd  RLu
RnL
RLu   Ta4,K
RLd  9.35 106  Ta6,K
 = Stefan-Boltzmann constant (5.67 x 10-8 W m-2 K-4)
p = surface albedo (typically 0.15)
Ta,K4 = air temperature (K)
b = 0.2
45
Direct and diffuse solar radiation



Direct (dr) component
 from a single direction
 causes shadows
Diffuse (df) component
 from all directions
 does not cause shadows
Need to distinguish the two
 interception by plants is different for the two
components
46
Daily direct and diffuse radiation
A set of empirical equations:
I df , d I t , d  1
I df , d I t , d  1  2.3  I t , d I et , d  0.07 
for I t , d I et , d  0.07
2
for 0.07  I t , d I et , d  0.35
I df , d I t , d  1.33  1.46 I t , d I et , d
for 0.35  I t , d I et , d  0.75
I df , d I t , d  0.23
for 0.75  I t , d I et , d
Idr,d = It,d – Idf,d
47
Hourly direct and diffuse radiation
A set of empirical equations:
I df I t  1
I df I t  1  6.4  I t I et  0.22 
for I t I et  0.22
2
for 0.22  I t I et  0.35
I df I t  1.47  1.66 I t I et
for 0.35< I t I et  K
I df I t  R
for K  I t I et
Idr = It – Idf
48
RH
ea  es (Ta )
100
RH = relative humidity (%)
Ta = air temperature (C)
saturated vapor pressure (mbar)
Air vapour pressure
100
80
60
40
20
0
0
10
20
30
40
50
air temperature (oC)
Relationship between saturated vapor
pressure and air temperature


Ta
es Ta   6.1078exp 17.269

T

237.3
a


49
Serdang RH and vapour pressure
100
es
50
80
40
30
ea
20
RH (%)
vapor pressure (mbar)
60
60
40
20
10
0
0
0
2
4
6
8
10 12 14 16 18 20 22 24
local solar hour
Air vapor pressure (ea) and saturated air
vapor pressure (es) for Serdang town
(3.0333 N; 101.7167 E), Malaysia on
31 October 2004
0
2
4
6
8 10 12 14 16 18 20 22 24
local solar hour
Relative humidity for Serdang on
31 October 2004
50
wind speed
Wind speed
0
2
4
6
8
10 12 14 16 18 20 22 24
local solar hour
Idealized daily trend for mean hourly wind speed
actual hourly wind speed can be erratic and difficult to simulate
51
Air temperature
air temperature (C)
35
I
II
III
30
25
20
sunrise
sunset
15
0
2
4
6
8
10 12 14 16 18 20 22 24
local solar hour, th
Measured air temperature for Serdang on 31 October 2004
52

Pattern of change
 before sunrise, air. temp gradually drops due
to heat loss from ground
 when sun rises, air temp. still drops because
heat gain from sun is not enough to overcome
heat loss from ground
 1-2 hours after sunrise, air temp. begins to
rise as heat supplied from sun increases

that’s why it is often the coldest just after dawn
53
 1-2

hours after solar noon (about 14:00
hours), air temp. reaches maximum then
starts to drop because ground radiation (heat
loss) now exceeds solar radiation
For simulation, divide the day into 3 sections
 Section I – before (sunrise + 1.5 h)
 Section II – from (sunrise + 1.5 h) to sunset
 Section III – after sunset
54

Tmin  Tset  24  th  tss 
Section I
Tset 
 tsr  1.5   24  tss 


   th  tsr  1.5  

Ta  Tmin  Tmax  Tmin  sin 
 Section II
tss  tsr




T T
t t
Tset   min set  h ss 
Section III

 tsr  1.5   24  tss 
Tmin & Tmax: min. and max. air temperature (C)
Tset
: air temperature at sunset (C)
– determine from Section II (th = tss)
tsr & tss
: sunrise and sunset time (hour)
th
: local solar hour
55
Serdang air temperature
air temperature, T a (° C)
35.0
Mean error 2%
30.0
25.0
20.0
measured
simulated
15.0
0
2
4
6
8 10 12 14 16 18 20 22 24
local solar hour, th
Comparison between actual and simulated air temperature
56
Part 3: Plant-radiation
regime
57
Interception
Irradiance above (I0) and below (I) the plant canopies
Intercepted = I0 - I
Intercepted radiation potentially available for transpiration and photosynthesis
58
Hypothetical plant canopy



A randomly placed leaf of area (a) over a ground
area (A)
 probability light intercepted is a/A
 probability light not intercepted is (1 - a/A)
A second randomly placed leaf (same area a)
over ground area (A)
 probability light not intercepted is (1 - a/A)2
So, N randomly placed leaves (all having area a)
over ground area (A)
 probability light not intercepted is (1 - a/A)N
59

For small leaf area a << A,
 (1 - a/A)N  exp(-Na/A)  exp(-L)
where L is the leaf area index (m2 m-2) or the total
leaf area in a unit ground area
 exp(-L) is known as the penetration function


But exp(-L) is for horizontal leaves only
 leaves are in all angles
 horizontal leaves, maximum light interception
 more vertical leaves, light interception
decreases
60
Interception of solar radiation depends on the solar
direction and leaf angle. Note: a is the area of the leaf
shadow on the ground, and aL the area of the leaf (one
side)
61

so we reduce light penetration by
 exp(-kdrL), where kdr is a value between 0 and
typically less than 1
 kdr is known as the canopy extinction
coefficient for direct light
 kdr = 1 means horizontal leaves, kdr < 1 means
leaves are not horizontal
 the smaller the kdr, the smaller the leaf angle

90 horizontal leaf; 0 erect leaf
62
kdr 
a
A*
where a is shadow area on ground; A* is exposed canopy area (sunlit leaves)

Most canopies have random (spherical) leaf distribution
 leaves are facing all directions equally (like the surface of a sphere)
63
Extinction coefficient is calculated as the ratio
between the area of canopy shadow on the ground
and the exposed surface area of the canopy
Thus,
 r2
sin 
2 r 2
0.5
0.5
kdr 
or kdr 
sin 
cos 
kdr 
64
Direct light
Direct light below canopies:
I p,dr  I dr exp  kdr L 
Direct light intercepted:
I i , dr  I dr  I dr exp  kdr L   I dr 1  exp  k dr L  
I
I0
I = I0 exp(-kL)
Attenuation of irradiance through a
canopy according to Beer’s law
0
L
65
Scattering
incoming
scattering = reflected + transmitted
reflected
scattering contributes to radiation
regime within canopies

 dr ,  exp   kdr L
transmitted
(through the leaf)
LEAF

where  is the scattering coefficient;
0.8 for PAR, 0.2 for NIR (near infrared),
and 0.5 for total solar radiation
(mean of both PAR and NIR).
66
Canopy reflection
incoming
reflected out of canopy

I p , dr  1  p  I dr exp   kdr L
into the
canopy
CANOPY



Ii , dr  1  p  I dr 1  exp   kdr L 


where p is the canopy reflection
coefficient, with it being equal to 0.04,
0.25 and 0.11 for PAR, NIR and
total solar radiation, respectively
67
Diffuse light
1.0
kdf 
0.9
1+0.1174 L
1+0.3732 L
kdf
0.8
0.7

I p, df  (1  p) I df exp   kdf L
0.6

0.5
0.4
0
2
4
6
8
10


Ii , df  (1  p) I df 1  exp   kdf L 


L
Canopy extinction coefficient for diffuse solar
radiation kdf at leaf area index (L) from 0.01 to
10 (random leaf distribution only).
Constant kdf at a given L.
68
Discontinuous canopies


Beer’s law assumes closed, homogenous
canopies
When early growth periods or sparse planting,
canopies are opened
 violates Beer’s law
Discontinuous canopies violate
one of the assumptions of
Beer’s law which require a
uniform distribution of canopies
69
For discontinuous canopies, modify Beer’s law by introducing a clump factor :
 dr ,  exp  kdr L 
  012 
0   ln  b  (1   b ) exp  kdr L (1   b )  kdr L
 L 3 for L  3
(1   b )  
 1 for L  3
70
PAR absorption



In photosynthesis modelling, we are interested
in the PAR irradiance incident on leaves, rather
than PAR intercepted
When direct solar beams enter the canopy, a
fraction of it will be intercepted by the leaves and
be scattered. Thus, the direct solar component
within the canopies is segregated into a
component that is scattered and the other that
remained direct.
The other component is the diffuse component
71

So, within the canopies, there are 3 components
 total direct component of PAR, Qp,dr
(unintercepted beam plus scattered beam)
Qp , dr  (1  p) dr , Qdr

 dr ,  exp   kdr L

 direct
component of the total direct PAR
component, Qp,dr,dr (unintercepted beam
without scattering)
Q p , dr , dr  (1  p ) dr Qdr
 dr  exp  kdr L 
 diffuse
PAR component, Qp,df
Q p , df  (1  p ) df Qdf

 df  exp   kdf L

72

Hence, the scattered component only is: Qp , dr ,  Q p , dr  Q p , dr , dr

2
Average diffuse component within canopies:
Qp , df


(1  p)Qdf 1  exp   kdf L 



 kdf L
73

Sunlit leaves absorption:
Qsl    kdr Qdr  Q p , df  Q p , dr , 

Shaded leaves absorption:
Qsh    Qp , df  Qp , dr , 
where  is the leaf absorption coefficient (0.8 for PAR)
74
Sunlit and shaded leaves
Recall:
a
kdr 
A*
where a is shadow area on ground; A* is exposed canopy area (sunlit leaves),
so applied to canopies,
kdr  a Lsl
 Lsl 
a
kdr
since we are taking ground area as 1 m2, a is 1 - exp(-kdrL) => fraction of
ground covered by shadows
Sunlit LAI:
1  exp(kdr L)
Lsl 
kdr
Shaded LAI: Lsh  L  Lsl
75
Conversion

W m-2 convert to mol (photons) m-2 s-1
 1 W m-2 is 4.55 mol (photons) m-2 s-1
76
Example

Determine the sunlit and shaded leaves PAR
absorption for a canopy with a spherical leaf
distribution and LAI of 3.0. The incident total
solar radiation above the canopies is 800 W m-2,
with the diffuse and direct solar components
comprising 40% and 60% of the total solar
radiation, respectively. Solar inclination is 40º.
77
Solution

PAR is typically 50% of total solar radiation
 so PAR flux density is half of the total 800 W m-2; that
is, 400 W m-2

or 400 x 4.55 = 1820 mol (photons) m-2 s-1.
 Of

this total PAR, diffuse and direct components are
0.4 x 1820 = 728 and 0.6 x 1820 = 1092 mol
(photons) m-2 s-1, respectively.
The three flux components within the canopies that must
be calculated are: the total direct component Qp,dr; the
direct component of the total direct component, Qp,dr,dr
and the mean diffuse component
78
kdr  0.5 cos 40  0.65


Qp , dr  (1  0.04) 1092  exp  0.8  0.65  3  183.2
Qp,dr ,dr  (1  0.04) 1092  exp  0.65  3  149.1
Qp,dr ,  183.2  149.1 2  17.1

 1+0.3732 3   0.73
(1  0.04)  728 1  exp   0.8  0.73  3 

  306.5

kdf  1+0.1174 3
Qp , df
0.8  0.73  3
Qsl  0.8  0.65 1092  306.5  17.1  826.7
Qsh  0.8  306.5  17.1  258.9
1  exp(0.65  3)
 1.3
0.65
Lsh  3.0  1.3  1.7
Lsl 
79
Part 4: Plant water
uptake and soil
evaporation
80
Energy balance

Rn = H + ET + G + M
 Rn = net radiation

(all in W m-2)
main energy supplier
H
= sensible heat density
 ET = latent heat flux density
 G = ground heat flux density

M
main energy
consumers
energy into or out of soil sub-surface
= miscellany energy density
small term (usually less than 5% of Rn)
 often neglected (M = 0)

81

Latent heat (ET)
 all energy supplied is to break bonds and
phase change water

no temperature change
 energy
to convert 1 kg liquid water to vapour
is 2454000 J
latent heat of vapourisation of water ()
 ET (kg water m-2 ground s-1) x  (J kg-1 water)
gives ET (J m-2 s-1 or W m-2)
 ET also known as evapotranspiration

82

Evapotranspiration
 water loss from both soil and plant
soil = evaporation
 plant = transpiration

 often

equals plant water uptake
so knowing ET, we know water uptake
 plants
can reduce transpiration
conditions of water stress, stomata openings are
reduced or closed
 bad in long term, photosynthesis reduced

83

Potential vs. actual ET
 PET

maximum ET that can occur under current
conditions if there was no water stress
 AET
actual ET occurring under current conditions
 may equal or be less than PET due to water stress

84

Sensible heat (H)
 energy supplied is to raise temperature


thus, can be “sensed”; can be measured by
thermometer
Heat transfer
 Rn = radiative
 ET, H and G = non-radiative

by conduction and convention only
 +ve
for energy flow from surface to air
 -ve for energy flow from air to surface
85
Energy balance for day and night
Day = ground gains heat
Night = ground loses heat
+ET = evaporation
-ET = condensation
86
K-theory transport
Vertical transport of a generic property 
F  
 A  B


or F  
or F   K 
z A  zB
z
z
87
Latent heat transport equation
ET  


or  ET  
z
z
 is the absolute humidity of air, which is defined as the mass of water vapor
contained in a given volume of air (kg m-3)
The problem with using absolute humidity is that the volume of air is
sensitive to changes in both the air temperature and pressure.
Absolute humidity changes when the volume changes, even though the
mass of water vapor has not changed.
For instance, a 1 m3 of air parcel which contains 2 g of water has an
absolute humidity of 2 g m-3. But if that air parcel is expanded to double its
volume (2 m3), this means the absolute humidity is now 1 g m-3 even though
the air parcel still contains the same weight of water in it (2 g). Given the way
absolute humidity is calculated it appears the amount of water in the air
parcel has decreased
88
so express it using air vapour pressure ea:
cp
e
 ET  
K ET a

z
where KET is the atmospheric transfer coefficient for sensible heat (m2 s-1);
 is the density of moist air (1.209 kg m-3); cp is the specific heat capacity
of moist air which is the amount of heat per unit mass of air required to
raise its temperature by one Kelvin (1010 J kg-1 K-1); and  is known as
the psychometric constant, and it has a value of 0.658 mbar K-1.
cp is known as the volumetric heat capacity: the amount of heat required
to raise the temperature of a unit volume of air by one K (1221.09 J m-3 K-1)
89
Sensible heat transport equation
H 
T
T
or H    c p K H
z
z
where KH is the atmospheric transfer coefficient for sensible heat (m2 s-1);
and cp gives the volumetric heat capacity of air.
Multiplication by cp is so that the above equation gives the amount of heat
transferred per unit area ground area per unit time (which is the heat flux
density).
90
Penman-Monteith equation
Uses the electrical network analogy to explain heat transfers
Penman-Monteith evapotranspiration (potential) model. Key: ET and H are
the latent and sensible heat fluxes, respectively; Tr and To are the temperatures for
the reference height and canopy, respectively; er and e0 are the vapor pressure at the
reference height and canopy, respectively; ra is aerodynamic resistance; rc is the
canopy (or soil surface) resistance.
91
cp
ea
 ET  
K ET

z
OR
cp
1
ea   ET 
z

K ET
which, incidentally, has the same form as Ohm’s law used to describe
electrical current flow: Potential difference (V) = Current (I) x Resistance (R).
So analogously,
cp
V
ea

I   ET
R
1
z
K ET
where the current (latent heat flux density) is driven by the potential difference
between two points (their vapor pressure difference) but is opposed by
resistance (the distance between the two points, and the reciprocal of the
atmospheric transfer coefficient KET). Recall that the atmospheric transfer
coefficient KET is the ease of atmospheric transfer, or its atmospheric
“conductance”. So taking its reciprocal 1/KET denotes the opposite: the
atmospheric resistance to transfer.
92
 c p er  e0
T  T0
 ET  
and H    c p r
 ra  rc
ra
Evaporation from a saturated environment (within stomata):
Rn  G   ET  H
 c p es T0   er
T T
Rn  G 
 cp 0 r

ra  rc
ra
93
But T0 as well as e0 are unknown. To eliminate them from calculations, a
vapour pressure deficit D is introduced which is defined as the difference
(deficit) between the current amount of moisture in the air and the maximum
amount of moisture the air can hold (i.e., saturation), or
D  es Tr   er
where es(Tr) is the saturated vapor pressure (mbar) at temperature Tr (C).
94
saturated vapor pressure (mbar)
 is the slope of the saturated vapor pressure curve (mbar K-1)
100

80
es Tr   es T0 
Tr  T0
60
for small differences between Tr and T0:
40
20

0
0
10
20
30
40
50
air temperature (oC)
Relationship between saturated vapor
pressure and air temperature

des Tr 
dTr
25029.4 exp 17.269Tr Tr  237.3  
Tr  237.3
2


Ta
es Ta   6.1078exp 17.269

T

237.3
a


95
Rn  G   ET  H
 c p es T0   er
T T
Rn  G 
 cp 0 r

ra  rc
ra
So introducing D and , and after some algebraic manipulations:
  Rn  G  ra  rc    c p D
H
ra    ra  rc 
 ET 
  Rn  G    c p D ra
    ra  rc  ra
96

Problem with PM equation
 assumes either evaporation or transpiration
but not both occurring simultaneously
 not applicable for open canopies (e.g., early
growth periods or sparse planting density)
97
Shuttleworth-Wallace equation

SW equation:
 extension of the PM equation
 ET occurs from both soil and plant
simultaneously
 good for early growth periods or for sparse
planting densities
98
Shuttleworth-Wallace evapotranspiration (potential) model. raa is the
aerodynamic resistance between the mean canopy flow and reference height; rsa
is the aerodynamic resistance between the soil and mean canopy flow; rca is the
bulk boundary layer resistance; rcs and rss are the canopy and soil surface
resistance, respectively.
99
Entire energy balance of the system can be described in 8 equations:
 c p e0  er
 ET 

raa
T T
H  cp 0 a r
ra
Hc  cp
 c p es T f  - e0
 ETc 

rac  rsc
T f  T0
rac
 c p es Ts   e0
 ETs 

ras  rss
Ts  T0
H s  cp
ras
Ac   ETc  H c
As   ETs  H s
or
Ac  1   dr ,  Rn
or
As   dr , Rn  G
100
After some algebraic manipulations:
PM c 
PM s 
 ET  Cc PM c  Cs PM s
A    c p D  rac As   raa  rac 
   1  rsc
r
a
a
 rac 
A    c p D  ras Ac   raa  ras 
   1  rss
r
a
a
 ras 
Cc  1  Rc Ra Rs  Rc  Ra 
1
Cs  1  Rs Ra Rc  Rs  Ra 
1
Ra       raa
Rc       rac   rsc
Rs       ras   rss
101
Once we know total ET, we determine its components:
D0  es T0   e0
 ETs 
Hs 
OR
As   c p D0 ras
   r  r
s
s
s
a

s
a
r
 As  rss  ras    c p D0
r    r  r
s
a
s
s
s
a

raa
D0  D 
 A        ET 
cp
and  ETc 
and H c 
Ac   c p D0 rac
    rsc  rac  rac
 Ac  rsc  rac    c p D0
rac    rsc  rac 
102
Soil heat flux (G):
G  0.35  dr , Rn  cos 
G is 35% of net radiation reaching the ground, and this fraction varies according
to the cosine of the solar inclination

103
Aerodynamic resistances
Wind speed decreases exponentially with height, and equals zero at a certain
height above ground/surface. How high above the ground/surface the wind
speed falls to zero is a measure of how rough the surface is: roughness length (z0)
Roughness length (z0) and zero-plane displacement (d)
In the presence of objects (e.g., canopies) the roughness length is displaced by
d (zero-plane displacement)
104
Roughness lengths z0 for some surface types (Hansen, 1993)
Surface
z0 (m)
Grass, closely mowed
0.001
Bare soil, tilled
0.002-0.006
Thick grass, 0.5 m tall
0.09
Forest, level topography
0.70-1.20
Coniferous forest
1.10
Alfalfa
0.03
Potato, 0.6 m tall
0.04
Cotton, 1.3 m tall
1.30
Citrus orchard
0.31-0.40
105
For crops with crop height h:
d  0.64h
z0  0.13h
u*  z  d 
ln 
;
k  z0 
Wind speed above canopy:
u( z) 
zh
Wind speed below canopy:
u ( z )  u (h) exp   1  z h   ; z  h
k is the von Karman constant (0.40)
u* is the friction velocity (m s-1): the effectiveness of air turbulence transfer
 is the wind speed attenuation coefficient (unitless)
106
Wind attenuation coefficients α for some vegetation types (Cionco, 1972)
Vegetation
α
Vegetation
α
Immature corn
2.8
Sunflower
1.3
Oats
2.8
Pine trees
2.4
Wheat
2.5
Larch trees
1.0
Corn
2.0
Citrus orchard
0.4
107
ras 
h  exp(n) 
 zs 0
exp
n


nK (h) 
h


 z0  d  
-1

exp

n
s
m



h  


  z0  d   
1
h 
 zr  d 


-1
r 
ln 

exp
n
1


1
s
m

 
 
ku*  h  d  nK (h) 
h

 



a
a
h = crop height (m); zs0 = roughness length of soil surface (m);
z0 = roughness length of crop (m); d = zero-plane displacement
(m); zr = reference height (m); n = eddy diffusivity coefficient
(unitless, n=2 to 3)
K(h) = eddy diffusivity transfer coefficient (m2 s-1)
= ku*h
108
Boundary layer resistance

Every surface has a thin boundary layer of still
air
 thicker the layer, the more resistance to
transfer of heat or vapour
 flow can be laminar, turbulent or mixed
 when turbulence is suppressed, transfer
occurs solely due to molecular diffusion (very
slow)
109
rac 

0.012 L 1  exp    2   u (h) w
s m-1
where L = leaf area index; u(h) = wind speed at canopy top of height h;
w = mean leaf width (m);  = wind speed attenuation coefficient
Equation includes turbulent effects
110
Stomatal resistance
stomatal resistance (s m -1)
rst 
a1  I PAR
s m -1
a2 I PAR
Canopy resistance is:
PAR (W m-2)
Stomatal resistance decreases with
increasing PAR (photosynthetically
active radiation) irradiance, following
the relationship by Jarvis (1976)
 rst L
r 
rst 0.5L cr
c
s
for L  0.5Lcr
for L  0.5Lcr
where Lcr is critical LAI, typically
taken as maximum LAI (=4)
111
Soil surface resistance
soil surface
water
dry layer
water
thickness, l
wet layer
assume that a soil is always made up of two layers: a thin upper layer that is
completely dry, and a thicker, lower layer that is wet.
water vapor traverses from the lower wet layer through the upper dry layer to
reach the soil-atmosphere boundary. This traversal by vapour through the dry
layer is by molecular diffusion.
vapor flux in the soil is controlled by four factors:
• vertical vapor pressure gradient between the dry and wet soil layers
• molecular diffusion coefficient of vapor in the soil
• soil porosity (fraction of soil that is made up of pores)
• soil tortuosity (ratio of the actual path length to the straight path length of
flow) (  1)
112
rss
rss (v )

rss,dry
rss, dry 

v
exp  

v, sat





rssdry
l
 p Dm,v
  1/ 
v / v,sat
Dm,v is the molecular diffusion coefficient of vapor in the soil (24.7 x 10-6 m2 s-1); 
is the tortuosity for soils (2); l is the upper dry layer depth (0.15 m); p is the soil
porosity; and  is the pore size distribution index
113
Pore size distribution index is the slope of the linear line of relative saturation
to soil suction


ln v v, sat   ln  e    ln  
ln (v / v,sat)
 ln e

higher the suction, drier the soil
and smaller the relative saturation

114
Conversion

W m-2 to mm (water) day-1
 (ET / ) x (60 x 60 x 24)
 e.g., (120 / 2454000) x 86400 = 4.2 mm day-1
 1 mm water is equivalent to 1 kg or 1 liter of
water in a 1-m2 ground area
115
Part 5: Water balance
116
Expressions of soil water content


usually expressed as depth of water (mm) or
volumetric water content (m3 m-3)
depth of water (mm):
 1 mm water depth is equivalent to 1 kg m-2
ground area or 1 liter m-2 ground area
117
Volumetric water content is the volume of water per unit volume of soil:
v 
volume of water
volume of soil
v 
area  depth of water depth of water

area  depth of soil
depth of soil
depth of water (mm) = v (m3 m -3 )  depth of soil (m)  1000
118
Water balance
R + I + CR = P + OF + ETa + 
(all in mm day-1)
WATER INPUT:
R = rainfall; I = irrigation; CR = capillary rise
WATER OUTPUT:
P = percolation; OF = overland flow; ETa = actual evapotranspiration;
 = change in soil water content
119

equation looks deceptively simple, but in
practice, the individual components can be
difficult to determine/measure
 use some assumptions:
1.
2.
3.

no irrigation supplied, so I = 0
deep water table (> 1 m deep), so CR = 0
flat, levelled land, so OF = 0
therefore water balance equation becomes:

R = P + ETa +  or  = R - P - ETa
120
Two-layered soil
Downward flow of water out of the top layer (i=1) is denoted by P1,t, and this
component subsequently becomes the percolation of water into the root zone
below (i=2). In other words, the infiltration of water into the second layer is P1,t
which is the percolation of water from the top layer. Within the root zone, water
will still flow downward, and any water leaving this zone is denoted by the
component P2,t which is regarded as deep percolation or drainage
121
Percolation

drainage (loss) of water from a soil layer/zone
 consists of two components:
1.
2.
percolation due to excess water pe
percolation due to redistribution pd
 P  pe  pd
122
Excess water percolates below if the amount of water in soil and amount of
water (due to rainfall R) received exceed the soil saturation level:
0

pe  
v  R  v , sat
if v  R  v , sat
if v  R  v , sat
pour 150 ml
pe = 40 + 150 – 100 = 90 ml
amount overflow, pe?
initial amount = 40 ml
max can hold = 100 ml
123

Redistribution occurs due to gravity and matric
potentials, as defined by Darcy’s Law
 gravity potential / energy

flow due to gravity (downward)
 matric


potential / energy
flow due to differences in water content (wet to dry)
Darcy’s Law
 flow is proportional to differences in potential
(or head) and inversely proportional to
distance
124

Flow, q  H/L or q = K H/L
 where L is distance (m); H is potential
difference (m); and K is hydraulic conductivity
(m day-1)
 H is total head which is the sum of matric and
gravity heads
 flow is faster if the difference in potentials is
larger, or the distance to flow is smaller
125
  Hm  H g 
HT
qK
K
z
z
If the depth difference between two soil layers is z, then Hg = z, and
qK
  Hm  z
z
 H m

 K
 1
 z

Assuming uniformly wetted soil means no differences in matric potential
any where in that soil layer, so
H m
0
z
which gives:
qK
Thus, flow is controlled only by the soil’s K
126
K depends on soil texture, soil structure and soil water content
K increases with increasing water content until maximum at soil saturation
hydraulic conductivity (m s-1)

v , sat  v
K  K sat exp  

v , sat

Ksat



where  is 13-16 for most soils
v,sat
volumetric water content (m 3 m-3)
127
Soil texture
Ksat (m day-1)
v,sat (m3 m-3)
Sand
15.21
0.43
Loamy sand
13.51
0.41
Sandy loam
2.99
0.41
Silty loam
0.62
0.45
Loam
0.60
0.43
Sandy clay loam 0.55
0.39
Clay loam
0.21
0.41
Sandy clay
0.19
0.38
Silty clay loam
0.15
0.43
Clay
0.11
0.38
Silty clay
0.09
0.36
Silt
0.06
0.46
128
Law of mass conservation
v
q

z
t
q/z is the change of water flux density q over the vertical distance z.
If q/z increases then this is the same as saying that qin < qout, and that
water storage in the volume element must decrease because more
water is lost from outflow qout than that gained by inflow qin.
Stated more specifically: the rate of increase of q with z must equal the
rate of decrease of volumetric water content v with time t.
129
If we take the soil layer thickness as L, then q   L
Earlier, we established q = K, so K   L
v
t
 v
t
v
t   L
K
re-arranging:
t2
 t 
t1

v 2


v 1
v 2

v 1

L
v
K
L


 v
K sat exp   v , sat
v , sat




v
130
So at time t2, the volumetric water content is:
v ,t2  v , sat 

 
 K sat  t2  t1 
ln 
 exp 
v , sat  v ,t1

 v , sat

 Lv , sat
v , sat





 

Therefore, percolation due to redistribution is
t2 - t1 = R – (pe + pd)
pd = t2 - t1 - R - pe
t2 is now available for evapotranspiration ETa
131
Actual ET


When water is limiting, evapotranspiration is not
at maximum but is reduced to a rate known as
actual ET
PET is scaled down to AET by a reduction factor
dependent on the amount of water in the soil
132
Actual soil evaporation
Ea  ETs  RD ,e
1.0
RD ,e 
1
1  3.6073  v v , sat  
-9.3172
reduction factor
0.8
0.6
0.4
0.2
Ea (1,t )  0.26 Ea and Ea (2,t )  0.74 Ea
0.0
0.0
0.2
0.4
0.6
0.8
1.0
relative soil water content
Potential soil evaporation is reduced to
actual evaporation by a reduction factor
that is dependent on the relative water
content
relative water content  v v , sat
133
Actual transpiration
1.0
Ta  ETc  RD ,t
RD ,t   v  v , wp   v ,cr  v , wp 
v , cr  v , wp  p  v , sat  v , wp 
reduction factor
0.8
C4
0.6
C3
0.4
critical point
0.2
all transpiration is from water in the
second soil layer only
0.0
0.0
v,wp
0.2
0.4
0.6
0.8
1.0
relative soil water content
Potential plant transpiration is
reduced to actual transpiration by a
reduction factor that is dependent on
the relative water content
relative water content   v  v , wp   v , sat  v , wp 
134



Plant cannot use the water below the soil wilting point
level
Most agricultural crops are C3 plants; only three are C4:
sugar cane, maize and sorghum
 C3 plants photosynthesize to produce a 3-C
compound (3-phosphoglyceric acid) and C4 a 4carbon compound (oxaloacetic acid). C4 are more
efficient in using water and solar radiation to convert
into biomass.
Critical water point for C3 and C4 plants are 50% and
30% of relative water content, respectively. C4 more
efficient in using water.
135
Photosynthesis (C3)
136
Empirical approaches
Assimilation rate (µmol m -2 s-1)
20.0
Lmax
15.0
Goudriaan and
van Laar (1978)
10.0
de Wit (1965)
de Wit (1965):
L max I PAR
L
I PAR  hI PAR
Goudriaan and van Laar (1978):
L  L max 1  exp   I PAR hI PAR  
5.0
0.0
0
20
40
60
PAR (W m-2)
80
100
Gross photosynthesis as a function of absorbed PAR, determined using the
equations by de Wit (1965), and Goudriaan and van Laar (1978)
137
Total dry matter (g m-2)
L n , canopy  e  f  I t , d
800
Ln,canopy is the net canopy
assimilation rate of CO2; e is
the RUE; It,d is the total solar
radiation incident on the
canopy; and f is the fraction
of It,d intercepted by the
canopy.
y = 1.32x
600
400
200
0
0
100
200
300
400
500
600
-2
Cumulative radiation intercepted (MJ m )
Total dry matter as a function of cumulative
total solar radiation intercepted (Monteith,
1977)
RUE for C3 and C4 plants
are typically measured at
about 1.9 and 2.5 g MJ-1 of
intercepted PAR
138
Mechanistic approaches


Today, increasingly more mechanistic photosynthesis
models are being used.
 Over the years numerous mathematical models have
been developed that take into account the underlying
mechanism of photosynthesis such as the diffusion of
CO2 into the chloroplast, enzyme kinetics, and the
biochemical reactions in the carbon-reduction cycle.
Two most significant work to build upon this progress are
by Farquhar et al. (1980) and Collatz et al. (1991)
139
Light and dark reactions

Photosynthesis equation:
light
6CO2 +12H 2 O   CH 2 O 6 +6O 2 +6H 2 O


Photosynthesis involves two chemical reaction steps:
light and dark reactions.
The light reactions involve the absorption of sunlight by
chlorophyll and carotenoids to produce the energy
carriers, NADPH and ATP, via a process called
photophosphorylation
 occur in the thylakoid membranes inside the
chloroplast, and NADPH and ATP are subsequently
transferred from there to the stroma for use in the
dark reactions
140



Term “dark reactions” is misleading because
these reactions do not occur in the dark
 so-called only because they do not require
light for their reactions.
Dark reactions are the biochemical reduction of
CO2 to carbohydrates using the high energy
carriers NADPH and ATP.
 occur in the stroma inside the chloroplast.
Depending on the plant type, there are three
pathways of dark reactions: C3, C4 and CAM
(Crassulacean acid metabolism)
141
Photosynthesis occurs in the chloroplast, an organelle in the mesophyll cells
142




The C3 pathway
 converts atmospheric carbon into a chemical compound with
three carbon atoms (3-phosphoglyceric acid or PGA),
The C4 pathway
 produces a compound with four carbon atoms (oxaloacetic acid)
The CAM pathway
 so-called after the plant family in which it was first found
(Crassulaceae)
 CO2 is stored in the form of an acid (malic acid) before use in
photosynthesis.
 Unlike C3 and C4 plants, CAM plants open their stomata at night
and close them during the day.
Most agricultural crops are C3 types, whereas sugar cane, maize
and sorghum are C4 types. CAM plants include pineapple and
dragon fruit (pitaya), and many succulents such as cacti.
143
Calvin cycle. Shaded area denotes the carboxylation cycle.
144
The major processes and their equations in the Calvin cycle
a) Carboxylation of RuBP: RuBP + CO2 + H2O  2PGA
energy
b) Regeneration of RuBP: 2PGA 
 RuBP+CH 2 O+O2
(NADPH+ATP)
c) Oxygenation of RuBP: RuBP + O2  PGA + PGIA
d) Regeneration of PGA: PGIA + 0.25O2  0.5PGA + 0.5CO2
145


Cycle begins by the enzyme Rubisco (ribulose
biphosphate-carboxylase/oxygenase) fixing one
mol of CO2 to one mol of RuBP (ribulose 1,5biphosphate) to produce two mol of PGA. This
step is the start of the carboxylation cycle.
Regeneration of RuBP requires two mol of PGA
and energy from NADPH and ATP to yield one
mol of RuBP and one mol each of a
carbohydrate compound and oxygen.
146





Rubisco is an enzyme that catalyzes both the fixation of
CO2 and O2 with RuBP.
CO2 competes with O2 for RuBP.
The fixation of O2 to RuBP is known as photorespiration
(or oxygenation), and it is regarded as a wasteful
process because about 10% of assimilated carbon is
lost.
In photorespiration, Rubisco fixes one mol of oxygen
with one mol of RuBP to give one mol each of PGA and
PGIA (2-phosphoglycolate).
PGA is regenerated when one mol of PGIA reacts with a
quarter mol of oxygen to yield half mol each of PGA and
CO2. So the final result of oxygenation is the release of
CO2 which is a loss of assimilated carbon in making
carbohydrates.
147


Carboxylation and oxygenation are highly temperature
dependent.
Moreover, the ratio between carboxylation and
oxygenation depends primarily on concentrations of CO2
and O2 at the carboxylation site and temperature.
 With increasing temperature, for example, the
solubility of CO2 relative to O2 decreases; thus, this
favors oxygenation and results in a greater loss of
assimilated CO2.
 Oxygenation is also favored over carboxylation in
high solar irradiance.
148



Carbon losses, however, are not solely due to photorespiration, but
also due to dark respiration.
“dark respiration”
 occurs all the time regardless of the presence or absence of
light. This is in contrast to photorespiration which occurs only in
light.
 occurs in the cell organelle mitochondria
 it is a process whereby energy is released from the oxidation of
organic compounds, and this energy is used by the plants for
their living maintenance (upkeep of cell activities), and growth
(synthesis of structural compounds).
Both photorespiration and dark respiration are components of plant
respiration, and both result in losses of assimilated carbon
149
Enzyme kinetics: Michaelis-Menten equation
k
k2
1


E  C 
EC

P  E
k 
1
Velocity (µmol s-1)
200
Vmax
150
v
Vmax [C ]
[C ]  K c
100
50
Kc is defined as the substrate
concentration at half the maximum
velocity
Kc
0
0
200
400
600
800
1000
[C] (µmol mol-1)
150
Competitive inhibition
k
k2
1


E  C 
EC

P  E
k 
1
k
and
o 

E  O 
EO
k 
o
In competition with O, velocity of C reaction is
v
Vmax [C ]
K c 1  [O ] K o   [C ]
151
Rubisco-limited
In plants, there is a minimum amount of internal CO2 below which there is no
CO2 assimilation. This critical or minimum amount of CO2 is known as the CO2
compensation point (*).
applied to photosynthesis:
vc 
Vc ,max  Ci   *
Kc 1  Oa Ko   Ci
vc is the carboxylation velocity (per unit leaf area) (mol m-2 s-1); Vc,max is the
maximum carboxylation velocity (per unit leaf area) (mol m-2 s-1); Ci is the
concentration of internal (intercellular) CO2 in air (mol mol-1); Oa is the
concentration of ambient O2 in air (mol mol-1); and Kc and Ko are the
Michaelis-Menten constants for CO2 and O2, respectively (mol mol-1)
* is the CO2 compensation point (mol mol-1) so that when Ci = *, vc = 0
Oa
* 
2
152
Kc(25)
Michaelis-Menten constant for CO2 (300 mol mol-1)
Ko(25)
Michaelis-Menten constant for O2 (300000 mol mol-1)
(25)
CO2 / O2 specificity factor (2600 mol mol-1)
Vc,max(25) Rubisco capacity rate (per unit leaf area) (200 mol m-2 s-1)
em
Quantum efficiency (0.06 mol mol-1)
α
Leaf absorption for PAR (0.8)
Oa
Ambient concentration of O2 in air (210000 mol mol-1)
Ci
Intercellular CO2 concentration in air (C3 plant) (245 mol mol-1)
 reflects the ability of Rubisco to discriminate between CO2 and O2. The
higher the value, the better the discrimination and the higher productivity for
CO2.
153
Sensitivity to temperature
Kc, Ko, , and Vc,max are highly sensitive to temperature. It is assumed that these
parameters depend on temperature in the same way as the activity of enzyme
depends on temperature.
The reaction rates of enzyme typically increase by a factor of 2 for every 10 C
increase in temperature, or
  (25) 2(T  25) /10
where  and (25) are the enzyme reaction rates at temperatures T and 25 C,
respectively. To generalize
(T 25)/10
  (25) Q10 f
where (25) is the model parameter value at 25 ºC (Kc(25), Ko(25), (25) or Vc,max(25));
Tf is the actual leaf temperature (ºC); and Q10 is the relative change in the
parameter  for every 10 ºC change.
154
Q10 values
Kc(25)
Michaelis-Menten constant for CO2
(2.1)
Ko(25)
Michaelis-Menten constant for O2
(1.2)
(25)
CO2 / O2 specificity factor
(0.57)
Vc,max(25) Rubisco capacity rate (per unit leaf area)
(2.4)
155
Additionally, Vc,max is sensitive to inhibition at temperatures exceeding 35
ºC. Vc,max needs a high temperature cutoff, after which it decreases rapidly.
Studies have shown that the critical temperature value for Vc,max is
actually at about 40 ºC. The high temperature responses of Vc,max is
determined empirically by
Vc ,max 
Vc ,max(25) 2.4
(T f  25) /10
1  exp 0.128 T f  40  
156

Photosynthesis can be limited by amount of light
and amount of sink (e.g., sucrose) already
produced
 light can be limited, so photosynthesis can be
limited by light
 too much sink may have already been
produced, so photosynthesis becomes limited
by the sink
157
Light-limited
Ci   *
vq  em Qt
Ci  2 *
Qt is the PAR flux density based on per unit leaf area, and not per unit
ground area. Additionally, Qt is expressed in the unit mol (photons) m-2 s-1,
rather than the usual unit W m-2 for solar radiant energy
em is the intrinsic quantum efficiency (also known as quantum yield) for CO2
uptake (mol mol-1), which is the number of moles of CO2 fixed per unit mole
of absorbed PAR
C4 plants = 0.053 mol mol-1
C3 plants: em  0.081  0.000053Tf  0.000019Tf
2
158
Sink limited
vs 
Vc ,max
2
vs is the sink-limited rate of CO2 assimilation (per unit leaf area) (mol m-2 s-1)
159
Gross assimilation rate
What we have now is three equations to calculate the gross CO2 assimilation
rate. Which equation to select depends on which factor (Rubisco, light or sink)
that is most limiting to gross assimilation:
L  MIN  vc , vq , vs 
where MIN() denotes the minimum of the enclosed values.
160
Gross assimilation rate (µmol m-2 s-1)
60
50
Qp = 1800
40
Qp = 1200
30
20
Qp = 600
10
0
0
10
20
30
Leaf temperature (ºC)
40
Photosynthesis response to temperature at three levels of PAR
flux densities (Qp, mol m-2 s-1)
161
Gross assimilation rate (µmol m-2 s-1)
60
30 ºC
50
40
20 ºC
30
10 ºC
20
10
0
0
500
1000
1500
-2 -1
PAR (µmol m s )
2000
Photosynthesis response to PAR at three levels of leaf temperature
162
Gross assimilation rate (µmol m-2 s-1)
90
Qp = 1800
60
Qp = 1200
Qp = 600
30
0
-30
-60
0
100
200
300
400
500
600
-1
C i (µmol mol )
Photosynthesis response to CO2 concentration at three levels of PAR
flux densities (Qp, mol m-2 s-1). Leaf temperature is fixed at 30 ºC.
163
Leaf temperature Tf
Solve energy balance first to obtain total sensible heat (H):
Hraa
T0 
 Tr
cp
Determine the sensible heat for crop (Hc), then calculate Tf:
Hc  cp
T f  T0
rac
H c rac  Hraa
Tf 
 Tr
cp
164
Scaling up to canopy photosynthesis
Divide the canopy into sunlit and shaded leaf classes and calculate the
assimilation rate for each leaf class. The gross assimilation rate for each leaf
class is then summed based on the fraction of leaf area for each class:
Lcanopy  Lsl Lsl  Lsh Lsh
165
Example

Determine the gross canopy photosynthetic rate
at 20 ºC (leaf temperature) for a canopy with a
spherical leaf distribution and LAI of 3.0. The
incident total solar radiation above the canopies
is 800 W m-2, with the diffuse and direct solar
components comprising 40% and 60% of the
total solar radiation, respectively. Solar
inclination is 40º.
166
Solution

PAR is typically 50% of total solar radiation
 so PAR flux density is half of the total 800 W m-2; that
is, 400 W m-2

or 400 x 4.55 = 1820 mol (photons) m-2 s-1.
 Of

this total PAR, diffuse and direct components are
0.4 x 1820 = 728 and 0.6 x 1820 = 1092 mol
(photons) m-2 s-1, respectively.
The three flux components within the canopies that must
be calculated are: the total direct component Qp,dr; the
direct component of the total direct component, Qp,dr,dr
and the mean diffuse component
167
kdr  0.5 cos 40  0.65


Qp , dr  (1  0.04) 1092  exp  0.8  0.65  3  183.2
Qp,dr ,dr  (1  0.04) 1092  exp  0.65  3  149.1
Qp,dr ,  183.2  149.1 2  17.1

 1+0.3732 3   0.73
(1  0.04)  728 1  exp   0.8  0.73  3 

  306.5

kdf  1+0.1174 3
Qp , df
0.8  0.73  3
Qsl  0.8  0.65 1092  306.5  17.1  826.7
Qsh  0.8  306.5  17.1  258.9
1  exp(0.65  3)
 1.3
0.65
Lsh  3.0  1.3  1.7
Lsl 
168
Determine the gross leaf assimilation rate for the sunlit and shaded leaves
that have absorbed Qsl = 826.7 and Qsh = 258.9 mol m-2 s-1, respectively.
Referring to the chart, Lsl = 27.8 mol m-2 s-1 and Lsh = 8.7 mol m-2 s-1.
So, gross canopy assimilation rate of CO2 (per unit ground area) is:
Lcanopy   27.8 1.3  8.7 1.7   50.9  mol m-2 s-1
Homework: Use the equations as shown earlier to determine Lsl and Lsh.
169
CO2 to carbohydrates


Every gram of CO2 is equal to 30/44 g of CH2O.
This is because one mol of CH2O and CO2 are
equivalent to their molecular weights of 30 and
44 g of CH2O and CO2, respectively. Likewise,
one mol of CO2 is 44  10-6 g of CO2
So this translates to every one mol CO2 m-2
day-1 being equivalent to 30  10-6 g CH2O m-2
day-1.
170
Part 7: Respiration
171



Plant respiration is separated into two
components: photorespiration and dark
respiration
Photorespiration is the oxygenase reaction
between O2 and Rubisco, resulting in the loss of
assimilated carbon
Dark respiration is the oxidation of
carbohydrates (plant food) to release energy,
which is used for living maintenance and growth
172
Conceptual model of plant respiration
173



A maintenance tax (RM) is subtracted from the total
substrate supply (i.e., gross photosynthesis, Lcanopy)
 leftover (Lcanopy - RM) is used for tissue production
with a growth efficiency Yg
Yg is the plant’s ability to convert substrate into new plant
structures.
 the plant’s ability to convert substrate into new plant
structures.
What is actually used for the production of new plant
materials is Yg(Lcanopy - RM), and the leftover RG = (1Yg)(Lcanopy - RM) is lost via growth respiration
174
Maintenance respiration


Maintenance respiration is required to sustain
living tissues. Even the maintenance of electrical
potentials across the cell membranes requires
energy.
Maintenance respiration varies widely,
depending on the plant species, and even differs
among different parts of the same plant
175
RM (25) 

i  plant parts
k M ,iWi
  k M , greenleavesWgreenleaves    kM , stemWstem    kM ,rootsWroots    k M , storageWstorage 
R’M is the maintenance respiration rate (g CH2O m-2 ground area day-1); kM is the
maintenance respiration coefficient (g CH2O g-1 dry matter day-1); and W is the
plant weight (g dry matter m-2 ground area). Maintenance respiration is taken to
be proportional to the plant weight W
Maintenance respiration coefficient (at 25 C air temperature) for various plant parts
Plant part
g CH2O g-1 dry matter day-1
green leaves, kM,greenleaves
0.030
stem, kM,stem
0.015
roots, kM,roots
0.015
storage organs, kM,storage
0.010
176
Maintenance respiration rates, however, are highly dependent on temperature. It
is assumed that the maintenance respiration rates depend on temperature in the
same way as the enzyme activities depend on temperature, where the Q10 value
for maintenance respiration is taken as 2.
RM  RM (25) Q10(Ta  25) /10
Maintenance rates additionally need to be corrected for plant age. As the plant
ages, its protein content decreases but the amount of stable components such
as support tissues and reserve compounds increases. This, in turn, decreases
the maintenance respiration requirement.
RM  RM Wgreenleaves Wleaves
Wleaves is the weight of both green and dead leaves (g dry matter m-2 ground
area). As the plant ages, the proportion of dead leaves to green leaves
increases (due to leaf death). This lowers the fraction of green leaves to total
leaves (Wgreenleaves/Wleaves) and decreases the maintenance respiration rate
accordingly.
177
Growth respiration


The amount of glucose required to synthesize a
new material depends on the chemical make up
and its amount in the material.
It has been found that this production process,
unlike plant maintenance, is independent of
environmental conditions, and dependent only
on the nature of the plant component formed.
178
Glucose requirement G and the carbon content for major biochemical groups in
plant tissues
G (g CH2O g-1
dry matter)
Carbon content
(fraction)
Carbohydrates
1.242
0.450
Proteins
2.700
0.532
Lipids
3.106
0.773
Lignin
2.174
0.690
Organic acids
0.929
0.375
Minerals
0.050
0.000
Biochemical group
179
Fractions of major biochemical groups in several plant parts, and the
calculated glucose requirement for the production of the plant parts
Group
Young
leaf
Wheat
seed
Broad
bean
Oil-rich Wood
seed
stem
Sugar
beet
roots
Carbohydrates
0.53
0.79
0.54
0.15
0.49
0.78
Proteins
0.25
0.12
0.33
0.30
0.02
0.05
Lipids
0.05
0.02
0.01
0.48
0.01
0.00
Lignin
0.05
0.03
0.04
0.03
0.38
0.05
Organic acids
0.06
0.02
0.04
0.02
0.05
0.06
Minerals
0.06
0.02
0.04
0.02
0.05
0.06
G
1.656
1.452
1.719
2.572
1.569
1.271
Gleaves   0.53 1.242    0.25  2.700    0.05  3.106  
 0.05  2.174    0.06  0.929    0.06  0.05
 1.656
180
Total glucose requirement for growth:
G

i  plant parts
Fi Gi
  Fgreenleaves Ggreenleaves    Fstem Gstem    Froots Groots    Fstorage Gstorage 
G is the total glucose requirement (g CH2O g-1 dry matter); Gi is the glucose
requirement for each plant part i (i.e., green leaves, stem, roots and storage
organs) (g CH2O g-1 dry matter); and Fi is the fraction of dry matter of the
individual plant parts
Shortcut: nitrogen content is equal to the protein content multiplied by 0.16
(the average nitrogen N content of proteins is about 16%):
G  3.736C  6.136 N  0.251
only need to analyse total C and N content – faster, cheaper and simpler
181
Since (Lcanopy – RM) is the amount of assimilates potentially available for growth
(expressed in g CH2O m-2 day-1), (Lcanopy - RM) / G is the total weight of dry
matter actually produced in a unit ground area per day (expressed as g dry
matter m-2 day-1). Thus, the weight of a plant part is incremented by
Wi ,t 1
 L canopy  RM
 Wi ,t  Fi 
G


 t

where Wi,t and Wi,t+1 are the weights (g m-2 ground area) of a given plant part i
(e.g., green leaves, stem, roots or storage organs) at the current time step t
and next time step t+1, respectively; t is the interval for each time step
(days); and Fi is the fraction of dry matter of plant part i
182
Typical glucose requirement for the synthesis of various plant parts
Plant part
g CH2O g-1 dry matter
green leaves, Ggreenleaves
1.463
stem, Gstem
1.513
roots, Groots
1.444
storage organs, Gstorage
1.415*
* highly variable
183
Growth development stage


Phenology is the study of the timing of life cycle
events, and how they respond to their
environment, in particular to weather
The current phase in the growth of a plant is
known as the growth development stage (s),
and it is defined by its physiological age and the
formation of its various organs and their
appearance
184


The growth development of a plant is
punctuated by several milestones or points of
significance such as the point of seed
emergence, flowering, tuber initiation, bulking
and maturity, ripening, plant maturity, and
senescence
Very often, the most important point in the
growth development stage is the switch from
vegetative to reproductive stage
185

Adopt a one-dimensional and irreversible scale
to denote the growth development stage
 no “standard” to follow – up to us to decide
 seed emergence (s = 0), flowering (s = 1),
and plant maturity (s = 2)
186
Scale for the growth development stage for barley and wheat (as
defined by the Agrometeorological Centre of Excellence, Canada)
Scale
Description
0
Planting
1
Emergence (more than half the plants are visible)
2
Jointing (earliest date for 1st internode elongation,
appearance of first leaf)
3
Heading (base of head reaches the same height as the
base of the short blade)
4
Soft dough (kernel deforms easily but no “milk”
exudes)
5
Ripe (kernels can no longer be deformed)
187
Growth development rate



The progress rate of plant growth is known as
the growth development rate (r; day-1)
The rate at which the growth development stage
advances
 The higher the rate, the earlier the next
milestone in the development stage is
reached
Temperature is often the most important factor
that determines the growth development rate
 Other factors: vernalization and day length,
depending on the plant species
188


Growth development rate is not possible to measure
directly
Determined indirectly by measuring the duration (in
days) between two growth development stages
 setup an experiment in a controlled environment
where the air temperature is set constant at a certain
value, and the time it takes for a plant to reach a
certain growth development stage is recorded
 then repeat experiment a few more times but using a
different constant air temperature for each experiment
189
0.08
60
0.06
days
1/days
40
0.04
20
T b = 2.6
0.02
0
0
0
10
20
30
40
temperature (deg. C)
Dependence of growth duration
on air temperature
0
10
20
30
40
temperature (deg. C)
Dependence of growth
development rate on air
temperature
Tb = base temperature, below which there is no growth
(note: no –ve growth)
190
Base temperature for some crops
Crop
Field pea, lentils, linseed, oats, spinach
Base temperature (ºC)
1-2
Barley, rape, wheat
3
Lettuce
4
Asparagus, peas
3-6
Canola and forages
5
General plant growth
5
Potatoes
6-7
Safflower, sunflower
7-8
Beans, cucumbers, maize, soybean
10
Millets
Cowpea, sorghum
Pumpkins, tomatoes
8-14
11
10-13
Castor, peanut, pigeon pea
13
Guar
15
Sesame
16
Melons
15-18
191
For a given air temperature T, the growth development rate r (day-1) can
be determined by
r 
 s ,i 1   s ,i
tT
where s,i and s,i+1 are the growth development at stage i and i+1,
respectively; and tT is the time period (days) between s,i and s,i+1 at
constant growing air temperature T ºC.
But if we keep every successive milestone in the growth development
stage one unit apart, then
r 
1
tT
192
 s ,t 1   s ,t   r ,t t
where the subscripts t and t+1 represent the time step at t and t+1,
respectively; t represents the time interval between two time steps (days);
and r t gives the advance in growth stage that had occurred from the time t
to t+1.
193
Leaf area index growth


The increase in leaf area index is determined
from the current weight of green leaves and
specific leaf area (SLA)
SLA is the ratio of leaf area to leaf weight, and it
is an important indicator of how much biomass is
allocated to the expansion of leaf area. The leaf
area index for the next time step t+1is
determined by
Lt 1  Wgreenleaves ,t  SLAt
SLA is typically 0.022 m2 g-1
194
Leaf death
There are two possible reasons for leaf death: 1) leaf age, and 2) self-shading
of leaves. Leaf death due to age typically occurs after flowering (post-anthesis):
Dage
0


  r

 2  s

 r
 0.1
for
 2   s   1.0
for 0.1   2   s   1.0
for
 2   s   0.1
where Dage is the leaf death rate due to leaf age (day-1). Dage is proportional to
the growth development rate as well as to leaf age. With increasing leaf age,
s approaches 2 (reaching maturity) so that 2- s becomes increasingly small,
so that leaf death rate becomes increasingly large. A minimum of 0.1,
however, is set as the difference between 2 and s to avoid any excessive leaf
death rates at the later growing stages.
195
Leaf death is also due to self-shading where the shading from the upper parts
of the canopy diminishes the solar irradiance within the lower plant canopy;
thus, causing leaf deaths in the lower canopy parts. A critical LAI (leaf area
index) of 4.0 is typically chosen as the point where self-shading becomes a
significant effect. Leaf death rate due to self-shading is determined by
Dshade
for L  Lcr

0


 MIN 0.03, 0.03  L  Lcr  Lcr  for L  Lcr
where Dshade is the leaf death rate due to leaf self-shading (day-1); L and Lcr
are the LAI and critical LAI (m2 m-2), respectively; and MIN[] is the minimum
of the enclosed values. Here, it is assumed that leaf death rate due to selfshading begins after LAI exceeds a critical value (typically Lcr = 4.0), after
which death rate increases linearly with increasing LAI until a maximum value
of 0.03 day-1. This maximum value is set to avoid any excessive leaf death
rates at the later growing stages
196
The actual death rate of leaves Dleaves (day-1) is then the larger of the two
rates Dage and Dshade:
Dleaves  MAX  Dage , Dshade 
where MAX() the maximum of the enclosed values. The weight of dead
leaves can then be calculated as
Wdeadleaves ,t 1  Wdeadleaves ,t  Wgreenleaves ,t Dleaves ,t  t 
where Wdeadleaves,t+1 and Wdeadleaves,t are the weights of dead leaves for the
next time step t+1 and current time step t; and (Wgreenleaves Dleaves t) gives
the weight of green leaves that have died within the time period between t
and t+1.
197
The weight of green leaves is calculated differently from that for stem, roots and
storage organs because the death rate of green leaves must be subtracted from
the growth rate of green leaves.
Wgreenleaves ,t 1  Wgreenleaves ,t

 Lcanopy  RM
  Fgreenleaves 
G




  Wgreenleaves ,t Dleaves ,t  t


198
Plant height growth
ht 1  ht 
dh
t
dt
height (m)
b0b1hm exp  b1Tts 
dh
 Tts 
2
dt
1  b0 exp  b1Tts  
accumulated temperature sum (deg. C day)
Typical plant height growth
following the logistic function
where Tts is the accumulated
temperature sum (ºC day); ht is the
current height (m) at time t; hm is the
maximum possible height of the plant
(m); and b0 and b1 are the intercept
(unitless) and slope (ºC-1 day-1)
coefficients, respectively
199
Temperature sum

A plant must accumulate a certain number of
temperature sum (or heat units) to advance to
the next milestone in the development stage
Tts  Tts ,t    H Tavg ,t  Tb   Tavg ,t  Tb 
t
Unit function:
t
0 for x  0
H ( x)  
1 for x  0
Increment temperature sum only when mean temperature is greater
than base temperature
200
Example
An example showing the daily temperature sum Tts and accumulated
temperature sum Tts (base temperature Tb is 5 ºC)
Day
1
2
3
4
5
6
7
8
9
Tavg
4
5
8
10
12
10
8
5
8
Tts
0
0
3
5
7
5
3
0
3
Tts
0
0
3
8
15
20
23
23
26
Tavg = average of Tmax and Tmin
201
Root elongation



Roots grow to a certain maximum depth,
provided that they are not limited by soil
conditions such as a compacted layer
The maximum depth depends very much on the
plant species and ranges from 0.5 to 1.5 m or
more
Root growth generally stops at about the
flowering stage, and root elongation rate is
surprisingly quite independent of root weight
202
d r ,t 1
for v ,t  v , wp or  s  1

 d r ,t

 MIN  d m ,  d r ,t  d g  t   for v ,t  v , wp and  s  1

where dr,t and dr,t+1 are the rooting depth (m) at time step t and t+1,
respectively; dm is the maximum rooting depth (m); dg is a constant root
elongation rate, denoting rooting depth increase per day (m day-1); t is
the time step interval (days); s is the growth development stage; and
v,t is the volumetric soil water content at time t (m3 m-3); and v,wp is
volumetric soil water content at permenant wilting point (m3 m-3).
Assume that root elongation growth only occurs before flowering (preanthesis or s < 1) and when the soil water content exceeds the
permenant wilting point (i.e., v,t > v,wp). No growth occurs if these two
conditions are violated (i.e., dr,t+1 = dr,t). Moreover, root elongation cannot
exceed the maximum possible root depth (dm).
203
Water stress effects
When water supply is insufficient, plant growth is additionally limited to water.
First: we need to determine the level of water stress being experienced. This is
equivalent to the ratio between actual and potential plant transpiration:
RD ,t  Ta ETc
Then reduce plant height growth rate, root elongation rate and gross
assimilates produced:
dh dh

 RD ,t
dt
dt
d g  d g  RD ,t
Lcanopy  L canopy  RD ,t
204