1

PRT4301:Modelling and computer simulations in agriculture

Dr. Christopher Teh (Room C202)

chris@agri.upm.edu.my

Tel: 8946 6976

2

Pre-amble

Objectiveunderstand how mathematics is applied in

agriculture, in particular in crop growthunderstand how computer models are built

and used 2+1 credits

lab is completely in the computer lab

3

Exams20% Test 120% Test 210% Lab work50% FinalAlways in essay format (never in multiple

choice @ objective format)

4

Reading materialsKropff, 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.

5

Reading materialsMonteith, 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.

6

Reading materialsTeh, C. 2006. Introduction to mathematical

modeling of crop growth: how the equations are derived and assembled into a computer model. BrownWalker Press, Florida, US.

7

Part 1:What is mathematical modelling?

8

Definition of a modelsimplified representation of real systems

Types of modelspictorial, conceptual/verbal, physical,

mathematical Definition of a mathematical model

represents a real system in a mathematical form (one or more equations)

9

Uses of mathematical modelshelp us to understand, predict and control a

system identify areas of deficient knowledge less experimentation by trial-and-erroranswer various “what if?” scenariosadd value to experimentsmay replace experiments (in rare cases)encourage collaboration among researchers

from various disciplines

10

Characteristics of models incomplete description of real systemsmodels built from assumptionsmodel simplicity vs. model accuracyno one best model for all circumstancesnot about computers or ICT

11

Modelling methodology: an example

How to determine the number of leavesin this tree (or any in trees)?

Most accurate method: manually counteach and every leaf

Problem: tedious and time-consuming

Alternative: N = nl x nb

where N is no. of leaves; nl is averageno. of leaves per branch; and nb is no.of branches

nl = 153, nb = 99, so N = 15,147 leaves

12

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)

argl esmall

small

VN

V

So max. no. of small boxes that canfit into the large box is:

13

So applied to canopy:

The geometrical shape of an ellipsoid is used to represent the canopy volume of a tree and the volume a single leaf

42 2 23 6c

WH LV HWL

342 2 23 6l

l l lV l

c lN V V

63 3

6

c

l

V HWL HWLN

V l l

Vc = vol. of canopyVl = average vol. of one leaf

So no. of leaves in canopy (N) is:

14

H = 3.5m; W = L = 2.5m; l = 0.14m, so N is 7,185 leaves

vs. 15,147 leavesmodel error of 47% (large!)

Why the error?check our assumptions

15

Revised model:

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.

42 2 23

42 2 23

16

o c e

H W L

H W L

V V V

WH L

WH Lf f f

HWL f f f

Vol. of canopy occupied by leaves is:

16

So number of leaves (N) is now:

o lN V V

Vo = vol. of canopy occupied by leavesVl = average vol. of one leaf

But, let’s account for leaf density (how closely packed are the leaves):

31

6ll

V l

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.

17

Finally, we have:

3

3

1 161

6

H W Ll H W Lo

l

l

HWL f f f HWL f f fVN

V ll

Measuring, we get: fH = fW = fL = 0.5; and pl = 2, so N is 13,951 leaves.Error of only 8% (acceptable).

18

Our model advantages:no manual leaf counting faster and less tedious

Our model disadvantages:accurate determination of pl difficult

assumes uniform distribution of leavesassumes ellipsoidal volumetric canopy and

leaves

19

REVIEW OFREVIEW OFMODELLINGMODELLING

METHODOLOGYMETHODOLOGY

20

Types of mathematical modelsMechanistic (process-based) and EmpiricalStatic and DynamicDiscrete and continuousDeterministic and stochastic

21

Static modelsno time factor

Dynamic modelshas time factor

Discrete models time is an integer (1, 2, 3, …)

Continuous models time is a real value (1.1, 2.5, 3.0, …)

22

Deterministic modelsno element of randomness

Stochastic modelshas elements of randomness (probabilities)

23

Mechanistic modelsprocess-based modelsdescribes and explains the processesmore difficult to build

Empirical modelscorrelative or statistical modelsdescribes but does not explain the processeseasier to build

24

Properties of empirical models1) easier to build; curve-fitting exercise

x

y

y = b0 + b1x

x

y y = b0 + b1x + b2x2

x

y

y = b0 + b1 log(x)

x

y

y = b0 eb

1x

x

y y = A sin (2 x/P)

P

A

0

linear logarithmic sine

quadratic exponential

25

mechanistic models difficult to build because we need to know which and how the factors interact with one another to produce the system process

26

2) empirical models cannot imply causality (cause-and-effect) describes how variable are related but does not

explain why

50

60

70

80

100 150 200 250 300

x, number of storks

y, p

opul

atio

n (x

10

3 )

y = 139.1x + 38183.7

R2 = 0.9

Relationship between the population of Oldenburg city, Germany with the number of bird storks in 1930-36

Other examples:- sale of ice cream and reservoir level- electricity bill and weather

27

3) empirical models are highly environment-specific 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

28

Crop production levels

Four levels of production:Level 1: potential growth, limited only by solar

radiationLevel 2: additionally limited by waterLevel 3: additionally limited by nitrogenLevel 4: additionally limited by other nutrients

Helps us to focus on developing our models

29

Overview of the Gg (Generic crop growth) model

30

31

Part 2:Meteorology

32

Importance

Agriculture strongly affected by meteorology (weather)

Water (rain)photosynthesis, respiration, nutrient supply

RH40-80% desirable; too high, high disease and

pests low crop yield because high vegetative growth

33

Solar radiationmajor energy supplier; photosynthesis, water

uptake length of day: flowering

soybean (short day); groundnut (long day) Air temperature

every plant has an optimum temp. rangeregulates chemical reactions in

photosynthesis, flowering, germination, transpiration, respiration

34

Solar positionSolar position with respect to the observer

= azimuth (-ve, before solar noon) = elevation (or solar height)

asin sin sin cos cos cos

sin sin sinacos

cos cos

= site latitude = solar declination = hour angle

35

Solar position with respect to the Earth. Np and Sp are the North Pole and South Pole, respectively.

Location of observer on the Earth’s surface. Np and Sp are the North Pole and South Pole, respectively.

36

-30

-20

-10

0

10

20

30

0 50 100 150 200 250 300 350

day of year

sun

decl

inat

ion

(deg

.)

Solar declination varies depending on the Earth’s position in orbit around the Sun. Np is the North Pole.

Cyclical change in solar declination with the day of year

0.4093cos 2 10 365dt

td = day of year (1=Jan 1, …, 365=Dec 31)

37

Hour angle, : Earth rotates 360 every 24 hours, so every 1 hour, Earth rotates 15

1212 ht

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)

38

Therefore,

asin sin sin cos cos cos

sin sin sinacos

cos cos

solar elevation(from horizontal)

solar azimuth(from South)

solar inclination(from vertical)

solar azimuth(from North in aneastward direction)

acos sin sin cos cos cos

39

Daylength and sunrise and sunset time

Sunset time

Sunrise time

Daylengthnumber of hours between sunrise and sunset

12 sin sin12 acos

cos cossst

12 sin sin24 12 acos

cos cossr sst t

24 sin sin2 12 acos

cos cosssDL t

40

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

41

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)

42

Daily irradiance

, , 0 1t d et d

sI I b b

DL

s = sunshine hours (no. of hours when solar irradiance >120 W m-2)DL = daylengthIet,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 temperate0.18 0.55

Dry tropical 0.25 0.45

Humid tropical 0.29 0.42

43

, 03600 1370 sinss

sr

t

et d h

t

I dt

1212 ac os /

1212 ac os /

2

sin cos 2 12 / 24

24 acos 1

ss

sr

a bt

h h h

t a b

dt a b t dt

a a b b a b

0 1 0.033cos 2 10 / 365dt

wherea = sinsin b = coscos

01370 sinetI Hourly ET irradiance:

Daily ET irradiance:

44

Hourly irradiance

0

200

400

600

800

1000

0 2 4 6 8 1012 1416 1820 2224

local solar hour

sola

r irra

dian

ce (

W m

-2)

Typical diurnal trend for solar irradiance

cos 2 / 24t hI A t B

,

2

and

86400where

acos 1

t d

A b B a

I

a a b b a b

a = sinsinb = coscos

45

Net radiation

1n t nLR p I R

1nL nL

sR R b b

DL

nL Ld LuR R R

4,Lu a KR T

6 6,9.35 10Ld a KR T

= Stefan-Boltzmann constant (5.67 x 10-8 W m-2 K-4)p = surface albedo (typically 0.15)Ta,K

4 = air temperature (K)b = 0.2

net = incoming - outgoing

46

Direct and diffuse solar radiation

Direct (dr) component from a single directioncauses shadows

Diffuse (df) component from all directionsdoes not cause shadows

Need to distinguish the two interception by plants is different for the two

components

47

Daily direct and diffuse radiation

, , , ,

2

, , , , , ,

, , , , , ,

, , , ,

1 for 0.07

1 2.3 0.07 for 0.07 0.35

1.33 1.46 for 0.35 0.75

0.23 for 0.75

df d t d t d et d

df d t d t d et d t d et d

df d t d t d et d t d et d

df d t d t d et d

I I I I

I I I I I I

I I I I I I

I I I I

Idr,d = It,d – Idf,d

A set of empirical equations:

48

Hourly direct and diffuse radiation

Idr = It – Idf

A set of empirical equations:

2

1 for 0.22

1 6.4 0.22 for 0.22 0.35

1.47 1.66 for 0.35<

for

df t t et

df t t et t et

df t t et t et

df t t et

I I I I

I I I I I I

I I I I I I K

I I R K I I

49

Air vapour pressure

( )100a s a

RHe e T

6.1078exp 17.269237.3a

s aa

Te T

T

0

20

40

60

80

100

0 10 20 30 40 50

air temperature (oC)

satu

rate

d va

por

pre

ssu

re (

mba

r)

Relationship between saturated vapor pressure and air temperature

RH = relative humidity (%)Ta = air temperature (C)

50

Serdang RH and vapour pressure

0

10

20

30

40

50

60

0 2 4 6 8 10 12 14 16 18 20 22 24

local solar hour

vapo

r pr

essu

re (

mba

r)

es

ea

0

20

40

60

80

100

0 2 4 6 8 10 12 14 16 18 20 22 24

local solar hour

RH

(%

)Air vapor pressure (ea) and saturated air

vapor pressure (es) for Serdang town

(3.0333 N; 101.7167 E), Malaysia on 31 October 2004

Relative humidity for Serdang on 31 October 2004

51

Wind speed

0 2 4 6 8 10 12 14 16 18 20 22 24

local solar hour

win

d sp

eed

Idealized daily trend for mean hourly wind speed

actual hourly wind speed can be erratic and difficult to simulate

52

Air temperature

Measured air temperature for Serdang on 31 October 2004

15

20

25

30

35

0 2 4 6 8 10 12 14 16 18 20 22 24

local solar hour, th

sunset

I II III

sunrise

air temperature (C)

53

Pattern of changebefore sunrise, air. temp gradually drops due

to heat loss from groundwhen 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

54

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 sectionsSection I – before (sunrise + 1.5 h)Section II – from (sunrise + 1.5 h) to sunsetSection III – after sunset

55

min

min max min

min

Section I

Section II

Section III

24

1.5 24

1.5sin

1.5 24

set h ssset

sr ss

h sra

ss sr

set h ssset

sr ss

T T t tT

t t

t tT T T T

t t

T T t tT

t t

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

56

Serdang air temperature

15.0

20.0

25.0

30.0

35.0

0 2 4 6 8 10 12 14 16 18 20 22 24

local solar hour, th

air

tem

pera

ture

, Ta (

° C

)

measured simulated

Comparison between actual and simulated air temperature

Mean error 2%

57

Part 3: Plant-radiation regime

58

Interception

Irradiance above (I0) and below (I) the plant canopies

Intercepted = I0 - I

Intercepted radiation potentially available for transpiration and photosynthesis

59

Hypothetical plant canopy A randomly placed leaf of area (a) over a ground

area (A)probability light intercepted is a/Aprobability 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

60

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 angleshorizontal leaves, maximum light interceptionmore vertical leaves, light interception

decreases

61

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)

62

so we reduce light penetration byexp(-kdrL), where kdr is a value between 0 and

typically less than 1kdr is known as the canopy extinction

coefficient for direct lightkdr = 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

63

*dr

ak

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)

64

Extinction coefficient is calculated as the ratio between the area of canopy shadow on the ground and the exposed surface area of the canopy

2

2

sin

20.5 0.5

or sin cos

dr

dr dr

rk

r

k k

Thus,

65

Direct light

, expp dr dr drI I k L

, exp 1 expi dr dr dr dr dr drI I I k L I k L

Direct light below canopies:

Direct light intercepted:

L

I

0

I0

I = I0 exp(-kL)

Attenuation of irradiance through a canopy according to Beer’s law

66

Scattering

transmitted(through the leaf)

reflected

incomingscattering = reflected + transmitted

scattering contributes to radiationregime within canopies

, expdr drk L

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).

LEAF

67

Canopy reflection

incoming reflected out of canopy

into thecanopy

, 1 expp dr dr drI p I k L

, 1 1 expi dr dr drI p I k L

CANOPY

where p is the canopy reflectioncoefficient, with it being equal to 0.04,0.25 and 0.11 for PAR, NIR andtotal solar radiation, respectively

68

Diffuse light

0.4

0.5

0.6

0.7

0.8

0.9

1.0

0 2 4 6 8 10

L

k df

Canopy extinction coefficient for diffuse solar radiation kdf at leaf area index (L) from 0.01 to

10 (random leaf distribution only).

1+0.1174 L

1+0.3732 Ldfk

, (1 ) expp df df dfI p I k L

, (1 ) 1 expi df df dfI p I k L

Constant kdf at a given L.

69

Discontinuous canopies

Beer’s law assumes closed, homogenous canopies

When early growth periods or sparse planting, canopies are openedviolates Beer’s law

Discontinuous canopies violate one of the assumptions of Beer’s law which require a uniform distribution of canopies

70

, expdr drk L

01 2

0 ln (1 )exp (1 )b b dr b drk L k L

For discontinuous canopies, modify Beer’s law by introducing a clump factor :

3 for 3(1 )

1 for 3b

L L

L

71

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

72

So, within the canopies, there are 3 components total direct component of PAR, Qp,dr

(unintercepted beam plus scattered beam)

direct component of the total direct PAR component, Qp,dr,dr (unintercepted beam without scattering)

diffuse PAR component, Qp,df

, ,(1 )p dr dr drQ p Q , expdr drk L

, , (1 )p dr dr dr drQ p Q expdr drk L

, (1 )p df df dfQ p Q expdf dfk L

73

, , , , , 2p dr p dr p dr drQ Q Q Hence, the scattered component only is:

,

(1 ) 1 expdf df

p df

df

p Q k LQ

k L

Average diffuse component within canopies:

74

Sunlit leaves absorption:

Shaded leaves absorption:

, , ,sl dr dr p df p drQ k Q Q Q

, , ,sh p df p drQ Q Q

where is the leaf absorption coefficient (0.8 for PAR)

75

Sunlit and shaded leaves

*dr

ak

A

where a is shadow area on ground; A* is exposed canopy area (sunlit leaves),so applied to canopies,

dr sl

sldr

k a L

aL

k

1 exp( )drsl

dr

k LL

k

sh slL L L

Sunlit LAI:

Shaded LAI:

Recall:

since we are taking ground area as 1 m2, a is 1 - exp(-kdrL) => fraction of ground covered by shadows

76

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

77

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.

78

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

79

0.5 cos 40 0.65drk

, (1 0.04) 1092 exp 0.8 0.65 3 183.2p drQ

, , (1 0.04) 1092 exp 0.65 3 149.1p dr drQ

, , 183.2 149.1 2 17.1p drQ

1+0.1174 3 1+0.3732 3 0.73dfk

,

(1 0.04) 728 1 exp 0.8 0.73 3306.5

0.8 0.73 3p dfQ

0.8 0.65 1092 306.5 17.1 826.7slQ

0.8 306.5 17.1 258.9shQ

1 exp( 0.65 3)1.3

0.65slL

3.0 1.3 1.7shL

80

Part 4: Plant water uptake and soil evaporation

81

Energy balance Rn = H + ET + G + M (all in W m-2)

Rn = net radiation main energy supplier

H = sensible heat densityET = latent heat flux densityG = ground heat flux density

energy into or out of soil sub-surfaceM = miscellany energy density

small term (usually less than 5% of Rn) often neglected (M = 0)

main energyconsumers

82

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

83

Evapotranspirationwater 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

84

Potential vs. actual ETPET

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

85

Sensible heat (H)energy supplied is to raise temperature

thus, can be “sensed”; can be measured by thermometer

Heat transferRn = radiativeET, 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

86

Energy balance for day and night

Day = ground gains heatNight = ground loses heat+ET = evaporation-ET = condensation

87

K-theory transport

Vertical transport of a generic property

or or A B

A B

F F F Kz z z z

88

Latent heat transport equation

or ET ETz 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

89

so express it using air vapour pressure ea:

p aET

c eET K

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 requiredto raise the temperature of a unit volume of air by one K (1221.09 J m-3 K-1)

90

Sensible heat transport equation

or p H

T TH H c K

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).

91

Penman-Monteith equation

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.

Uses the electrical network analogy to explain heat transfers

92

1pa

ET

ce ET z

K

pa

cV e

I ET1

ET

R zK

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,

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.

p aET

c eET K

z

OR

93

0 0 and p r rp

a c a

c e e T TET H c

r r r

0 0

n

p s r rn p

a c a

R G ET H

c e T e T TR G c

r r r

Evaporation from a saturated environment (within stomata):

94

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

s r rD e T e

where es(Tr) is the saturated vapor pressure (mbar) at temperature Tr (C).

95

6.1078exp 17.269237.3a

s aa

Te T

T

0

20

40

60

80

100

0 10 20 30 40 50

air temperature (oC)

satu

rate

d v

apo

r pr

ess

ure

(m

bar)

Relationship between saturated vapor pressure and air temperature

2

25029.4exp 17.269 237.3

237.3

s r

r

r r

r

de T

dT

T T

T

0

0

s r s

r

e T e T

T T

for small differences between Tr and T0:

is the slope of the saturated vapor pressure curve (mbar K-1)

96

n a c p

a a c

n p a

a c a

R G r r c DH

r r r

R G c D rET

r r r

So introducing D and , and after some algebraic manipulations:

0 0

n

p s r rn p

a c a

R G ET H

c e T e T TR G c

r r r

97

Problem with PM equationassumes either evaporation or transpiration

but not both occurring simultaneouslynot applicable for open canopies (e.g., early

growth periods or sparse planting density)

98

Shuttleworth-Wallace equation

SW equation:extension of the PM equationET occurs from both soil and plant

simultaneouslygood for early growth periods or for sparse

planting densities

99

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 rs

s are the canopy and soil surface

resistance, respectively.

100

Entire energy balance of the system can be described in 8 equations:

0 rp a

a

T TH c

r

0fc p c

a

T TH c

r

0ss p s

a

T TH c

r

0p ra

a

c e eET

r

0-s fpc c c

a s

e T ecET

r r

0p s ss s s

a s

c e T eET

r r

c c cA ET H

s s sA ET H

,1c dr nA R

,s dr nA R G

or

or

101

After some algebraic manipulations: c c s sET C PM C PM

1

c a cp a s a a

c c a cs a a

A c D r A r rPM

r r r

1

s a sp a c a a

s s a ss a a

A c D r A r rPM

r r r

11c c a s c aC R R R R R

11s s a c s aC R R R R R

aa aR r

c cc a sR r r

s ss a sR r r

102

Once we know total ET, we determine its components:

0 0 0sD e T e 0

aa

p

rD D A ET

c

0 0 and

s cs p a c p a

s cs s s c c cs a a s a a

A c D r A c D rET ET

r r r r r r

0 0 and

s s c cs s a p c s a p

s cs s s c c ca s a a s a

A r r c D A r r c DH H

r r r r r r

OR

103

,0.35 cosdr nG R

Soil heat flux (G):

G is 35% of net radiation reaching the ground, and this fraction varies accordingto the cosine of the solar inclination

104

Aerodynamic resistances

Roughness length (z0) and zero-plane displacement (d)

Wind speed decreases exponentially with height, and equals zero at a certainheight above ground/surface. How high above the ground/surface the windspeed falls to zero is a measure of how rough the surface is: roughness length (z0)

In the presence of objects (e.g., canopies) the roughness length is displaced by d (zero-plane displacement)

105

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

106

0

0.64

0.13

d h

z h

For crops with crop height h:

*

0

( ) ln ; u z d

u z z hk z

( ) ( ) exp 1 ; u z u h z h z h

Wind speed above canopy:

Wind speed below canopy:

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)

107

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

108

-10 0exp( )exp exp s m

( )s s

a

z z dh nr n n

nK h h h

-10

*

1ln exp 1 1 s m

( )a r

a

z dz d hr n

ku h d nK h h

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

109

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 mixedwhen turbulence is suppressed, transfer

occurs solely due to molecular diffusion (very slow)

110

-1 s m

0.012 1 exp 2 ( )c

arL u h w

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

111

Stomatal resistance

PAR (W m-2)

stomatal resistance (s m-1)

Stomatal resistance decreases with increasing PAR (photosynthetically active radiation) irradiance, following the relationship by Jarvis (1976)

-11

2

s mPARst

PAR

a Ir

a I

for 0.5

0.5 for 0.5st crc

sst crcr

r L L Lr

r L L L

Canopy resistance is:

where Lcr is critical LAI, typicallytaken as maximum LAI (=4)

112

Soil surface resistance

dry layer

wet layer

soil surface

thickness, lwater water

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)

113

,,

( ) exps s vs v s dry

v sat

r r

v / v,sat

rss

rssdry

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

,,

ss dry

p m v

lr

D

1/

114

,ln ln lnv v sat e

Pore size distribution index is the slope of the linear line of relative saturationto soil suction

ln (v / v,sat)

ln e

higher the suction, drier the soiland smaller the relative saturation

115

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

116

Part 5: Water balance

117

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

118

volume of water

volume of soilv

area depth of water depth of water

area depth of soil depth of soilv

3 -3depth of water (mm) = (m m ) depth of soil (m) 1000v

Volumetric water content is the volume of water per unit volume of soil:

119

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

120

equation looks deceptively simple, but in practice, the individual components can be difficult to determine/measure use some assumptions:

1. no irrigation supplied, so I = 0

2. deep water table (> 1 m deep), so CR = 0

3. flat, levelled land, so OF = 0

therefore water balance equation becomes: R = P + ETa + or = R - P - ETa

121

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

122

Percolation drainage (loss) of water from a soil layer/zone

consists of two components:1. percolation due to excess water pe

2. percolation due to redistribution pd

e dP p p

123

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 if

if v v sat

ev v sat v v sat

Rp

R R

initial amount = 40 mlmax can hold = 100 ml

pour 150 ml

amount overflow, pe?

pe = 40 + 150 – 100 = 90 ml

124

Redistribution occurs due to gravity and matric potentials, as defined by Darcy’s Lawgravity 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

125

Flow, q H/L or q = K H/Lwhere 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

126

m gTH HH

q K Kz z

If the depth difference between two soil layers is z, then Hg = z, and

1m m

H z Hq K K

z z

Assuming uniformly wetted soil means no differences in matric potential any where in that soil layer, so

0mH

z

q K

Thus, flow is controlled only by the soil’s K

which gives:

127

K depends on soil texture, soil structure and soil water contentK increases with increasing water content until maximum at soil saturation

volumetric water content (m3 m-3)

hydraulic conductivity (m s-1)

v,sat

Ksat ,

,

exp v sat vsat

v sat

K K

where is 13-16 for most soils

128

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

129

Law of mass conservation

vq

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.

130

If we take the soil layer thickness as L, then vq Lt

Earlier, we established q = K, so vK Lt

2

1

2

1

2

1

,

,

exp

v

v

v

v

v

t

v

t

v

v sat vsat

v sat

t LK

Lt

K

L

K

re-arranging:

131

2 1

, 2 1, , , ,

, ,

ln expv sat satv t v sat v sat v t

v sat v sat

K t t

L

So at time t2, the volumetric water content is:

Therefore, percolation due to redistribution is

t2 - t1 = R – (pe + pd) pd = t2 - t1 - R - pe

t2 is now available for evapotranspiration ETa

132

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

133

Actual soil evaporation

,a s D eE ET R

0.0

0.2

0.4

0.6

0.8

1.0

0.0 0.2 0.4 0.6 0.8 1.0

relative soil water content

redu

ctio

n fa

ctorPotential 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

, -9.3172

,

1

1 3.6073D e

v v sat

R

(1, ) (2, )0.26 and 0.74a t a a t aE E E E

134

Actual transpiration

,a c D tT ET R

, , , ,D t v v wp v cr v wpR

, , ,relative water content v v wp v sat v wp

, , , ,v cr v wp v sat v wpp

0.0

0.2

0.4

0.6

0.8

1.0

0.0 0.2 0.4 0.6 0.8 1.0

relative soil water content

redu

ctio

n fa

ctor

C4 C3

v,wp

critical point

Potential plant transpiration is reduced to actual transpiration by a reduction factor that is dependent on the relative water content

all transpiration is from water in thesecond soil layer only

135

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 4-carbon 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.

136

Photosynthesis (C3)

137

Empirical approaches

0.0

5.0

10.0

15.0

20.0

0 20 40 60 80 100PAR (W m-2)

de Wit (1965)

Goudriaan andvan Laar (1978)

Lmax

Assimilation rate (µmol m-2 s-1)

Gross photosynthesis as a function of absorbed PAR, determined using the equations by de Wit (1965), and Goudriaan and van Laar (1978)

max PAR

PAR PAR

I

I hI

LL

max 1 exp PAR PARI hIL L

de Wit (1965):

Goudriaan and van Laar (1978):

138

y = 1.32x

0

200

400

600

800

0 100 200 300 400 500 600

Cumulative radiation intercepted (MJ m-2)

Total dry matter (g m-2)

Total dry matter as a function of cumulative total solar radiation intercepted (Monteith, 1977)

, ,n canopy t de f IL

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.

RUE for C3 and C4 plants are typically measured at about 1.9 and 2.5 g MJ-1 of intercepted PAR

139

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)

140

Light and dark reactions Photosynthesis equation:

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

2 2 2 2 26

light6CO +12H O CH O +6O +6H O

141

Term “dark reactions” is misleading because these reactions do not occur in the darkso-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)

142

Photosynthesis occurs in the chloroplast, an organelle in the mesophyll cells

143

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.

144

Calvin cycle. Shaded area denotes the carboxylation cycle.

145

The major processes and their equations in the Calvin cycle

a) Carboxylation of RuBP: RuBP + CO2 + H2O 2PGA

b) Regeneration of RuBP:

c) Oxygenation of RuBP: RuBP + O2 PGA + PGIA

d) Regeneration of PGA: PGIA + 0.25O2 0.5PGA + 0.5CO2

2 2

energy

(NADPH+ATP)2PGA RuBP+CH O+O

146

Cycle begins by the enzyme Rubisco (ribulose biphosphate-carboxylase/oxygenase) fixing one mol of CO2 to one mol of RuBP (ribulose 1,5-biphosphate) 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.

147

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.

148

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.

149

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

150

Enzyme kinetics: Michaelis-Menten equation

max[ ]

[ ] c

V Cv

C K

0

50

100

150

200

0 200 400 600 800 1000

[C] (µmol mol-1)

Kc

Velocity (µmol s-1)

Vmax

Kc is defined as the substrate concentration at half the maximum velocity

1 2

1

k k

kE C EC P E

151

Competitive inhibition

1 2

1

k k

kE C EC P E

k o

k oE O EO

and

max[ ]

1 [ ] [ ]c o

V Cv

K O K C

In competition with O, velocity of C reaction is

152

Rubisco-limited

,max *

1c i

cc a o i

V Cv

K O K C

applied to photosynthesis:

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

In plants, there is a minimum amount of internal CO2 below which there is noCO2 assimilation. This critical or minimum amount of CO2 is known as the CO2

compensation point (*).

*2

aO

153

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.

154

Sensitivity to temperatureKc, 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)( 25) /102 T

where and (25) are the enzyme reaction rates at temperatures T and 25 C, respectively. To generalize

(25) 10

( 25) /10fTQ

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.

155

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)

Q10 values

156

,max(25)

,max

( 25) /102.4

1 exp 0.128 40

cc

f

fTV

VT

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

157

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

158

Light-limited

*

2 *i

q m ti

Cv e Q

C

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

20.081 0.000053 0.000019m f fe T T

C4 plants = 0.053 mol mol-1

C3 plants:

159

Sink limited

,max

2c

s

Vv

vs is the sink-limited rate of CO2 assimilation (per unit leaf area) (mol m-2 s-1)

160

Gross assimilation rate

, ,c q sMIN v v vL

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:

where MIN() denotes the minimum of the enclosed values.

161

0

10

20

30

40

50

60

0 10 20 30 40Leaf temperature (ºC)

Gross assimilation rate (µmol m-2 s-1)

Qp = 600

Qp = 1800

Qp = 1200

Photosynthesis response to temperature at three levels of PAR flux densities (Qp, mol m-2 s-1)

162

0

10

20

30

40

50

60

0 500 1000 1500 2000PAR (µmol m

-2 s

-1)

10 ºC

20 ºC

30 ºC

Gross assimilation rate (µmol m-2 s-1)

Photosynthesis response to PAR at three levels of leaf temperature

163

-60

-30

0

30

60

90

0 100 200 300 400 500 600

C i (µmol mol-1)

Qp = 600

Qp = 1800

Qp = 1200

Gross assimilation rate (µmol m-2 s-1)

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.

164

Leaf temperature Tf

0

aa

rp

HrT T

c

0fc p c

a

c ac a a

f rp

T TH c

r

H r HrT T

c

Solve energy balance first to obtain total sensible heat (H):

Determine the sensible heat for crop (Hc), then calculate Tf:

165

Scaling up to canopy photosynthesis

canopy sl sl sh shL LL L L

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:

166

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.

167

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

168

0.5 cos 40 0.65drk

, (1 0.04) 1092 exp 0.8 0.65 3 183.2p drQ

, , (1 0.04) 1092 exp 0.65 3 149.1p dr drQ

, , 183.2 149.1 2 17.1p drQ

1+0.1174 3 1+0.3732 3 0.73dfk

,

(1 0.04) 728 1 exp 0.8 0.73 3306.5

0.8 0.73 3p dfQ

0.8 0.65 1092 306.5 17.1 826.7slQ

0.8 306.5 17.1 258.9shQ

1 exp( 0.65 3)1.3

0.65slL

3.0 1.3 1.7shL

169

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.

-2 -127.8 1.3 8.7 1.7 50.9 mol m scanopy L

So, gross canopy assimilation rate of CO2 (per unit ground area) is:

Homework: Use the equations as shown earlier to determine Lsl and Lsh.

170

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.

171

Part 7: Respiration

172

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

173

Conceptual model of plant respiration

174

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 = (1-Yg)(Lcanopy - RM) is lost via growth respiration

175

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

176

(25) ,

, , , ,

M M i ii plant parts

M greenleaves greenleaves M stem stem M roots roots M storage storage

R k W

k W k W k W k W

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

177

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.

(25) 10( 25) /10a

M MTR R Q

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.

M M greenleaves leavesR R W W

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.

178

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.

179

Glucose requirement G and the carbon content for major biochemical groups in plant tissues

Biochemical groupG (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

180

Fractions of major biochemical groups in several plant parts, and the calculated glucose requirement for the production of the plant parts

GroupYoung leaf

Wheat seed

Broad bean

Oil-rich seed

Wood 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

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

leavesG

181

i ii plant parts

greenleaves greenleaves stem stem roots roots storage storage

G FG

F G F G F G F G

Total glucose requirement for growth:

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

3.736 6.136 0.251G C N

Shortcut: nitrogen content is equal to the protein content multiplied by 0.16 (the average nitrogen N content of proteins is about 16%):

only need to analyse total C and N content – faster, cheaper and simpler

182

, 1 ,canopy M

i t i t i

RW W F t

G

L

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

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

183

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

184

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

185

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

186

Adopt a one-dimensional and irreversible scale to denote the growth development stageno “standard” to follow – up to us to decideseed emergence (s = 0), flowering (s = 1),

and plant maturity (s = 2)

187

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)

188

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

advancesThe 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 rateOther factors: vernalization and day length,

depending on the plant species

189

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

190

0

20

40

60

0 10 20 30 40

temperature (deg. C)

days

0

0.02

0.04

0.06

0.08

0 10 20 30 40

temperature (deg. C)

1/d

ays

T b = 2.6

Dependence of growth duration on air temperature

Dependence of growth development rate on air temperature

Tb = base temperature, below which there is no growth (note: no –ve growth)

191

Base temperature for some crops

Crop Base temperature (ºC)

Field pea, lentils, linseed, oats, spinach 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 8-14

Cowpea, sorghum 11

Pumpkins, tomatoes 10-13

Castor, peanut, pigeon pea 13

Guar 15

Sesame 16

Melons 15-18

192

, 1 ,s i s ir

Tt

1r

Tt

For a given air temperature T, the growth development rate r (day-1) can be determined by

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

193

, 1 , ,s t 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.

194

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

1 ,t greenleaves t tL W SLA

SLA is typically 0.022 m2 g-1

195

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):

0 for 2 1.0

for 0.1 2 1.02

for 2 0.10.1

s

r sage

s

r s

D

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.

196

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

0 for

0.03, 0.03 for cr

shadecr cr cr

L LD

MIN L L L L L

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 self-shading 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

197

The actual death rate of leaves Dleaves (day-1) is then the larger of the two rates Dage and Dshade:

,leaves age shadeD MAX D D

where MAX() the maximum of the enclosed values. The weight of dead leaves can then be calculated as

, 1 , , ,deadleaves t deadleaves t greenleaves t leaves tW W W D 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.

198

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.

, 1 , , ,canopy M

greenleaves t greenleaves t greenleaves greenleaves t leaves t

RW W F W D t

G

L

199

Plant height growth

accumulated temperature sum (deg. C day)

hei

gh

t (m

)

Typical plant height growth following the logistic function

1t t

dhh h t

dt

0 1 1

2

0 1

exp

1 exp

m tsts

ts

b b h b TdhT

dt b b T

where Tts 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

200

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

, , ,ts ts t avg t b avg t bt t

T T H T T T T

0 for 0( )

1 for 0

xH x

x

Unit function:

Increment temperature sum only when mean temperature is greaterthan base temperature

201

Example

An example showing the daily temperature sum Tts and accumulated

temperature sum Tts (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

Tts 0 0 3 8 15 20 23 23 26

Tavg = average of Tmax and Tmin

202

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

203

, , ,

, 1, , ,

for or 1

, for and 1

r t v t v wp s

r tm r t g v t v wp s

dd

MIN d d d t

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 (pre-anthesis 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).

204

Water stress effects

,D t a cR T ET

,canopy canopy D tRL L ,D t

dh dhR

dt dt

,g g D td d R

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:

Then reduce plant height growth rate, root elongation rate and grossassimilates produced: