dynamic plant uptake modeling stefan trapp. steady-state considerations: simple & small data...
TRANSCRIPT
Dynamic plant uptake modeling
Stefan Trapp
Steady-state considerations: simple & small data need
However: often emission pattern is non-steady, e.g.: non-steady plant growth (logistic) pesticide spraying application of manure and sewage sludge on
agricultural fields
In these scenarios: steady-state solutions are not valid, and dynamic simulation is required.
Steady-state vs. dynamic models
• Three different types of input, namely• pulse input (pesticide spraying)
• constant input (deposition from air)• irregular input (this we cannot solve --> use numerical integration)
Dynamic input patterns
Dynamic models
Designed for
- repeated input- dynamic growth- pesticides- manure or sewage appl. - 1 year or 10 year - easy to handle (excel)
1.0E-06
1.0E+00
2.0E+00
3.0E+00
4.0E+00
5.0E+00
0 10 20 30 40 50 60
Real time (d)
C(t
)
C3 Stem C2 root C4 Fruit
C1 soil bioavailable C4a Leaves
● Repeated pulse input from soil or air or constant emission
● Logistic growth of plants (here: summer wheat)
● n variable periods (30 in practice)
Dynamic models
Dynamic Model
Differential equation system
In words
soil: change of mass = + Input - degradation - uptake into plants
roots: change of mass = + uptake from soil - loss to stem - degradation
stem: change of mass = + uptake from roots - loss to leaves (fruits) - deg
leaves: change of mass = +uptake from stem ± exchange air - degradation
fruits: change of mass = +uptake from stem ± exchage air - degradation
Differential equation system
soil
roots
(stem)
leaves
fruits
inputQKM
CCk
dt
dC
SWSoil
SoilSoil
Soil deg
RRRRWRRSW
SoilR CkMQKCMQK
C
dt
dC ///
StStStStWStRW
RSt
St CkMQKCK
CMQ
dt
dC ///
LLLLLA
LA
L
depLSt
StWL
LL CkCMK
mLgAC
M
vAC
KM
Q
dt
dC
31000
FFFFFA
FA
F
FSt
StWF
FF CkCMK
mLgAC
M
gAC
KM
Q
dt
dC
31000
Cascade of compartments
111 mk
dt
dm 2211
2 mkmkdt
dm
Mass balance:
"The change of mass in tank 2 is what flows out of tank 1 minus what flows out of tank 2"
Differential equation system
1 soil
2 roots
3 stem
4a leaves
4b fruits
1111 bCk
dt
dC
2221122 bCkCk
dt
dC
3332233 bCkCk
dt
dC
4443344 bCkCk
dt
dC
4443344 bCkCk
dt
dC
The system written in a schematic way:
Each DE always relates to the DE before, but not to any other DE
transfer rate constants kij (d-1)
loss rate constant ki (d-1)
constant external input b (mg kg-1 d-1).
Structure of the multi-cascade crop model
Dynamic Model
bC
kk
kk
kk
k
dt
Cd
434
323
212
1
00
00
00
000
Same processes, same differential equations, but formulated as matrix
1 is soil k1 loss rate k12 transfer rate2 is roots k2 loss rate k23 transfer rate 3 is stem k3 loss rate k34 transfer rate 4 is leaves or fruits k4 loss rate
b is the input vector
Steady-state solution
Set dC/dt (left hand) to zero. Then
Cascade
etc.
1
1
11
11 )(
k
b
Mk
ItC
)()( 12
12
22
22 tC
k
k
Mk
ItC
)()( 23
23
33
33 tC
k
k
Mk
ItC
Conc. = Input / loss
tkeCtC 1)0()( 11
tktktk
eCkk
e
kk
eCktC 2
21
)0()()(
)0()( 22112
1122
tk
tktk
tktktk
eC
kk
e
kk
eCk
kkkk
e
kkkk
e
kkkk
eCkktC
3
32
321
)0(
)()()0(
))(())(())(()0()(
3
3223223
231332123121123123
etc. …
Analytical solution for pulse input, i.e. C(0) ≠ 0
tktk eCek
btC 11 01 1
1
11
tktktktk eCeBeeAtC 2221 01 22
tktktktktktk eCeFeeEeeDtC 333231 01 33
112
1121121 0
kkk
bkkkCA
21
21112
kk
bkbkB
13
23
kk
kAD
23
223 0
kk
BACkE
3
323
k
bBkF
Cascade with constant input
Analytical solution for all t
That's what you always wanted to know
about math, wasn't it?
Questions?
Principle of superposition
Concentrations are additive
We can thus calculate several subsequent periods with different values, and the output from one period is the input to the next.
This allows to simulate non-constant conditions.
Our "cascade model" has by default 24 periods to 5 days (= 120 days, i.e. one vegetation period), but this is variable.
Principle of superposition
Figure: Concentrations are additive
Most annual crops show a logistic growth curve initial growth is exponential
towards ripening, growth slows down
and finally stops
Change of plant mass M [kg]:
Plant growth
max
1M
MMk
dt
dM k First-order rate constant (for exponential growth)
[1/d]
Mmax Maximum plant mass [kg]
tke
MM
MtM
11
0
max
max
Plant mass as a function of time
M0 Initial plant mass [kg]
Growth and transpiration of plants are related by
the water use efficiency (kg plant / L water) or the
transpiration coefficient TC (L water / kg plant).
Typical values range between 200 and 1000 L/kg dry weight
Default value for TC is 100 L/kg fresh weight.
Plant growth and transpiration
max
1M
MMkT
dt
dMTQ CC
Q Transpiration [L/d]
TC Transpiration coefficient [L/kg dw]
In our model, transpiration takes place only when plants are growing
Data obtained from agricultural handbooks (summer wheat)
Annual seed plant
- Initial mass 10-4 kg (0.1 g for seeds)- Growth rate constant k = 0.1 d-1 (doubling time ≈ 1 week)- Final mass 1 kg
data related to 1 m2
Transpiration coefficient TC = 50 L/kg fw (water content green plans ≈ 90%)
Plant growth and transpiration
Standard scenario: summer wheat
tk
eMM
MtM
11
0
max
max
max
1M
MMkT
dt
dMTQ CC
Maximum transpiration Qmax is at ½ Mmax (inflection point)
with maxmax 41 MkTQ C
at time
1
1ln
1
0max MMkt
Plant growth and transpirationPlant growth and transpiration
Annual seed plant
Plant growth and transpiration
Growth is exponentialfor t < 70 d
Absolute growth & transpiration peak at t = 92 d
Growth almost stops for t > 135 d
= phase in which fruit or corn ripe leaves decay and plants dry out
Biomass M and transpiration Q of summer wheat
Dynamic model: Default scenario
Reading:
0.0
0.2
0.4
0.6
0.8
0 5 10 15 20 25 30 35
Time (d)
C(t)
C3 Stem C2 root C4 Fruit C1 soil
Example simulation for a repeated pesticide application
Repeated application of insecticide by drip irrigation to soil
Comparison to measured data
Model result before calibration
0
0.05
0.1
0.15
0.2
0.25
0 5 10 15 20 25 30 35
Time (d)
C Fr
uit
Measured Model
Comparison to measured data
After fit of two parameters (temperature, soil depth)
00.05
0.10.15
0.20.25
0.3
0 5 10 15 20 25 30 35
Time (d)
C Fr
uit
Measured Model
More reading:
Lucky You - you survived this part.
Questions?