Week 9 Flashcards
qr =
rainfall
qin =
canopy interception
Ea =
evapotranspiration
qro =
runoff
qi =
infiltration
qd =
drainage
qvp =
groundwater recharge
Soil water balance IGNORES
Snow
d(H)/dt =
qr - qin - qro - qd - Ea
(H) =
depth of water stored in soil
Found by integration of theta (moisture content) with respect to elevation above soil layer (z) between 0-H
H = thickness of soil layer
Infiltration =
flux of water passing into soil
Infiltration capacity =
maximum flux that can pass into soil at given time
Runoff =
excess rainfall that can’t infiltrate
Infiltration rate (y) vs time (x) progression
- add water at constant rate
- initially all infiltrates
- ponding
- decrease infiltration as near-surface pore space used
- eventually steady state related to soil permeability
Horton’s empirical curve
i(t) = ic + (io-ic)exp(-kt)
Shows that following ponding, infiltration capacity declines exponentially with time
i = CURRENT infiltration capacity ic = FINAL io = INITIAL k = decay rate t = time after infiltration event started
How do we find cumulative infiltration? What does this allow us to find?
- Measure with an infiltrometer diagram
- diameter = ~30cm
- maintain near zero pressure head
- measure supply rate
- measure amount of runoff
- difference = infiltration - F(t) = integration of i(t)dt
THEREFORE CAN FIT 2 TO 1 DATA TO FIND PARAMETERS FROM HORTON’S EQUATION
Units of infiltration rate
mm/min
Rainfall>infiltration capacity…
RUNOFF
If PE = Ea…
Is this the case?
Then discharge would = rainfall - PE
BUT PE > Ea
Field capacity
S = Smax
= amount held after XS drained and rate of movement materially decreased
Permanent wilting point
S = 0
= state of soil water when plants wilt and don’t recover
Soil moisture deficit
Smax-S
= extent soil drier than field capacity
dS/dt =
qi - qd - Ea
qd =
0 ; S=Smax
qi-Ea ; S>Smax
- b/c S is higher than the straw, any excess water will just drain
- dS/dt = 0
Ea =
0 ; S=<0
Ep ; S >0
qi =
qr - qin - qro
Ep =
potential evapotranspiration
Using the finite difference method to find S(n+1) =
- dS/dt \n = (S(n+1) - S(n)) / /\t
- dS/dt \n = q(i,n) - q(d,n) - E(a,n)
S(n+1) = S(n) + /\t(q(i,n) - q(d,n) - E(a,n))
= if we know storage in a bucket at time n we can look into future and predict and therefore qd and Ea too!!!
S =
Water level in bucket
How to constrain 0 < S < Smax
- Calculate auxiliary variable
F = Sn + /\t(q(i,n) - E(p,n))
i. e. F = everything but qd, and E(a,n)=E(p,n)
- if 0 < S < Smax
Why do we need to constrain 0 < S < Smax
-ve values not physically possible
Can’t go over Smax due to ‘straw’/tube
Constraining 0 < S < Smax; Sn+1
0 ; F=<0
F ; 0 < F<= Smax
Smax ; F>Smax
Constraining 0 < S < Smax; q(d,n)
0 ; F=Smax
Constraining 0 < S < Smax; E(a,n)
q(i,n) + Sn//\t ; Fs =<0
E(p,n) ; F>0
Bob Moore’s Probability Distributed Model (PDM) CONCEPT
A = area As = waterlogged area c = spatially random variable, storage capacity
If largest value of c within waterlogged area of C and F(C) is the probability of c not exceeding C in the catchment, then As=AF(C)
qro= F(C)(qr-qin)
Local water storage level within waterlogged areas = c
Outside = C
S = $C/0 (1-F(c))dc F(c) = 1-exp(-c/Smax)
qro = (S/Smax)(qr-qin)
Allows you to account for extra drainage
PDM qin =
canopy interception rate
Implementation of the PDM
Difficult to distinguish qro/qd at catchment scale –> lumped as qd
Same for canopy interception/evapotranspiration –> lumped as Ea
qi therefore = qr
dS/dt = qi - qro - Ea
qd =
(S/Smax)qi ; S=Max
Ea =
0 ; S=<0
Ep ; S>0
PDM with the finite difference method
- Auxiliary variable
G = (Sn/Smax)qi,n
F = Sn +/\t(qi,n - Ep,n - G)
Because 0 < S < Smax
S(n+1) =
0 ; F=<0
F ; 0 < F =Smax
q(d,n)
G ; F=Smaç
E(a,n)
q(i,n)+Sn/\t-G ; F=<0
E(p,n) ; F>0
