Week 9

Question 1

qr =

Answer 1

rainfall

Question 2

qin =

Answer 2

canopy interception

Question 3

Ea =

Answer 3

evapotranspiration

Question 4

qro =

Answer 4

runoff

Question 5

qi =

Answer 5

infiltration

Question 6

qd =

Answer 6

drainage

Question 7

qvp =

Answer 7

groundwater recharge

Question 8

Soil water balance IGNORES

Answer 8

Snow

Question 9

d(H)/dt =

Answer 9

qr - qin - qro - qd - Ea

Question 10

(H) =

Answer 10

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

Question 11

Infiltration =

Answer 11

flux of water passing into soil

Question 12

Infiltration capacity =

Answer 12

maximum flux that can pass into soil at given time

Question 13

Runoff =

Answer 13

excess rainfall that can’t infiltrate

Question 14

Infiltration rate (y) vs time (x) progression

Answer 14

• add water at constant rate
• initially all infiltrates
• ponding
• decrease infiltration as near-surface pore space used
• eventually steady state related to soil permeability

Question 15

Horton’s empirical curve

Answer 15

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

Question 16

How do we find cumulative infiltration? What does this allow us to find?

Answer 16

1. Measure with an infiltrometer diagram
   - diameter = ~30cm
   - maintain near zero pressure head
   - measure supply rate
   - measure amount of runoff
   - difference = infiltration
2. F(t) = integration of i(t)dt

THEREFORE CAN FIT 2 TO 1 DATA TO FIND PARAMETERS FROM HORTON’S EQUATION

Question 17

Units of infiltration rate

Answer 17

mm/min

Question 18

Rainfall>infiltration capacity…

Answer 18

RUNOFF

Question 19

If PE = Ea…

Is this the case?

Answer 19

Then discharge would = rainfall - PE

BUT PE > Ea

Question 20

Field capacity

Answer 20

S = Smax

= amount held after XS drained and rate of movement materially decreased

Question 21

Permanent wilting point

Answer 21

S = 0

= state of soil water when plants wilt and don’t recover

Question 22

Soil moisture deficit

Answer 22

Smax-S

= extent soil drier than field capacity

Question 23

dS/dt =

Answer 23

qi - qd - Ea

Question 24

qd =

Answer 24

0 ; S=Smax

qi-Ea ; S>Smax

• b/c S is higher than the straw, any excess water will just drain
• dS/dt = 0

Question 25

Ea =

Answer 25

0 ; S=<0 Ep ; S >0

Question 26

qi =

Answer 26

qr - qin - qro

Question 27

Ep =

Answer 27

potential evapotranspiration

Question 28

Using the finite difference method to find S(n+1) =

Answer 28

1. dS/dt \n = (S(n+1) - S(n)) / /\t 2. 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!!!

Question 29

S =

Answer 29

Water level in bucket

Question 30

How to constrain 0 < S < Smax

Answer 30

1. 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

Question 31

Why do we need to constrain 0 < S < Smax

Answer 31

-ve values not physically possible | Can't go over Smax due to 'straw'/tube

Question 32

Constraining 0 < S < Smax; Sn+1

Answer 32

0 ; F=<0 F ; 0 < F<= Smax Smax ; F>Smax

Question 33

Constraining 0 < S < Smax; q(d,n)

Answer 33

0 ; F=Smax

Question 34

Constraining 0 < S < Smax; E(a,n)

Answer 34

q(i,n) + Sn//\t ; Fs =<0 E(p,n) ; F>0

Question 35

Bob Moore's Probability Distributed Model (PDM) CONCEPT

Answer 35

``` 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

Question 36

PDM qin =

Answer 36

canopy interception rate

Question 37

Implementation of the PDM

Answer 37

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

Question 38

PDM with the finite difference method

Answer 38

1. 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