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Hydrology > Week 9 > Flashcards

Week 9 Flashcards

1
Q

qr =

A

rainfall

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2
Q

qin =

A

canopy interception

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3
Q

Ea =

A

evapotranspiration

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4
Q

qro =

A

runoff

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5
Q

qi =

A

infiltration

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6
Q

qd =

A

drainage

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7
Q

qvp =

A

groundwater recharge

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8
Q

Soil water balance IGNORES

A

Snow

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9
Q

d(H)/dt =

A

qr - qin - qro - qd - Ea

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10
Q

(H) =

A

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

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11
Q

Infiltration =

A

flux of water passing into soil

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12
Q

Infiltration capacity =

A

maximum flux that can pass into soil at given time

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13
Q

Runoff =

A

excess rainfall that can’t infiltrate

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14
Q

Infiltration rate (y) vs time (x) progression

A
  • add water at constant rate
  • initially all infiltrates
  • ponding
  • decrease infiltration as near-surface pore space used
  • eventually steady state related to soil permeability
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15
Q

Horton’s empirical curve

A

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
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16
Q

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

A
  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

17
Q

Units of infiltration rate

A

mm/min

18
Q

Rainfall>infiltration capacity…

A

RUNOFF

19
Q

If PE = Ea…

Is this the case?

A

Then discharge would = rainfall - PE

BUT PE > Ea

20
Q

Field capacity

A

S = Smax

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

21
Q

Permanent wilting point

A

S = 0

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

22
Q

Soil moisture deficit

A

Smax-S

= extent soil drier than field capacity

23
Q

dS/dt =

A

qi - qd - Ea

24
Q

qd =

A

0 ; S=Smax

qi-Ea ; S>Smax

  • b/c S is higher than the straw, any excess water will just drain
  • dS/dt = 0
25
Q

Ea =

A

0 ; S=<0

Ep ; S >0

26
Q

qi =

A

qr - qin - qro

27
Q

Ep =

A

potential evapotranspiration

28
Q

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

A
  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!!!

29
Q

S =

A

Water level in bucket

30
Q

How to constrain 0 < S < Smax

A
  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

31
Q

Why do we need to constrain 0 < S < Smax

A

-ve values not physically possible

Can’t go over Smax due to ‘straw’/tube

32
Q

Constraining 0 < S < Smax; Sn+1

A

0 ; F=<0

F ; 0 < F<= Smax

Smax ; F>Smax

33
Q

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

A

0 ; F=Smax

34
Q

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

A

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

E(p,n) ; F>0

35
Q

Bob Moore’s Probability Distributed Model (PDM) CONCEPT

A 
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

36
Q

PDM qin =

A

canopy interception rate

37
Q

Implementation of the PDM

A

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

38
Q

PDM with the finite difference method

A
  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

Hydrology (9 decks)

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