Wim Cornelis, Greet Oltenfreiter, Donald Gabriels & Roger Hartmann

Wim Cornelis, Greet Oltenfreiter,Donald Gabriels & Roger Hartmann

WEPP-WEPS workshop, Ghent-Wageningen, 2003

Splash-saltation of sand due towind-driven rain

Outline of presentation

• Introduction: some theory

• Materials and methods

• Results

• Conclusions

Introduction – some theory

Rainless conditions Saltation

c*t

c*

c* uuuCQ

,

e.g. Owen (1964)Lettau & Lettau (1977)

Rainless conditions Saltation


Windfree conditions Splash

detachment


td EEKD

Windfree conditions Splash

e.g. Sharma & Gupta (1989)

or Qr


Wind-driven rain conditions Rainsplash-saltation


Wind-driven rain conditions Rainsplash-saltation


Total sediment transport rate

wwr

r

QQQ

QQ

'

'

0for

0for

*

*

u

u



Objectives:

• Determine sediment mass flux qx and qz (kg m-2 s-1)

and express them as function of x and z resp.under wind-driven rain (and rainless wind) conditions

• Determine sediment transport rate Qwr (kg m-1 s-1)and relate them to rain and wind erosivity (KE or M and u*)

1. )(F~ xqx )(F~ zqz

Vertical deposition flux in kg m-2 s-1

Horizontal mass flux in kg m-2 s-1

ICE wind-tunnel experiments(dune sand, under different u* and KE

or M)

wind-tunnel wall

w ind

z

y (m )

x (m )6 7 8 9 10 11 12

trough

0.4

0.8

x = 0

tray w ith test m ateria l

(a)

wind

z

y (m )

x (m )6 7 8 9 10 11 12

0.4

0.8

x = 0

W ilson and Cooke catcher

(b)

Kinetic energy KEz or Momentum Mz

splash cups

Shear velocity u*

5 vane probes

Mass flux qx

23 troughs

Mass flux qz

4 W&C bottles

Materials and methods

Shear velocity u*

wind-velocity profiles5 vane probes


Shear velocity

Eq. [7]

wind velocity u (m s-1)

0 2 4 6 8 10 12 14

hei

gh

t z

(m)

10-5

10-4

10-3

10-2

10-1

100

101

u* = 0.27 m s-1

u* = 0.39 m s-1u*= 0.50 m s-1


Shear velocity

0

* lnz

zuu

uref (m s-1)

6 8 10 12

u* (

m s

-1)

0.2

0.3

0.4

0.5

0.6Observed dataEq. [9]; R² = 0.999


Shear velocityrefuu 037.0050.0*


Kinetic energy or Momentum

2

2

1vmKE

vmM

v from nomograph of Laws (1941)

S (rainsplash from cup)


Kinetic energy or Momentum

SKE z 141.0010.0 SM z 04200030 ..

rainsplash from splash cups S (g m-2 s-1)

0 2 4 6 8

kin

etic

en

erg

y K

Ez

(J m

-2 s

-1)

0.0

0.5

1.0

1.5

mo

men

tum

Mz

(kg

m-1

s-2

)

0.0

0.1

0.2

0.3

0.4Observed dataEq. [10] or [11]; R² = 0.857


SaltiphoneSensit “KE of rain field sensor”

Did not work properly under given circumstances

2.

),(F *uEQ Mass transport rate in kg m-1 s-1

xqQ

x

xx max

0

dCalibration

Contribution ofE (KEz or Mz)u*

ValidationzqQ

z

zz max

0

d



shear velocity u* (m s-1)

0.2 0.3 0.4 0.5 0.6

mea

sure

d r

ain

fall

inte

nsi

ty I

(mm

h-1

)

0

50

100

150p = 75 kPap = 100 kPap = 150 kPaEq. [8]; R² = 0.995

43.2387119 *uI

Results – wind-driven rain

Vertical deposition flux qx (g m-2 s-1)

x (m)

0 1 2 3 4 5

qx

(g m

-2 s

-1)

0.001

0.01

0.1

1

10

100

1000u

* = 0.27 m s-1; KEz

= 0.250 J m-2 s-1

u* = 0.39 m s-1; KEz

= 0.455 J m-2 s-1

u* = 0.50 m s-1; KEz

= 0.591 J m-2 s-1

x (m)

0 1 2 3 4 5

qx

(g m

-2 s

-1)

0.001

0.01

0.1

1

10

100

1000u

* = 0.27 m s-1; KEz

= 0.250 J m-2 s-1

u* = 0.39 m s-1; KEz

= 0.455 J m-2 s-1

u* = 0.50 m s-1; KEz

= 0.591 J m-2 s-1

Eq. (8.9)


xxq ΔΔ ee


R2 > 0.99

z (m)

0.0 0.1 0.2 0.3

qz

(g m

-2 s

-1)

0.01

0.1

1

10

100

1000u

* = 0.27 m s-1; KEz

= 0.250 J m-2 s-1

u* = 0.39 m s-1; KEz

= 0.455 J m-2 s-1

u* = 0.50 m s-1; KEz

= 0.591 J m-2 s-1

Horizontal flux qz (g m-2 s-1)



z (m)

0.0 0.1 0.2 0.3

qz

(g m

-2 s

-1)

0.01

0.1

1

10

100

1000u

* = 0.27 m s-1; KEz

= 0.250 J m-2 s-1

u* = 0.39 m s-1; KEz

= 0.455 J m-2 s-1

u* = 0.50 m s-1; KEz

= 0.591 J m-2 s-1

Eq. (8.12)

zbaq e


R2 > 0.98

xqQ

x

xx max

0

dCalibration

Contribution ofE (KEz or Mz)u*

ValidationzqQ

z

zz max

0

d

Transport rate Q (g m-1 s-1)


(KEz - KEzt) u*0.4 (-)

0.1 0.2 0.3 0.4 0.5

Q (

g m

-1 s

-1)

0

1

2

3Qx data

Eq. (9.11); R² = 0.956

4.0*

3105.4 uKEKEQ ztz



4.0*

3105.4 uKEKEQ ztz


(KEz - KEzt) u*0.4 (-)

0.1 0.2 0.3 0.4 0.5

Q (

g m

-1 s

-1)

0

1

2

3Qx data

Qz data

Eq. (9.11); R² = 0.956



2.1ztzd EEKQ

R2 = 0.96 4.0*uEEKQ ztzd

ztzd EEKQ

R2 = 0.93

R2 = 0.92


u* and KEz or Mz


Results – rainless wind (control)

(d)

x (m)

0 1 2 3 4 5

qx

(g m

-2 s

-1)

0.001

0.01

0.1

1

10

100

1000u

* = 0.33 m s-1

u* = 0.36 m s-1

u* = 0.39 m s-1

u* = 0.50 m s-1

Eq. [13]


xxq ΔΔ ee



(d)

x (m)

0.0 0.1 0.2 0.3

qx

(g m

-2 s

-1)

0.01

0.1

1

10

100

1000u

* = 0.33 m s-1

u* = 0.36 m s-1

u* = 0.39 m s-1

u* = 0.50 m s-1

Eq. [5]

zbaq e



3**

3106.18 tuuQ

(u* - u*t)3 (-)

0 1 2 3 4 5 6

sed

imen

t tr

ansp

ort

rat

e Q

(g

m-1

s-1

)

0

20

40

60

80

100

120

140Eq. [8]Qx data



3**

3106.18 tuuQ

(u* - u*t)3 (-)

0 1 2 3 4 5 6

sed

imen

t tr

ansp

ort

rat

e Q

(g

m-1

s-1

)

0

20

40

60

80

100

120

140Qz data

Eq. [8]Qx data

Results – wind-driven rain vs. rainless wind

wind-driven rain rainless wind

Q (g m-1 s-1) u* (m s-1) KEz (J m2 s-1) Q (g m-1 s-1) u*

(m s-1) KEz (J m2 s-1)

0.44 0.27 0.185 0.16 0.33 0

2.08 0.5 0.653 168.32 0.5 0

• Vertical deposition flux of sand was described with double exponential equation, q = f(x).

• Horizontal flux of sand was described with single exponential equation, q = f(z).

• Same expressions (and same equipment) can be used for wind-driven rain and rainless wind conditions.But model coefficients are different.

Conclusions

• Sediment transport rate Q relates well to normal component of KE or M (R2 = 0.93).

• Observed variation is better explained if u* is considered as well (R2 = 0.96).

• Qwr > Qw at low shear velocitiesQw >> Qwr at high shear velocities

Conclusions

Documents

Wim Cornelis, Greet Oltenfreiter, Donald Gabriels & Roger Hartmann