Transcript Slide 1

DIMENSIONLESS BANKFULL HYDRAULIC RELATIONS
FOR EARTH AND TITAN
European Space Agency
Gary Parker
Dept. of Civil & Environmental Engineering and Dept. of Geology
University of Illinois
1
UNTIL RECENTLY TITAN WAS SHROUDED IN MYSTERY
What we knew or could reasonably infer:
1. Larger than Mercury
2. Atmospheric pressure ~ 1.5 Earth atmospheres near surface
3. ~ 95 K near surface
4. Atmosphere of nitrogen (mostly), methane, ethane
5. Crustal material of water/ice
6. Near triple point of methane/ethane: possibility of
a. methane/ethane oceans
b. methane/ethane precipitation as liquid/solid
7. Possibility of rivers of liquid methane carrying sediment of
solid water ice!
But a thick shroud of smog produced by the
breakdown of methane under ultraviolet light
prevented any surface visualization.
2
AND THEN JANUARY 14, 2005 ARRIVED!
This and other images of Titan courtesy European Space Agency and NASA
Cassini/Huygens Mission:
very strong evidence for
rivers of liquid methane carrying sediment of water ice
I was glued to the internet! I had waited for years!
3
MARS VERSUS TITAN
Mars shows evidence of ancient
rivers of flowing water that
carried sediment similar to that
4
of the Earth’s crust.
MARS VERSUS TITAN contd.
But the era of flowing rivers was a long time ago, as evidenced by the
fairly intense impact cratering of Mars, and may not has lasted very
5
long as compared to Earth.
MARS VERSUS TITAN contd.
Tectonic ridges?
Titan shows evidence of active tectonics, vulcanism, aeolian and fluvial
reworking, and has very few impact craters: so its surface is likely active 6in
modern geological time!
MARS VERSUS TITAN contd.
Volcano?
Titan shows evidence of active tectonics, vulcanism, aeolian and fluvial
reworking, and has very few impact craters: so its surface is likely active 7in
modern geological time!
MARS VERSUS TITAN contd.
Aeolian dunes?
Titan shows evidence of active tectonics, vulcanism, aeolian and fluvial
reworking, and has very few impact craters: so its surface is likely active 8in
modern geological time!
MARS VERSUS TITAN contd.
River drainage basin?
Titan shows evidence of active tectonics, vulcanism, aeolian and fluvial
reworking, and has very few impact craters: so its surface is likely active 9in
modern geological time!
MARS VERSUS TITAN contd.
Impact crater
Titan shows evidence of active tectonics, vulcanism, aeolian and fluvial
reworking, and has very few impact craters: so its surface is likely active10in
modern geological time!
ALLUVIAL GRAVEL-BED RIVERS ON TITAN?
The evidence suggests that at least near where Huygens touched down,
there is a plethora of alluvium in the gravel and sand sizes. The gravel
presumably consists of water ice and appears to be fluvially rounded. 11
CAN OUR KNOWLEDGE OF ALLUVIAL GRAVEL-BED
RIVERS ON EARTH HELP US MAKE INFERENCE ABOUT
TITAN?
12
IF WE KNEW THE PHYSICS BEHIND RELATIONS FOR
BANKFULL GEOMETRY HERE ON EARTH
• Bankfull Depth
Hbf ~ (Qbf)0.4
• Bankfull Width
Bbf ~ (Qbf)0.5
• Bed Slope
S ~ (Qbf)-0.3
where Qbf = bankfull discharge
we might be able to extend the relations to Titan.
13
WE BEGIN WITH EARTH
The Parameters:
Qbf
QbT,bf
Bbf
Hbf
S
D

s
R
=
=
=
=
=
=
=
=
=
g

=
=
bankfull discharge (m3/s)
volume bedload transport rate at bankfull discharge (m3/s)
bankfull width (m)
bankfull depth (m)
bed slope (1)
surface geometric mean or median grain size (m)
density of water (kg/m3)
density of sediment (kg/m3)
(s/ ) – 1 = submerged specific gravity of sediment ~ 1.65
(1)
gravitational acceleration (m/s2)
kinematic viscosity of water (m2/s)
The forms sought: dimensionless versions of
n
H bf ~ Q bfh
,
n
B bf ~ Q bfb
,
n
S ~ Q bfs
,
n
Q bT ,bf ~ Q bfbT
Why dimensionless?
In order to allow scaling between Earth and Titan!
14
Meet my friends the DIMENSIONLESS PARAMETERS
RgD D
Re p 
Qˆ 
Particle Reynolds number

Q bf
gD D
Dimensionless bankfull discharge
2
1/ 5
~ g H bf
H
2/5
Q bf
Dimensionless bankfull depth
1/ 5
~ g B bf
B
2/5
Q bf
Dimensionless bankfull width
S
Down-channel bed slope
Q bT ,bf
ˆ
QT 
2
gD D
Dimensionless bankfull bedload
transport rate

 bf 

c
H bf S
Bankfull Shields number
RD
  0 . 5 [ 0 . 22 Re
 0 .6
p
 0 . 06  10
 0 .6
(  7 . 7 Re p
)
]
Shields number at
threshold of motion
15
DATA SETS FOR GRAVEL-BED RIVERS ON EARTH
1.
2.
3.
4.
Alberta streams, Canada1
Britain streams (mostly Wales)2
Idaho streams, USA3
Colorado River, USA (reach averages)
1
Kellerhals, R., Neill, C. R. and Bray, D. I., 1972, Hydraulic and
geomorphic characteristics of rivers in Alberta, River Engineering
and Surface Hydrology Report, Research Council of Alberta, Canada,
No. 72-1.
2 Charlton, F. G., Brown, P. M. and Benson, R. W., 1978, The
hydraulic geometry of some gravel rivers in Britain, Report INT 180,
Hydraulics Research Station, Wallingford, England, 48 p.
3 Parker, G., Toro-Escobar, C. M., Ramey, M. and Beck S., 2003,
The effect of floodwater extraction on the morphology
of mountain streams, Journal of Hydraulic Engineering, 129(11),
2003.
4 Pitlick, J. and Cress, R., 2002, Downstream changes in the channel of a
large gravel bed river, Water Resources Research 38(10), 1216,
doi:10.1029/2001WR000898, 2002.
16
WHAT THE DATA SAY: WIDTH, DEPTH, SLOPE
The four independent sets of data form a coherent set!
100
~
B
Britain width
Alberta width
Idaho width
Colorado width
Britain depth
Alberta depth
Idaho depth
Colorado depth
Britain slope
Alberta slope
Idaho slope
Colorado slope
Btilde, Htilde, S
10
~ ~
B, H, S
1
~
H
0.1
S
0.01
0.001
0.0001
1.0E+02
1.0E+03
1.0E+04
1.0E+05
Qhat
ˆ
Q
1.0E+06
1.0E+07
17
REGRESSION RELATIONS BASED ON THE DATA
~
ˆ 0 .00004
H  0 . 3785 Q
,
~
ˆ 0 .0661
B  4 . 698 Q
,
ˆ  0 .3438
S  0 . 1003 Q
To a high degree of approximation,
~ ~
H  H c  0 . 3785
Remarkable, no?
1.E+02
Bdimtilde, Hdimtilde, S
y = 4.6977x0.0661
~ ~
B, H, S
1.E+01
y = 0.3785x4E-05
1.E+00
Bdimtilde
Hdimtilde
S
Power (Hdimtilde)
Power (Bdimtilde)
Power (S)
1.E-01
y = 0.1003x-0.3438
1.E-02
1.E-03
1.E-04
1.E+02
1.E+03
1.E+04
1.E+05
Qdim
Qˆ
1.E+06
1.E+07
18
WHAT DOES THIS MEAN?
0 .4
H bf ~ Q bf or
0 . 3785
H bf 
0 . 4661
B bf ~ Q bf
B bf
g
1/ 5
Q
0 .4
bf
or
  4 . 698

 0 . 3438

S ~ Q bf
or
S   0 . 1001


gD
5/2
gD

5/2
 0 . 0661
g

0 . 3438
1 / 5
 Q 0 .4661
 bf
 Q  0 . 3438
 bf
19
WHAT THE DATA SAY: BANKFULL SHIELDS NUMBER

 bf ~ 0 . 0486
( average )

 bf Threshold for Significant

Shields Diagram
with Threshold for Motion,
 c  and
0 . 03
, r   for1 Gravel-bed
. 62
Suspension
Bankfull Shields Number
Streams
c
threshold for significant
suspension
10
*
1
suspension
motion
Alta
Brit
Ida
Colo
Average
0.1
0.01
threshold of motion
(modified Shields curve)

c
  0 .0.001
5 [ 0 . 22 Re
1
 0 .6
p
 0 . 06  10
10
100
 0 .6
(  7 . 7 Re p
1000
Rep
)
]
10000
100000 1000000
20
THE PHYSICS BEHIND IT ALL
100
Assume the following relations.
Q bf
Cz 
B bf H bf
gH bf
H 
 3 . 732  bf 
 D 
Cz
Cz
Manning-Strickler resistance relation
Cz
Fit
10
0 . 2645
1
1
10
Parker-Einstein bedload relation
B bf
 
 11 . 2 
RgD D
 3/2
bf


c 
1   



bf



0 . 0562
 bf  0 . 02301 Qˆ
1000
1
4 .5
Relation for bankfull Shields number
100
0.1

tausbf
Q bT ,bf
H bfHhat/ D
 bf
tausbf
FitQ
0.01
0.001
1.E+02
1.E+03
1.E+04
1.E+05
1.E+06
1.E+07
Qˆ
Qhat
Channel form relation of type of Parker (1978)

 bf


c
 r  1 . 62
“Gravel yield” relation
ˆ  0 . 003176 Q
ˆ 0 .5504
Q
T
21
THE RELATIONS OF THE PREVIOUS SLIDE YIELD
PRECISELY THE OBSERVED DIMENSIONLESS
RELATIONS!
~
ˆ 0 .0661
B  4 . 698 Q
ˆ  0 .3438
S  0 . 1003 Q
y = 4.6977x0.0661
Bdimtilde, Hdimtilde, S
~ ~
H  H o  0 . 3785
1.E+02
1.E+01
y = 0.3785x4E-05
1.E+00
~ ~
B, H, S
1.E-01
y = 0.1003x-0.3438
1.E-02
1.E-03
1.E-04
1.E+02
1.E+03
1.E+04
1.E+05
1.E+06
1.E+07
Qdim
ˆ
Q
22
GENERALIZATION FOR OTHER PLANETS/SATELLITES
Manning-Strickler resistance relation
Q bf
Cz 
B bf H bf
gH bf
H 
 3 . 732  bf 
 D 
0 . 2645
Parker-Einstein bedload relation
Q bT ,bf
B bf
 

RgD D
 11 . 2  bf
3/2


c 
1   

 bf 

The presence of
g and R allow us
to go from
4 .5
Relation for bankfull Shields number

 0 . 02301 

RD

H bf S


2 
gD D 
0 . 0562
Q bf
Channel form relation of type of Parker (1978)

 bf
 r  1 . 623

c
to
“Gravel yield” relation (volume to mass)
Q bT ,bf
gD D
2
0 . 00841 


1  R 


2 
gD D 
0 . 5504
Q bf
23
BACK-CALCULATED DIMENSIONALLY HOMOGENEOUS
BANKFULL HYDRAULIC RELATIONS FOR
ALLUVIAL GRAVEL RIVERS ON
H bf  0 . 1751
B bf 
S 
(1  R )
g
The presence of g and R
allow us to go from
0 . 7908
0 . 4661
15 . 992
R (1  R ) g
0 . 1314 R g
(1  R )
Q
1/ 5
0 .4
bf
Q bf
0 . 2331
0 . 1719
0 . 7908
D
0 . 1653
D
0 . 8595
Q
0 . 3438
bf
to
24
FROM
Pressure E-atmo
Temperature K
Grav. accel. m/s2
Fluid dens. kg/m3
Sed. Dens. kg/m3
(s/) - 1
Kin. Viscosity m2/s
TO
Parameter Earth
p
1
Titan
1.5
T
g

~ 293
9.81
1000
~ 95
1.40
446
s
R

2650
1.65
1.00x10-6
931
1.09
4.04x10-7
25
CONSIDER A STREAM WITH THE SAME BANKFULL
DISCHARGE Qbf AND CHARACTERISTIC GRAIN SIZE D
HOW SHOULD TITAN COMPARE WITH EARTH?
E = Earth, T = Titan
H bf ,T
H bf ,E
B bf ,T
B bf ,E
ST
SE
 gT
 
 gE
 gT
 
 gE
 gT
 
 gE









1/ 5
 (1  R T ) 


 (1  R E ) 
 0 . 2331
0 . 1719
RT

 RE
 RT

 RE



1/ 2
From
0 . 7908
 (1  R T ) 


(
1

R
)
E 

  (1  R T ) 
 

  (1  R E ) 
1
to
 0 . 7908
26
CONSIDER A STREAM WITH THE SAME BANKFULL
DISCHARGE Qbf AND CHARACTERISTIC GRAIN SIZE D
HOW SHOULD TITAN COMPARE WITH EARTH?
E = Earth, T = Titan
H bf ,T
H bf ,E
B bf ,T
B bf ,E
ST
SE
 gT
 
 gE
 gT
 
 gE
 gT
 
 gE









1/ 5
 (1  R T ) 


 (1  R E ) 
 0 . 2331
0 . 1719
RT

 RE
 RT

 RE



0 . 7908
1/ 2
= 1.48 x 0.83 = 1.23
1
 (1  R T ) 

 = 1.57 x 1.56 = 2.46
 (1  R E ) 
  (1  R T ) 
 

  (1  R E ) 
 0 . 7908
= 0.72 x 0.80 = 0.57
27
SO FOR THE SAME BANKFULL DISCHARGE Qbf AND
CHARACTERISTIC GRAIN SIZE D
A gravel-bed river on
might be
1.23 x the bankfull depth,
2.46 x the bankfull width and
0.57 x the down-channel slope
of a gravel-bed river on
Could braiding be more
common on Titan?
28
BUT WAIT A MINUTE!
IS “GRAVEL” ON TITAN GRAVEL ON EARTH?
For dynamic similarity in grain Reynolds number
Re p ,T  Re p ,E
or
R T g TD T D T

T
R E g ED E D E
E
or
DT
DE
 T 

 
 E 
2/3
 RT

 RE



1/ 3
 gT

 gE



1/ 3
 1 . 20
So the answer is “yes” to a reasonable approximation!
29
GRAIN REYNOLDS INVARIANCE
Besides, the dynamics of sediment transport becomes
approximately invariant to particle Reynolds number for
Re p  ~ 3320
or
D >~ 8.8 mm on Earth
or
D >~ 10.6 mm on Titan
based on the condition c*/c,asymp*  0.90 using


c
 0 . 5 [ 0 . 22 Re
 0 .6
p
 0 . 06  10
 0 .6
(  7 . 7 Re p
)
]
30
WHAT ABOUT AEOLIAN PROCESSES ON TITAN?
Let Ua = wind velocity, a = atmospheric density, Cf = drag coefficient, s =
sediment density, D = grain size. Scaling for mobility of grain size D:
  a C f U a2

  gD
 s

  a C f U a2
  

  gD
T
 s



E
Atmospheric density
Earth 293K 1 E-atmo,
a = 1.21 kg/m3
Titan (nitrogen) 95K 1.5 E-atmo,
a = 5.39 kg/m3
Assuming Reynolds invariance (Cf  constant), critical velocity Uac to blow
around size D scales as:
1/ 2
1/ 2
1/ 2
  a ,T 
  s ,T   g T 


 


 0 . 16





U ac ,E
  a ,E 
  s ,E   g E 
Much easier to blow sediment around on Titan!
But much less solar heating to drive meteorology!
U ac ,T
31
QUESTIONS OR COMMENTS?
32