Chapter 3
Flow - stability
Introduction
focus on stability of loose non-cohesive grains rock: important material for protection
• • • grains may vary in size from μc (sand) to m (rock)
Uniform flow – Horizontal bed Forces on a grain in flow
Drag force :
2
= D
D
C
F
A
ρ u D w
2
Shear force :
=
2
S
S
F
C
F
A
ρ u w
2 ρ u d w
Lift force :
2
= L
L
C
F
A
ρ u L w
1 2 1 2 1 2
⎫ ⎪ ⎪ ⎪ F ≈ ⎬ ⎪ ⎪ ⎪ ⎭
Balance equations
: 0 = H
Σ
=
FW x f = F
3
( ∝
ρ
d g )
2 2 d u cw
- ρρ s w
⋅
Σ
SD,
F SD, W = F : 0 = V Σ L )( )( dO W = dO F : 0 = M ⋅
⎫ ⎪ ⎬ ⎪ ⎭
Relation between load and strength
d g
d g
Δ
→
Δ
2 c
2 u c
K = u
- ρρ s w ρ w
⎞ ⎟⎟ = d g ⎠
⎛ ⎜⎜ ∝ ⎝
Isbash (1930)
d g
or
1.7 =
or
0.7 = d
Δ
Δ
2 1.2 = u c
2 u c g 2
u c d g Δ
used for first approximation when:
(e.g a jet entering a body of water)
• relation between velocity and waterdepth not clear
Shields (1936)
f
f
=
=
=
=
(
)
ψ c
*Re
g d
2 u c * g d Δ
u d c * υ
⎛ ⎜ ⎝
⎞ ⎟ ⎠
(
)
τ c − ρ ρ w
s
:ψ Mobility parameter (when actual u used)
=
*Re
Re≠
:cψ Shields parameter (stability parameter)
cu d * υ *Re
Note:
Critical shear stress Shields Van Rijn
f
f
=
=
=
=
(
)
ψ c
Re *
dg
du * c υ
2 u * c dg Δ
)
⎛ ⎜ ⎝
⎞ ⎟ ⎠
τ c ( ρρ − w
s
Example
What is u*c for sand with d = 2 mm?
Shields
1500
Re
=
=
=
*
du c * ν
002 6 − 10
.01 × 33.1 ×
Ψc = 0.055
gd
.0
055
65.1
81.9
002.0
042.0
/ sm
=
Δ
=
×
×
×
=
ψ c
u =→ * c
ψ c
u * c gd Δ
Re
63
04.0
=
=
=→=
ψ c
*
002 6 −
cdu * ν
.0 042 33.1
.0 10
× ×
Guess: u*c = 1 m/s • •
* c
u gd 04.0 65.1 81.9 .0 002 .0 036 / sm Δ = × × × = =ψ c
Example (cont.)
Van Rijn
d = 2 mm
3
3
42
d
d
.0
002
=
=
=
*
65.1 33.1(
g Δ 2 ν
81.9 × 26 )10 ×
0.04
cΨ =
u
gd
04.0
65.1
81.9
.0
002
.0
036
/ sm
Δ
=
×
×
×
=
* c
=ψ c
•
Relative protrusion
Load and strength distribution
0. no movement at all 1. occasional movement at some locations 2. frequent movement at some locations 3. frequent movement at several locations 4. frequent movement at many locations 5. frequent movement at all locations 6. continuous movement at all locations 7. general transport of the grains
Shields
Videos on stability of rock on a bed with current only
u = 0.60 m/s, Ψ = 0.03
u = 0.70 m/s, Ψ = 0.04
u = 0.83 m/s, Ψ = 0.05
u = 0.92 m/s, Ψ = 0.06
u = 0.97 m/s, Ψ = 0.07
Threshold of motion
(Extrapolation of transport to zero)
18
q
(for
0.05)
=
6.56 10 ⋅
<
16 ψ
ψ
q
* s
with
q
=
* s
2.5
3
q
13
(for
0.05)
=
>
ψ
ψ
s g d
Δ
* s
⎫ ⎪ ⎬ ⎪⎭
Shields Paintal
Stone dimension
Nominal diameter:
3
d
3 V M =
/ ρ
n =
d≠
nd
50
50
50
0.84
≈
nd d
50
Influence of waterdepth
u
=
Uniform flow:
u *
ψ = c
g C 2 u *c g d Δ
C
ψ
u
c
=
c g d
g
Δ
n
5 0
R
1.7
=
C
18log
=
gd
icu Δ
12 k
r
Isbash:
u≠
u c
ic
Attention:
roughness kr = 2*d50 or kr = 3*d50
Influence waterdepth on critical velocity
Practical application
2
C
ψ
u
u
c
d
=
→
=
n
50
2
c g d
g
Δ
C
c Δ
ψ
n
50
c
Roughness and threshold of motion
dn50=0.146m
Lammers, 1997
Choose Ψ:
do we select Ψ on the safe side or do we use the expected value of Ψ ??
Angles of repose for non-cohesive materials
1:1
1:1.5
1:2
1:3
Influence of slope on stability
φ = 40ο
Case b: slope parallel to flow Case c: slope perpendicular to flow
Slope parallel to current
)
W
sin α
K(
= )
=
=
α //
F( α // F(0)
cos tan W - φα tan W φ
sin
α
cos αφ
sin αφ
=
=
cos − sin φ
sin ( - ) αφ sin φ
Slope perpendicular to current
2
2 α
cos
sin
- 1
- 1 =
cos
α
=
= =
= )K( α
2 α 2 φ
2 α 2 φ
tan tan
sin sin
F( ) α F(0)
2 - φα tan 2 φ tan
Stability on top of sill
Experiments: first damage at downstream crest
use velocity on top of the sill
Stability and head difference
Shields is useless here because Shields contains waterdepth
waterlevel downstream is below the top of the dam Shields for flow over sill
(0.5
+
0.04
2 = u 1
2 (h g 2 μ u
= )h - d
(h g 2 ) u
)h - d
h d d
n
50
discharge coefficient
Vertical constriction Stability with flow under weir
Shields in horizontal constriction
2
h
= C
1 - 4
4.5log
=
ψ c
ψ c g
α 2 φ
u gap gd Δ
sin sin
n
50
3 d
50
n
⎛ ⎜ ⎝
⎞ ⎟ ⎠
Correction
General formula
h
C
18log
18log
=
=
h 3 d
50
n
50
(horizontal closure with trucks)
Damage at half depth: 12 1/ 2 × 2 n d α = 30o; φ = 40o α slope of construction ϕ angle of repose (internal stability)
Stability on head of dam
Deceleration
cu
u
structure
=
K = v
c
cs
without with
c u
structure
u u
ucu: vertically averaged critical velocity in uniform flow ucs: velocity in case with a structure
Effect of flow field
Relation between Kv and turbulence level
K
)
=
⎯ →⎯
=
=
r 1 3 ( + cu
u cu
cs
v
b 1 3 +
g r u cs
u cu u cs
r 1 3 + cs r 1 3 + cu
ucu : vertically averaged critical velocity in uniform flow ucs : velocity in case with a structure rcu : turbulence intensity in uniform flow rcs : vertically averaged turbulence intensity
Stability downstream of a sill
no dam
Kv in vertical constriction
)
(
u
K
=
−
→
≈
huDhu 1 22
2
u =→ 1
2
v
h 2 Dh − 2
h 2 Dh − 2
high dam
Damage after some time
Stone stability downstream of a hydraulic jump
Peak velocities and incipient motion in horizontal constriction
damage after constriction
Kv - factors for various structures
Structure Groyne
K vG 1.3 - 1.7
K vM 1.1 - 1.2
K v0 b0*K vG/bG
1.2
1
b0*K vbG
Abutment
1.3 - 1.7
1.2
b0*K v/bG
Shape Rect- angular Trape- zoidal Rect- Angular Round
1.2
1.2 - 1.3
b0*K v/bG
1 - 1.1
1 - 1.1
b0*K v/bG
Stream Lined Round
Pier
1.2 - 1.4 ⊗ 1 - 1.1
1.4 - 1.6 ⊗ 1.2 - 1.3
Rect- Angular Abruptly
Outflow
--
b0*K v/bG ⊗ 2*K v b0*K v/bG ⊗ 2*K v --
1
--
--
0.9
Stream Lined Top
Sill
Section 3.6.1 Fig 3.13
Section 3.6.1 Fig 3.13
Section 3.6.1 Fig 3.13
Down Stream
⊗ For many piers in a river the first expression for K v is appropriate. The second is valid for a detached pier in an infinitely wide flow, where K G is not defined.
Definition of velocities
groyne vertical pole
Combined equation
2
= d
2
2 u * K v c K C Δψ s c
Kv : reduction for constriction, etc. Ks : reduction for slope (parallel, perpendicular))
Coherent material
Relative density:
n
−
( 1
)
s
=
Δ
)( − ρρ w ρ w
Gabions
