Density Currents and Stratified Flow
Environmental Hydraulics
Density Currents and Stratification
Already small differences in density give rise to
stratification in receiving waters. Stable conditions
develop, associated with water motion.
⇒ Reduced exchange between layers
⇒ Consequences for water quality
Causes of density differences:
• Temperature
• Salinity
• Suspended matter
1
Characterization of density differences:
σ=
ρ − ρo
ρo
Example:
0 oC – 1 oC
=>
Δσ=0.06
10oC – 11oC
=>
Δσ=-0.10
20oC – 21oC
=>
Δσ=-0.21
Density varies approximately linearly with salinity:
0.1%
=> Δσ=0.80
(independent of temperature)
Classification of Stratified Flow
• surface flow
• bottom flow
• inter flow
Interface between layers:
Thermocline
-
temperature
Halocline
-
salinity
Lutocline
-
suspended solids
Saltwater
wedge
2
Net Force due to Density Difference
Water volume with density r1 in an ambient with density r2.
ρ1 gV
Gravity:
Buoyancy:
ρ 2 gV
Net force:
( ρ1 − ρ2 ) gV = ρ1
Reduced acceleration
due to gravity:
g'=
( ρ1 − ρ2 )
( ρ1 − ρ2 )
ρ1
ρ1
gV = ρ1 g 'V
g
Froude Densimetric Number
U
Fd =
g
( ρ1 − ρ2 ) y
=
momentum
buoyancy
ρ1
U and y velocity and length scale, respectively
Example: Jet discharge
U = exit velocity
y = D = exit diameter
3
Example of Stratified Flows
• wind setup and associated displacement of
density interface (e.g., lake)
• saltwater wedge in a river mouth
Inclination of Water Surface due to Wind
Force balance:
ρw g
2
h
− ρw g
2
(h +
dh
dx) 2
dx
+ (τ0 + τb ) dx = 0
2
4
Resulting in:
dh τ0 + τb
=
dx ρ w gh
Receiving water with constant depth and length:
Δh =
τ 0 + τb
L
ρ w gh
Assume that
τ0 τb and
Δh =
τo = CDρair w102
CDρair w102 L
ρw gh
Numerical Example
h = 10 m
w10 = 10 m/s
CD = 1.3 · 10-3
L = 10 km
Which gives a maximum wind setup of:
Δh = 1.3 ⋅10−3 ⋅
1.3
1 102 ⋅104
⋅
⋅
= 0.016 m
1000 9.81
10
5
Inclination of Density Interface
Surface inclination is small but a compensating tilt in the
density interface becomes large.
Force balance (neglecting
water movement and
associated shear stresses):
dh2
dh ρ1
=− 1
dx
dx ρ 2 − ρ1
Numerical Example
ρ1 = 998.2 kg/m3 (20 0C)
ρ2 = 1000 kg/m3 (4 0C)
dh1
= 10 −6
dx
These values yield:
dh2
1000
= −10 −6 ⋅
= −0.5 ⋅10−3
dx
1000 − 998.2
With L = 1 km one obtains Δh2 ≈ 0.5 m, whereas the water
surface only rises Δh1 = 0.001 m
6
Saltwater Wedge
Assume no mixing at interface.
Constant velocity along the
verticals.
Slope of the energy line:
S =−
d ⎛
U2 ⎞
z
+
y
+
y
+
2
1
⎜
⎟
dx ⎝
2g ⎠
Definitions:
U=
q
q2
dz
, F12 = 3 , So = −
y1
gy1
dx
Introduce definitions in the equation yields:
dy2
dy
+ (1 − F12 ) 1 = So − S
dx
dx
Steady-state conditions are assumed => net force is zero
τi − y 2
dpo
+ ρ2 gy2 sin φ = 0
dx
po = ρ1 gy1 + ρ2 gy2
(pressure at bottom)
7
Shear stress at interface:
τi = ρ1 gSy1
Introduce into force balance:
S
y1 dy1 ρ2 dy2 ρ2
=
+
− So
y2 dx ρ1 dx ρ1
ρ1 y1
dy1
ρ 2 y2
= −S
ρ
dx
1 − 1 − F12
ρ2
1+
Combine equations:
(Equation to determine the shape of the
saltwater wedge)
y2 is small at the front and dy1/dx < 0. Salt water
penetration may only occur if:
F12 <
ρ2 − ρ1
ρ2
If y2 -> 0, F1 -> Fo and:
1/ 2
⎛ρ −ρ ⎞
Fo < ⎜ 2 1 ⎟
⎝ ρ2 ⎠
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Numerical example:
r1 = 1000 kg/m3
r2 = 1029 kg/m3
=>
⎛ 1029 − 1000 ⎞
Fo < ⎜
⎟
⎝ 1029
⎠
1/ 2
= 0.17
Thus, if Fo > 0.17 there is no risk of a salt water
wedge forming.
In order to determine the shape of the salt water
wedge using the previous equation, a boundary
condition is needed at the river mouth:
U = g ' y1
Density Interface
Large density variation over short vertical distance.
Prevents vertical exchange of water.
Mixing across interface needs supply of energy (mixing implies
increase in the potential energy of the system).
The Richardson number characterizes the ability of the
interface to prevent mixing:
∂u
∂ρ
Ri = − g / ρ ⎛⎜ ⎞⎟
∂z ⎝ ∂z ⎠
2
(simpler definition of Ri yields:
Ri =
1
Fo2
)
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Density Interface Development
0
Water Depth (m)
-5
-10
Month
FEB
-15
APR
JUN
AUG
OCT
-20
DEC
-25
4
8
12
16
20
24
Water Temperature (degrees)
Many laboratory experiments
on mixed-layer depening
Temperature
distribution Nainital
Lake
rinitial
rmixed
Internal Waves
Similar governing equations as for surface
waves, if g is replaced by g’.
Additional considerations:
• effect of the upper layer (inertia etc)
• shear stresses at the interface
• thickness of the interface
Lake Biwa,
Japan
10
Internal Waves in the Ocean
Straits of Gibraltar
Sulu Sea
Breaking of Internal Waves
Breaking generates
substantial mixing
In the atmosphere
Laboratory
experiments
11