6 Initial Mixing
Introduction
Integral Analysis
Dimensional Analysis
Multi-port Diffusers
Gravitational spreading, intrusion & mixing
Multi-port Diffusers in Shallow Water
Buoyant Surface Jets
Combined Near and Far Field Analysis
Submerged Discharge
Mixing by turbulent
entrainment rather
than exchange
Dilution
xs
Q
Ua
c
H
cb
Qo
co
„
„
S = Q/Qo
S = (Co-Cb)/(C-Cb)
Mixing zones
„
„
Hydrodynamic
Regulatory
Dilution a solution to
pollution?
Biodegradable contaminant?
High ambient concentration of
contaminant?
Toxics?
Pure Jet
y
A
b
x
umax
b
Momentum driven
Bell-shaped velocity
distribution (in jet)
Irrotational flow
(entrainment field)
Properties
„
B
„
Figure by MIT OCW.
Daily and Harleman, (1966)
„
b~x
u ~ x-1
Q ~ ub2 ~ x
Buoyant Jet
Buoyancy driven
„
∆T
r
λb
~
u
Z
~x
b
θo
Uo, ∆To, ∆ρo
D
X
Figure by MIT OCW.
„
Temperature
Dissolved/Suspended
solids
Bell-shaped velocity
& scalar distributions
Linear spread
Finite initial size
(ZOFE)
Equation of State
(Gill, 1982)
ρ = ρ (T ) + ∆ρ ( S ) + ∆ρ (TSS )
⎡
ρ (T ) = 1000⎢1 −
⎤
T + 288.9414
(T − 3.9863) 2 ⎥
508929.2(T + 68.12963)
⎦
⎣
∆ρ ( S ) = AS + BS 3 / 2 + CS 2
A = 0.824493 − 4.0899 x10 −3 T + 7.6438 x10 −5 T 2 − 8.2467 x10 −7 T 3 + 5.3875 x10 −9 T
B = −5.72466 x10 −3 + 1.0227 x10 − 4 T − 1.6546 x10 −6 T 2
C = 4.8314 x10 − 4
1 ⎤
⎡
−3
∆ρ (TSS ) = TSS ⎢1 −
10
x
⎣ SG ⎥⎦
ρ = kg/m3, T in oC, S in PSU (g/kg), TSS in mg/L
Seawater Density ( σ1 Units )
44
-8
40
-6
σt = 1000*(ρ-1)
-4
36
Temperature (oC)
Fischer, et al. (1979)
-10
48
-2
(ρ in g/cm3)
0
32
2
4
28
6
8
10
24
12
14
20
16
18
16
20
22
12
24
26
8
28
30
4
0
0
4
8
12
16
20
24
28
32
36
40
SALINITY (% )
Figure by MIT OCW.
Model Types
Computational Fluid Dynamics (3-D)
Integral Analysis (1-D)
Dimensional Analysis (0-D)
Integral Analysis: Self-Similarity
∆T
r
λb
~
u
Z
~x
b
θo
Uo, ∆To, ∆ρo
D
u~
= f ( r / b)
~
uc
∆ρ
∆T
∆c
= g ( r / b)
=
=
∆cc ∆Tc ∆ρ c
X
Figure by MIT OCW.
f ( r / b) = e
g ( r / b) = e
−r
−r
2
b2
2
( λb ) 2
Integrated Fluxes
Volume
∞
∞
−∞
0
~dA = u~ f 2πrdr = 2πI u~ b 2
u
1 c
∫
∫ c
Q≅
Momentum* M ≅
Mass
J≅
∞
∞
−∞
0
~ 2 dA = u~ 2 f 2 2πrdr = 2πI u~ 2 b 2
u
2 c
∫
∫ c
∞
∞
−∞
0
~∆cdA = u~ ∆c fg 2πrdr = 2πI 3u~ ∆ b 2
u
c c
∫
∫ c
Neglects turbulent momentum fluxes
Conservation Statements
Continuity
Longitudinal
Momentum
Horizontal
Momentum
Contaminant
mass
Geometry 1
Geometry 2
dQ
~
=
π
=
π
α
b
v
b
u
2
2
e
c
d~
x
∞
dM
2
=
π
∆
ρ
θ
=
π
∆
ρ
g
rdr
I
gb
2
sin
2
sin θ
4
∫
~
dx
0
d ( M cosθ )
=0
~
dx
dJ
=0
~
dx
dx
= cosθ
~
dx
dy
= sin θ
~
dx
Solution Technique
Initial Value Problem
6 equations in 6 unknowns
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢⎣
A
⎤ ⎧ u& c ⎫ ⎡ ⎤
⎥ ⎪∆ρ& ⎪ ⎢ ⎥
⎥⎪ c ⎪ ⎢ ⎥
⎥ ⎪⎪ b& ⎪⎪ ⎢ ⎥
⎥ ⎨ & ⎬ = ⎢C⎥
⎥⎪ θ ⎪ ⎢ ⎥
⎥ ⎪ x& ⎪ ⎢ ⎥
⎥⎪
⎪ ⎢ ⎥
⎥⎦ ⎪⎩ y& ⎪⎭ ⎢⎣ ⎥⎦
Results
Output as function of
Densimetric
Froude Number
Fo =
Dimensionless
Distance, Height
x / Do
Limiting Conditions
uo
g (∆ρ o ρ ) Do
z / Do
Fo = ∞
Pure jet
Fo = 1
Pure plume
2.5
1.0
140
5
10
120
15
20
RNN
Temperature Chart
o
θ = 0.0
30
=
.01
40
c/
∆T
o
50
∆T
100
Vertical Distance Z/D
.015
80
75
.02
.025
60
.03
100
.04
40
.06
20
0
0
150
.08
.1
.15
.5
200
.3
20
40
60
80
100
120
140
160
180
F = 600
200
Horizontal Distance X/D
Shirazi & Davis, 1974
Figure by MIT OCW.
1.0
Centerline Excess Temperature
∆Tc
∆To
RNN
Temperature/ Width Chart
o
θ = 90
0.1
W/D = 10
F = 600
50
20
0.01
30
10
40
20
5
50
2.5
0
15
1.0
0
40
80
120
Vertical Distance Z/D
Shirazi & Davis, 1974
Figure by MIT OCW.
160
200
Example Calculations (WE 6-1)
Qo = 0.00125 m3/s
Do = 0.1 m
∆ρο/ρ = 0.025
u=Qo/(πD2/4)=0.16m/s
Fo=uo/(∆ροg/ρD)0.5 = 1
z/Do = 70
c/co = 0.008
Z (m)
Qm cm
hs
Qo
co
8
7
0
2.5
1.0
140
5
10
120
15
20
RNN
Temperature Chart
o
θ = 0.0
30
=
.01
40
c/
∆T
o
50
∆T
100
Vertical Distance Z/D
.015
80
75
.02
.025
60
.03
100
.04
40
.06
20
0
0
150
.08
.1
.15
.5
200
.3
20
40
60
80
100
120
Horizontal Distance X/D
Shirazi & Davis, 1974
Figure by MIT OCW.
140
160
180
F = 600
200
1.0
Centerline Excess Temperature
∆Tc
∆To
RNN
Temperature/ Width Chart
o
θ = 90
0.1
W/D = 10
F = 600
50
20
0.01
30
10
40
20
5
50
2.5
0
15
1.0
0
40
80
120
Vertical Distance Z/D
Shirazi & Davis, 1974
Figure by MIT OCW.
160
200
Example Calculations (cont’d)
Base Case
Increased
Momentum
Increased
Flow
Do
0.1
0.05
0.1
Qo
0.00125
0.00125
0.0025
uo
0.16
0.64
0.032
∆ρο/ρ
0.025
0.025
0.0125
Fo
1
5.7
2.8
z/Do
70
140
70
c/co
0.008
0.008
0.016
In deep water behavior depends mainly on buoyancy—not momentum,
flow rate, port size or orientation
2.5
1.0
140
5
10
120
15
20
RNN
Temperature Chart
o
θ = 0.0
30
=
.01
40
c/
∆T
o
50
∆T
100
Vertical Distance Z/D
.015
80
75
.02
.025
60
.03
100
.04
40
.06
20
0
0
150
.08
.1
.15
.5
200
.3
20
40
60
80
100
120
Horizontal Distance X/D
Shirazi & Davis, 1974
Figure by MIT OCW.
140
160
180
F = 600
200
1.0
Centerline Excess Temperature
∆Tc
∆To
RNN
Temperature/ Width Chart
o
θ = 90
0.1
20
W/D = 10
F = 600
50
0.01
30
40
10
15
20
5
50
2.5
0
1.0
0
40
80
120
Vertical Distance Z/D
Shirazi & Davis, 1974
Figure by MIT OCW.
160
200
Dimensional Analysis
Identify important independent and
dependent variables
Arrange in dimensionally consistent
manner
Determine coefficients empirically
Buckingham Π Theorem
Number of dimensionless parameters equals
number of independent plus dependent
variables minus number of dimensions used
to describe these variables
Example: D = ½ gt2
„
„
„
3 variables (g, t, D)
2 dimensions (length, time)
1 dimensionless variable (D/gt2)
“Empirical” coefficient (1/2)
Axi-symmetric Plume
Neglect ambient current, stratification
Assume deep water (initial momentum,
flow rate, nozzle size, discharge angle
less important than buoyancy)
Kinematic buoyancy flux
„
Bo = Qo g∆ρο/ρ
[L4T-3)
Axi-symmetric Plume (cont’d)
Q = f(B, z)
3 variables – 2 dimensions = 1 nondimensional parameter (c1)
c1 =
Q
α
Bo z β
α
Q = c1 Bo z
β
Axi-symmetric Plume (cont’d)
α
Q ~ Bo z β
L3 L4α β
= 3α L
T T
3 = 4α + β
1 = 3α
∴α = 1 / 3, β = 5 / 3
S = Q / Qo
1/ 3
c1 Bo z 5 / 3
Sc =
Qo
c1 ≅ 0.1
Integral vs Dim Anal (WE 6-2)
Input variables
„
„
„
„
Qo = 0.00125 m3/s
Do = 0.1 m
z=7m
∆ρο/ρ = 0.025 (salt water-fresh water)
Derived variables
„
„
„
Bo = Qo g ∆ρο/ρ = 0.00031 m4/s3
Fo = uo/(g ∆ρο/ρ Do)0.5 = 1
z/Do = 70
Integral vs Dim Anal (cont’d)
1.0
2.5
140
5
10
120
15
20
RNN
Temperature Chart
θ = 0.0o
30
Integral Analysis
=
.01
40
c/
∆T
o
50
∆T
100
Vertical Distance Z/D
.015
80
„
75
.02
„
.025
60
.03
Dimensional Analysis
100
.04
40
.06
20
0
0
150
.08
.1
.15
.5
200
.3
20
40
60
80
100
120
Horizontal Distance X/D
Figure by MIT OCW.
140
160
180
∆cc/∆co = 0.008
Sc = 125
F = 600
200
„
Sc=0.1Bo1/3z5/3/Qo=138
Blockage at surface (or trap
elevation)
xs
Qm c m
Ua
cn
cb
Qo
co
hs
H
Prevents entrainment of
ambient water near top of
trajectory
Mixing & extra entrainment
as jet “turns the corner”
“Near field” dilution
„
„
Sn = 0.26Bo1/3H
Xn/H = 2.8
5/3/Q
o
hs/H ~ 0.11 (horizontal
discharge)
Ambient Stratification
Stratification
frequency N
Ua
ρ(z)
Qm cm
cb
h
Qo
co
g∂ρ
N =
ρ∂z
2
H
Plume traps at level
of neutral buoyancy
with reduced dilution
ht = 2.8Bo 1/4/N3/4
Sm = 0.9 Bo3/4/QoN5/4
Ambient Current
xs
Qm c m
Ua
H
cb
Qo
co
Deflects plume
downstream
Augments dilution if
strong
Sm = 0.32uaH2/Qo
xs=0.3Qo(g∆ρο/ρ)/ua3
Dense plumes
Typical applications:
„
„
Ua
ht
co
Sm
Cold water from LNG
terminals
Brine from desal
plants, sol’n mining
of salt domes
ht=2.3Mo3/4/Bo1/2
Sm=2.8Mo5/4/QoBo1/2
Example: solution mining of
salt domes
Strategic Petroleum
Reserve
„
„
„
Dates from 1970’s
~700x106 bbl stored in
4 domes in LA & TX
Salinity gradients in
GoM confuse shrimp
Also used for
„
„
„
Salt production
Compressed gas
storage
Waste isolation
Multi-phase Plumes
Bubble plumes
„
„
„
„
Reservoir
destratification
Aeration
Ice prevention
Pollutant containment
Droplet plumes
„
Deep oil spills
Sediment plumes
„
„
Dredged mat’l disposal
CO2 ocean storage
Figure by MIT OCW.
Deep Oil-well Blowout
Figure by MIT OCW.
CO2Sequestration
Adapted from Heroz et al. (2000).
Figure by MIT OCW.
What are gas hydrates?
“Filled ice”
Example: methane hydrate
Cage structures of gas hydrates
⎯→ CO2 ⋅ nH 2O
CO2 + nH 2O ←⎯
T, P
n ≈ 5.75
ρ h = 1100 − 1140 kg/m3
CO2/seawater phase diagram
200
CO2(g)
(float)
Depth (m)
400
Stable Hydrates
(sink)
600
CO2(l)
(float)
800
0
4
8
o
Temperature ( C)
12
16
Figure by MIT OCW.
Laboratory studies
(Oak Ridge National Lab)
Liquid CO2
ORNL SPS
Details of mixing zone
(Seafloor process simulator)
Water
d
Hydrate–liquid CO2–water
composite extrusion
T
Water
Water
Liquid CO2
West et al., 2003; Lee et al 2003
P
Two-phase plume model
(100 kg/s CO2, 1 cm diameter spheres, release depth 800 m, Qc/Qw = λ = 0.49)
Neutrally
buoyant
particles:
only solute
density effect
Plume
depth
Intrusions
Pure hydrate
particles
Multi-port diffusers
Construction:
„
„
Cut and cover
Bored tunnel
Ports
„
H
„
l~ 0.3H (or 0.3 h)
Often 2 or more per riser
Line source approx.
hT
„
l
L
qo = Qo/L, bo = Bo/L
No current; no strat
„
Sm=0.42Hbo1/3/qo
No current; strat
„
Sm = 0.97bo2/3/qoN
Current; strat
„
Sm = 2.2ua1/2bo1/2/Nqo
Single vs Multiport (WE 6-3)
Boston Outfall
„
„
„
„
„
Diffuser Length L = 2000 m
No ports Np = 440
Flow rate Qo = 20 m3/s
Water depth H = 30 m
g∂ρ
2
Stratification frequency N =
ρ∂z
Š N2 = (9.8)(25-22)/(1025)(30) = 0.001 s-2
Single vs Multiport (cont’d)
As single port
„
„
„
„
„
Qo = 20/440 = 0.045 m3/s
Bo = 0.045*9.8*0.025 = 0.011 m4/s3
ht = 2.8Bo1/4/N3/4 = 12 m
l = L/Np = 2000/440 = 4.5 m
Š l > 0.3 ht => no merging
Sm = 0.9 Bo3/4/QoN5/4 =
0.9(0.011)3/4/(0.045)(0.0013)5/8 = 51
Single vs Multi-port (cont’d)
As multi-port diffuser (line source of
buoyancy)
„
„
„
„
qo = 20/2000 = 0.01 m2/s
bo = 0.01*0.025*9.8 = 0.0025 m3/s3
ht = 2bo1/3/N = 2(0.0025)1/3/(0.001)½ =9 m
Sm = 0.97bo2/3/qoN =
0.97(0.0025)2/3/(0.01)(0.001)½ = 56
Numerical modeling of sewage outfalls?
MWRA, 1999
Numerical modeling of sewage outfalls?
MWRA, 1999
Gravitational spreading,
intrusion, mixing
Surface spreading layer
xs
ua
hs = ua / g N '
2
QN cN
hs
H
cb
xs = 0.3QN g N ' / ua
Qo
co
bs = 0.8QN g N ' / ua
Internal spreading layer
3
H
ua
hs
ρ(z)
Qm cm
cb
h
Qo
co
3
hs = 1.2ua / N
xs = 0.25QN N / ua
2
bs = 0.65QN N / ua
2
Neutrally buoyant jet in
stratification
r
hm
ρ(z)
x
xm = 3.5M o
1/ 4
S m = 0.63M o
hm = 0.95M o
/ N 1/ 2
3/ 4
/ N 1/ 2Qo
1/ 4
/ N 1/ 2
Wachusetts Reservoir Algae
Occasional taste and odor
problems
„
„
Synura (left)
Chrysosophaerella
Algal locations
„
„
„
Hypolimnion
Metalimnion
Under ice
Conventional treatment
(surface algae) with CuSO4
from boat
How to efficiently treat (place
algaecide in proper stratum)
under ice & at depth?
Layout of Treatment System
(potential system being discussed)
Figure by MIT OCW.
CDM, 2005
Mid-Depth Air Driven Circulator
3 meters
10 meters
Chemical
Injection
Air Diffusers
CDM, 2005
Application at Depth
T
r
x
hm
z1
zo
z2
Elevation view
z
Figure 1
Length, thickness and dilution
(hence required operation time)
depend on reservoir stratification
and discharge momentum
Plan view
Application under Ice
T(z)
Algaecide
Introduced
...
...
..
a
T(z)
ζ
b
T(z)
Relies on bubble
plume to transport
algaecide to surface
c
T(z)
d
Figure 5
Multi-port diffusers in shallow water
Alternating
Staged
Typical for power
plant (thermal)
discharges
0.26 g ' H 2 L4 / 3 ⎛ ua HL ⎞
⎟⎟
Sa =
+ ⎜⎜
4/3
Q
Qo
⎝ o ⎠
2/3
0.5ua HL
+
Ss =
Qo
Tee
2
2
⎛ 0.5ua HL ⎞ 0.19 HLuo
⎜⎜
⎟⎟ +
Q
Qo
o
⎝
⎠
2
Co-Flowing
Figure by MIT OCW.
HLu a
St =
2
2Qouo + 10ua HL
0.5ua HL
+
Sc =
Qo
2
⎛ 0.5ua HL ⎞ 0.5 HLuo
⎜⎜
⎟⎟ +
Qo
⎝ Qo ⎠
Buoyant surface discharges
Plan View
Unsteady
Spreading
(Far Field)
t1
2bo
x
y
t2
Fo ' =
Buoyant Jet
(Near Field)
Lateral
Entrainment
xt
z
x
Vertical
Entrainment
uo
g ( ∆ρ o ρ ) l o
l o = ho bo
Transition
Line
uo, go'
ho
Time
Thermal plumes and
river discharges
Independent variables
t1
Side View
t2
Dependent variables
„
„
Figure by MIT OCW.
S = 1.4Fo’
Lengths ~Fo’ l o
Combined near and far field analysis
(accounting for background build-up)
co − c a
c F − ca
Far Field Dilution
SF =
Near Field Dilution
c − cF
SN = o
cN − cF
Total Dilution
co − c a
ST =
c N − ca
xs
Qm cm
Ua
H
cb
Qo
co
1
1
1
1
1
1
=
+
−
≅
+
ST S N S F S N S F S N S F
Total dilution less than either near field or far field
dilution and controlled by the smaller of the two
Example
Far field dilution
SF = 50 to 100
Near Field
dilution SN = 50
to 100
Total Dilution
ST = 25 to 33 to
50
MWRA, 1999