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Buoyant Jets and Plumes:Theory and Models
Environmental Hydraulics
Definitions
Plume = boundary layer flow originating from a source of buoyancy
Buoyant jet (forced plume) = boundary layer flow originating from a source of momentum and buoyancy
2
Positive and Negative Buoyancy
Positive Negative
Negative/positive
Buoyant Jet Evolution
Zone of jet evolution:
1. jet development (ZFE)2. fully developed jet (ZEF)3. final vertical elevation of jet4. horizontal spreading
3
Densimetric Froude Number
00
00
0
r
uFg D
=ρ −ρρ
00
0
0
0
'
' r
uFg D
g
=
ρ −ρ=
ρ
Related is the Richardson number:
21
o
RiF
=
(in oceanography: )2
/ uRi gz z∂ρ ∂⎛ ⎞= − ρ⎜ ⎟∂ ∂⎝ ⎠
Model of Buoyant Jet
Assumptions:
• hydrostatic pressure
• small variations in density
• symmetric jet
• self-similar properties
• no influence from boundaries
• density linearly related to concentration
4
Geometrical conditions:
sindzds
= θ
cosdxds
= θ
Continuity (water):
max0
( 2 ) 2 sd u rdr u rds
∞
ρ π = α ρ π∫
Continuity (pollutant):
0 00
2cu rdr c Q∞
π =∫
Momentum equation (x-direction):
20 0 0
0
cos 2 cosu rdr Q u∞
θ ρ π = θ ρ∫
5
Momentum equation (z-direction):
20 0 0
0 0 0
sin 2 sin ( 2 )s
u rdr Q u g rdr ds∞ ∞
θ ρ π = θ ρ + Δρ π∫ ∫ ∫
Self-similarity:
2
max
2
max
exp( ( ) )
exp( ( ) )
u rku s
c rkc s
= − ⋅
= −μ ⋅ ⋅
Buoyant Jet Trajectories
00
00
0
0
,
r
mm
o o
uFg D
cSc
x zD D
=ρ −ρρ
=
Governing parameters:
6
Typical values on self-similarity coefficients:
Fr0 k μ
0 96 0.74∞ 77 0.80
Homogenization before Horizontal Spreading
max 1.4after
cc
≈
Regard temperature as a ”pollution” with:
0
* r
r
T TTT T−
=−
7
Example of Discharge of Wastewater
Wastewater, at a flow rate of 0.1 m3/s and with a density of 1005 kg/m3, is discharged horizontally through a circular pipe at a depth of 20 m below the sea surface. The density of the seawater is assumed to be constant, ρr = 1025 kg/m3. What is the dilution at the jet axis immediately below the water surfaceand how far from the discharge point will the jet reach the water surface? Make calculations for an initial jet diameter of D0 = 0.25 m and D0 = 0.20 m.
01 2 20
0.1 0.1 2.04 m/s/ 4 0.25 / 4
uD
= = =π π⋅
Initial velocity:
02 20.1 3.18 m/s
0.20 / 4u = =
π⋅
Density difference: ρr – ρ0 = 1025 – 1005 = 20 kg/m3
8
Densimetric Froude number:
012.04 9.2320 0.25
1005
Fg
= =⋅ ⋅
023.18 16.120 0.20
1005
Fg
= =⋅ ⋅
z/Do + x/Do and Fo => solution through Figure 40
Water surface at: z/Do=20/0.25=80 + Fo=9.23
⇒ x/Do=45 x=11 m
Sm=50
Water surface at: z/Do=20/0.20=100 + Fo=16.1
⇒ x/Do=70 x=14 m
Sm=57