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Interpreting profiles of pore water solutes
First, solute transport (simple)
1. Diffusive Transport:
€
Flux =−φDseddCdx
2. Sediment BurialGenerally: Assume a constant mass flux (…not always true)
€
massvolume
=ρsolid 1 −φ( ) ingsolid
cmsed3
Mass accumulation rate:
€
MAR =wρsolid 1 −φ( ) ingsolid
cmsed2 ⋅y
Solute burial, cont.
Below the “compacting layer”
€
MAR =w∞ρsolid 1 −φ( )
Which must be the same as MAR at shallower depths…
€
wρsolid 1 −φ( ) =w∞ρsolid 1 −φ∞( )
w x( ) =w∞
1 −φ∞( )
1 −φ x( )( )
That’s for solids…
For solutes…
The burial rate for solutes is slightly smaller than for solids…
€
v x( ) =φ∞φ x( )
w∞ AND
€
F =φvC
Interpreting pore water profiles…
In general: we consider rates of change of concentration over space and time
We derive descriptive expressions by considering the mass balance in a layer of sediment
x1
x2
R
F1
F2
In the box:
€
x2 −x1( )∂φC∂t
=F1 −F2 + φR x2 −x1( )
€
x1 −x2( )
Interpreting, cont.
(x2 - x1) --> small…
€
∂φC∂t
=−∂F∂x
+ φR
Since…
€
F =−φDsed∂C∂x
+ φvC
€
−∂F∂x
=∂∂x
φDsed∂C∂x
⎧ ⎨ ⎩
⎫ ⎬ ⎭−
∂∂x
φvC{ }
Interpreting, cont…
€
∂∂t
φC{ } =∂∂x
φDsed∂C∂x
⎧ ⎨ ⎩
⎫ ⎬ ⎭−
∂∂x
φvC{ } + φR
Then…
Simplify: steady state ; constant ; constant D ; burial << transport:
€
0 =Dsedd2C
dx2+ R
==> A simple interpretation of pw profiles
€
d2C
dx2=−
RDsed
60
50
40
30
20
10
0
403020100
NO3- μ /mol l
- CVP Site M 3 NO data : & .Fit A S stoich
Interpretation of profile shapes : quantitative
Steady-state mass balance in a sediment layer: Rate of reaction within the layer = net flux out of the layer
€
R = Fout − Fin
Diffusive flux :
€
F = −φDseddC
dxoxic
denitrification
12
Flux at pt. 1 (x=0) : gives total, net NO3 Production in sediment column
Flux at pt. 2 : gives rate of NO3 consump.By denitrification
Sum of absolute values of Flux at 1 + Flux at 2:Gives rate of NO3 production by oxicDecomposition of organic matter
But we can get more information…
€
0 =Dsedd2C
dx2+ R
What else do we need to solve this equation?
But we can get more information…
€
0 =Dsedd2C
dx2+ R
Boundary conditions!
At sediment-water interface (x=0)
€
C x =0( ) =Cbw
At depth in the sediments:
€
dCdx
⎛
⎝ ⎜
⎞
⎠ ⎟x=x max
=0 or Cx = xmax( ) =0
What about R ?
Example : organic matter oxidation by O2
€
CH 2O( )106NH 3( )16
H 3PO4( ) +138O2 ⇒ 106HCO3
− +16NO3
− + H 2PO4
− +124H + +16H 2O
What solutes could you measure to define reaction rate?
For O2, it has been convenient to use
R = P(x)i.e., with no dependence on [O2], even though it’s a reactant
Can that be justified?
Devol (1978) DSR 25, 137-146
Cultured marine bacteria from low-O2 waters…Found O2 consumption followedMichaelis-Menten kinetics:
€
dO2
dt= Vm
O2[ ]Ks + O2[ ]
And found a “Critical O2 Concentration” belowWhich rate depended on [O2]Of ~ 2.4 µmol/l
What about R ?
Pore water profiles :O2 all done by in situ microelectrode profiling
2
1
0
Depth (cm)
200150100500
O2 (µmol/l)
BBay August 2003 dep_22 dep_12 dep_13
4
3
2
1
0
-1
Depth (cm)
2001000O2 (µmol/l)
616m JSL II Dive 2949-a JSL II Diev 2949 -c JSL II Dive 2949 -d Fit Range of x*
7
6
5
4
3
2
1
0
Depth (cm)
250200150100500
O2 (µmol/l)
Ceara Rise3200 m
(Hales et al., 1996)
TotalCorg ox.Rate(µmol/cm2/y)
14 45 350
Continental margin sediments:* large organic matter flux
* electron acceptors other than O2
Let’s consider a sediment dominated by sulfate reduction:
€
CH 2O( )106NH 3( )16
H 3PO4( ) + 53SO4
2− →106HCO3
− +16NH 4
+ + HPO4
2− + 53HS− + 39H +
Defining P as the production rate of a solute,
€
PNH 4
PTCO 2
=16
106
€
PHPO 4
PTCO 2
=1
106
What would we predict pore water profiles of these 3 solutes to look like?
€
0 = Dd 2Cdx 2
+ P0e− p1x
C x = 0( ) = Cbw
dCdx
x → ∞( ) = 0
Solution:
€
C = Cbw +P0
p1
2D1− e− p1x( )
Solve the equation for each solute:
€
PTCO 2 = P0e− p1x
€
PNH 4 =16106
PTCO 2
€
PHPO 4 =1
106PTCO 2
Assume porosity = 0.8 and Dsed = Dsw x (porosity)2 … then
DHCO3- = 323 cm2/y , DNH4+ = 543 cm2/y, DHPO42- = 208 cm2/y
Assume P0 = 100,p1 = 0.2
25
20
15
10
5
0
800050002000TCO2 (µmol/l)
5000NH4 (µmol/l)
1000PO4 (µmol/l)
Plotting the concentration of one solute vs. another…
600
400
200
0
8000600040002000TCO2 (µmol/l)
s = 0.08979
100
80
60
40
20
0
8000600040002000TCO2 (µmol/l)
s = 0.01465
Interpreting the slopes:At any depth,
€
FTC = −DTC
dTCdx
FN = −DN
dNdx
€
FTC
FN
=DTC
dTCdx
DNdN
dx~
FTC
FN
=DTC
DN
×ΔCΔN
Therefore, the slopes imply
€
PN
PTC
=16106
€
PP
PTC
=1
106
A mineral,undersaturated in seawater
apparently simple dissolution kinetics…
What do we expect [Si(OH)4] in pore water to look like?
Depth
ConcentrationCbw Csat
Diagenesis of a solid Undersaturated in bottom water Asymptotic approach to saturation in pore water
CBW CSATConcentration
Diagenesis of a solid, undersaturated in bw Asymptotic approach to
Saturation in pore water
Csat = 100-120
Csat = 600
Csat = 550-830Csat = 500-750
N. Atlantic (Bermuda)
PeruMargin
Southern Ocean
Observations: Si(OH)4
In pore waters