Lecture 13 Tracers for Gas Exchange Examples for gas exchange using: 222 Rn 14 C E&H Sections 5.2...
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Transcript of Lecture 13 Tracers for Gas Exchange Examples for gas exchange using: 222 Rn 14 C E&H Sections 5.2...
Lecture 13 Tracers for Gas Exchange
Examples for gas exchange using:222Rn14C
E&H Sections 5.2 and 10.2
Rates of Gas ExchangeStagnant Boundary Layer Model.
Depth (Z)
ATM
OCN
Cg = KH Pgas = equil. with atm
CSW
ZFilm
Stagnant BoundaryLayer – transport by molecular diffusion
well mixed surface SW
well mixed atmosphere
0
Z is positive downward
C/ Z = F = + (flux into ocean)see:
Liss and Slater (1974) Nature, 247, p181Broecker and Peng (1974) Tellus, 26, p21Liss (1973) Deep-Sea Research, 20, p221
Expression of Air -Sea CO2 Flux
k = piston velocity = D/Zfilm
From wind speed
From CMDLCCGG network
S – Solubility
From Temperature & Salinity
From measurements
F = k s (pCO2w- pCO2a) = K ∆ pCO2
pCO2apCO2w
Need to calibrate!
222Rn Example Profile fromNorth Atlantic
226Ra
222RnDoes Secular Equilibrium Apply?
t1/2 222Rn << t1/2 226Ra
(3.8 d) (1600 yrs)
YES!A226Ra = A222Rn
Why is 222Rn activity less than 226Ra?
222Rn is a gas and the 222Rn concentration in the atmosphere is much less than in the ocean mixed layer ( mixed layer).
Thus there is a net evasion of 222Rn out of the ocean.
The 222Rn balance for the mixed layer, ignoring horizontal advection and vertical exchange with deeper water, is:
ml 222Rn [222Rn]/t = ml 226Ra [226Ra] – 222Rn [222RnML] + D/Zfilm { [222Rnatm] – [222RnML]}
Knowns: 222Rn, 226Ra, DRn
Measure: ml, A226Ra, A222Rn, d[222Rn]/dt
Solve for Zfilm
222Rn/dt = sources – sinks = decay of 226Ra – decay of 222Rn - gas exchange to atmosphere
ml 222Rn d[222Rn]/dt = ml 226Ra [226Ra] – ml 222Rn [222Rn] + D/Zfilm { [222Rnatm] – [222RnML]}
ml δA222Rn/ δt = ml (A226Ra – A222Rn) + D/Z (CRn, atm – CRn,ML)
for SS = 0 atm Rn = 0
Then
-D/Z ( – CRn,ml) = ml (A226Ra – A222Rn)
+D/Z (ARn,ml/Rn) = ml (A226Ra – A222Rn)
+D/Z (ARn,ml) = ml Rn (A226Ra – A222Rn)
ZFILM = D (A222Rn,ml) / ml Rn (A226Ra – A222Rn)
ZFILM = (D / ml Rn) ( )226
222
1
1Ra
Rn
A
A
Z = DRn / 222Rn (1/A226Ra/A222Rn) ) - 1
Average Zfilm = 28 m
Stagnant Boundary Layer Film Thickness
Histogram showing results of film thicknesscalculations from many stations.
Organized by Ocean and by Latitude
Q. What are limitations of this approach?1. unrealistic physical model2. steady state assumption
Cosmic Ray Produced Tracers – including 14C
Cosmic ray interactions produce a wide range of nuclides in terrestrial matter, particularly in the atmosphere, and in extraterrestrial material accreted by the earth.
Isotope Half-life Global inventory3H 12.3 yr 3.5 kg14C 5730 yr 54 ton10Be 1.5 x 106 yr 430 ton7Be 54 d 32 g26Al 7.4 x 105 yr 1.7 ton32Si 276 yr 1.4 kg
Carbon-14 is produced in the upper atmosphere as follows:
Cosmic Ray Flux Fast Neutrons Slow Neutrons + 14N* 14C (thermal)
The reaction is written:
14N + n 14C + p(7n, 7p) (8n, 6p)
(5730 yrs)
From galactic cosmic rayswhich are more energetic thansolar wind. So these are not from the sun.
Tritium (3H) is produced from cosmic ray interactions with N and O.
After production it exists as tritiated water ( H - O -3H ), thus it is an ideal tracer for water.
Tritium concentrations are TU (tritium units) where1 TU = 1018 (3H / H)
Thus tritium has a well defined atmospheric input via rain and H2O vapor exchange.
Its residence time in the atmosphere is on the order of months.
In the pre-nuclear period the global inventory was only 3.5 kg which means there was very little 3H in the ocean at that time. The inventory increased by 200x and was at a maximum in the mid-1970s
Tritium is a conservative tracer for water (as HTO) – thermocline penetration
Meridional Section in the Pacific
Eq
Time series of northern hemisphere atmospheric concentrationsand tritium in North Atlantic surface waters
Atmospheric Record of Thermocline Ventilation TracersConservative, non-radioactive tracers (CFC-11, CFC-12, CFC13, SF6)
Bomb Fallout Produced TracersNuclear weapons testing and nuclear reactors (e.g. Chernobyl) have been an extremely important sources of nuclides used as ocean tracers.
In addition to 3H and 14C the main bomb produced isotopes have been:
Isotope Half Life Decay90Sr 28 yrs beta238Pu 86 yrs alpha239+240Pu 2.44 x 104 yrs alpha
6.6 x 103 yrs alpha137Cs 30 yrs beta, gamma
Nuclear weapons testing has been the overwhelmingly predominant source of 3H, 14C, 90Sr and 137Cs to the ocean.
Nuclear weapons testing peaked in 1961-1962.
Fallout nuclides act as "dyes"
Another group of man-made tracers that fall in this category but are not bomb-produced and are not radioactive are the chlorofluorocarbons (CFCs).
The bomb spike: surface ocean and atmospheric Δ14C since 1950
• Massive production in nuclear tests ca. 1960 (“bomb 14C”)
• Through air-sea gas exchange, the ocean took up ~half of the bomb 14C by the 1980s
bomb spike in 1963data: Levin & Kromer 2004; Manning et al 1990; Druffel 1987; Druffel 1989; Druffel & Griffin 1995
Example – Use 14C to calculate ZFILM using the Stagnant Boundary Layer
Use Pre-bomb 14C – assume steady state
source = sink14C from gas exchange = 14C lost by decay
14Catm
14C decay
Assume [CO2]top = [CO2]bottom = [CO2]surface ocean (e.g. no CO2 gradient, only a 14C gradient)
[14C]
1-box model
AssumeD = 3 x 10-2 m2 y-1
h = 3800m1 = 8200 y[CO2]surf = 0.01 moles m-3
[DIC]ocean = 2.4 moles m-3
14CO2/CO2 = 1.015 (14C-CO2 is more soluble than CO2)( equals solubility constant)(14C/C) surf = 0.96 (14C/C)atm(14C/C)deep = 0.84 (14C/C)atm
Then:Zfilm = 1.7 x 10-5 m = 17 m
Rearrange andSolve for Vmix
Use pre-nuclear 14C data when surface 14C > deep 14C(14C/C)deep = 0.81 (14C/C)surf
Vmix = (200 cm y-1) A A = ocean areafor h = 3200m
thus age of deep ocean box (t)t = 3200m / 2 my-1 = 1600 years
Example:What is the direction and flux of oxygen across the air-sea interface given?
PO2 = 0.20 atmKH,O2 = 1.03 x 10-3 mol kg-1 atm-1
O2 in mixed layer = 250 x 10-6 mol l-1 (assume 1L = 1 kg)The wind speed (U10) = 10 m s-1
Answer:O2 in seawater at the top of the stagnant boundary layer = KH PO2 = 1.03 x 10-3 x 0.20 = 206 x 10-6 mol l-1
So O2 ml > O2 atm and the flux is out of the ocean.
What is the flux?With a wind speed = 10 m s-1, the piston velocity (k) = 5 m d-1
C = (250 – 206) x 10-6 = 44 x 10-5 mol l-1 Flux = 5 m d-1 x 44 x 10-6 mol l-1 x 103 l m-3 = 5 x 44 x 10-6 x 103 = 220 x 10-3 mol m-2 d-1
ExampleThe activity of 222Rn is less than that of 226Ra in the surface water of theNorth Atlantic at TTO Station 24 (western North Atlantic). Calculate the thickness of the stagnant boundary layer (ZFILM).
A226Ra = 8.7 dpm 100l-1
A222Rn = 6.9 dpm 100L-1
Assume:222Rn = 2.1 x 10-6 s-1
D222Rn = 1.4 x 10-9 m2 s-1
ml = 40m
Answer: ZFILM = 40 x 10-6 m