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Lecture 8 Tracers for Gas Exchange

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Lecture 8 Tracers for Gas Exchange

Examples for calibration of gas exchange using:

222Rn – short term

14C - long term

E&H Sections 5.2 and 10.2

Stagnant Boundary Layer Model.

well mixed atmosphere

Cg = KH Pgas = equil. with atm

ATM

0

OCN

Stagnant Boundary

Layer –

transport by

molecular diffusion

ZFilm

Depth (Z)

CSW

well mixed surface SW

Z is positive downward

C/ Z =

F = + (flux into ocean)

see:

Liss and Slater (1974) Nature, 247, p181

Broecker and Peng (1974) Tellus, 26, p21

Liss (1973) Deep-Sea Research, 20, p221

Expression of Air -Sea CO2 Flux

Need to calibrate!

S – Solubility

From Temperature & Salinity

k = piston velocity = D/Zfilm

From wind speed

F = k s (pCO2w- pCO2a) = K ∆ pCO2

pCO2a

pCO2w

From measurements

From CMDL

CCGG network

Gas Exchange and Environmental Forcing: Wind

Wanninkhof, 1992

from 14C

Liss and Merlivat,1986

from wind tunnel exp.

~ 5 m d-1

20 cm hr-1 = 20 x 24 / 102 = 4.8 m d-1

U-Th Series Tracers

Analytical Method for 222Rn and 226Ra

5 half-lives

charcoal

liquid N2

222Rn

Activity is what

is measured. Not

concentration!

SW

226Ra

Apply the principle of secular equilibrium!

226Ra in Atlantic and Pacific

Q. What controls the ocean distributions of 226Ra?

226Ra source from the sediments

222Rn Example Profile from

North Atlantic

Does Secular Equilibrium Apply?

t1/2222Rn << t1/2 226Ra

(3.8 d) (1600 yrs)

YES!

A226Ra = A222Rn

222Rn

226Ra

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 (Zmlmixed layer).

Thus there is a net evasion of 222Rn out of the ocean.

The simple 1-D 222Rn balance for the mixed layer, with thickness Zml,

ignoring horizontal advection and vertical exchange with deeper water, is:

d222Rn/dt = sources – sinks

= decay of 226Ra – decay of 222Rn - gas exchange to atmosphere

Zmll222Rnd[222Rn]/dt = Zmll226Ra [226Ra] – Zmll222Rn [222RnML]

- D/Zfilm{ [222Rnatm] – [222RnML]}

Knowns: l222Rn, l226Ra, DRn

Measure: Zml,A226Ra, A222Rn, d[222Rn]/dt

Solve for Zfilm

Zmlλ222Rnd[222Rn]/dt = Zmlλ226Ra[226Ra] – Zmlλ222Rn[222Rn]

- D/Zfilm { [222Rnatm] – [222RnML]}

ZmlδA222Rn/ δt = Zml(A226Ra – A222Rn) + D/Z (CRn, atm – CRn,ML)

Note: diffusion is

expressed in terms of

concentrations not

activities

atm Rn = 0

for SS = 0

Then

-D/Z ( – CRn,ml) = Zml(A226Ra – A222Rn)

+D/Z (ARn,ml/λRn) = Zml(A226Ra – A222Rn)

+D/Z (ARn,ml) =ZmlλRn(A226Ra – A222Rn)

ZFILM = D (A222Rn,ml) / ZmlλRn(A226Ra – A222Rn)

ZFILM = (D /ZmlλRn) ()

Stagnant Boundary Layer Film Thickness

Z = DRn / Zfilm l222Rn

(1/A226Ra/A222Rn) ) - 1

Histogram showing results of film thickness

calculations from many stations.

Organized by ocean and by latitude

Average Zfilm = 28 mm

- Q. What are limitations of
- this approach?
- unrealistic physical model
- steady state assumption
- short time scale

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.

IsotopeHalf-lifeGlobal inventory (pre-nuclear)

3H 12.3 yr 3.5 kg

14C 5730 yr 54 ton

10Be 1.5 x 106yr 430 ton

7Be 54 d 32 g

26Al 7.4 x 105yr 1.7 ton

32Si 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)

(5730 yrs)

From galactic cosmic rays

which are more energetic than

solar wind. So these are not from

the sun.

The overall reaction is written:

14N + n 14C + p

(7n, 7p) (8n, 6p)

So the production rate from cosmic rays can be calculated

Nuclear weapons testing and nuclear reactors (e.g. Chernobyl) have been an extremely important sources of nuclides used as ocean tracers.

The main bomb produced isotopes have been:

IsotopeHalf LifeDecay

3H 12.3 yrs beta

14C 5730 yrs beta

90Sr 28 yrs beta

238Pu 86 yrs alpha

239+240Pu 2.44 x 104yrs alpha

6.6 x 103yrs alpha

137Cs 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).

Atmospheric 14CO2 in the second half of the 20th century.

The figure shows the 14C / 12C ratio relative to the natural level in the atmospheric

CO2 as a function of time in the second half of the 20th century.

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

data: Levin & Kromer 2004; Manning et al 1990; Druffel 1987;

Druffel 1989; Druffel & Griffin 1995

bomb spike in 1963

Example – Use 14C to calculate ZFILM using the Stagnant Boundary Layer

14Catm

Use Pre-bomb 14C – assume steady state

1-box model

source = sink

14C from gas exchange = 14C lost by decay

[14C]

14C decay

Assume [CO2]top = [CO2]bottom = [CO2]surface ocean (e.g. no CO2 gradient,

only a 14C gradient)

D = 3 x 10-2 m2 y-1

h = 3800m

l-1 = 8200 y

[CO2]surf = 0.01 moles m-3

[DIC]ocean = 2.4 moles m-3

a14CO2/aCO2 = 1.015 (14C-CO2 is more soluble than CO2)(a 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 mm

Solve 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 area

for h = 3200m

thus age of deep ocean box (t)

t = 3200m / 2 my-1 = 1600 years

What is the direction and flux of oxygen across the air-sea interface given?

PO2 = 0.20 atm

KH,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 O2ml > O2atm 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

DC = (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

The activity of 222Rn is less than that of 226Ra in the surface water of the

North Atlantic at TTO Station 24 (western North Atlantic).

Calculate the thickness of the stagnant boundary layer (ZFILM).

A226Ra = 8.7 dpm100 L-1

A222Rn = 6.9 dpm100 L-1

Assume:

λ222Rn= 2.1 x 10-6 s-1

D222Rn = 1.4 x 10-9 m2 s-1

Zml= 40m

Answer: ZFILM = 40 x 10-6 m

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) where

1 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 (3H) in rain

and surface SW

Tritium is a conservative tracer for water (as HTO)

– thermocline penetration

Eq

Meridional Section in the Pacific

Atmospheric Record of Thermocline Ventilation Tracers

Conservative, non-radioactive tracers (CFC-11, CFC-12, CFC13, SF6)

Time series of northern hemisphere atmospheric concentrations

and tritium in North Atlantic surface waters

222Rn as a tracer for gas exchange

d222Rn/dt = sources – sinks

= decay of 226Ra – decay of 222Rn - gas exchange to atmosphere

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