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Lecture 5 Principles of Mass Balance

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Lecture 5 Principles of Mass Balance. Simple Box Models The modern view about what controls the composition of sea water. Two main types of models used in chemical oceanography. - Box (or reservoir) Models - Continuous Transport-reaction Models In both cases: - PowerPoint PPT Presentation
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Lecture 5 Principles of Mass Balance Simple Box Models The modern view about what controls the composition of sea water.
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Page 1: Lecture 5 Principles of Mass Balance

Lecture 5Principles of Mass Balance

Simple Box Models

The modern view about what controls the composition of sea water.

Page 2: Lecture 5 Principles of Mass Balance

Two main types of models used in chemical oceanography.

-Box (or reservoir) Models

-Continuous Transport-reaction Models

In both cases:

Change in Sum of Sum ofMass with = Inputs - OutputsTime

Page 3: Lecture 5 Principles of Mass Balance

At steady state the dissolved concentration (Mi) does not change with time:

(dM/dt)ocn = dMi / dt = 0

Sum of sources must equal sum of sinks

Page 4: Lecture 5 Principles of Mass Balance

Box Models

How would you verify that this 1-Box Ocean is at steady state?

Page 5: Lecture 5 Principles of Mass Balance

For most elements in the ocean:

(dM/dt)ocn = Fatm + Frivers - Fseds + Fhydrothermal

The main balance is even simpler:

Frivers = Fsediment + Fhydrothermal

all elements all elements source: Li, Rb, K, Ca, Fe, Mn sink: Mg, SO4, alkalinity

Page 6: Lecture 5 Principles of Mass Balance

Residence Time

= mass / input or removal flux = M / Q = M / S

Q = input rate (e.g. moles y-1)S = output rate (e.g. moles y-1)[M] = total dissolved mass in the box (moles)

Page 7: Lecture 5 Principles of Mass Balance

d[M] / dt = Q – S

input = Q = Zeroth Order flux (e.g. river input) not proportional to how much is in the ocean

sink = S = many are First Order (e.g. Radioactive decay, plankton uptake, adsorption by particles)

If steady state, then inflow equals outflow

Q = S

Page 8: Lecture 5 Principles of Mass Balance

First order removal is proportional to how much is there.

S = k [M]

where k (sometimes ) is the first order removal rate constant (t-1)and [M] is the total mass.

Then:

d[M] / dt = Q – k [M]

at steady state

[M] / Q = 1/k = and [M] = Q / k

Page 9: Lecture 5 Principles of Mass Balance

Dynamic Box Models

If the source (Q) and sink (S) rates are not constant with time or they may have been constant and suddenly change.

Examples: Glacial/Interglacial; Anthropogenic Inputs to Ocean

Assume that the initial amount of M at t = 0 is Mo. The initial mass balance equation is:

dM/dt = Qo – So = Qo – k Mo

The input increases to a new value Q1.

The new balance at the new steady state is:dM/dt = Q1 – k M1

and the solution for the approach to the new equilibrium state is:M(t) = M1 – (M1 – Mo) exp ( -k t )

M increases from Mo to the new value of M1 (= Q1 / k) with a response time of k-1 or see Emerson and Hedges: Appendix 2.2

Page 10: Lecture 5 Principles of Mass Balance

Dynamic Box Models

The response time is defined as the time it takes to reduce the imbalance to e -1 or 37% of the initial imbalance (e.g. M1 – Mo). This response time-scale is referred to as the “e-folding time”. If we assume Mo = 0, after one residence time (t =) we find that: Mt / M1 = (1 – e-1) = 0.63 (Remember that e = 2.7.). Thus, for a single box with a sink proportional to its content, the response time equals the residence time. Elements with a short residence time will approach their new value faster than elements with long residence times.

=

Page 11: Lecture 5 Principles of Mass Balance

Broecker two-box model (Broecker, 1971)

v is in m y-1

Flux = VmixCsurf = m yr-1 mol m-3 = mol m-2y-1see Fig. 2 of Broecker (1971)Quaternary Research“A Kinetic Model of Seawater”

Page 12: Lecture 5 Principles of Mass Balance

Vs dCs/dCt = VrCr + VmCd – VmixCs – B

B = VrCr + VmixCd - VmixCs

Page 13: Lecture 5 Principles of Mass Balance

How large is the transport term:If the residence time of the deep ocean is 1000 yrs (from 14C)and = Vold / Vthen:

V = (3700m/3800m)(1.37 x 1018 m3) / 1000 y = 1.3 x 1015 m3 y-1

If River Inflow = 3.7 x 1013 m3 y-1

Then River Inflow / Deep Box Exchange = 3.7 x 1013/1.3 x 1015

= 1 / 38

This means water circulates on average about 40 times through the ocean (surface to deep exchange) before itevaporates.

volumefraction of total depth that is deep ocean

Page 14: Lecture 5 Principles of Mass Balance

Broecker (1971) defines some parameters for the 2-box model

g = B / input = (VmixCD + VrCr – VmixCs) / VmixCd + VrCr

f = VrCr / B = VrCr / (VmixCd + VrCr - VmixCs)

f x g

In his model Vr = 10 cm y-1

Vmix = 200 cm y-1

so Vmix / Vr = 20

Here are some values:g f f x g

N 0.95 0.01 0.01P 0.95 0.01 0.01C 0.20 0.02 0.004Si 1.0 0.01 0.01Ba 0.75 0.12 0.09Ca 0.01 0.12 0.001

Q. Explain these values andwhy they vary the way they do.

because fB = VrCr

fraction of element removed to sediment per visit to the surface

fraction of inputremoved as B

See Broecker (1971) Table 3

Page 15: Lecture 5 Principles of Mass Balance

Why is this important for chemical oceanography?What controls ocean C, N, P?g ≈ 1.0

Mass Balance for whole ocean:C/ t = VRCR – f B

CS = 0; CD = CD

VU = VD = VMIX

Negative Feedback Control:if VMIX ↑VUCD ↑B ↑f B ↑ (assumes f will be constant!)assume VRCR then CD ↓ (because total ocean balanceVUCD ↓ has changed; sink > source)B ↓

CS

CD

if VMIX = m y-1 and C = mol m-3

flux = mol m-2 y-1

The nutrient concentration of the deep ocean will adjust so thatthe fraction of B preserved in thesediments equals river input!

Page 16: Lecture 5 Principles of Mass Balance
Page 17: Lecture 5 Principles of Mass Balance

sw

Reactivity andResidence Time Cl

Al,Fe

A parameterization of particle reactivityWhen the ratio is small elements mostly on particles

Elements with small KY have short residence times.

Page 18: Lecture 5 Principles of Mass Balance

Multi-Box Models

Vt – total ocean volume (m3)Vs = surface ocean volumeVu,Vd = water exchange (m3 y-1)R = river inflow (m3 y-1)C = concentration (mol m-3)P = particulate flux fromsurface box to deep box (mol y-1)B = burial flux from deep box (mol y-1)

Page 19: Lecture 5 Principles of Mass Balance

1. Conservation of water R = evap – precip Vu = Vd = V2. Surface Box mass balance (units of mol t-1) Vols dCs/dt = R[CR] + V [Cd] – V ([Cs]) - P

Vols dCs/dt = R[CR] – V ([Cs] – [Cd]) - P 3. Deep Box mass balance Vold d[Cd] / dt = V [Cs] – V[Cd] + P - B

Vold d[Cd] / dt = V ([Cs] – [Cd]) + P - B4. At steady state d[Ct] / dt = 0 and R [CR] = B

Page 20: Lecture 5 Principles of Mass Balance

Example: Global Water Cycle

103 km3

103 km3 y-1

Q. Is the water content of the Atmosphere at steady state?

Residence time of water in the atmosphere = 13 x 103 km3 / 495 x 103 km3 y-1 = 0.026 yr = 9.6 d

Residence time of water in the ocean with respect to rivers = 1.37 x 109 km3 / 37 x 103 km3 y-1 = 37,000 yrs

Page 21: Lecture 5 Principles of Mass Balance

CO2,atm = 590/130 = 4.5 yC,biota = 3/50 = 0.06 y

C,export = 3/11 = 0.29 y

export/tbiota = 0.27/0.06 = 4.5 times recycled

Example: Global Carbon Cycle

Page 22: Lecture 5 Principles of Mass Balance

Summary

Salinity of seawater is determined by the major elements.

Early ideas were that the major composition was controlled by equilibrium chemistry.

Modern view is of a kinetic ocean controlled by sources and sinks.

River water is main source – composition from weathering reactions.Evaporation of river water does not make seawater.

Reverse weathering was proposed – but the evidence is weak.

Sediments are a major sink. Hydrothermal reactions are a major sink.Still difficult to quantify!


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