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Lecture 11 Principles of Mass Balance. Simple Box Models The modern view about what controls the composition of sea water. Four Main Themes Global Carbon Cycle Are humans changing the chemistry of the ocean? 3.What are chemical controls on biological production?

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Lecture 11

Principles of Mass Balance

Simple Box Models

The modern view about what controls

the composition of sea water.

Four Main Themes
  • Global Carbon Cycle
  • Are humans changing the
  • chemistry of the ocean?
  • 3.What are chemical controls
  • on biological production?
  • 4. What is the fate of organic matter
  • made by biological production?
Example: Global Carbon Cycle

tC,biota = 3/50 = 0.06 y

tC,export = 3/11 = 0.29 y

texport/tbiota = 0.27/0.06 = 4.5 times recycled

tCO2,atm = 590/130 = 4.5 y

No red export!

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 of

Mass with = Inputs - Outputs


At steady state the dissolved concentration (Mi)

does not change with time:

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

Sum of sources must equal sum of sinks at steady state

Box Models

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

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

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)

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 inflow equals outflow

Q = S


d[M] / dt = 0 or steady state

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.


d[M] / dt = Q – k [M]

at steady state when d[M] / dt = 0 Q = k[M]

[M] / Q = 1/k =  and [M] = Q / k

inverse relationship

Reactivity and

Residence Time




Elements with small KY have

short residence times.

When t < tsw not evenly mixed!

A parameterization of particle reactivity

When the ratio is small elements mostly on particles

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 M

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 

Dynamic Box Models

t =

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 = 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.

e = Σ 1/n!

Broecker two-box model (Broecker, 1971)

v is in m y-1

Flux = VmixCsurf = m yr-1 mol m-3 = mol m-2y-1

see Fig. 2 of Broecker (1971)

Quaternary Research

“A Kinetic Model of Seawater”

Mass balance for surface box

VsdCs/dCt= VrCr + VmCd – VmixCs – B

At steady state:

B = VrCr + VmixCd– VmixCsand fB= VrivCriv

How large is the transport term:

If the residence time of the deep ocean is 1000 yrs (from 14C)

and t = Vold / Vmix


Vmix= (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 it

evaporates and returns as river flow.

fraction of total depth

that is deep ocean


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

fraction of input

to surface box

removed as B

Fraction of B flux

preserved in sediments

because fB = VrCr

fraction of element removed to

sediment per visit to the surface

Here are some values:

g f f x g

N 0.95 0.01 0.01

P 0.95 0.01 0.01

C 0.20 0.02 0.004

Si 1.0 0.01 0.01

Ba 0.75 0.12 0.09

Ca 0.01 0.12 0.001

Q. Explain these values and

why they vary the way they do.

See Broecker (1971) Table 3

Why is this important for

chemical oceanography?

What controls ocean C, N, P?

assume g ≈ 1.0

Mass Balance for whole ocean:

C/ t = VRCR – f B

CS = 0; CD = CD


Negative Feedback Control:




B ↑ (assumes g is constant!)

f B ↑ (assumes f will be constant!)

assume VRCR 

then CD↓ (because total ocean balance

VUCD↓ has changed; sink > source)

B ↓

The nutrient concentration of

the deep ocean will adjust so that

the fraction of B preserved in the

sediments equals river input!



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

flux = mol m-2 y-1

Multi-Box Models

Vt – total ocean volume (m3)

Vs = surface ocean volume

Vu,Vd = water exchange (m3 y-1)

R = river inflow (m3 y-1)

C = concentration (mol m-3)

P = particulate flux from

surface box to deep box (mol y-1)

B = burial flux from deep box

(mol y-1)

1. Conservation of water

R = evap – precip

Vu = Vd = V

2. 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 - B

4. At steady state

d[Ct] / dt = 0 and R [CR] = B

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

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!