1. Principles of Mass Balance
Lecture 5
Principles of Mass Balance
Simple Box Models
The modern view about what controls
the composition of sea water.

2. -Box (or reservoir) Models -Continuous Transport-reaction Models
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
Time

3. 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

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

5. 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

6.  = mass / input or removal flux = M / Q = M / S
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)

7. 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

8. 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

9. 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

10. 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.

11. 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”

12. Vs dCs/dCt = VrCr + VmCd – VmixCs – B
B = VrCr + VmixCd - VmixCs

13. How large is the transport term:
If the residence time of the deep ocean is 1000 yrs (from 14C)
and t = Vold / V
then:
V = (3700m/3800m)(1.37 x 1018 m3) / 1000 y
= 1.3 x 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.
fraction of total depth
that is deep ocean
volume

14. 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
removed as B
because fB = VrCr
fraction of element removed to
sediment per visit to the surface
Here are some values:
g f f x g N P C Si Ba Ca
Q. Explain these values and
why they vary the way they do.
See Broecker (1971) Table 3

15. 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 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! CS CD
if VMIX = m y-1 and C = mol m-3
flux = mol m-2 y-1

17. sw Reactivity and Residence Time Cl Al,Fe Elements with small KY have
short residence times.
A parameterization of particle reactivity
When the ratio is small elements mostly on particles

18. 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)

19. 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

20. 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 = yr = 9.6 d
Residence time of water in the ocean with respect to rivers
= x 109 km3 / 37 x 103 km3 y-1 = 37,000 yrs

21. texport/tbiota = 0.27/0.06 = 4.5 times recycled
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

22. 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!