Principles of Mass Balance Lecture 12 Principles of Mass Balance Box Models: The modern view about what controls the composition of sea water

We have already seen examples of a few different types of box models, and have seen how they can be useful – simple models are tools to better understand complex natural systems. Can determine fluxes, reservoir sizes, residence times, whether or not a system is in steady state, etc. Can draw simple conclusions, identify gaps in knowledge, etc.

Change in Mass with Time Two main types of models used in chemical oceanography: Box (or Reservoir) Models Continuous Transport-Reaction Models In both cases: Change in Mass with Time Sum of Inputs Sum of Outputs Two types of models: Box Models CTRM: box models that take into account transport and mixing processes and production/consumption of chemical species. Either way, we can model the change in the amount of mass in the reservoir as a balance between the mass inputs and mass outputs.

Change in Mass with Time = 0 At steady state the dissolved concentration (Mi) does not change with time: (dM/dt)ocn = ΣdMi / dt = 0 (i.e. the sum of the sources must equal the sum of the sinks at steady state) Change in Mass with Time = 0 Sum of Inputs Sum of Outputs At steady state the mass of the reservoir does not change with time. The sum of the sources must equal the sum of the sinks. Sum of Inputs Sum of Outputs

How could we verify that this 1-Box Ocean is in steady state? We have drawn a one box model of the ocean with the main fluxes indicated: River input, glacial input Atmospheric exchange, Hydrothermal exchange, sediment exchange, radioactive decay (internal source/sink) (notice these are shown as both inputs and outputs – these processes are sinks and sources, depending on what element you are looking at) How could we determine whether this system is in steady state? We could measure M over time/composition of species over time We could try to measure all of the sources and sinks over time, to see if they are balanced. Both of these things are hard to do. And there is a lot of uncertainty in both types of measurements.

For most elements in the ocean: (dM/dt)ocn = Fatm + Frivers - Fseds + Fhydrothermal If we assume steady state, and assume atmospheric flux is negligible (safe for most elements)… the main balance is even simpler: Frivers = Fsediment + Fhydrothermal all elements all elements source: Li, Rb, K, Ca, Fe, Mn sink: Mg, SO4, alkalinity For most elements, the main sources are atmospheric deposition and river input. The main sink is deposition in ocean sediments. For some elements, hydrothermal vents are a source (+) and for some HT vents are a sink (-). If we assume steady state (dM/dt = 0) and that atmospheric deposition is negligible, then we get an even simpler balanced equation.

Residence Time   = mass / input or output flux = M / Q = M / S We have seen this figure before. This is our very simple box model with 1 input (Q), 1 output (S), and the reservoir of our constituent M. We know we can calculate the residence time with respect to either flux. This represents how long it would take to fill the reservoir (wrt Q) or drain the reservoir (wrt S) if the other flux were zero. Q = input rate (e.g. moles y-1) S = output rate (e.g. moles y-1) [M] = total dissolved mass in the box (moles)

Source = Q = e.g. river input flux d[M] / dt = Q – S Source = Q = e.g. river input flux = Zeroth Order flux (flux is not proportional to how much M is present in the ocean) Sink = S = many removal mechanisms are First Order (the flux is proportional to how much M is there) (e.g. radioactive decay, plankton uptake, adsorption by particles) We could also look at the time rate of change of M: this is the difference between the source and the sink. In most cases, the input is independent of how much M is in the box (zeroth order flux) The removal flux, however, is often proportional to how much M is there. These are first order removal mechanisms.

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, we can rewrite d[M]/dt = Q – S, to include the first order sink: d[M] / dt = Q – k [M] At steady state when d[M]/dt = 0, Q = k[M] Rearrange: [M]/Q = 1/k =  * and [M] = Q / k *inverse relationship between first order removal constant and residence time These fluxes can be described as k[M], or some first order removal rate constant times how much M is present. So, we can rewrite our time rate of change equation for M in our reservoir. At steady state Q = k[M] (input flux equals the removal rate constant times the total mass). If we rearrange this equation, we see that this equation predicts an inverse relationship between k and tau. K can also be thought of as elemental reactivity (if reactivity is high, the removal rate constant is high). SO, if elemental reactivity is high, by this equation, residence time will be short. As k increases, tau decreases. Elements will have a short residence time if they are readily removed from the ocean via particle reactivity, biological uptake, etc.

Reactivity vs. Residence Time sw Elements with small KY have Cl Al, Fe In an analysis by Whitfield and Turner (1979), we see a robust correlation between mean ocean residence time of an element and the partitioning of that element between seawater and crustal rock *Note that log plots tend to hide a lot of scatter* Y-axis = log residence time X-axis = log of seawater/crust partitioning (i.e. elements with short residence time are readily incorporated into particles that are removed from ocean water) This is essentially a parameterization of particle reactivity (when the ratio is small, elements are mostly adsorbed onto particles. Notice that Cl- is mostly in seawater and has a long residence time. Al and Fe are at the lower left – they are mostly in the crust and have a shorter residence time. *When  < sw the element is not evenly mixed! Tau cannot be strictly interpreted as a residence time if tau of the element is less than the tau of seawater. This indicates that the element is not homogenously distributed in the ocean. Elements with small KY have short residence times. When  < sw the element is not evenly mixed!

Dynamic Box Models In some instances, the source (Q) and sink (S) rates are not constant with time OR they may have been constant, but suddenly change. Examples: Glacial/Interglacial cycles, 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 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)e-kt] “M increases from Mo to the new value of M1 (Q1/k) with a response time of k-1 or ”

Dynamic Box Models M(t) = M1 – [(M1 – Mo)e-kt] t = This response time is defined as the time it takes to reduce the imbalance (M1 – Mo). to e-1 (or 37%) of the initial imbalance ((1/e)*( 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 This is 37% reduced, t = e-folding time! For a single box model with a 1st order sink, response time = residence time. Elements with a short residence time will approach their new value faster than elements with long residence times. E-folding time here is defined as the time interval over which is takes something to decrease by a factor of e. (the timescale for M to reach a new value that is 1/e its original value). Walk through solution of Mo=0. Remember e = 2.7 If the sink is proportional to the amount of your species in the reservoir ([M]), then the e-folding response time of that species will be equal to the residence time of that species. If tau is longer, e-folding time will be longer. It tau is shorter, e-folding time will be shorter.

Controls on Atmospheric CO2 Remarkable consistency for glacial/interglacial concentrations of CO2. A main Control on atm CO2 is the B flux! We need to understand B… PI CO2 = 280 ppm PI CO2 w/o B = 970 ppm!

How do we get from the marine food web to a global assessment of CO2 flux??? With great difficulty!

Broecker two-box model (Broecker, 1971) B = VmCd + VrCr - VmCs B (1-f)B fB see Fig. 2 of Broecker (1971) v is in m y-1 then flux is mol m-2y-1 Flux = Vmix Csurf = m yr-1 mol m-3 = mol m-2 y-1

Appliation of the 2-Box model Nutrients (Si, P, C, N) and trace metals Internal cycling can be described by the simple 2-box ocean model The main balance is input from rivers and removal as biological debris to sediments Sources = Sinks Input from rivers = removal to sediments VrCr = f B where f is the fraction of biogenic flux that is buried (escapes remineralization) Q. What are B and fB for C in Sarmiento and Gruber?

Broecker (1971) defines some parameters for the 2-box model Two important parameters are g and f: g = the fraction of an element put in at the surface that is removed as B (the efficiency of bioremoval of an element from the surface – how efficiently it sinks as a particle (as B flux) out of the surface ocean) = B / surface ocean input = (VmixCD + VrCr – VmixCs) / VmixCd + VrCr f = the fraction of particles that are buried (the efficiency of ultimate removal from the water column) = VrCr / B = VrCr / (VmixCd + VrCr - VmixCs) g = efficiency of bioremoval from the surface as particles = bioremoval/inputs = B/(VrCr + VmixCdeep) f = fraction of particles that are buried = the efficiency of ultimate removal fB = VrCr Walk through equation derivations.

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 this model Vr = 10 cm y-1 Vmix = 200 cm y-1 so Vmix / Vr = 20 From 14C mass balance (next slide) fraction of input to surface 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 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

How large is the transport term: If the residence time of the deep ocean is 1850 yrs (from 14C) and t = Vold / Vmix then: Vmix = (3700m/3800m)(1.37 x 1018 m3) / 1850 y = 0.72 x 1015 m3 y-1 If River Inflow = 3.7 x 1013 m3 y-1 Then River Inflow / Deep Box Exchange = 3.7 x 1013/72 x 1013 = 1 / 19.5 This means water circulates on average about 19.5 times through the ocean (surface to deep exchange) before it evaporates and returns as river flow. fraction of total depth that is deep ocean volume

Example – 14C Deep Ocean Residence Time substitute for B vmix in cm yr-1; vC in cm yr-1 x mol cm-3

Rearrange and 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 = 3700m thus age of deep ocean box (t) t = 3700m / 2 my-1 = 1850 years

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

Example: Perturbation analysis – Mass Balance Control Double Upwelling Rate Double rate of ocean mixing sequence of events Paleo record VrCr = fB at the beginning and at the end! The deep concentration (Cd) is cut in half

Example: Perturbation Analysis 1. Double River Input 2. CaCO3 burial increases and Carbonate Compensation Depth (CCD) deepens