Lecture 11 Principles of Mass Balance

1. Lecture 11 Principles of Mass Balance Simple Box Models The modern view about what controls the composition of sea water.

2. 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?

3. 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!

4. 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

5. 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

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

7. 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

8. 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)

9. 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 then d[M] / dt = 0 or steady state

10. 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 when d[M] / dt = 0 Q = k[M] [M] / Q = 1/k =  and [M] = Q / k inverse relationship

11. Reactivity and Residence Time Cl sw Al,Fe 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

12. 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 

13. 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!

15. 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”

16. Mass balance for surface box VsdCs/dCt= VrCr + VmCd – VmixCs – B At steady state: B = VrCr + VmixCd– VmixCsand fB= VrivCriv

17. How large is the transport term: If the residence time of the deep ocean is 1000 yrs (from 14C) and t = Vold / Vmix then: 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 volume

18. 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

20. 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 VU = VD = VMIX Negative Feedback Control: if VMIX↑ VUCD↑ 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! CS CD if VMIX = m y-1 and C = mol m-3 flux = mol m-2 y-1

21. 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)

22. 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

23. 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

24. 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!