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

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Presentation on theme: "Lecture 11 Principles of Mass Balance Simple Box Models The modern view about what controls the composition of sea water."— Presentation transcript:

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 1.Global Carbon Cycle 2.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 CO2,atm = 590/130 = 4.5 y C,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 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 = dM i / 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 = F atm + F rivers - F seds + F hydrothermal The main balance is even simpler: F rivers = F sediment + F hydrothermal all elements all elements source: Li, Rb, K, Ca, Fe, Mn sink: Mg, SO 4, 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 sw Reactivity and Residence Time Cl Al,Fe A parameterization of particle reactivity When the ratio is small elements mostly on particles Elements with small K Y have short residence times. When < sw not evenly mixed!

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 M o. The initial mass balance equation is: dM/dt = Q o – S o = Q o – k M o The input increases to a new value Q 1. The new balance at the new steady state is: dM/dt = Q 1 – k M and the solution for the approach to the new equilibrium state is: M(t) = M 1 – (M 1 – M o ) exp ( -k t ) M increases from M o to the new value of M 1 (= Q 1 / k) with a response time of k -1 or

13 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. M 1 – M o ). This response time-scale is referred to as the e-folding time. If we assume M o = 0, after one residence time (t = ) we find that: M t / M 1 = (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!

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15 Broecker two-box model (Broecker, 1971) v is in m y -1 Flux = V mix C surf = m yr -1 mol m -3 = mol m -2 y -1 see Fig. 2 of Broecker (1971) Quaternary Research A Kinetic Model of Seawater

16 Mass balance for surface box V s dC s /dC t = V r C r + V m C d – V mix C s – B At steady state: B = V r C r + V mix C d – V mix C s and fB= V riv C riv

17 How large is the transport term: If the residence time of the deep ocean is 1000 yrs (from 14 C) and = Vol d / V mix then: V mix = (3700m/3800m)(1.37 x m 3 ) / 1000 y = 1.3 x m 3 y -1 If River Inflow = 3.7 x m 3 y -1 Then River Inflow / Deep Box Exchange = 3.7 x /1.3 x = 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. volume fraction of total depth that is deep ocean

18 Broecker (1971) defines some parameters for the 2-box model g = B / input = (V mix C D + V r C r – V mix C s ) / V mix C d + V r C r f = V r C r / B = V r C r / (V mix C d + V r C r - V mix C s ) f x g In his model V r = 10 cm y -1 V mix = 200 cm y -1 so V mix / V r = 20 Here are some values: gff x g N P C Si Ba Ca Q. Explain these values and why they vary the way they do. Fraction of B flux preserved in sediments because fB = V r C r fraction of element removed to sediment per visit to the surface fraction of input to surface box removed as B See Broecker (1971) Table 3

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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 = V R C R – f B C S = 0; C D = C D V U = V D = V MIX Negative Feedback Control: if V MIX V U C D B (assumes g is constant!) f B (assumes f will be constant!) assume V R C R then C D (because total ocean balance V U C D has changed; sink > source) B CSCS CDCD if V MIX = m y -1 and C = mol m -3 flux = mol m -2 y -1 The nutrient concentration of the deep ocean will adjust so that the fraction of B preserved in the sediments equals river input!

21 Multi-Box Models V t – total ocean volume (m 3 ) V s = surface ocean volume V u,V d = water exchange (m 3 y -1 ) R = river inflow (m 3 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 V u = V d = V 2. Surface Box mass balance (units of mol t -1 ) Vol s dC s /dt = R[C R ] + V [C d ] – V ([Cs]) - P Vol s dC s /dt = R[C R ] – V ([C s ] – [C d ]) - P 3. Deep Box mass balance Vol d d[C d ] / dt = V [Cs] – V[Cd] + P - B Vol d d[C d ] / dt = V ([C s ] – [C d ]) + P - B 4. At steady state d[C t ] / dt = 0 and R [C R ] = B

23 Example: Global Water Cycle 10 3 km km 3 y -1 Q. Is the water content of the Atmosphere at steady state? Residence time of water in the atmosphere = 13 x 10 3 km 3 / 495 x 10 3 km 3 y -1 = yr = 9.6 d Residence time of water in the ocean with respect to rivers = 1.37 x 10 9 km 3 / 37 x 10 3 km 3 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!


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