UW OCEAN 400 - Principle of mass balance

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Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Lecture 11Principles of Mass BalanceSimple Box ModelsThe modern view about what controls the composition of sea water.Four Main Themes1.Global Carbon Cycle2.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?tCO2,atm = 590/130 = 4.5 ytC,biota = 3/50 = 0.06 ytC,export = 3/11 = 0.29 ytexport/tbiota = 0.27/0.06 = 4.5 times recycledExample: Global Carbon CycleNo red export!Two main types of models used in chemical oceanography.-Box (or reservoir) Models-Continuous Transport-reaction ModelsIn both cases:Change in Sum of Sum ofMass with = Inputs - OutputsTimeAt steady state the dissolved concentration (Mi) does not change with time: (dM/dt)ocn = SdMi / dt = 0Sum of sources must equal sum of sinks at steady stateBox ModelsHow 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 + Fhydrothermalall elements all elements source: Li, Rb, K, Ca, Fe, Mn sink: Mg, SO4, alkalinityResidence 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 – Sinput = Q = Zeroth Order flux (e.g. river input) not proportional to how much is in the oceansink = S = many are First Order (e.g. Radioactive decay, plankton uptake, adsorption by particles)If inflow equals outflowQ = Sthend[M] / dt = 0 or steady stateFirst 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 / kinverse relationshipswReactivity andResidence TimeClAl,FeA parameterization of particle reactivityWhen the ratio is small elements mostly on particlesElements with small KY have short residence times.When t < tsw not evenly mixed!Dynamic Box ModelsIf 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 OceanAssume 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 Mand 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 ModelsThe 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. t =e = Σ 1/n!Broecker two-box model (Broecker, 1971)v is in m y-1Flux = VmixCsurf = m yr-1 mol m-3 = mol m-2y-1see Fig. 2 of Broecker (1971)Quaternary Research“A Kinetic Model of Seawater”Mass balance for surface boxVs dCs/dCt = VrCr + VmCd – VmixCs – BAt steady state:B = VrCr + VmixCd – VmixCs and fB= VrivCrivHow large is the transport term:If the residence time of the deep ocean is 1000 yrs (from 14C)and t = Vold / Vmixthen: Vmix = (3700m/3800m)(1.37 x 1018 m3) / 1000 y = 1.3 x 1015 m3 y-1If River Inflow = 3.7 x 1013 m3 y-1Then River Inflow / Deep Box Exchange = 3.7 x 1013/1.3 x 1015 = 1 / 38This means water circulates on average about 40 times through the ocean (surface to deep exchange) before itevaporates and returns as river flow.volumefraction of total depth that is deep oceanBroecker (1971) defines some parameters for the 2-box modelg = B / input = (VmixCD + VrCr – VmixCs) / VmixCd + VrCrf = VrCr / B = VrCr / (VmixCd + VrCr - VmixCs)f x gIn his model Vr = 10 cm y-1 Vmix = 200 cm y-1so Vmix / Vr = 20Here are some values:g f f x gN 0.950.010.01P 0.950.010.01C 0.200.020.004Si 1.0 0.010.01Ba 0.750.120.09Ca 0.010.120.001Q. Explain these values andwhy they vary the way they do.Fraction of B fluxpreserved in sedimentsbecause fB = VrCrfraction of element removed to sediment per visit to the surface fraction of inputto surface box removed as BSee Broecker (1971) Table 3Why is this important for chemical oceanography?What controls ocean C, N, P?assume g ≈ 1.0Mass Balance for whole ocean:C/ t = VRCR – f BCS = 0; CD = CDVU = VD = VMIXNegative Feedback Control:if VMIX ↑VUCD ↑B ↑ (assumes g is constant!)f B ↑ (assumes f will be constant!)assume VRCR then CD ↓ (because total ocean balanceVUCD ↓ has changed; sink > source)B ↓CSCDif VMIX = m y-1 and C = mol m-3flux = mol m-2 y-1The nutrient concentration of the deep ocean will adjust so thatthe fraction of B preserved in thesediments equals river input!Multi-Box ModelsVt – total ocean volume (m3)Vs = surface ocean volumeVu,Vd = water exchange (m3 y-1)R = river inflow (m3 y-1)C = concentration (mol m-3)P = particulate flux fromsurface 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 = V2. 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 - B4. At steady state d[Ct] / dt = 0 and R [CR] = BExample: Global Water Cycle103 km3103 km3 y-1Q. Is the water content of the Atmosphere at steady state? Residence time of water in the atmospheret = 13 x 103 km3 / 495 x 103 km3 y-1 = 0.026 yr = 9.6 dResidence time of water in the ocean with respect to


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