In dynamic-R/RTM: Reaction Transport Modelling: Course Tutorials and Exercises

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Exercise 1 ANSWER. Decay of particulate organic matter due to bacterial activity

Based on the conceptual diagram in Figure 1, the mass balance equations are:

$$\frac{dPOC}{dt} = FluxPOC + Bact_Mortality - Hydrolysis_POC$$ $$\frac{dHMWC}{dt} = Hydrolysis_POC - Hydrolysis_HMWC$$ $$\frac{dLMWC}{dt} = Hydrolysis_HMWC - Bact_uptake$$ $$\frac{dBACT}{dt} = Bact_uptake - Growth_respiration - Basal_respiration - Bact_Mortality$$

Exercise 2 ANSWER. Methane dynamics in wetlands

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Based on the following reactions $$Reaction~1:\qquad CH_3COOH \rightarrow CH_4 + CO_2$$ $$Reaction~2:\qquad CO_2 + 4H_2 \rightarrow CH_4 + 2H_2O$$ $$Reaction~3:\qquad CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$$ and the additional fluxes $CH_3COOH_burial$, $CH_3COOH_input$, $CH4_emission$, $CH_4_burial$, $CO_2_emission$, $CO_2_burial$, $O_2_influx$, the mass balance equations can be written as follows:

$$\frac{dCH_3COOH}{dt} = -2\cdot R_1 - CH_3COOH_burial + CH_3COOH_input$$ $$\frac{dCH_4}{dt} = R_1 + R_2 - R_3 - CH_4_emission - CH_4_burial$$ $$\frac{dCO_2}{dt} = R_1 - R_2 + R_3 - CO_2_emission - CO_2_burial$$ $$\frac{dO_2}{dt} = - 2\cdot R_3 + O_2_influx$$ Units of the state variables and fluxes are:

• $CH_3COOH$, $CH_4$, $CO_2$: $mol~C~m^{-2}$
• $O_2$: $mol~O_2~m^{-2}$
• all carbon fluxes: $mol~C~m^{-2}~d^{-1}$
• $O_2$ flux: $mol~O_2~m^{-2}~d^{-1}$

Model application

To estimate the fluxes, we use the linear equations: $$41 = R_1+R_2-R_3, \qquad 31 = R_1-R_2+R_3, \qquad 20 = 2\cdot R_3.$$ Solving these equations yields $R_1=36$, $R_2=15$, and $R_3=10~mmol~m^{-2}~d^{-1}$.

Exercise 3 ANSWER. Methane removal by anoxic processes in marine sediments

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Based on the conceptual diagram shown in Figure 3, we can write the following mass balance equations:

$$\frac{dCH_{4,diss}}{dt} = Methanogenesis + rHyd2Dis - rDis2Hyd + rGas2Dis - rDis2Gas - AOM - Diffusion$$ $$\frac{dCH_{4,gas}}{dt} = rHyd2Gas - rGas2Hyd + rDis2Gas - rGas2Dis - Ebullition$$ $$\frac{dCH_{4,hyd}}{dt} = -rHyd2Gas + rGas2Hyd + rDis2Hyd - rHyd2Dis $$ Units of the state variables and fluxes are:

• $CH_{4,diss}$, $CH_{4,gas}$, $CH_{4,hyd}$: $mol~C~m^{-2}$
• all fluxes: $mol~C~m^{-2}~d^{-1}$

Exercise 4 ANSWER. The coupled cycles of carbon, nitrogen and oxygen in marine sediments

First, we reproduce here the reactions:

aerobic mineralization: $$OxicMin:\quad (CH_2O)1(NH_3){NC}(H_3PO_4){PC} + O_2 \rightarrow CO_2 + {NC} \cdot NH_3 + {PC} \cdot H_3PO_4 + H_2O $$ denitrification: $$Denitri:\quad (CH_2O)_1(NH_3){NC}(H_3PO_4){PC} + \frac{4}{5}\cdot HNO_3 \rightarrow CO_2 + \frac{2}{5}\cdot N_2 + {NC} \cdot NH_3 + {PC} \cdot H_3PO_4 + \frac75\cdot H_2O$$ anoxic mineralization: $$AnoxicMin:\quad (CH_2O)_1(NH_3){NC}(H_3PO_4)_{PC} + XO_2 \rightarrow CO_2 + X + {NC} \cdot NH_3 + {PC} \cdot H_3PO_4 + H_2O$$ nitrification: $$Nitri:\quad NH_3 + 2\cdot O_2 \rightarrow HNO_3 + H_2O$$ aerobic reoxidation: $$Reox:\quad X + O_2 \rightarrow XO_2$$

By taking into account net influxes for all state variables, we have the following mass balance equations:

$$\frac{dCorg}{dt} = Org_Depo - oxicMin - Denitri - anoxicMin$$ $$\frac{dO_2}{dt} = O_2_influx - oxicMin - 2 \cdot Nitri - Reox$$ $$\frac{dCO_2}{dt} = CO_2_influx + oxicMin + Denitri + anoxicMin$$ $$\frac{dHNO_3}{dt} = HNO_3_influx - 0.8 \cdot Denitri + Nitri$$ $$\frac{dNH_3}{dt} = NH_3_influx + \frac{16}{106} \cdot (oxicMin + Denitri + anoxicMin) - Nitri$$ $$\frac{dX}{dt} = X_influx + anoxicMin - Reox$$ $$\frac{dXO_2}{dt} = XO_2_influx - anoxicMin + Reox$$

• Suitable units for the state variables in the sediment are:
• $Corg$ and $CO_2$: $mol~C~m^{-2}$
• $O_2$: $mol~O_2~m^{-2}$
• $HNO_3$ and $NH_3$: $mol~N~m^{-2}$

• $XO_2$ and $X$: $mol~X~m^{-2}$

• Units of the time derivatives and of the process rates and influxes will have an extra $d^{-1}$ after the corresponding $mol~m^{-2}$ part.

• Note that each influx corresponds to the net flux. We use the following sign convention: a positive value of the influx corresponds to a net flux of a substance from the overlying water into the sediment, whereas a negative value corresponds to a net flux from the sediment into the overlying water.

If we denote by $h$ the height of the water column, the mass balance equations for the water column components are: $$\frac{dO_{2,wat}}{dt} = - \frac{O_2_influx}{h}$$

$$\frac{dNH_{3,wat}}{dt} = - \frac{NH_3_influx}{h}$$

and similarly for the other components.

• Suitable units for the state variables in the water column are $mol~m^{-3}$, units of the time derivatives are $mol~m^{-3}~d^{-1}$. By dividing the influx with $h$, where the latter is in $m$, we obtain the units of rate on the right-hand side of the water-column components in $mol~m^{-3}~d^{-1}$, as required.

Exercise 5 ANSWER. Benthic and pelagic primary producers in shallow seas

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Based on the conceptual diagram shown in Figure 4, we can write the following mass balance equations:

$$\frac{dPhytoplankton}{dt} = Nup_p - Death_p - \frac{Sinking_p}{Depth}$$ $$\frac{dDetitrus}{dt} = Death_p - Min_p - \frac{Sinking_det}{Depth}$$ $$\frac{dNH_4}{dt} = Min_p - Nup_p + \frac{Diffusion}{Depth}$$ $$\frac{dSediment_detritus}{dt} = Sinking_det + Sinking_p + Death_b - Min_b$$ $$\frac{dSediment_{NH4}}{dt} = Min_b - Nup_b - Diffusion$$ $$\frac{dPhytobenthos}{dt} = Nup_b - Death_b$$ Regarding units, water column constituents are in $mol~N~m^{-3}$. Rates are in $mol~N~m^{-3}~d^{-1}$, except for the rates of sinking and diffusion, which are in $mol~N~m^{-2}~d^{-1}$.

dynamic-R/RTM documentation built on Feb. 28, 2025, 1:23 p.m.