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ANSWERS
Exercise 1 ANSWER. Carbon cycling in a Canadian National Park
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- Model domain: entire river and riparian ecosystem
- Boundary: sea
- Time-scale: months, the length of the migration
- State variables: BEAR, SALMON, CARCASS, SCAVENGERS
Suitable units are: $mol~carbon$ (for the entire area) or $mol~C~m^{-2}$.
The mass balance equations based on the conceptual scheme in Figure 1 are:
$$\frac{dSALMON}{dt} = Migration - BearFeeding - SalmonLoss$$ $$\frac{dBEAR}{dt} = BearFeeding-CarcassProd-BearLossIngest-BearLoss$$ $$\frac{dCARCASS}{dt} = CarcassProd -ScavFeeding - CarcassDecay$$ $$\frac{dSCAVENGERS}{dt} = ScavFeeding -ScavLossIngest - ScavLoss$$
Exercise 2 ANSWER. Lake biomanipulation
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- Possible forcing function could be f1, the inflow of DIP.
- Suitable units: $mol~P~m^{-3}$
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The effects of biomanipulation can be investigated by changing f10.
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Mass balance equations are:
$$\frac{dDIP}{dt} = f1+f4+f6-f2-f7$$ $$\frac{dPHYTO}{dt} = f2-f3-f8$$ $$\frac{dZOO}{dt} = f3-f4-f5-f9$$ $$\frac{dFISH}{dt} = f5-f6-f10$$
Exercise 3 ANSWER. Population dynamics in the Serengeti National Park
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- Scale of the model: entire park
- Time scale: season
- Units: $kg~km^{-2}$
- State variables: WILDEBEEST, LIONS
- Reaction processes: death (f3, f5), predation (f4)
- Transport: immigration (f1) and emigration (f2)
The mass balance equations are:
$$\frac{dWILDEBEEST}{dt} = f1-f2-f4-f3$$ $$\frac{dLIONS}{dt} = a \times f4-f5$$
Exercise 4 ANSWER. Oysters in the Oosterschelde
- Scale of the model: Oosterschelde, mean depth = D.
- Time scale: year
- State variables: NITRATE, ALGAE, CREUSEOYS, FLATOYS, BOTTOMDETRITUS
- Units:
- $mol~N~m^{-3}$ (NITRATE, ALGAE)
- $mol~N~m^{-2}$ (OYSTERS, BOTTOM DETRITUS)
- Reaction processes:
- algal growth (f1)
- oyster filtration (f2a, f2b)
- pseudofaeces production (f3a, f3b)
- faeces production (f4a, f4b)
- oyster mortality (f5a, f5b)
- harvesting (f6a, f6b)
- detritus mineralisation (f7)
- Transport: import of NITRATE (f8) and ALGAE (f9)
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The mass balance equations are :
$$\frac{dNITRATE}{dt} = f8-f1+f7/D$$ $$\frac{dALGAE}{dt} = f9+f1-f2a/D-f2b/D$$ $$\frac{dFLAT.OYSTER}{dt} = f2a-f3a-f4a-f5a-f6a$$ $$\frac{dCREUSE.OYSTER}{dt} = f2b-f3b-f4b-f5b-f6b$$ $$\frac{dBOTTOMDETRITS}{dt} = f3a+f3b+f4a+f4b+f5a+f5b-f7$$ Units of f1, f8 and f9 are $mol~N~m^{-3}d^{-1}$, while the units of the other processes are $mol~N~m^{-2}d^{-1}$.
We need to take into account the system’s depth (D) when implementing the effect of the bottom detritus mineralisation (in $mol~N~m^{-2}~d^{-1}$) on the rate of change of nitrate (in $mol~N~m^{-3}~d^{-1}$). This assumes that the flux of nitrate from the bottom to the water column is immediately homogenised over the entire water depth (i.e., the water column is well-mixed).
The same applies for the ingestion by the oysters (f2a, f2b), when taking into account their effects on algae.
