Lecture 04
August 30, 2023
A mathematical model of a system is a set of equations which maps inputs (forcings, decisions) to outputs (outcomes, metrics).
Mathematical models of systems can be:
Text: VSRIKRISH to 22333
Diagram of CRUD release from two factories
Two factories are discharging a chemical, chlororadiated ureadicarboxyl (CRUD), into the Riley River.
Diagram of CRUD release from two factories
Environmental authorities have sampled water from the river and CRUD concentrations exceed the legal standard (1 mg/L).
Diagram of CRUD release from two factories
We want to design a CRUD removal plan to get back in compliance.
Diagram of CRUD release from two factories
CRUD decays in the river with first-order decay coefficient \(k=0.45 \ \text{d}^{-1}\).
Consider the problem at each point of release.
\[\begin{align*} \text{(recall that } 1 \ \text{mg/L} &= 1 \ \text{g/m}^3 \\ &= 10^{-3} \ \text{kg/m}^3\text{)} \end{align*}\]
Factory 1 releases \(1000 \ \text{kg/d}\) without treatment.
What is the outflow for a given treatment fraction \(E_1\)?
Factory 1 releases \(1000 \ \text{kg/d}\) without treatment.
What is the outflow for a given \(E_1\)?
Total CRUD after factory 1 release: \(\color{blue}100 + \color{red} 1000(1-E_1) \color{black} \ \text{kg/d}\)
Is this the inflow at Release 2?
Given first-order decay rate of \(0.45 \ \text{d}^{-1}\): \[\frac{dM}{dt} = -0.45 M \Rightarrow \frac{dM}{M} = -0.45 dt\]
\[\int_{M(0)}^{M(T)} \frac{dM}{M} = -0.45 \int_0^T dt\]
\[\ln\left(\frac{M(T)}{M(0)}\right) = -0.45 T\]
So, after \(t\) days: \[M(t) = M(0) \exp\left(-0.45 t\right)\]
Is this what we need?
No! We need \(M(x)\), where \(x\) is some distance downstream.
Since the velocity of the river is \(25 \ \text{km/d}\):
\[M(x) = M_0 \exp\left(-\frac{0.45 x}{25}\right), \quad x \leq 10 \ \text{km}.\]
Simplifying and plugging in \(x = 10\) and \(M_0\), the inflow of CRUD at the factory 2 release is:
\[M(10) = (1100 - 1000E_1) \exp(-0.18) \ \text{kg/d}.\]
This means that after factory 2 releases CRUD, the mass is:
\[\color{red}(1100 - 1000E_1) \exp(-0.18) \color{black}+ \color{blue}1200(1 - E_2) \ \color{black}\text{kg/d}.\]
We can use this as an initial condition for \(M(x), x > 10\):
\[M(x) = M_1 \exp\left(-\frac{0.45x}{25}\right), \quad x > 10,\] where \[M_1 = (1100 - 1000E_1) \exp(-0.18) + 1200(1 - E_2).\]
Therefore the concentration of CRUD \(C(x)\) (in \(\text{mg/L}\)) at any point \(x \ \text{km}\) downstream of the first release is:
\[ C(x) = \begin{cases} M_0\exp\left(-0.45 x/25\right) / 600, & x \leq 10 \ \text{km} \\ M_1 \exp\left(-0.45x/25\right) / 660, & x > 10 \ \text{km} \\ \end{cases} \] where \[ \begin{align*} M_0 &= 1100 - 1000E_1,\\ M_1 &= (1100 - 1000E_1) \exp(-0.18) + 1200(1 - E_2). \end{align*} \]
Looking at these equations:
The downstream concentration (past Release 2) depends on both \(E_1\) and \(E_2\).
This means whether the regulatory limit is achieved depends on both factories!
We don’t know what needed levels of \(E_2\) are without knowing \(E_1\).
One option: We could leave \(E_1\) up to the owners of Factory 1 and require \(E_2\) be set based on this level.
Friday: Lake Eutrophication Example
Wednesday: Box Models and Simulation