Ch 2.6 Case Study: Lake Burley Griffin

Lake Burley Griffin

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Background

  • Lake Burley Griffin is located in Canberra, Australia's capital.

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Background

  • Lake Burley Griffin was created artificially in 1962

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Background

  • Serves recreational and aesthetic purposes.

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Background

  • Queanbeyan is in the state of New South Wales (NSW)
  • Lake Burley Griffin is in the Australian Capital Territory
  • Safety levels/standards are different between territories.

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Background

  • Extensive measurements of pollution levels (1970s)
  • Upstream sewage plants (Queanbeyan) contribute pollution

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Background

  • Significant also was rural and urban runoff during summer rainstorms.
  • These contributed to dramatic increases in pollution levels
  • Responsible for lifting concentration levels above safety limits.

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Background

  • In 1974 the mean concentration of the bacteria faecal coliform count was approximately \( 10^7 \) bacteria per \( \mathrm{m}^3 \) at the point where the river feeds into the lake.

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Background

Safety threshold for faecal coliform count in water for contact recreational sports:

  • No more than 10% of total samples over a 30-day period should exceed \( 4 \times 10^6 \) bacteria per \( \mathrm{m}^3 \).

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Use Model to Investigate

  • If sewage management were improved, would lake would flush out?
  • Would pollution levels would drop below the safety threshold?

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General compartmental model

  • The lake is the only compartment.

  • Applying the balance law there is an input of polluted water from the river(s) flowing into the lake, or due to a pollution dump into the lake, and an output as water flows from the lake carrying some pollution with it.

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Assumptions, Model & Solution

  • Lake is continuously well mixed
  • Pollution is uniform throughout
  • Lake has a constant volume \( V = 28 \times 10^6 \mathrm{m}^3 \).
  • Lake has constant flow F, with \( F_{in} = F_{out} \).
  • IVP and solution:

\[ \begin{align*} \frac{dC}{dt} & = \frac{F}{V}c_{in} - \frac{F}{V}C(t) , \,\, C(0)= c_0 \\ \\ C(t) & = c_{in} - (c_{in} - c_0) e^{-Ft/V } \end{align*} \]

Lake Pollution Model

  • How long will it take for the lake's pollution level to reach a specified level \( C_L \) if only fresh water flows into the lake?
  • Only fresh water coming in: \( c_{in} = 0 \)

\[ \begin{align*} C(t) & = c_{in} - (c_{in} - c_0) e^{-Ft/V } \\ & = c_0 e^{-Ft/V } \end{align*} \]

  • Solve for \( t \), using \( C = C_L \):

    \[ t = - \frac{V}{F} \ln \left( \frac{C_L}{c_0} \right) \]

Lake Pollution Model: Predictions

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  • Fresh water entering the lake: \( c_{in} = 0 \)
  • Volume: \( V = 28 \times 10^6 \mathrm{m}^3 \).
  • Mean monthly flow: \( F = 4 \times 10^6 \mathrm{m}^3/ \mathrm{month} \)
  • Initial faecal coliform count: \( c_0 = 10^7\mathrm{bacteria}/ \mathrm{m}^3 \)
  • \( C_L = ? \)

Lake Pollution Model: Predictions

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  • Fresh water entering the lake: \( c_{in} = 0 \)
  • Volume: \( V = 28 \times 10^6 \mathrm{m}^3 \).
  • Mean monthly flow: \( F = 4 \times 10^6 \mathrm{m}^3/ \mathrm{month} \)
  • Initial faecal coliform count: \( c_0 = 10^7\mathrm{bacteria}/ \mathrm{m}^3 \)
  • \( C_L = 4 \times 10^6 \mathrm{bacteria}/ \mathrm{m}^3 \)

Lake Pollution Model: Predictions

  • Parameter Values
  • \( V = 28 \times 10^6 \mathrm{m}^3 \)
  • \( F = 4 \times 10^6 \mathrm{m}^3/ \mathrm{month} \)
  • \( C_L = 4 \times 10^6\mathrm{bacteria}/ \mathrm{m}^3 \)
  • \( c_0 = 10^7\mathrm{bacteria}/ \mathrm{m}^3 \)

  • Formula \[ t = - \frac{V}{F} \ln \left( \frac{C_L}{c_0} \right) \]

  • Use R for calculations:
V <- 28*10^6 
F <- 4*10^6 
CL <- 4*10^6
c0 <- 10^7
t <- - V/F*log(CL/c0)

t

[1] 6.414035

Lake Pollution Model

  • However, pure water entering the lake is not a very realistic scenario with three sewage plants and much farmland upstream.
  • Thus including the entrance of polluted river water into the lake model is essential.

\[ C(t) = c_{in} - (c_{in} - c_0) e^{-Ft/V } \]

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Numerical Solutions: Maple and MATLAB

  • Including the entrance of polluted river water into the lake model is essential.

\[ \frac{dC}{dt} = \frac{F}{V} \left[ c_{in} - C(t) \right] , \,\, C(0)= c_0 \]

  • Case 1: \[ \begin{align*} c_{in} & = 3 \times 10^6 \mathrm{bacteria}/ \mathrm{m}^3 \\ F & = (4*12) \times 10^6 \mathrm{m}^3/ \mathrm{year} \end{align*} \]
  • Case 2: \[ \begin{align*} c_{in} & = 10^6 \left[ 10 + 10 \cos(2 \pi t) \right] \mathrm{bacteria}/ \mathrm{m}^3 \\ F(t) & = 10^6 \left[ 10 + 6 \sin(2 \pi t) \right] \mathrm{m}^3/ \mathrm{year} \end{align*} \]

Case 1: Maple (Ch2.5 Listing 2.7, Part 1)

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Case 1: Maple (Ch2.5 Listing 2.7, Part 2)

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Case 1: Maple (Ch2.5 Listing 2.7)

\[ \lim_{t \rightarrow \infty} C(t) = \lim_{t \rightarrow \infty} \left( c_{in} - (c_{in} - c_0) e^{-Ft/V }\right) = c_{in} \]

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Case 1: MATLAB (Ch2.5 Listing 2.8)

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\[ \begin{align*} c_{in} & = 3 \times 10^6 \mathrm{bacteria}/ \mathrm{m}^3 \\ F & = (4*12) \times 10^6 \mathrm{m}^3/ \mathrm{year} \end{align*} \]

Case 1: MATLAB (Ch2.5 Listing 2.8)

  • ch25Lake

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Case 2: Maple, Part 1 (Ch2.6 DrG)

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Case 2: Maple, Part 2 (Ch2.6 DrG)

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Case 2: MATLAB (Ch2.6 DrG)

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Case 2: MATLAB (Ch2.6 DrG)

  • ch26Lake2

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