Activating project at `~/Teaching/BEE4750/fall2023/slides`
Lecture 20
October 20, 2023
Activating project at `~/Teaching/BEE4750/fall2023/slides`
Variable | Meaning |
---|---|
\(C\) | Concentration (g/m\(^3\)) |
\(Q\) | Emissions Rate (g/s) |
\(H\) | Effective Source Height (m) |
\(u\) | Wind Speed (m/s) |
\(y, z\) | Crosswind, Vertical Distance (m) |
Gaussian Plume Distribution
\[\begin{aligned} C(x,y,z) = &\frac{Q}{2\pi u \sigma_y \sigma_z} \exp\left(\frac{-y^2}{2\sigma_y^2} \right) \times \\\\ & \quad \left[\exp\left(\frac{-(z-H)^2}{2\sigma_z^2}\right) + \exp\left(\frac{-(z+H)^2}{2\sigma_z^2}\right) \right] \end{aligned}\]
Text: VSRIKRISH to 22333
We could use a Gaussian plume to simulate the effect of a single source, but often we:
Take three sources of SO2 (air quality standard 150 \(\mu\text{g/m}^3\)):
Source | Emissions (kg/day) | Effective Height (m) | Removal Cost ($/kg) |
---|---|---|---|
1 | 34,560 | 50 | 0.20 |
2 | 172,800 | 200 | 0.45 |
3 | 103,680 | 30 | 0.60 |
and five receptors at ground level with \(u = 1.5\) m/s.
Our goal:
Minimize cost of removing SO2 from the plume sources to ensure all receptors are not exposed beyond the 150 \(\mu\text{g/m}^3\) standard.
Need to know relationship between source emissions (\(Q_i\)) and receptor exposure.
\[ C_\text{total} = \frac{M_1 + M_2 + M_3}{V} = \frac{M_1}{V} + \frac{M_2}{V} + \frac{M_3}{V} = C_1 + C_2 + C_3 \]
What is the key set of decision variables?
Fraction of SO2 removed at source \(i\): \(R_i\).
Alternatively, can reframe as level of emissions: \[Q_i = (1-R_i) \times E_i,\] where \(E_i\) is the emissions level (given in problem).
What is our main constraint?
Need to ensure compliance with the air quality standard:
\[\text{Exp}_j \leq .00015 \text{g/m}^3\]
This means that we need to express \(\text{Exp}_j\) as a linear function of the \(R_i\).
Write \(C_i(x,y) = Q_it_i(x,y)\), where the \(t_i\) is the transmission factor from the Gaussian dispersion model:
\[t_i(x,y) = \frac{1}{2\pi u \sigma_y \sigma_z} \exp\left(\frac{-y^2}{2\sigma_y^2} + \frac{H^2}{\sigma_z^2}\right)\]
This lets us write the exposure constraints as a linear function of \(R_i\):
For a receptor \(j\) (with fixed location \((x_j, y_j)\)),
\[ \begin{align} \text{Exp}_j &= \sum_i Q_i t_i(x_j, y_j) \\ &= (1-R_i)E_it_i(x_j, y_j) \leq 0.00015\ \text{g/m}^3 \end{align} \]
Write \(t_{ij} = t_i(x_j, y_j)\).
However, we still need to do this analysis given a particular atmospheric stability class (or can test across all). Let’s assume we’re in stability class C.
Dispersion Coefficients
Using the equations:
\[\begin{align} \sigma_y &= ax^{0.894} \\[0.5em] \sigma_z &= cx^d + f, \end{align}\]
we have \[\sigma_y = 104x^{0.894}, \qquad \sigma_z = 61x^{0.911}.\]
# delta_x, delta_y should be in m
function transmission_factor(Δx, Δy, u, H)
if Δx <= 0 # check if source is upwind of receptor
tf = 0.0 # ensure this is a Float
else
σy = 104 * (Δx / 1000)^0.894
σz = 61 * (Δx / 1000)^0.911
tf_coef = 1/(2 * pi * u * σy * σz)
tf = tf_coef * exp((-0.5 * (Δy / σy)^2) + (H / σz)^2)
end
return tf
end
For example, from Source 1 to Receptor 1:
Or from Source 1 to Receptor 2:
What is our objective?
Minimize cost:
\[ \begin{align} \min_{R_i} & \sum_i RemCost_i \times (E_i \times R_i) \\[0.5em] &= (0.20) (34560) R_1 + (0.45)(172800) R_2 + (0.60)(103680) R_3 \\[0.5em] &= 6912R_1 + 77760R_2 + 62208R_3 \end{align} \]
\[ \begin{align} \min_{R_i} \quad & \sum_i RemCost_i \times (E_i \times R_i) & \\ \text{subject to:} & \\[0.5em] & \sum_{i=1}^3 E_i (1-R_i)t_{ij} \leq 0.00015 & \forall j \in 1:5 \\[0.5em] & R_i \geq 0 & \forall i \in 1:3 \\[0.5em] & R_i \leq 1 & \forall i \in 1:3 \end{align} \]
Source | SO2 Emissions (g/m^3) | Removal Percentage (%) |
---|---|---|
1 | 400 | 85.0 |
2 | 2000 | 100.0 |
3 | 1200 | 75.0 |
Does this make sense?
In the absence of our treatment plan (exposure in \(\mu\)g/m\(^3\)):
Source | R1 | R2 | R3 | R4 | R5 |
---|---|---|---|---|---|
1 | 0.0 | 1008.0 | 0.0 | 51.0 | 5.0 |
2 | 0.0 | 0.0 | 0.0 | 278.0 | 1000.0 |
3 | 0.0 | 0.0 | 0.0 | 58.0 | 603.0 |
Need to reduce Source 1’s emissions to bring Receptor 2’s exposure down. What about Sources 2 and 3?
We could reduce emissions from source 2 by ~85% to comply at receptor 5, but then would also need to reduce source 3’s emissions by 100%.
This plan involves eliminating all of source 2’s emissions, but only 75% of source 3’s.
Since removal at source 2 ($0.45/kg) is cheaper than source 3 ($0.60/kg), eliminating all of source 2’s is cheaper, at a cost of $124,416.
Source | R1 | R2 | R3 | R4 | R5 |
---|---|---|---|---|---|
1 | 0.0 | 150.0 | 0.0 | 8.0 | 1.0 |
2 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 |
3 | 0.0 | 0.0 | 0.0 | 14.0 | 149.0 |
Monday: Mixed Integer Linear Programming
Wednesday/Friday: Applications (network models, unit commitment)
HW4: Due next Friday (10/27) at 9pm.
Regulatory Report: Due next Friday also.