diff --git a/concept/allocation.html b/concept/allocation.html index 1bfee7247..7bbc77fd5 100644 --- a/concept/allocation.html +++ b/concept/allocation.html @@ -574,7 +574,7 @@

4.4 Example

The following is an example of an optimization problem for the example shown here:

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using Ribasim
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println(p.allocation.allocation_models[1].problem)
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Min F[(Basin #2, UserDemand #3)]² + F[(Basin #5, UserDemand #6)]²
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Min F[(Basin #5, UserDemand #6)]² + F[(Basin #2, UserDemand #3)]²
 Subject to
- flow_conservation[Basin #2] : -F[(Basin #2, LinearResistance #4)] + F[(LinearResistance #4, Basin #2)] - F[(Basin #2, UserDemand #3)] + F[(FlowBoundary #1, Basin #2)] + F[(UserDemand #3, Basin #2)] = 0
- flow_conservation[Basin #5] : F[(LinearResistance #4, Basin #5)] - F[(Basin #5, LinearResistance #4)] - F[(Basin #5, UserDemand #6)] - F[(Basin #5, TabulatedRatingCurve #7)] + F[(UserDemand #6, Basin #5)] = 0
+ flow_conservation[TabulatedRatingCurve #7] : -F[(TabulatedRatingCurve #7, Terminal #8)] + F[(Basin #5, TabulatedRatingCurve #7)] = 0
+ flow_conservation[LinearResistance #4] : -F[(LinearResistance #4, Basin #5)] + F[(Basin #5, LinearResistance #4)] + F[(Basin #2, LinearResistance #4)] - F[(LinearResistance #4, Basin #2)] = 0
  flow_conservation[Terminal #8] : F[(TabulatedRatingCurve #7, Terminal #8)] = 0
- flow_conservation[LinearResistance #4] : F[(Basin #2, LinearResistance #4)] - F[(LinearResistance #4, Basin #2)] - F[(LinearResistance #4, Basin #5)] + F[(Basin #5, LinearResistance #4)] = 0
- flow_conservation[TabulatedRatingCurve #7] : F[(Basin #5, TabulatedRatingCurve #7)] - F[(TabulatedRatingCurve #7, Terminal #8)] = 0
- F[(Basin #2, LinearResistance #4)] ≥ 0
- F[(LinearResistance #4, Basin #2)] ≥ 0
+ flow_conservation[Basin #5] : F[(UserDemand #6, Basin #5)] - F[(Basin #5, UserDemand #6)] + F[(LinearResistance #4, Basin #5)] - F[(Basin #5, LinearResistance #4)] - F[(Basin #5, TabulatedRatingCurve #7)] = 0
+ flow_conservation[Basin #2] : F[(UserDemand #3, Basin #2)] + F[(FlowBoundary #1, Basin #2)] - F[(Basin #2, UserDemand #3)] - F[(Basin #2, LinearResistance #4)] + F[(LinearResistance #4, Basin #2)] = 0
+ F[(TabulatedRatingCurve #7, Terminal #8)] ≥ 0
+ F[(UserDemand #3, Basin #2)] ≥ 0
+ F[(UserDemand #6, Basin #5)] ≥ 0
+ F[(Basin #5, UserDemand #6)] ≥ 0
+ F[(FlowBoundary #1, Basin #2)] ≥ 0
  F[(LinearResistance #4, Basin #5)] ≥ 0
  F[(Basin #5, LinearResistance #4)] ≥ 0
  F[(Basin #2, UserDemand #3)] ≥ 0
- F[(Basin #5, UserDemand #6)] ≥ 0
+ F[(Basin #2, LinearResistance #4)] ≥ 0
+ F[(LinearResistance #4, Basin #2)] ≥ 0
  F[(Basin #5, TabulatedRatingCurve #7)] ≥ 0
- F[(TabulatedRatingCurve #7, Terminal #8)] ≥ 0
- F[(FlowBoundary #1, Basin #2)] ≥ 0
- F[(UserDemand #3, Basin #2)] ≥ 0
- F[(UserDemand #6, Basin #5)] ≥ 0
  source[(FlowBoundary #1, Basin #2)] : F[(FlowBoundary #1, Basin #2)] ≤ 172800
  source_user[UserDemand #3] : F[(UserDemand #3, Basin #2)] ≤ 0
  source_user[UserDemand #6] : F[(UserDemand #6, Basin #5)] ≤ 0
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@@ -472,7 +472,7 @@ 

 
1.2.1 Interpolation

 
At the given timestamps the values are set in the simulation, such that the timeseries can be seen as forward filled.

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 Code
 
import numpy as np
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1.4.1.1 Level to area

 
The level to area relationship is defined with the Basin / profile data using linear interpolation. An example of such a relationship is shown below.

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 Code
 
fig, ax = plt.subplots()
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\[
     S(h) = \int_{h_0}^h A(h')\text{d}h'.
 \]

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 Code
 
storage = np.diff(level) * area[:-1] + 0.5 * np.diff(area) * np.diff(level)
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1.4.1.3 Interactive basin example

 
The profile data is not detailed enough to create a full 3D picture of the basin. However, if we assume the profile data is for a stretch of canal of given length, the following plot shows a cross section of the basin.

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 Code
 
import plotly.graph_objects as go
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fig.show()

 

 

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