Code
using Ribasim
@@ -600,27 +600,27 @@ println(p.allocation.allocation_models[1].problem)
-
Min F[(Basin #2, UserDemand #3)]² + F[(Basin #5, UserDemand #6)]²
+Min F[(Basin #5, UserDemand #6)]² + F[(Basin #2, UserDemand #3)]²
Subject to
- flow_conservation[Basin #2] : -F[(Basin #2, UserDemand #3)] + F[(FlowBoundary #1, Basin #2)] + F[(UserDemand #3, Basin #2)] - F[(Basin #2, LinearResistance #4)] + F[(LinearResistance #4, Basin #2)] = 0
+ flow_conservation[Basin #2] : -F[(Basin #2, LinearResistance #4)] + F[(LinearResistance #4, Basin #2)] + F[(UserDemand #3, Basin #2)] + F[(FlowBoundary #1, Basin #2)] - F[(Basin #2, UserDemand #3)] = 0
+ flow_conservation[Basin #5] : -F[(Basin #5, UserDemand #6)] + F[(UserDemand #6, Basin #5)] - F[(Basin #5, TabulatedRatingCurve #7)] + F[(LinearResistance #4, Basin #5)] - F[(Basin #5, LinearResistance #4)] = 0
flow_conservation[Terminal #8] : F[(TabulatedRatingCurve #7, Terminal #8)] = 0
- flow_conservation[Basin #5] : -F[(Basin #5, UserDemand #6)] + F[(UserDemand #6, Basin #5)] + F[(LinearResistance #4, Basin #5)] - F[(Basin #5, LinearResistance #4)] - 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[TabulatedRatingCurve #7] : -F[(TabulatedRatingCurve #7, Terminal #8)] + F[(Basin #5, TabulatedRatingCurve #7)] = 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
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
- F[(Basin #2, UserDemand #3)] ≥ 0
+ source_user[UserDemand #3] : F[(UserDemand #3, Basin #2)] ≤ 0
F[(Basin #5, UserDemand #6)] ≥ 0
F[(UserDemand #6, Basin #5)] ≥ 0
- F[(FlowBoundary #1, Basin #2)] ≥ 0
- F[(TabulatedRatingCurve #7, Terminal #8)] ≥ 0
- F[(LinearResistance #4, Basin #5)] ≥ 0
- F[(Basin #5, LinearResistance #4)] ≥ 0
F[(Basin #5, TabulatedRatingCurve #7)] ≥ 0
- F[(UserDemand #3, Basin #2)] ≥ 0
F[(Basin #2, LinearResistance #4)] ≥ 0
F[(LinearResistance #4, Basin #2)] ≥ 0
+ F[(UserDemand #3, Basin #2)] ≥ 0
+ F[(LinearResistance #4, Basin #5)] ≥ 0
+ F[(Basin #5, LinearResistance #4)] ≥ 0
+ F[(TabulatedRatingCurve #7, Terminal #8)] ≥ 0
+ F[(FlowBoundary #1, Basin #2)] ≥ 0
+ F[(Basin #2, UserDemand #3)] ≥ 0
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.
-
+
Code
import numpy as np
@@ -675,7 +675,7 @@
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.
-
+
Code
fig, ax = plt.subplots()
@@ -758,7 +758,7 @@ \[
S(h) = \int_{h_0}^h A(h')\text{d}h'.
\]
-
+
Code
storage = np.diff(level) * area[:-1] + 0.5 * np.diff(area) * np.diff(level)
@@ -803,7 +803,7 @@
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.
-
+
Code
import plotly.graph_objects as go
@@ -918,9 +918,9 @@ fig.show()
-
Code
import numpy as np
@@ -675,7 +675,7 @@
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.
-
+
Code
fig, ax = plt.subplots()
@@ -758,7 +758,7 @@ \[
S(h) = \int_{h_0}^h A(h')\text{d}h'.
\]
-
+
Code
storage = np.diff(level) * area[:-1] + 0.5 * np.diff(area) * np.diff(level)
@@ -803,7 +803,7 @@
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.
-
+
Code
import plotly.graph_objects as go
@@ -918,9 +918,9 @@ fig.show()
-