From b3e198a668e4e79c55e0227789932085842393e4 Mon Sep 17 00:00:00 2001 From: Bart de Koning Date: Mon, 29 Jan 2024 16:32:54 +0100 Subject: [PATCH] Remove using S for storage (use u) --- docs/core/equations.qmd | 24 ++++++++++++------------ docs/core/numerics.qmd | 2 +- 2 files changed, 13 insertions(+), 13 deletions(-) diff --git a/docs/core/equations.qmd b/docs/core/equations.qmd index cc238d8a6..92d5d258f 100644 --- a/docs/core/equations.qmd +++ b/docs/core/equations.qmd @@ -42,7 +42,7 @@ $\mathbf{u}(t)$ is given by all the states of the model, which are (currently) t Given a single basin with node ID $i \in B$, the equation that dictates the change of its storage over time is given by $$ -\frac{\text{d}S_i}{\text{d}t} = +\frac{\text{d}u_i}{\text{d}t} = \sum_{(i',j') \in E | j' = i} Q_{i',j'} - \sum_{(i',j') \in E | i' = i} Q_{i',j'} + F_i(p,t). $$ @@ -74,15 +74,15 @@ $$ Most of these terms are always $0$, because a flow over an edge only depends on a small number of states. Therefore the matrix $J$ is very sparse. -For many contributions to the Jacobian the derivative of the level $l(S)$ of a basin with respect to its storage $S$ is required. To get an expression for this, we first look at the storage as a function of the level: +For many contributions to the Jacobian the derivative of the level $l(u)$ of a basin with respect to its storage $u$ is required. To get an expression for this, we first look at the storage as a function of the level: $$ -S(l) = \int_{l_0}^l A(\ell)d\ell. +u(l) = \int_{l_0}^l A(\ell)d\ell. $$ -From this we obtain $S'(l) = A(l)$ and thus +From this we obtain $u'(l) = A(l)$ and thus $$ -\frac{\text{d}l}{\text{d}S} = \frac{1}{A(S)}. +\frac{\text{d}l}{\text{d}u} = \frac{1}{A(u)}. $$ :::{.callout-note} @@ -149,11 +149,11 @@ plt.show() The precipitation term is given by $$ - Q_P = P \cdot A(S). + Q_P = P \cdot A(u). $$ {#eq-precip} Here $P = P(t)$ is the precipitation rate and $A$ is the wetted area. $A$ is a -function of the storage $S = S(t)$: as the volume of water changes, the area of the free water +function of the storage $u = u(t)$: as the volume of water changes, the area of the free water surface may change as well, depending on the slopes of the surface waters. ## Evaporation @@ -161,13 +161,13 @@ surface may change as well, depending on the slopes of the surface waters. The evaporation term is given by $$ - Q_E = E_\text{pot} \cdot A(S) \cdot \phi(d;0.1). + Q_E = E_\text{pot} \cdot A(u) \cdot \phi(d;0.1). $$ {#eq-evap} -Here $E_\text{pot} = E_\text{pot}(t)$ is the potential evaporation rate and $A$ is the wetted area. $\phi$ is the [reduction factor](equations.qmd#sec-reduction_factor) which depends on the depth $d$. It provides a smooth gradient as $S \rightarrow 0$. +Here $E_\text{pot} = E_\text{pot}(t)$ is the potential evaporation rate and $A$ is the wetted area. $\phi$ is the [reduction factor](equations.qmd#sec-reduction_factor) which depends on the depth $d$. It provides a smooth gradient as $u \rightarrow 0$. -A straightforward formulation $Q_E = \mathrm{max}(E_\text{pot} A(S), -0)$ is unsuitable, as $\frac{\mathrm{d}Q_E}{\mathrm{d}S}(S=0)$ is then not well-defined. +A straightforward formulation $Q_E = \mathrm{max}(E_\text{pot} A(u), +0)$ is unsuitable, as $\frac{\mathrm{d}Q_E}{\mathrm{d}u}(u=0)$ is then not well-defined.