Add max_flow_rate to LinearResistance

[visr]

This is needed to be able to model structures that exchange flow based on a water level difference, but only up to a certain capacity.

Ref Deltares/Ribasim-NL#62

The updated test model now looks like this:

[visr]

@SouthEndMusic it would be great if you could help updating this test value.

[Huite]

Solve for Q(level) == max_flow_rate

Find out at which times it switches from max_flow_rate to dynamic flow rate

Compute from there

[SouthEndMusic]

DE for $u(t)$:

$$ u'(t) = \text{clamp}\left(\frac{L - \frac{u(t)}{A}}{R}, -q_{\max}, q_{\max}\right), \quad R, A, q_{\max} > 0. $$

For simplicity we assume a constant area here, which is not the case in the current linear resistance test model but is now supported.

Initial condition:

$$ u(0) = u_0 > 0. $$

We assume $u_0$ is big enough such that

$$ \frac{L - \frac{u_0}{A}}{R} < -q_{\max}. $$

Hence for $0 &lt; t &lt; t_\text{shift}$ we have that

$$ u(t) = u_0 - q_{\max} t. $$

$t_\text{shift}$ is given by

$$ \frac{L - \frac{u(t_\text{shift})}{A}}{R} = -q_{\max}. $$

Substituting the above expression for $u(t)$ yields

$$ t_\text{shift} = \frac{u_0 - A(L + Rq_{\max})}{ q_{\max}}, $$

and so

$$ u(t_\text{shift}) = A(L + Rq_{\max}). $$

Now we solve from here with

$$ u'(t) = \frac{L - \frac{u(t)}{A}}{R}, $$

which yields

$$ AL = ARu'(t) + u(t) = AR \exp\left(-\frac{t}{AR}\right)\left[u(t)\exp\left(\frac{t}{AR}\right)\right]' $$

and so

$$ u(t)\exp\left(\frac{t}{AR}\right) = \frac{L}{R}\int \exp\left(\frac{t}{AR}\right) \text{d} t = AL \exp\left(\frac{t}{AR}\right) + C, $$

which gives

$$ u(t) = AL + C \exp\left(-\frac{t}{AR}\right). $$

This makes sense; $u(t)$ will converge to $AL$, the volume in the basin at the level of the level boundary.

Filling in our initial condition:

$$ u(t_\text{shift}) = AL + C \exp\left(-\frac{t_\text{shift}}{AR}\right) = A(L + Rq_{\max}) $$

yields

$$ C = ARq_{\max} \exp\left(\frac{t_\text{shift}}{AR}\right). $$

We conclude:

$$ u(t) = \begin{cases} u_0 - q_{\max} t \quad \text{if} \quad 0 \leq t \leq t_\text{shift}, \\ AL + ARq_{\max} \exp\left(-\frac{t - t_\text{shift}}{AR}\right) \quad \text{if} \quad t > t_\text{shift} \end{cases} $$

and

$$ u'(t) = \begin{cases} - q_{\max}\quad \text{if} \quad 0 \leq t \leq t_\text{shift}, \\ -q_{\max} \exp\left(-\frac{t - t_\text{shift}}{AR}\right) \quad \text{if} \quad t > t_\text{shift} \end{cases}, $$

where

$$ t_\text{shift} = \frac{u_0 - A(L + Rq_{\max})}{ q_{\max}}. $$

[SouthEndMusic]

Looking good, I trust you can update the analytical test with the maximum flow 👍

[visr]

The storage in black nicely goes from your u pre-shift in green to u post-shift in blue.