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Chapter 3 : Reservoir models - KU Leuven

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If the concentration time is not taken into account as a variable parameter with<br />

the runoff, the extra variation in smoothing and peak shift can be modelled using<br />

the dynamic storage. In that way the smoothing module and the dynamic storage<br />

module overlap to some extent, although they cannot replace each other completely.<br />

The differential equation that describes a linear static reservoir model is :<br />

dV<br />

dt<br />

= Q − Q<br />

(3.8)<br />

in<br />

through<br />

with :<br />

Q<br />

through<br />

V<br />

=<br />

V<br />

Q m<br />

(3.9)<br />

m<br />

In this set of equations, V is the instantaneous storage volume in the reservoir,<br />

V m the maximum storage, Q in the inflow, Q m the maximum throughflow and<br />

Q out the instantaneous throughflow. The analytical solution for this set of equations<br />

at time t+dt as a function of the instantaneous storage V t in the previous time step t is :<br />

V Q V ⎛<br />

m<br />

Qm<br />

Q<br />

V dt V Qm<br />

t+ dt = in − ⎛ ⎞⎞<br />

⎜−<br />

⎟<br />

⎜<br />

t<br />

V<br />

dt<br />

m ⎝ ⎝ m ⎠<br />

⎟<br />

+ ⎛ ⎞<br />

1 exp exp ⎜−<br />

⎟ (3.10)<br />

⎠<br />

⎝ m ⎠<br />

When the maximum storage V m is reached, the overflow starts spilling and the overflow<br />

discharge is equal to the difference between inflow Q in and maximum throughflow Q m .<br />

The reservoir modelling system Remuli however uses piecewise linear relationships for<br />

the storage/throughflow relationships, which also makes the inclusion of extra storage<br />

during the overflow event possible (see paragraph 3.3.2).<br />

The dynamic storage and the static storage are fitted together as shown in figure 3.15<br />

using the following equations :<br />

Vm<br />

V k Q k Q<br />

Q Q Vdm<br />

= stat through + dyn in = through +<br />

Q<br />

Q in<br />

(3.11)<br />

in which V dm is the maximum dynamic storage corresponding with the maximum<br />

inflow Q dm .<br />

Combined with the continuity equation this leads to :<br />

Or solved :<br />

dV<br />

dt<br />

⎛ Vdm<br />

V<br />

Vt+ dt = Qin<br />

⎜ +<br />

⎝ Q Q<br />

dm<br />

⎛ Vdm<br />

Qm<br />

⎞ Qm<br />

= Qin<br />

⎜1 + ⎟ −<br />

⎝ Q V ⎠ V<br />

V<br />

(3.12)<br />

m<br />

m<br />

dm<br />

m<br />

m<br />

m<br />

⎞ ⎛ Qm<br />

V dt V Qm<br />

⎟ − ⎛ ⎞⎞<br />

⎜−<br />

⎟<br />

t<br />

⎠<br />

⎜<br />

⎝<br />

V<br />

dt<br />

⎝<br />

m ⎠<br />

⎟ + ⎛ − ⎞<br />

1 exp exp ⎜ ⎟ (3.13)<br />

⎠ ⎝ m ⎠<br />

dm<br />

<strong>Chapter</strong> 3 : <strong>Reservoir</strong> <strong>models</strong> 3.29

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