Calculating with Storage Elements
Let us look at the following numerical example:
\[
\begin{gathered}
A = 0.1\,\mathrm{m}^2 \\[4pt]
\text{Filling time 10 s: } t_{\mathrm{Start}} = 0\,\mathrm{s};\ t_{\mathrm{Ende}} = 10\,\mathrm{s} \\[4pt]
\dot{v} = 1.5\,\frac{\mathrm{l}}{\mathrm{s}} = 0.0015\,\frac{\mathrm{m}^3}{\mathrm{s}}
\end{gathered}
\]
The bucket is hmax = 30 cm high. So the fill level is limited to the range between hmin = 0 cm and hmax = 30 cm; beyond that it overflows. Let the bucket initially be half full with h0 = 15 cm = 0.15 m. The fill level after 10 seconds is then
\[
\begin{gathered}
h(t_{\mathrm{Ende}}) = \frac{1}{A} \cdot \int_{t_{\mathrm{Start}}}^{t_{\mathrm{Ende}}} \dot{v}(t)\,dt + h_0(t_{\mathrm{Start}}) \\[6pt]
h(10\,\mathrm{s}) = \frac{1}{0.1\,\mathrm{m}^2} \cdot \int_{0\,\mathrm{s}}^{10\,\mathrm{s}} 0.0015\,\frac{\mathrm{m}^3}{\mathrm{s}}\,dt + 0.15\,\mathrm{m}
\end{gathered}
\]
Since the inflow is constant over time, we use the antiderivative of a constant.
\[
\begin{gathered}
\text{In general: } y = \int_{x_1}^{x_2} k\,dx = k \cdot x\Big|_{x_1}^{x_2} = k(x_2 - x_1) \\[6pt]
\text{For } \dot{v}(t) = \text{constant over } t \text{ between } t_{\mathrm{Start}} \text{ and } t_{\mathrm{Ende}}\text{:} \\[6pt]
\int_{t_{\mathrm{Start}}}^{t_{\mathrm{Ende}}} \dot{v}(t) \cdot dt = \dot{v} \cdot (t_{\mathrm{Ende}} - t_{\mathrm{Start}}) = 0.0015\,\frac{\mathrm{m}^3}{\mathrm{s}} (10\,\mathrm{s} - 0\,\mathrm{s}) = 0.015\,\mathrm{m}^3 \\[6pt]
h(10\,\mathrm{s}) = \frac{1}{0.1\,\mathrm{m}^2} \cdot 0.015\,\mathrm{m}^3 + 15\,\mathrm{cm} = 0.15\,\mathrm{m} + 15\,\mathrm{cm} = 30\,\mathrm{cm}
\end{gathered}
\]
So the fill level at the end of the filling process is 30 cm.
After 10 seconds, the fill level of the bucket has risen by 15 cm to 30 cm.