Cylindrical Water Reservoir Model and System Behavior

Question

System Description

We're looking for the mathematical model of a cylindrical water reservoir. The reservoir is characterized by three variables:

The inlet flow, $Q_e(t)$, at time t The outlet flow, $Q_s(t)$, at time t The height of the liquid in the reservoir, $H(t)$, at time t The reservoir area is denoted by S.

Hypothesis

The outlet flow is governed by a turbulent gravity flow:

$Q_s(t) =a \sqrt{H(t)}$ where $a=s\sqrt{2g}$

​ (g: acceleration of gravity) (Eqn 1.1)

Reminder

The equation resulting from the balance of volumes entering, accumulating, and exiting during a time dt is of the form: $S \frac{d H(t)}{d t} = Q_e(t) - Q_s(t) \ \ $ ​ [Eqn 1.2]

Questions:

1. Determine the relationship corresponding to the steady state (static state) characterized by the nominal variables $(Q_0, H_0) \ \ $. From Eq 1.1, 1.2, we deduce this non-linear differential equation : $S \frac{d H(t)}{dt} + a\sqrt{H(t)} = Q_e(t)$ in the static state i think -this is where i could be wrong i want your insights- the derivates are zero, thus : $a \sqrt{H_0} = Q_0$ is the relation we are looking for.

2. We want to linearize equations (1.1) and (1.2) around a nominal operating point (static regime) characterized by the nominal variables $(Q_0, H_0)$. Here's the adopted notation:

$$Q_e(t) = Q_0 + q_e(t); Q_s(t) = Q_0 + q_s(t); H(t) = H_0 + h(t).$$

We need to find a linearized differential equation that describes the behavior of the system around a specific operating point, focusing on the variables $q_e(t)$ and $h(t)$.

For this question this is what i tried to do : from the non-linear version i found we can plug these substitutions $$S \frac{d h(t)}{dt} + a \sqrt{H_0 + h(t)} = Q_0 + q_e(t)$$ no idea how to make this a linear thing tbh

Answer 1

The nonlinear ode is

$$ \dot H + \frac aS \sqrt{H} = \frac 1S Q_e $$

and at the operating point

$$ 0 + \frac aS \sqrt{H_0} = \frac 1S Q_e^0 $$

subtracting we have

$$ \frac{d}{dt}(H-H_0) + \frac aS \sqrt{H}-\frac aS \sqrt{H_0}=\frac 1S Q_e-\frac 1S Q_e^0 $$

but

$$ \frac aS \sqrt{H}-\frac aS \sqrt{H_0}\approx \frac{a}{2S}\frac{1}{\sqrt{H_0}}(H-H_0) $$

then, around the operating point

$$ \dot\delta_h + c_1\delta_h = c_2 \delta_u,\ \ \ \ \delta_h = H-H_0, \delta_u = Q_e-Q_e^0 $$

Here

$$ \cases{ c_1 = \frac{a}{2S}\frac{1}{\sqrt{H_0}}\\ c_2 = \frac 1S } $$