Question 7.4.8: A Liquid-Level System with a Pump Figure 7.4.9 shows a liqui...

Let \Delta p = p_{1}  −  p_{a} . Denote the mass flow rates through each resistance as q_{m_{1}} and q_{m_{2}} . These flow rates are
q_{m_{1}} = \frac{1}{R_{1}} (p_{1}  −  ρgh  −  p_{a}) = \frac{1}{R_{1}} (\Delta p  −  ρgh)           (1)

q_{m_{2}} = \frac{1}{R_{2}} (ρgh + p_{a}  −  p_{a}) = \frac{1}{R_{2}}   ρgh              (2)
From conservation of mass,
ρ A \frac{d h}{d t} = q_{m_{1}}  −  q_{m_{2}} = \frac{1}{R_{1}} (\Delta p  −  ρgh)  −  \frac{1}{R_{2}} ρgh            (3)
At equilibrium, q_{m_{1}} = q_{m_{2}} , so from equation (3),
\frac{1}{R_{1}} (\Delta p  −  ρgh) = \frac{1}{R_{2}} ρgh
which gives
ρgh = \frac{R_{2}}{R_{1}  +  R_{2}} \Delta p           (4)
Substituting this into expression (2) we obtain an expression for the equilibrium value of the flow rate qm2 as a function of Δp:
q_{m_{2}} = \frac{1}{R_{1}  +  R_{2}} Δp          (5)
This is simply an expression of the series resistance law, which applies here because \dot{h} = 0 at equilibrium and thus the same flow occurs through R_{1} and R_{2}.
When equation (5) is plotted on the same plot as the pump curve, as in Figure 7.4.10, the intersection gives the equilibrium values of q_{m_{1}} and Δp. A straight line tangent to the pump curve and having the slope −1/r then gives the linearized model:
δ q_{m_{1}} = − \frac{1}{r} δ (\Delta p)             (6)
where δ q_{m_{1}} and δ(Δp) are the deviations from the equilibrium values.
From equation (4),
\Delta p = \frac{R_{1}  +  R_{2}}{R_{2}} ρgh
and thus
δ (\Delta p) = \frac{R_{1}  +  R_{2}}{R_{2}} ρg δ h

and from equation (6),
δ q_{m_{1}} = − \frac{1}{r} δ (\Delta p) = − \frac{1}{r} \frac{R_{1}  +  R_{2}}{R_{2}} ρg δ h            (7)
The linearized form of equation (3) is
ρ  A \frac{d}{d t} δ h = δ q_{m_{1}}  −  δ q_{m_{2}}
From equations (2) and (7),
ρ  A \frac{d}{d t} δ h = − \frac{1}{r} \frac{R_{1}  +  R_{2}}{R_{2}} ρg δ h  −  \frac{ρg}{R_{2}} δh
or
A \frac{d}{dt} δh = − \left(\frac{1}{r} \frac{R_{1}  +  R_{2}}{R_{2}} + \frac{1}{R_{2}} \right) g  δh
This is the linearized model, and it is of the form
\frac{d}{dt} δh = −b  δh
where
b = \left(\frac{1}{r} \frac{R_{1}  +  R_{2}}{R_{2}} + \frac{1}{R_{2}} \right) \frac{g}{A}
The equation has the solution δh(t) = δh(0)e^{−bt} . Thus if additional liquid is added to or taken from the tank so that δh(0) ≠ 0, the liquid height will eventually return to its equilibrium value.
The time to return is indicated by the time constant, which is 1/b.