Question 7.4: A Single-Tank Liquid-Level System with a Pump Consider the s...

We begin by applying the law of conservation of mass to the tank,

\frac{\mathrm{d} m}{\mathrm{~d} t}=q_{\mathrm{mi}}-q_{\mathrm{mo}}.

The fluid mass inside the tank is \rho A h. For constant fluid density and constant cross-sectional area,

\frac{\mathrm{d} m}{\mathrm{~d} t}=\rho A \frac{\mathrm{d} h}{\mathrm{~d} t}.

The mass flow rate into the tank is

q_{\mathrm{mi}}=\frac{p_{1}-p_{2}}{R}.

where p_{1}=p_{\mathrm{a}}+\Delta p and p_{2}=p_{\mathrm{a}}+\rho g h, which is equal to the hydrostatic pressure at the bottom of the tank. Thus,

q_{\mathrm{mi}}=\frac{p-\rho g h}{R}.

The mass flow rate out of the tank can be expressed in terms of the volume flow rate q_{0} as

q_{\mathrm{mo}}=\rho q_{\mathrm{o}} .

Substituting these expressions into the law of conservation of mass gives

\rho A \frac{\mathrm{d} h}{\mathrm{~d} t}=\frac{p-\rho g h}{R}-\rho q_{\mathrm{o}}.

Rearranging the equation gives

\rho A \frac{\mathrm{d} h}{\mathrm{~d} t}+\frac{\rho g}{R} h-\frac{p}{R}=-\rho q_{0}.

For a liquid-level system with two or more tanks, we apply the law of conservation of mass to each tank.