A Two-Tank Liquid-Level System Figure 7.12 shows a liquid-level system, in which two tanks have cross-sectional areas A1 and A2, respectively. A pump is connected to the bottom of tank 1 through a valve of linear resistance R1. The liquid flows from tank 1 to tank 2 through a valve of linear

Question

A Two-Tank Liquid-Level System

Figure 7.12 shows a liquid-level system, in which two tanks have cross-sectional areas [latex]A_{1}[/latex] and [latex]A_{2}[/latex], respectively. A pump is connected to the bottom of tank 1 through a valve of linear resistance [latex]R_{1}[/latex]. The liquid flows from tank 1 to tank 2 through a valve of linear resistance [latex]R_{2}[/latex], and leaves tank 2 through a valve of linear resistance [latex]R_{3}[/latex]. The density [latex]\rho[/latex] of the liquid is constant.

a. Derive the differential equations in terms of the liquid heights [latex]h_{1}[/latex] and [latex]h_{2}[/latex]. Write the equations in second-order matrix form.

b. Assume the pump pressure [latex]\Delta p[/latex] is the input and the liquid heights [latex]h_{1}[/latex] and [latex]h_{2}[/latex] are the outputs. Determine the state-space form of the system.

Accepted Answer

a. Applying the law of conservation of mass to tank 1 gives

[latex]\frac{\mathrm{d} m}{\mathrm{~d} t}=q_{\mathrm{mi}}-q_{\mathrm{mo}}.[/latex]

where

[latex]\frac{\mathrm{d} m}{\mathrm{~d} t}= ho A_{1} \frac{\mathrm{d} h_{1}}{\mathrm{~d} t}.[/latex]

The mass flow rates entering and leaving tank 1 can be written as

[latex]q_{\mathrm{mi}}=\frac{\left(p_{\mathrm{a}}+p ight)-\left(p_{\mathrm{a}}+ ho g h_{1} ight)}{R_{1}}=\frac{p- ho g h_{1}}{R_{1}}[/latex]

and

[latex]q_{\mathrm{mo}}=\frac{\left(p_{\mathrm{a}}+ ho g h_{1} ight)-p_{\mathrm{a}}}{R_{2}}=\frac{ ho g h_{1}}{R_{2}}[/latex]

Substituting these expressions results in

[latex] ho A_{1} \frac{\mathrm{d} h_{1}}{\mathrm{~d} t}=\frac{p- ho g h_{1}}{R_{1}}-\frac{ ho g h_{1}}{R_{2}},[/latex]

which can be rearranged as

[latex] ho A_{1} \frac{\mathrm{d} h_{1}}{\mathrm{~d} t}+ ho g h_{1}\left(\frac{1}{R_{1}}+\frac{1}{R_{2}} ight)=\frac{p}{R_{1}} .[/latex]

Applying the law of conservation of mass to tank 2 gives

[latex]\frac{\mathrm{d} m}{\mathrm{~d} t}=q_{\mathrm{mi}}-q_{\mathrm{mo}},[/latex]

where

[latex]\frac{\mathrm{d} m}{\mathrm{~d} t}= ho A_{2} \frac{\mathrm{d} h_{2}}{\mathrm{~d} t}.[/latex]

The mass flow rates entering and leaving tank 2 can be written as

[latex]q_{\mathrm{mi}}=\frac{\left(p_{\mathrm{a}}+ ho g h_{1} ight)-p_{\mathrm{a}}}{R_{2}}=\frac{ ho g h_{1}}{R_{2}}[/latex]

and

[latex]q_{\mathrm{mo}}=\frac{\left(p_{\mathrm{a}}+ ho g h_{2} ight)-p_{\mathrm{a}}}{R_{3}}=\frac{ ho g h_{2}}{R_{3}} .[/latex]

Substituting these expressions results in

[latex] ho A_{2} \frac{\mathrm{d} h_{2}}{\mathrm{~d} t}=\frac{ ho g h_{1}}{R_{2}}-\frac{ ho g h_{2}}{R_{3}}.[/latex]

which can be rearranged as

[latex] ho A_{2} \frac{\mathrm{d} h_{2}}{\mathrm{~d} t}-\frac{ ho g h_{1}}{R_{2}}+\frac{ ho g h_{2}}{R_{3}}=0.[/latex]

The system of differential equations in second-order matrix form is found to be

[latex] \left[\begin{array}{cc} ho A_{1} & 0 \ 0 & ho A_{2} \end{array} ight]\left\{\begin{array}{c} \frac{\mathrm{d} h_{1}}{\mathrm{~d} t} \ \frac{\mathrm{d} h_{2}}{\mathrm{~d} t} \end{array} ight\}+\left[\begin{array}{cc} \frac{ ho g}{R_{1}}+\frac{ ho g}{R_{2}} & 0 \ -\frac{ ho g}{R_{2}} & \frac{ ho g}{R_{3}} \end{array} ight]\left\{\begin{array}{l} h_{1} \ h_{2} \end{array} ight\}=\left\{\begin{array}{c} \frac{p}{R_{1}} \ 0 \end{array} ight\} . [/latex]

b. As specified, the state, the input, and the output are

[latex] \mathbf{x}=\left\{\begin{array}{l} x_{1} \ x_{2} \end{array} ight\}=\left\{\begin{array}{l} h_{1} \ h_{2} \end{array} ight\}, \quad u=p, \quad \mathbf{y}=\left\{\begin{array}{l} h_{1} \ h_{2} \end{array} ight\} . [/latex]

The state-variable equations are

[latex] \begin{aligned} & \dot{x}_{1}=\frac{\mathrm{d} h_{1}}{\mathrm{~d} t}=-\frac{g}{A_{1}}\left(\frac{1}{R_{1}}+\frac{1}{R_{2}} ight) h_{1}+\frac{1}{ ho A_{1} R_{1}} p, \ & \dot{x}_{2}=\frac{\mathrm{d} h_{2}}{\mathrm{~d} t}=\frac{g}{A_{2} R_{2}} h_{1}-\frac{g}{A_{2} R_{3}} h_{2}, \end{aligned} [/latex]

or in matrix form

[latex] \left\{\begin{array}{l} \dot{x}_{1} \ \dot{x}_{2} \end{array} ight\}=\left[\begin{array}{cc} -\frac{g}{A_{1}}\left(\frac{1}{R_{1}}+\frac{1}{R_{2}} ight) & 0 \ \frac{g}{A_{2} R_{2}} & -\frac{g}{A_{2} R_{3}} \end{array} ight]\left\{\begin{array}{c} x_{1} \ x_{2} \end{array} ight\}+\left[\begin{array}{c} \frac{1}{ ho A_{1} R_{1}} \ 0 \end{array} ight] u . [/latex]

The output equation is

[latex] \left\{\begin{array}{l} y_{1} \ y_{2} \end{array} ight\}=\left[\begin{array}{ll} 1 & 0 \ 0 & 1 \end{array} ight]\left\{\begin{array}{l} x_{1} \ x_{2} \end{array} ight\}+\left[\begin{array}{l} 0 \ 0 \end{array} ight] u .[/latex]

Question 7.5: A Two-Tank Liquid-Level System Figure 7.12 shows a liquid-le...

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