A liquid storage facility can be modeled by ̈y + 5̇y + y(t) = 8̇u + u(t) where y is the liquid level (m) and u is an inlet flow rate (m^3/s). Both are defined as deviations from the nominal steady-state values. Thus, y = u = 0 at the nominal steady state. Also, the initial values of all the

Question

A liquid storage facility can be modeled by

[latex]\ddot{y}+5 \dot{y}+y(t)=8 \dot{u}+u(t)[/latex]

where y is the liquid level (m) and u is an inlet flow rate [latex]\left( m ^{3} / s ight)[/latex]. Both are defined as deviations from the nominal steady-state values. Thus, y = u = 0 at the nominal steady state. Also, the initial values of all the derivatives are zero.

(a) If u(t) suddenly changes from 0 to [latex]1 m ^{3} / s ext { at } t=0[/latex], determine the liquid level response, y(t).

(b) If the tank height is 2.5 m, will the tank overflow?

(c) Based on your results for (b), what is the maximum flow change, [latex]u_{\max }[/latex], that can occur without the tank overflowing?

(Hint: Consider the Principle of Superposition.)

Accepted Answer

(a) Take the Laplace transform of each term, taking into account that all initial conditions are zero:

[latex]s^{2} Y-s y(0)-y^{\prime}(0)+5 s Y-5 y(0)+Y=8 s U-8 u(0)+U[/latex]

[latex]s^{2} Y+5 s Y+Y=U(8 s+1)[/latex]

[latex]U(s)=\frac{1}{s}[/latex]

[latex]Y\left(s^{2}+5 s+1\right)=\frac{8 s+1}{s}[/latex]

[latex]Y=\frac{8 s+1}{s\left(s^{2}+5 s+1\right)}[/latex]

Now use symbolic mathematical software (ex. Mathematica) to solve for y(t). InverseLaplaceTransform [latex]\left[\left(8^{*} s +1\right) /\left( s ^{*}\left( s ^{\wedge} 2+5^{*} s +1\right)\right), s , t \right][/latex]

[latex]g \left[ t _{-}\right]:=\frac{1}{41}\left(-21 e^{-\frac{5+\sqrt{21}}{2} t}-11 \sqrt{21} e^{-\frac{5+\sqrt{21}}{2} t}-21 e^{\frac{\sqrt{21-5}}{2} t}+11 \sqrt{21} e^{\frac{\sqrt{21}-5}{2} t}+42\right)[/latex]

Plot[g[t],{t,0,100},AxesLabel [latex]\rightarrow[/latex]{time,Y},PlotRange [latex]\rightarrow[/latex] {{0,100},{0 ,2}}]

(b) Define the time when y(t) reaches its maximum as [latex]t_{\max }[/latex]. This time occurs when [latex]y^{\prime}(t)=0[/latex]. Solve for this time using Mathematica and find that [latex]t_{\max }=0.877[/latex] and [latex]y\left(t_{\max }\right)=1.558 .[/latex] Therefore, the tank will not overflow.

(c) Now find the general solution for any input step size, M (the solution is denoted in this case as [latex]Y_{ M }(s)[/latex] and [latex]y_{ M }(t)[/latex] for clarity). The input U(s) is M/s.

[latex]U(s)=\frac{M}{s}[/latex]

[latex]Y_{M}\left(s^{2}+5 s+1\right)=\frac{M(8 s+1)}{s}[/latex]

[latex]Y_{M}=\frac{M(8 s+1)}{s\left(s^{2}+5 s+1\right)}=M Y[/latex]

[latex]Y_{ M }[/latex] is the previous Y, multiplied by the size of the step, M. Since M is a constant, taking the inverse Laplace transform gives:

[latex]y_{M}(t)=M y(t)[/latex]

Now solve the equation:

[latex]y_{M}\left(t_{\max }\right)=2.5=M y\left(t_{\max }\right)=M(1.558)[/latex]

[latex]M=\frac{2.5}{1.558}=1.605[/latex]

The maximum step change in the flow rate into the tank that will not overflow the tank is 1.605.

Question 3.19: A liquid storage facility can be modeled by ̈y + 5̇y + y(t) ...

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