Linearization of Water Tank Height and Outflow
Find the nonlinear differential equation describing the height of the water in the tank in Fig. 2.36. Assume that there is a relatively short restriction at the outlet and that $α = 2.$ Also linearize your equation about the operating point $h_0.$

Question

Linearization of Water Tank Height and Outflow
Find the nonlinear differential equation describing the height of the water in the tank in Fig. 2.36. Assume that there is a relatively short restriction at the outlet and that [latex] α = 2. [/latex] Also linearize your equation about the operating point [latex] h_0.[/latex]

Accepted Answer

Applying Eq. (2.72) yields the flow out of the tank as a function of the height of the water in the tank:

[latex] w_{out} = \frac{1}{R} (p_1 - p_a ) ^{1/2} . [/latex] (2.74)

Here,

[latex] p_1 = ρgh + p_a, [/latex] the hydrostatic pressure,
[latex] p_a = [/latex] ambient pressure outside the restriction.

[latex] w = \frac{1}{R} (p_1 - p_2)^{1/ \alpha } , [/latex] (2.72)

where

[latex] w = [/latex] mass flow rate,
[latex] p_1, p_2 = [/latex] pressures at ends of the path through which flow is occurring,
[latex] R, α = [/latex] constants whose values depend on the type of restriction.

Substituting Eq. (2.74) into Eq. (2.70) yields the nonlinear differential equation for the height:

[latex] \dot{h} = \frac{1}{A ho} \left( w_{in} - \frac{1}{R} \sqrt{p_1 - p_a} ight). [/latex] (2.75)

[latex] \dot{h} = \frac{1}{A ho} (w_{in} - w_ {out} ) , [/latex] (2.70)

where

[latex] A = [/latex] area of the tank,
[latex] ρ = [/latex] density of water,
[latex] h = m/Aρ = [/latex] height of water,
[latex] m = [/latex] mass of water in the tank.

Linearization involves selecting the operating point [latex] p_0 = ρgh_0 + p_a [/latex] and substituting [latex] p_1 = p_0 + Δp [/latex] into Eq. (2.74). Then we expand the nonlinear term according to the relation

[latex] (1 + ε)^β ≅ 1 + β ε , [/latex] (2.76)

where [latex] ε \ll 1. [/latex] Equation (2.74) can thus be written as

[latex] w_{out} = \frac{\sqrt{p_0 - p_a}}{R} \left( 1+ \frac{Δp}{p_0 - p_a } ight)^{1/2} \ \ \ \ \ \ \ \ ≅ \frac{\sqrt{p_0 - p_a}}{R} \left( 1+ \frac{1}{2} \frac{Δp}{p_0 - p_a } ight)^{1/2}.[/latex]

The linearizing approximation made in Eq. (2.77) is valid as long as [latex] Δp \ll p_0 − p_a ;[/latex] that is, as long as the deviations of the system pressure from the chosen operating point are relatively small.
Combining Eqs. (2.70) and (2.77) yields the following linearized equation of motion for the water tank level:

[latex] \Delta \dot{h} = \frac{1}{A ho } \left[w_{in} - \frac{\sqrt{p_0 - p_a}}{R} \left(1 + \frac{1}{2} \frac{\Delta p}{p_0 - p_a} ight) ight] . [/latex]

Because [latex] Δp = ρgΔh, [/latex] this equation reduces to

[latex] \Delta \dot{h} = - \frac{g}{2AR\sqrt{p_0 - p_a}} \Delta h + \frac{w_{in}}{A ho} - \frac{\sqrt{p_0 - p_a}}{ ho A R } , [/latex]

which is a linear differential equation for [latex] Δ\dot{h}. [/latex] The operating point is not an equilibrium point because some control input is required to maintain it. In other words, when the system is at the operating point [latex] (Δh = 0) [/latex] with no input [latex] (w_{in} = 0), [/latex] it will move from that point because [latex] Δ\dot{h} ≠ 0. [/latex] So if no water is flowing into the tank, the tank will drain, thus moving it from the reference point. To define an operating point that is also an equilibrium point, we need to require that there be a nominal flow rate,

[latex] \frac{w_{in_0}}{A ho} = \frac{\sqrt{p_0 - p_a}}{ ho A R } ,[/latex]

and define the linearized input flow to be a perturbation from that value.

Question 2.EX.18: Linearization of Water Tank Height and Outflow Find the nonl...

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