Question

For the cone water tank given in the figure, the relation between the flow rate \(q(t)\) and the water level \(h(t)\) is given as \(q(t)=0.005 h^{2}(t) \cdot q(t)\) is in \(\left[m^{3} / s\right]\) and \(h(t)\) in [m].

a. When \(t=0\), the valve is closed and the water level is \(2[\mathrm{~m}]\). Find the level of the water after 60 seconds of the valve opening. 

b. Find the linearized expression for water flow rate \(q(t)\) about initial conditions. 

Answer 1

For the cone water tank given in the figure, the relation between the flow rate q(t) and the water level h(t) is given as ?(?) = 0.005ℎ 2 (?). q(t) is in [m 3 /s] and h(t) in [m]. a. When t=0, the valve is closed, and the water level is 2 [m].

q(t) =0.005h2(t)

from the figure

\(\frac{s}{h}=\frac{1}{3} \Rightarrow s=h / 3\)

The capacitance of the system

\(C=\pi l^{2}=\pi(h / 3)^{2}\)

\(C=\pi h^{2} / 9\)

The relation between capacitance and head

\(C=\frac{d h}{d t}=-q(t)\)

\(\frac{\pi h^{2}}{9} \frac{d h}{d t}=-0.005 h^{2} \quad \frac{d h}{d t}=-\frac{-(0.005) \times 9}{\pi}=\)

\(\frac{d h}{d t}=-0.0143\)

Integrating

\(\int_{2}^{h} d h=-0.0143 \int_{0}^{60} d t\)

\((h-2)=-0.0143 \times 60=-0.858\)

\(h-2=-0.858=1.142\)

\(h=1.142 m\) at \(t=60\)

b)

Let \(\bar{H}\) =steady stste head before any change

h=small deviation of head from its steady state

\(\bar{Q}\) =steady state flow rate

small deviation of outflow rate from its steady state

Tank resistance

The rate of change in liquid stored in the tank in equal to flow in -flow out

$$ \begin{array}{l} q_{i n}=0 \\ e \frac{d h}{d t}=q_{i n}-q_{o u t}=-q_{o u t} \Rightarrow \frac{d h}{d t}=-q_{o u t} \\ \qquad_{\text {Resistance }} R=\frac{d H}{d Q}=\frac{h}{q_{0}} \Rightarrow q_{0}=\frac{h}{R} \end{array} $$

\(e \frac{d H}{d t}=\frac{-h}{R} \Rightarrow R_{C} \frac{d h}{d t}+h=0\) linearized mathematical model of system when 'h' is output

when \(\mathrm{q}_{0}\) is considered output then, \(\mathrm{h}=\mathrm{Rq}_{0}\)

\(R C \frac{d q_{0}}{d t}+q_{0}=0\) or \(R C \frac{d q(t)}{d t}+q(t)=0\)