Question

A liquid surge tank has \(F_{in}\) and \(F_{out}\) as the inlet and outlet flow rates respectively, as shown in the figure. \(F_{out}\) is proportional to the square root of the liquid level \(h\). The cross-sectional area of the tank is \(20~\text{cm}^2\). Density of the liquid is constant everywhere in the system. At steady state, \(F_{in}=F_{out}=10~\text{cm}^3\text{s}^{-1}\) and \(h=16~\text{cm}\). The variation of \(h\) with \(F_{in}\) is approximated as a first-order transfer function. Which one of the following is the CORRECT value of the time constant (in seconds) of this system?

• A. 20
• B. 32
• C. 64
• D. 128

Hint:

For tanks with outflow \(F_{out}(h)\), the linearized first-order time constant is \(\tau=A\big/\left(\dfrac{dF_{out}}{dh}\big|_{ss}\right)\).
Orifice-type outlets have \(F_{out}\propto \sqrt{h}\Rightarrow \dfrac{dF_{out}}{dh}=\dfrac{k}{2\sqrt{h}}\).
Use steady-state data to estimate \(k\) before linearization.

Answer 1

The Correct Option is C

Step 1: Dynamic balance on the tank (cross-sectional area \(A=20\)): \[ A\frac{dh}{dt}=F_{in}-F_{out},\qquad F_{out}=k\sqrt{h}. \] Use steady state \(F_{in}=F_{out}=10\) at \(h=16\) to find \(k\): \[ 10=k\sqrt{16}=4k \;\Rightarrow\; k=2.5~\frac{\text{cm}^3}{\text{s}\,\sqrt{\text{cm}}}. \] Step 2: Linearize about steady state. Let deviations be \(h'=h-h_{ss}\), \(f'_{in}=F_{in}-F_{in,ss}\). \[ A\frac{dh'}{dt}=f'_{in}-\left.\frac{dF_{out}}{dh}\right|_{ss} h', \qquad \left.\frac{dF_{out}}{dh}\right|_{ss}=\frac{k}{2\sqrt{h_{ss}}}=\frac{2.5}{2\cdot4}=0.3125~\frac{\text{cm}^3}{\text{s}\,\text{cm}}. \] Step 3: Put into standard first-order form \(A\frac{dh'}{dt}+\left.\dfrac{dF_{out}}{dh}\right|_{ss} h' = f'_{in}\). Hence the time constant is \[ \tau=\frac{A}{\left.\dfrac{dF_{out}}{dh}\right|_{ss}}=\frac{20}{0.3125}=64~\text{s}. \] \(\boxed{\tau=64~\text{s}}\).