The Equation of Continuity — Question 6

Question 6

A conical tank rises quickly near its tip

An open conical tank has its vertex downward, total height $H$ and top radius $R$. It is filled at constant volume rate $Q$ from a level $h_1>0$ to $h_2$. There is no outflow, and the liquid is incompressible.

1. Derive $dh/dt$ in terms of current height $h$ and derive the filling time from $h_1$ to $h_2$.

2. Evaluate the time for $H=1.0\,\mathrm{m}$, $R=0.50\,\mathrm{m}$, $Q=1.0\times10^{-3}\,\mathrm{m^3/s}$, $h_1=0.20\,\mathrm{m}$ and $h_2=0.60\,\mathrm{m}$. Explain the formal behavior as $h\to0$.

Question 6 - Solution

A conical tank rises quickly near its tip

(a) Derivation and calculation

Similar triangles give $r(h)=Rh/H$, hence$$A(h)=\pi R^2h^2/H^2,\quad Q=\frac{dV}{dt}=A(h)\frac{dh}{dt}.$$$$\boxed{\frac{dh}{dt}=\frac{QH^2}{\pi R^2h^2}}.$$$$\Delta t=\frac{\pi R^2}{QH^2}\int_{h_1}^{h_2}h^2\,dh=\boxed{\frac{\pi R^2(h_2^3-h_1^3)}{3QH^2}}.$$

(b) Result and interpretation

$$\boxed{\Delta t=\frac{\pi(.500)^2(.600^3-.200^3)}{3(.00100)(1.00)^2}=54.4543\,\mathrm s}.$$$$h\to0^+:\quad A(h)\to0,\quad dh/dt\to\infty,\quad Q=\text{finite}.$$The divergence concerns height rate, not incoming volume rate; finite inlet geometry and wetting limit the tip idealization.

Physical check

The same time follows from $[V(h_2)-V(h_1)]/Q$ with $V(h)=\pi R^2h^3/(3H^2)$.