Modelling and Analytic Solution

In class we modelled the rate at which a Big Gulp empties through a hole of area $a$ punched in its bottom. If $h$ is the height of the fluid above the hole, the pressure at the whole caused by the weight of the fluid bearing down, is

$$P=\rho g h,$$

where $\rho$ is the density of the fluid and $g$ is the acceleration due to gravty (so that $\rho h$ is the mass of fluid per area). Bernoulli's equation gives that the (ideal) speed, $u$, to which a pressure differential accelerates a fluid satisfies

$$P = \frac12 \rho u^2, \quad \mbox{and using the pressure/height relationship} \quad u = \sqrt{2 g h}.$$

Consequently, the volumetric flow ratethrough a hole of area $a$ is

$$\mbox{Volumetric Flow Rate } = a u = a \sqrt{2 g h}.$$

This gives a relationship between how volume changes and the height of the fluid above the hole,

$$\frac{dV}{dt} = - a \sqrt{2 g h}.$$

For regularly-shaped containers (like Big Gulps), the change in volume, $\Delta V$, over a small increment in time, $\Delta t$ can be approximated as $\Delta V = \Delta (A(h) h)$, where $A(h)$ is the area of the fluid at the top of the fluid (currently at height $h$). Thus,

$$\frac{dV}{dt} = \frac{d}{dt} \left[A(h) h \right] = \underbr... ...A(h) h \right]=- a \sqrt{2 g h}}_{\mbox{An ODE for $h(t)$!}} .$$

We now arrive at a (relatively) simple differential equation for the height of the fluid in the Big Gulp (neglecting ice):

$$\frac{dh}{dt} = - a \sqrt{2 g } \frac{\sqrt{h}}{\frac{d}{dh}\left[A(h) h \right]}.$$

The last part of the modelling for the Big Gulp is to determine how surface area at the top of the fluid varies with the height of the fluid. As indicated in the figure, the radius of a Gulp cross section varies linearly with height according to

$$r = r_b + \frac{r_T-r_b}{H} h,$$

where $r_T$ and $r_b$ are the radius at the top (at least of the fluid in) and bottom of the Gulp container and $H$ is the initial height of the fluid. Assuming that we have not sat upon or otherwise deformed the container the cross sections are circular, and thus

$$A(h) = \pi r^2 = \pi \left[ r_b + \frac{r_T-r_b}{H} h \right... ...r_b \frac{r_T-r_b}{H} h + \frac{(r_T-r_b)^2}{H^2} h^2 \right].$$

We need the derivative of $h A(h)$ to proceed.

$$h A(h) = \pi \left[ r_b^2 \ h + 2 r_b \frac{r_T-r_b}{H} h^2 + \frac{(r_T-r_b)^2}{H^2} h^3 \right],$$

so

$$\frac{d}{dh}\left[h A(h)\right]= \pi \left[ r_b^2 + 4 r_b \frac{r_T-r_b}{H} h + 3 \frac{(r_T-r_b)^2}{H^2} h^2 \right],$$

Thus,

$$\frac{dh}{dt} = - a \sqrt{2 g } \frac{\sqrt{h}}{\pi \left[ r... ... \frac{r_T-r_b}{H} h + 3 \frac{(r_T-r_b)^2}{H^2} h^2 \right]}.$$

Separating variables we arrive at

$$- a \sqrt{2 g } \ dt = \frac{\pi \left[ r_b^2 + 4 r_b \frac{... ...\frac12} + 3 \frac{(r_T-r_b)^2}{H^2} h^{\frac32} \right] \ dh.$$

Integrating both side gives

$$\begin{eqnarray*} C- a \sqrt{2 g } t &= & \pi \left[ r_b^2 \frac{h^{\frac12}}{\f... ...frac32} + \frac65 \frac{(r_T-r_b)^2}{H^2} \ h^{\frac52} \right]. \end{eqnarray*}$$

And actually, this is about as far as we can go analytically. The next step would normally be to solve for $h$ as a function of $t$, but given the expression for $h$ on the right hand side, we are out of luck. It is possible to determine the free constant of integration, $C$. Noting that at $t=0$ $h=H$ we can substitute these values into the equation above and get

$$\begin{eqnarray*} C & = & \pi \left[ 2 r_b^2 \ H^{\frac12} + \frac83 r_b \frac... ...rac{4}{3} r_b^2 + \frac{2}{5} r_T \ r_b + 3 r_T^2 \right]. \\ \end{eqnarray*}$$

Fortunately, this is enough to predict the time at which the Big Gulp should empty (at least according to the Torricelli model). Since the height of the fluid will be $h=0$ when $t=t_{\mbox{\tiny empty}}$, the right hand side of our integrated solution will be zero, giving

$$C-a\sqrt{2 g} t_{\mbox{\tiny empty}} = 0, \quad\mbox{ or }\q... ...\frac{4}{3} r_b^2 + \frac{2}{5} r_T \ r_b + 3 r_T^2 \right],$$

using $C$ as given above.

	EXERCISE 6: Even though we can't find an analytic form for $h(t)$, we can plot it! Using $H=16$cm, $r_b=3.5$cm, $r_T$=10cm and a small hole diameter of 4mm, solve for $t$ as a function of $h$. Using MATLAB pick an evenly spaced vector of $h$ values between 0 and $H$ using linspace, and calculate the corresponding values for $t$. Use plot(t,h) to plot these and verify the emptying time expression above.