A water tank of radius $r_1 = 1.75m$ with a vertical slit of width W and an outlet hole of radius $r_p = 0.05m$ placed at
a height $h_p = 0.3m$ is mounted in an elevator moving up and down causing a time dependent acceleration $g(t)$, see
Fig. 1. The slit width W can take any value between 0m and 0.04m.
Figure 1: Water tank inside an elevator
The height of water $h(t)$ in the tank can be modeled by the following ODE,
$\frac{dh}{dt} = \frac{f(t) - \rho \sqrt{g(t)} \left(W \sqrt[3]{h^3(t)} + \pi r_p^2 \sqrt{2} max(0, h(t) - h_p)\right)}{\rho \pi r_1^2}$,
where $\rho = 1000 kg/m^3$ and
$f(t) = c \left[1 + sin \left(2\pi f_1 t + cos(2\pi f_2 t)\right)\right]$.
models the inflow into the tank with $c = 37 kg/s$, $f_1 = 0.027Hz$, and $f_2 = 0.014Hz$. Note the cos is within the sin.
The mass flow rate $m(t)$ of water out of the tank is
$\dot{m}(t) = \rho \sqrt{g(t)} \left(W \sqrt[3]{h^3(t)} + \pi r_p^2 \sqrt{2} max(0, h(t) - h_p)\right)$.
To determine the non-periodic time dependent elevator acceleration $g(t) = \frac{dv}{dt}$, measurements of the elevator's
velocity $v(t)$ in m/s are taken every $\Delta t = 25s$ starting at $t = 0s$. The measured velocity data is available on Canvas
in the Matlab file vdat.mat. At $t = 0s$, the height of water in the tank is $h(t = 0s) = 0.4m$.
Task:
Find the slit width W such that the water mass leaving the tank from $0s \le t \le 2200s$ is $m = 80,000kg \pm 0.1kg$. For
this W, graph $\dot{m}(t)$ from $0s \le t \le 2200s$ and graph the power spectrum of $\dot{m}(t)$ for frequencies $f \le 0.1Hz$ using a
linear scale for frequency and logarithmic scale for power.
Required submission:
$\Box$ Concise report containing description of solution procedure, name of methods used, W, m and its accuracy; requested 2 graphs,
and justification/documentation that results have the required accuracy. Answers in code are insufficient
$\Box$ printout of any new script/functions not pan of prior homework/exam solutions as an appendix of the report;
