Question 4.2: Modeling a fluid-mechanical system A cylindrical water tank ......
Modeling a fluid-mechanical system
A cylindrical water tank has cross-sectional area A_1 and water level h_1(t) and is fed by an input volumetric flow of water f_1(t) with an orifice at height h_2 whose effective cross-sectional area is A_2, through which flows the output volumetric flow f_2(t) (Figure 4.5). Write a differential equation for the water level as a function of time and graph the water level versus time for a tank that is initially empty, under different assumptions of inflow.
Under certain simplifying assumptions, the velocity of the water flowing out of the orifice is given by Toricelli’s equation,
v_2 (t) = \sqrt{2g[h_1(t)-h_2 ]}where g is the acceleration due to earth’s gravity (9.8 m/s²). The rate of change of the volume A_1 h_1(t) of water in the tank is the volumetric inflow rate minus the volumetric outflow rate
\frac{d}{dt}(A_1 h_1(t)) = f_1(t) − f_2 (t)and the volumetric outflow rate is the product of the effective area A_2 of the orifice and the output flow velocity f_2(t) = A_2 v_2(t). Combining equations we can write one differential equation for the water level
A_1 \frac{d}{d t}\left(\mathrm{~h}_1(t)\right)+A_2 \sqrt{2 g\left[\mathrm{~h}_1(t)-h_2\right]}=\mathrm{f}_1(t). (4.1)
The water level in the tank is graphed in Figure 4.6 versus time for four constant volumetric inflows under the assumption that the tank is initially empty. As the water flows in, the water level increases and the increase of water level increases the water outflow. The water level rises until the outflow equals the inflow and after that time the water level stays constant. As first stated in Chapter 1, when the inflow is increased by a factor of two, the final water level is increased by a factor of four, a result of the fact that the differential equation (4.1) is nonlinear. A method of finding the solution to this differential equation will be presented later in this chapter.
