Consider a water tank with a hole of area a in its base. The tank will drain thr
ID: 2859955 • Letter: C
Question
Consider a water tank with a hole of area a in its base. The tank will drain through this hole, at a rate proportional to the height of the water (as this influences the pressure). Torricelli's Law gives a basic model that relates the volume of water to the height of the water: dV/dt = -a 2gh(t), where: V(t) = the volume of water in the tank at time t in seconds, and in m^3, h(t) = the height of water in the tank at time t in m, a = the area of the hole in the bottom of the tank in m^2, and g = 9.81 m/s^2, acceleration due to gravity. Note 1 m^3 = 1000 L c) If we add water to our tank at a rate of Q(t), then our equation becomes dV/dt =Q(t) - a 2gh(t). i) (*) As we do not have a technique for solving this form of equation exactly, linearise our equation in h(t) around our initial height (corresponding to 5000L). Note that you do not need to linearise Q(t). ii) (**) Solve the linearised ODE for Q(t) = 0 using the techniques you have been taught. Plot your solution with the solution you obtained in part (b) as a single figure. Where do the two solutions begin to differ significantly?) (***) Let Q(t) = 1 - e^-t. Substitute this into the linearised ODE from ii) and solve the resulting linearised ODE.Explanation / Answer
a