Consider a water tank with a hole of area n in its base. The tank will drain thr
ID: 1721449 • Letter: C
Question
Consider a water tank with a hole of area n 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 squareroot 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 Suppose we have a simple cylindrical tank with radius 1m, with a hole in the bottom of radius 5cm. By substituting in an expression for volume, convert Torricelli's Law into an ODE for h(t), i.e. dh/dt = By substitution into the equation, determine the value of A that will make h(t) = (At + B)^2 a solution to our ODE from a). Solve for B assuming that the tank has 5000L of water in it at time t = 0. Use this value to plot h(t) = (At + B)^2 in MATLAB. If we add water to our tank at a rate of Q(t), then our equation becomes dV/dt = Q(t) - a squareroot 2gh(t). 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 lineanse Q(t). Solve the linearised ODE for Q(t) = o 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> the volume of the cylinderical tank is :
V = pi*r^2h
r = 1m
=> dv/dt = pi*r^2dh/dt
we are given that dv/dt = -asqrt(2gh)
here h is a function of time t
=> -a*sqrt(2gh) = pi*r^2dh/dt
we are given that a = area of the circular hole = pi*r^2 = pi*(5/100)^2 = pi/400
=> -pi/400*sqrt(2*9.81h) = pi*(1)^2*dh/dt
dh/dt = -1/400*sqrt(19.62h)
b>
h(t) = (At+B)^2
=> dh/dt = 2(At+B)A
amd since h(t) is a solution of the above differential solution so it shall satisfy it.
=>
dh/dt = -1/400*sqrt(19.62h)
2(At+B)A = -1/400sqrt[19.62(At+B)^2]
=> A = -1/800*sqrt(19.62)
at t = 0
V = pi*r^2h = 5000
pi*r^2(At+B) = 5000
r = 1m
=> pi*(1)^2(0+B) =5000
=> B = 5000/pi