This problem corresponds to problem 1-11 in Applied Numerical Methods by Chapra

Question

This problem corresponds to problem 1-11 in Applied Numerical Methods by Chapra

A conical storage tank contains water at depth y. Water flows in at a sinusoidal rate of Q in = 3 sin2(t) and flows out according to Q out = 3(y - y out)1.5 y > y out Q out = 0 y le y out where flows has units of m3 / day and y = the elevation of the water surface above the bottom of the tank (m). Applying conservation of volume of water for the flow, dV/dt = 3 sin2(t) - 3(y - y out)1.5 Note that the volume of a cone is given by V = pir2y/3 and with the side slope s = y top/r top and r = y/s, the volume of water in the cone can be related to depth y as V = pi/3s2 y3 y = 3 3s2/piV Substitute Equation (3) into Equation (1), it becomes dV/dt = 3 sin2(t) - 3(3 3s2/piV - y out)1.5 For the case where the water level is below the outlet, outflow is zero and then it reduces to dV/dt = 3sin2(t) For the parameters r top = 2.5 m, y top = 4 m and y out = 1 m and assume that the water level is initially below the outlet pipe with y(0) = 0.7 m,

Explanation / Answer

clc
clear all
hold on
ytop=4;
rtop=2.5;
s=ytop/rtop;
yout=1;
f=@(V,t)(3*sin(t)^2-3*((3*s^2*V/pi)^1/3-yout))^1.5;
g=@(t)3*sin(t)^2;
dt=[1 0.5 0.1 0.05];
v(1)=0.7;
line_color=['r','g','b','m'];
for i=1:4
t=0:dt(i):10;
for n=1:(size(t,2)-1)
if (v(n) > (pi/(3*s^2)))
v(n+1)=v(n)+dt(i)*f(v(n),t(n));
else
v(n+1)=v(n)+dt(i)*g(t(n));
end
end
y=(3*s^2/pi .*v).^1/3;
plot(t,y,line_color(i))
clear y
end
xlabel('Time');
ylabel('y');
legend('dt=1','dt=0.5','dt=0.1','dt=0.05')

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