Integration 2. Nu Consider an industrial tank in the shape of an inverted cone.

Question

Integration 2. Nu Consider an industrial tank in the shape of an inverted cone. The radius of the tank at the top rim is 3 m, and the total height of the tank is 4 m. The volume of the tank in m3 is given by: V- (1/3)pi R2 H The volume of liquid in the tank when filled to a height h measured from the bottom vertex is: The Lab will consist of a single script, divided in two parts. In each part, the filling schedule will be different. A filling schedule is a function that provides flow rate, in m3/h, as a function of time. In Part I, your script will calculate the level of the liquid, h, after a two-hour filling schedule is completed. The filling schedule for Part I. Schedule I, is as follows During the first 30 minutes, the flow rate will increase linearly from zero to 10 m3/H . During the following 60 minutes, the flow rate will stay constant at 10 m3/h . During the last 30 minutes, the flow rate will decrease linearly from 10 m3/ h down to zero In Part II, your script will calculate the time it takes to completely fill the tank with a different filling schedule. Schedule Il. given by the equation: Flow Rate (m /h)-10 (1 e2t) m3/h where t is time in hours, and the exponent, 2t, is dimensionless

Explanation / Answer

clc;
clear;
R=3; %radius of tank
H=4; %tank height
% part I
q1=[20 0];    %polynomial representation for time 0-30min
q2=[20];      %polynomial representation for time 30-90min
q3=[-20 0];   %polynomial representation for time 90-120min
lm_1=[0 0.5]; %time for first schedule in hours
lm_2=[0.5 1.5]; %time for second schedule in hours
lm_3=[1.5 2]; %time for third schedule in hours
P_1=polyint(q1);
P_2=polyint(q2);
P_3=polyint(q3);
V_1=(polyval(P_1,lm_1(2))-polyval(P_1,lm_1(1)))+(polyval(P_2,lm_2(2))-polyval(P_2,lm_2(1)))+(polyval(P_3,lm_3(2))-polyval(P_3,lm_3(1)));
h=(3*V_1/(pi*(R/H)^2))^(1/3);
fprintf('The height of the liquid after Schedule I is %.3f meters ',h);

% part II
V=1/3*pi*R^2*H;   %total volume of tank
er=1;    %initial error for starting loop
a=0;     %initial limit
b=0.01; %final limit for iteration 1
Q =@(t) 10*(1-exp(-2*t)); %flow rate function
it=0;   %iteration count
V_t=0; %volume initial
while er>0.1;
    V_t=V_t+quad(Q,a,b);   %volume filled
    er=abs(V-V_t); %error
    t=b;       %new time
    a=b;       %new limits
    b=b+0.01; %new limits
    it=it+1;   %iteration number
end
fprintf('The time required to fill up the tank with Schedule II is %.3f hours. ',t);

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