// Example 2_13
clc;funcprot(0);
//Given data
m=[1 2 3 4 5 6 7 8 9 10 11 12];// Month
D=[500 200 1500 2500 3000 2400 2000 1500 1500 1000 800 600];// Discharge in millions of m^3 per month
H=80;// Available head in m
n_o=80/100;// Overall efficiency of the generation 
g=9.81;// The acceleration due to gravity in m/s^2

// Calculation
// (a)
Q_a1=(D(1)+D(2)+D(3)+D(4)+D(5)+D(6)+D(7)+D(8)+D(9)+D(10)+D(11)+D(12))/12;// The average monthly flow in millions of m^3/month
m_1=[0 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12];
D_1=[500 500 200 200 1500 1500 2500 2500 3000 3000 2400 2400 2000 2000 1500 1500 1500 1500 1000 1000 800 800 600 600 3200];
Q_a=[Q_a1,Q_a1];
m=[0,12];
xlabel('Month');
ylabel('Discharge in millions of m^3 per month');
subplot(2,1,1);
plot(m_1',D_1','b',m',Q_a','r-');
a=gca();
a.x_ticks.labels=["0","J","F","M","A","M","J","J","A","S","O","N","D"];
a.x_ticks.locations=[0;1;2;3;4;5;6;7;8;9;10;11;12];
legend('Hydrograph','Mean flow');
D=[200 500 600 800 1000 1500 2000 2400 2500 3000];
M=[12 11 10 9 8 7 4 3 2 1];// Total number of months during which flow is available
for(i=1:10)
    T(i)=(M(i)/12)*100;
end
subplot(2,1,2);
xlabel('Percentage of time');
ylabel('Discharge in millions of cu.m.month');
plot(T,D','b');
legend('Flow duration curve');

m=(Q_a1*10^6/(30*24*3600));// The average flow available in m^3/sec
P=(((m*1000*g*H)/1000)*(n_o/1000));// Average kW available in MW
printf('\nAverage kW available at the site=%0.3f MW',P);
// The answer provided in the textbook is wrong