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+// 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