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+//Example 4_19
+clc;funcprot(0);
+//Given data
+P_1=36000;// kW
+P_2=27000;// kW
+N_1=81.8;// r.p.m
+H_1=13;// m
+H_2=11;// m
+D_1=7.82;// m
+
+//Calculation
+//As the specific speeds are the same,using the definition of specific speed in terms of power,
+N_2=((N_1*sqrt(P_1)/(H_1^(5/4)))*((H_2^(5/4))/sqrt(P_2)));// rpm
+// As the unit speeds are same,
+D_2=(D_1*N_1*sqrt(H_2))/(sqrt(H_1)*N_2);// m
+// As the unit flow is same,Q=Q_2/Q_1
+Q=(D_2^2*H_2^(1/2))/((D_1^2*H_1^(1/2)));
+// By solving Q, it gives the relation,Q_2=0.886*Q_1;
+Q_r=(1-Q)*100;
+printf('\n Speed,N_2=%0.1f rpm \n Diameter,D_2=%0.2f m \n There is a reduction in flow by about %0.2f percentage.',N_2,D_2,Q_r);
+// The answer vary due to round off error