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+//Example 4_16
+clc;funcprot(0);
+//Given data
+P=5400;// kW
+N=200;// r.p.m
+D=3;// m
+H=240;// m
+n_t=0.82;
+rho=1000;// kg/m^3
+g=9.81;// m/s^2
+
+//Calculation
+//(a)
+Q=(P*1000)/(rho*g*H*n_t);// m^3/sec
+N_u=(N*D/sqrt(H));// Unit speed
+P_u=(P/(D^2*H^(3/2)));// Unit power
+Q_u=(Q/(D^2*(sqrt(H))));// Unit flow
+N_s=(2*%pi*N*sqrt(Q)*sqrt(n_t))/(60*(g*H)^(3/4));// Specific speed
+printf('\n(a)The flow rate,Q=%0.1f m^3/sec \n The unit speed,N_u=%0.1f \n The unit power,P_u=%0.3f \n The unit flow,Q_u=%0.2f \n The specific speed,N_s=%0.3f',Q,N_u,P_u,Q_u,N_s);
+//(b)
+// When the head is changed to 160 m,the diameter remains same.
+H=160;// m
+N=(N_u*sqrt(H))/D;// rpm
+P_1=(P_u*D^2*H^(3/2));// kW
+Q=(Q_u*D^2*sqrt(H));// m^3/sec
+printf('\n(b)Speed,N=%0.0f r.p.m\n Power,P=%0.0f kW\n The flow rate,Q=%0.2f m^3/sec ',N,P_1,Q);
+//(c)
+H=183;// m
+P=2850;// kW
+D_1=sqrt((P/(P_u*H^(3/2))));// m
+N_1=(N_u*sqrt(H))/(D_1);// r.p.m
+Q=Q_u*D_1^2*sqrt(H);// m^3/sec
+printf('\n(c)Diameter,D_1=%0.2f m\n Speed,N=%0.0f r.p.m\n The flow rate,Q=%0.2f m^3/sec ',D_1,N_1,Q);
+// The answer vary due to round off error