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+//Example 8_2
+clc;clear;funcprot(0);
+// Given values
+rho=62.42;//lbm/ft^3
+mu=1.038*10^-3;// lbm/ft.s
+D=0.01;// ft
+L=30;// ft
+V_avg=3;// ft/s
+g=32.2;// Ft/s^2
+
+//Calculation
+//(a)
+Re=(rho*V_avg*D)/mu;// Reynolds number
+f=64/Re;// Friction factor
+h_l=f*(L/D)*((V_avg^2)/(2*g));// ft
+printf('(a)The head loss,h_l =%0.1f ft\n',h_l);
+//(b)
+gradP_l=(f*(L/D)*rho*(V_avg^2/2))/32.2;// lbf/ft^2
+gradP_1=(gradP_l/144);// psi
+printf('(b)The pressure drop,gradP_l=%0.0f lbf/ft^2=%0.2f psi\n',gradP_1*144,gradP_1);
+//(c)
+A_c=(%pi*D^2)/4;// ft^2
+v=V_avg*A_c;// ft^3/s
+W_pump=v*gradP_l/0.737;// W
+printf('(c)The pumping power requirements,W_pump=%0.2f W\n',W_pump);
+// The answer vary due to round off error