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+//Example 8_3
+clc;clear;funcprot(0);
+// Given values
+rho=62.36;// lbm/ft^3
+mu=7.536*10^-4;// lbm/ft.s
+D=2/12;// ft
+v=0.2;// ft^3/s
+L=200;// ft
+g=32.2;// ft/s^2
+
+//Calculation
+A_c=(%pi*D^2)/4;//  ft^2
+V=v/A_c;// Average velocity in ft/s
+Re=(rho*V*D)/(mu);// Reynolds number
+// Re is greater than 4000. Therefore, the flow is turbulent. The relative roughness of the pipe is calculated using Table 8–2, (epsilon/D)=e
+E=0.000007;
+e=E/(D);
+//To avoid any reading error, we determine f from the Colebrook equation:(1/sqrt)=-2.0*log10*((e/3.7)+(2.51/(Re*sqrt(f)))
+// f=y(1)
+function[X]=frictionfactor (y)
+     X(1)=(-2.0*log10((0.000042/3.7)+(2.51/(126400*sqrt(y(1))))))-(1/sqrt(y(1)));
+endfunction
+y=[0.001];
+z=fsolve(y,frictionfactor);// Friction factor
+gradP_L1=(z*(L/D)*(rho*(V^2)/2))*(1/32.2);// lbm/ft^2
+gradP_L=gradP_L1/144;// psi
+printf('The pressure drop,gradP_L=%0.0f lbf/ft^2=%0.1f psi \n',gradP_L*144,gradP_L);
+h_L=(z*(L/D)*(V^2/(2*g)));// ft
+printf('The head loss,h_L=%0.1f ft\n',h_L);
+W_p=(v*gradP_L1)/0.737;// W
+printf('The required power input,W_pump=%0.0f W \n',W_p);
+// The answer vary due to round off error