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+//Example 8_1
+clc;clear;funcprot(0);
+// Given values
+P_i=745;
+P_o=97;//The pressure at the pipe inlet and outlet in kPa
+D=0.05;// m
+L=40;// m
+//Properties
+rho=888;//kg/m^3
+mu=0.800;// kg/m.s
+g=9.81;// m/s^2
+
+//Calculation
+gradP=P_i-P_o;//kPa
+A_c=(%pi*D^2)/4;// m^2
+//(a)
+//For the horizontal case, theta=0
+theta=0;// degree
+V=((((gradP*1000)-(rho*g*L*sind(theta)))*(%pi*D^4))/(128*mu*L));// m^3/s
+V_horiz=V;// m^3/s
+printf('(a)The flow rate for the horizontal case,theta=0,V_horiz=%0.5f m^3/s\n',V_horiz);
+
+//(b)
+// For uphill flow with an inclination of 15°, we have theta=+15°,
+theta_1=+15;// degree
+V=((((gradP*1000)-(rho*g*L*sind(theta_1)))*(%pi*D^4))/(128*mu*L));
+V_uphill=V;//m^3/s
+printf('(b)The flow rate for uphill flow with an inclination of 15°,V_uphill=%0.5f m^3/s\n',V_uphill);
+
+//(c)
+//For downhill flow with an inclination of 15°,we have theta=-15°,
+theta_2=-15;//degree
+V=((((gradP*1000)-(rho*g*L*sind(theta_2)))*(%pi*D^4))/(128*mu*L));
+V_downhill=V;//m^3/s
+printf('(c)The flow rate for downhill flow with an inclination of 15°,V_downhill=%0.5f m^3/s\n',V_downhill);
+V_avg=(V_downhill/A_c);
+Re=(rho*V_avg*D)/mu;
+disp("Re=100.Re<2300.Therefore, the flow is laminar for all three cases and the analysis is valid.");