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+//Example 6_2
+clc;clear;
+// Given values
+m=14;//Water flow rate in kg/s
+rho=1000;//The density of water in kg/m^3
+A_1=0.0113;// The cross sectional area of the elbow at inlet in m^2
+A_2=7*10^-4;// The cross sectional area of the elbow at outlet in m^2
+z_2=0.3;// m
+z_1=0;// m
+g=9.81;// The acceleration due to gravity in m/s^2
+theta=30;// degree
+b=1.03;// The momentum-flux correction factor
+
+// Calculation 
+//(a)
+v_1=m/(rho*A_1);
+v_2=m/(rho*A_2);//The inlet and the outlet velocities in m/s
+P_1g=(rho*g*(((v_2^2-v_1^2)/(2*g))+(z_2-z_1)))/1000;// kPa
+printf("The gage pressure at the center of the inlet of the elbow=%0.1f kPa\n",P_1g);
+//(b)   z
+F_Rx=b*m*(((v_2*cosd(theta))-v_1))-(P_1g*1000*A_1);// N
+F_Rz=b*m*v_2*sind(theta);// N
+printf("The anchoring force of the elbow be F_Rx=%0.0f N,F_Rz=%0.0f N\n",F_Rx,F_Rz);
+// The answer vary due to round off error