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+///Chapter No 7 Fluid Mechanics
+///Example 7.19 Page No:130
+///Find pipe inclined 30 degree,therefore Z2
+///Input data
+clc;
+clear;
+L1=36;                    //Length of pipe in m
+D11=0.15;                 //Diameter at upper side in m
+D12=0.3;                  //Diameter at lower side in m
+sin30=0.5;
+theta=sin(30);      //Pipe slope upward at angle in degree
+V11=2;                    //Velocity of water at smaller section in m/s  
+pi1=3.14;                 //Pi constant 
+rho=1000;                //Roh constant
+g1=9.81;                  //Gravity constant
+
+
+///Calculation
+//datum line is passing through the center of the low end,therefore
+Z1=0;
+Z2=Z1+L1*(0.5);           //pipe inclined 30 degree,therefore in m
+//Q=A1*V1=A2*V2  continuity eqation ,discharge
+V12=(pi1/4*(D11^2)*2)/(pi1/4*(D12^2));
+//Z=P1-P2 bernoulli's equation 
+Z=((((-V11^2)/(2*g1))+((V12^2)/(2*g1))-Z1+Z2)*(rho*g1))*10^-3;
+
+
+///Output
+printf('pipe inclined 30 degree,therefore Z2= %f m \n',Z2);
+printf('continuity eqation discharge V2= %f m/s \n',V12);
+printf('bernoullis equation Z=%f kpa \n',Z);