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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);
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