Question

If d=0.25m and D=0.40m. Assume headloss from the contraction to the end of the pipe can be found as hų = 0.9 (V is velocity in the pipe of diameter D). What height, H, will cause cavitation if atmospheric pressure is 100 kPa (abs)? What is the discharge when cavitation begins? Water T-20°C

Answer 1

Answer:

Height=14.25 m

Discharge=$0.8155 m^{3}/s$

Explanation:

Bernoulli’s equation for the two points, let’s say A and B is given by

$\frac {P_A}{\rho g}+\frac {(v_A)^{2}}{2g}+z_A=\frac {P_B}{\rho g}+\frac {(v_B)^{2}}{2g}+z_B+h_{L(A-B)}$

But since the elevations are the same then

$\frac {P_A}{\rho g}+\frac {(v_A)^{2}}{2g}=\frac {P_B}{\rho g}+\frac {(v_B)^{2}}{2g}+h_{L(A-B)}$

Since $h_{L(A-B)}=\frac {0.9v^{2}}{2g}$ then

$\frac {P_A}{\rho g}+\frac {(v_A)^{2}}{2g}=\frac {100}{\rho 9.81}+\frac {(v_B)^{2}}{2g}+\frac {0.9v^{2}}{2g}$

From continuity equation

$A_AV_A=A_BV_B=0.25\pi D^{2}V_A=0.25\pi d^{2}V_B hence [tex]D^{2}V_A= d^{2}V_B$ and substituting 0.4 for D and 0.25 for d

$V_B=\frac { D^{2}V_A}{d^{2}}=\frac { 0.4^{2}V_A}{0.25^{2}}=2.56V_A$

$\frac {P_A}{\rho g}=\frac {100}{\rho 9.81}+\frac {(v_B)^{2}}{2g}+\frac {0.9v^{2}}{2g}-\frac {(v_A)^{2}}{2g}$ and substituting $V_B$ with $2.56V_A$ we have

$\frac {P_A}{\rho g}=\frac {100}{\rho 9.81}+\frac {(2.56V_A)^{2}}{2g}+\frac {0.9v^{2}}{2g}-\frac {(v_A)^{2}}{2g}$

Since $\rho$ is taken as 1 then

$\frac {P_A}{9.81}=\frac {100}{9.81}+\frac {(2.56V_A)^{2}}{2*9.81}+\frac {0.9v^{2}}{2*9.81}-\frac {(v_A)^{2}}{2*9.81}$

v=6.49 m/s

Since discharge=$AV=0.25\pi 0.4^{2}*6.49=0.8155 m^{3}/s$

$H=\frac {P_B}{\rho g}+\frac {(v_B)^{2}}{2g}=0.24+14.01=14.25 m$