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

Height=14.25 m

Discharge=
`0.8155 m^(3)/s`

Step-by-step 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`