Question

A small submersible moves at velocity V , in fresh water at 20 8 C, at a 2-m depth, where ambient pressure is 131 kPa. Its critical cavitation number is known to be Ca 5 0.25. At what velocity will cavitation bubbles begin to form on the body? Will the body cavitate if V 5 30 m/s and the water is cold (5 8 C)?

Answer 1

Ca = 0.2892

Step-by-step explanation:

given data

fresh water temperature = 20°C

depth = 2-m

ambient pressure = 131 kPa

Ca = 0.25

V = 30 m/s

water cold = 5°C

solution

we know at 20°C here Pv = 2.337 kPa

so

Ca (critical) =
`(2(Pa-Pv))/(\rho v^2)`

0.25 =
`(2(131000-2337))/(998 * v^2)`

so

Vcrit = 32.11 m/s

and

when temperature decrese by 5

Pv = 863 pa

ϼ = 1000 kg/m³

so for this we know hee

v = 30 m/s

and

Ca =
`(2(131000-863))/(1000 * 30^2)`

Ca = 0.2892

here Ca > 0.25 so that this body will not cavitate for given condition