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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)?

Respuesta :

Answer:

Ca = 0.2892

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) = [tex]\frac{2(Pa-Pv)}{\rho v^2}[/tex]

0.25 = [tex]\frac{2(131000-2337)}{998 \times v^2}[/tex]

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 = [tex]\frac{2(131000-863)}{1000 \times 30^2}[/tex]

Ca = 0.2892

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

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