Question

A torpedo causes cavitation when it is traveling at 25 m/s at a water depth of 7 m. With seawater at 20C and the atmospheric pressure of 101KPa. Neglect Reynolds and Froude Number effects. a. At what speed will the torpedo cavitate at a water depth of 32 m

Answer 1

Using Bernoulli's Equation, we can calculate the speed at which the torpedo will cavitate at a water depth of 32 m. However, in this case, the velocity is found to be negative, indicating that the torpedo will not cavitate at this depth.

Step-by-step explanation:

To determine the speed at which the torpedo will cavitate at a water depth of 32 m, we need to use Bernoulli's Equation. Bernoulli's Equation relates the pressure, velocity, and height of a fluid in a system. With the given information, we can use Bernoulli's Equation to calculate the speed:

P1 + ½ ρ v12 + ρgh1 = P2 + ½ ρ v22 + ρgh2

Using this equation, we can write:

P1 + ½ ρ v12 + ρgh1 = P2 + ½ ρ v22 + ρgh2

At a water depth of 7 m, the cavitation speed is 25 m/s. Using this information, we can substitute the values into the equation and solve for v2:

P1 + ½ ρ v12 + ρgh1 = P2 + ½ ρ v22 + ρgh2

With the given values and solving the equation:

P1 + ½ ρ v12 + ρgh1 = P2 + ½ ρ v22 + ρgh2
(101000 Pa) + ½ (1000 kg/m3) (25 m/s)2 + (1000 kg/m3) (9.8 m/s2) (7 m) = (101000 Pa) + ½ (1000 kg/m3) v22 + (1000 kg/m3) (9.8 m/s2) (32 m)

Simplifying the equation and solving for v2:

312500 Pa + ½ (1000 kg/m3) (625 m2/s2) + (1000 kg/m3) (9.8 m/s2) (7 m) = (101000 Pa) + ½ (1000 kg/m3) v22 + (1000 kg/m3) (9.8 m/s2) (32 m)

Calculating the equation:

160000 Pa + ½ (1000 kg/m3)v22 = 132304 Pa

Simplifying the equation:

½ (1000 kg/m3)v22 = -27696 Pa

Simplifying the equation:

(1000 kg/m3)v22 = -55392 kg/m2/s2

Simplifying the equation:

v22 = -55.392 m2/s2

Simplifying the equation:

v2 = -7.44 m/s

Since the velocity cannot be negative, the torpedo will not cavitate at a water depth of 32 m.