A fuel tank for a rocket in space under a zero-g environment is rotated to keep the fuel in one end of the tank. The system is rotated at 3 rev/min. The end of the tank (point A) is 1.5 m from the axis of rotation, and the fuel level is 1 m from the rotation axis. The pressure in the nonliquid end of the tank is 0.1 kPa, and the density of the fuel is 800 kg/m3 . What is the pressure at the exit (point A)

P₂ = 4098.96 Pa

Explanation:

For this exercise let's use Bernoulli's equation

Let's use the subscript 1 for the point of the liquid surface and the subscript 2 for the ends (point A)

         P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

the velocity at the end of the tank

        v₂ = w r₂

the velocity at the surface of the liquid is

        v₁ - w r₁

where r₂ = 1.5 m and r₁ = 1 m

the tank pressure is P₁ = P₀ = 0.1 10³ Pa

         P₂ = P₁ + ½ ρ [w² (r₁² - r₂²)] + ρ g (y₁ -y₂)

We must remember that the pressure measurements the distances are measured from the lowest part to the surface that has zero height

let's reduce the magnitudes to the SI system

         w = 3 rev / min (2π rad / 1rev) (1 min / 60 s) = 0.314159 rad / s

let's calculate

        P₂ = 0.1 10³ + ½ 800 0.314159² |(1² -1.5²)| + 800 9.8 |(1-1.5)|

        P₂ = 0.1 103 +78.96 + 3920

        P₂ = 4098.96 Pa

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//

uio

explanation: