Anonymous asked in Science & MathematicsPhysics · 2 months ago

Water in cylinder physics problem?

A horizontal cylinder with volume V is filled with water and stopped up with a piston. On another end of the cylinder there is a small hole whose cross-section's area s is much smaller than the cylinder's cross-section's area. What will be the work done when I have to push out all the water out of the cylinder, affecting the piston with constant, horizontal force. Density of the water - ρ. Assume there is no friction and viscosity

1 Answer

  • NCS
    Lv 7
    2 months ago
    Favourite answer

    For a cylinder of volume V, the length and area are such that

    V = L*A

    By Bernoulli,

    p₁ + ρgh₁ + ½ρv₁² = p₂ + ρgh₂ + ½ρv₂²

    Let's take side 1 to be the outside of the cylinder. The pressure there is 0 (gauge). The velocity on side 2 can be taken as zero (since the hole's area "s is much smaller than the cylinder's area"). The h terms cancel (as the cylinder is horizontal). So

    ½ρv₁² = p₂

    p₂ = F/A = F/(V/L) = FL/V

    where F is the applied force


    ½ρv₁² = FL/V

    means that

    F = ρv₁²V / 2L

    work done = force * distance

    W = F*L = ½ρv₁²V

    This suggests, and I think quite reasonably, that the work done depends on the velocity with which the water is ejected. The greater the velocity, the greater the work required.

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