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
- NCSLv 72 months agoFavourite answer
For a cylinder of volume V, the length and area are such that
V = L*A
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
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.