Irodov 1.322: The Force That Outraces Its Own Piston

Problem Irodov 1.322

A horizontal cylinder full of water is closed by a piston at one end and has a small orifice of cross-sectional area $s$ at the other, with $s$ much smaller than the piston’s own cross-sectional area. A constant force pushes the piston until the entire volume $V$ of water has been squeezed out through the orifice, taking a time $t$ to do so. Neglecting friction and viscosity, find the work done by the force.

The horizontal cylinder: a constant force F drives the piston, squeezing the water of volume V out through the small orifice of area s. — The horizontal cylinder: a constant force $F$ drives the piston, squeezing the water of volume $V$ out through the small orifice of area $s$ .

Given

$V$
$s$
$t$
$s \ll S$ (piston area)

Find

Find the work

A

done by the force during the time

t

Solution

Step 1Why the flow is steady throughout

The force $F$ and the piston area $S$ are both constant, so the gauge pressure driving the water is constant for the whole process:

p = \frac{F}{S}.

Since $s \ll S$ , the piston itself advances far slower than the water leaves the orifice, so at every instant the flow is effectively the same steady jet driven by this fixed pressure: the exit speed $v$ it sets does not change with time.

Step 2Work done equals the kinetic energy carried out by the jet

The cylinder is horizontal, so no water changes height, and friction and viscosity are negligible by hypothesis. The piston’s own speed is smaller than the jet’s by the same factor $s/S \ll 1$ , so its kinetic energy is negligible next to the jet’s. By energy conservation, all the work $A$ done by the force converts into the kinetic energy of the ejected water:

A = \frac{1}{2}(\rho V) v^2,

where $\rho$ is the density of water and $\rho V$ the mass of the whole volume expelled.

A = \dfrac{1}{2}\rho V v^2

(1)

Step 3The exit speed from continuity

The orifice passes the whole volume $V$ over the time $t$ at the rate $sv$ , constant because the exit speed $v$ is, so

V = s v t \quad\Longrightarrow\quad v = \frac{V}{st}. \tag{2}

Step 4Combining

Substituting the exit speed (2) into (1) and squaring first,

A = \frac{1}{2}\rho V\left(\frac{V}{st}\right)^2 = \frac{1}{2}\rho V\cdot\frac{V^2}{s^2t^2}.

A = \dfrac{\rho V^3}{2 s^2 t^2}

(3)

Check

Dimensions: $[\rho][V]^3/([s]^2[t]^2) = (\text{kg/m}^3)(\text{m}^9)/(\text{m}^4\,\text{s}^2) = \text{kg}\,\text{m}^2/\text{s}^2 = \text{J}$ , correct for work. Also $A\propto 1/t^2$ : squeezing the same volume out twice as fast needs four times the work, since the exit speed doubles and the kinetic energy goes as $v^2$ ; and $A$ does not depend on the piston area $S$ at all, only on the orifice, which is the whole point of $s\ll S$ .

Remark

The piston does work on the water, but essentially none of it stays as the piston’s own kinetic energy: because $s\ll S$ the piston creeps while the jet flies, so all the work reappears as the jet’s kinetic energy. That is the same accounting used for a fire hose or a rocket nozzle, where a wide, slow reservoir is driven through a narrow, fast exit, and it is why $A$ never has to mention the piston’s own speed at all.