Irodov 1.329: Only the Two End Areas Matter – Amit & Deepali Sinha Foundation

Problem Irodov 1.329

A wide, open tank has a tube fixed to one of its side walls, through which water drains into the air. The tube’s own cross-sectional area narrows smoothly from $S = 3.0$ cm $^2$ , where it joins the tank, down to $s = 1.0$ cm $^2$ at its open outlet. The free surface of the water in the tank sits a height $h = 4.6$ m above the level of the tube itself. Neglecting the water’s viscosity, find the horizontal component of the force with which the flowing water tends to pull the tube out of the tank wall.

The tank drains through a tube narrowing from cross-section S at the wall to s at the tip (widths drawn proportional to sqrt(area)). The free surface sits a height h above the tube; the dashes are the emerging jet. — The tank drains through a tube narrowing from cross-section $S$ at the wall to $s$ at the tip (widths drawn proportional to $\sqrt{\text{area}}$ ). The free surface sits a height $h$ above the tube; the dashes are the emerging jet.

Given

$S = 3.0$ cm $^2$
$s = 1.0$ cm $^2$
$h = 4.6$ m

Find

The horizontal component of the force tending to pull the tube out of the tank

Solution

Step 1Exit speed, from the whole tank to the open tip

Applying Bernoulli’s equation between the free surface (atmospheric, essentially at rest in a wide tank) and the open outlet (also atmospheric), the intermediate shape of the tube never enters: only the height dropped, $h$ , sets the exit speed.

0 + 0 + \rho g h = \frac{1}{2}\rho v_2^2 + 0

v_2 = \sqrt{2gh}

(1)

Step 2The wide end’s own speed and the flow rate, from continuity

The same volume per second $Q$ crosses every section of the tube, so the speed $v_1$ at the wide end ( $S$ ) is fixed by the exit speed (1) at the tip ( $s$ ).

Q = v_2 s = v_1 S .

Solving that for the wide end’s speed,

v_1 = v_2\,\frac{s}{S}

(2)

Step 3Pressure at the wide end, from Bernoulli along the tube itself

The tube’s two ends sit at essentially the same height, so applying Bernoulli between the wide end (pressure $p_1$ , speed $v_1$ ) and the tip (atmospheric, speed $v_2$ ) isolates $p_1$ .

p_1 + \frac{1}{2}\rho v_1^2 = \frac{1}{2}\rho v_2^2 .

Moving the wide end’s kinetic term across leaves the pressure there alone:

p_1 = \frac{1}{2}\rho\bigl(v_2^2-v_1^2\bigr)

(3)

Step 4Momentum balance on the water inside the tube

Take the water between the two end sections as the system. The wide end feels the tank’s own pressure $p_1$ pushing it along the tube’s axis; the tip is open to the atmosphere and contributes nothing in gauge pressure; the tube walls exert the remaining force $F_w$ on the water. Newton’s second law for this steady flow, and the reaction $F=-F_w$ the water exerts on the walls by Newton’s third law, read

\begin{align*}p_1 S + F_w &= \rho Q\bigl(v_2-v_1\bigr) \\ F &= p_1 S - \rho Q(v_2-v_1).\end{align*}

Substituting (3) for $p_1$ and factoring,

\begin{align*}F &= \frac{1}{2}\rho S\bigl(v_2^2-v_1^2\bigr) - \rho Q(v_2-v_1) \\ &= \rho(v_2-v_1)\left[\frac{1}{2}S(v_2+v_1) - Q\right].\end{align*}

Since $Q=v_1 S$ , the bracket reduces to a single term,

\begin{align*}\frac{1}{2}S(v_2+v_1) - v_1S &= \frac{1}{2}S(v_2-v_1),\end{align*}

so the force collapses to

F = \frac{\rho S}{2}(v_2-v_1)^2.

Step 5Eliminate

v_1

and evaluate

The force above depends on the two speeds only through their difference, so substitute (2) for $v_1$ and take a common factor of $v_2$ :

\begin{align*}v_2-v_1 &= v_2 - v_2\frac{s}{S} \\ &= v_2\,\frac{S-s}{S}.\end{align*}

Squaring that inside the force,

\begin{align*}F &= \frac{\rho S}{2}\cdot v_2^2\frac{(S-s)^2}{S^2} \\ &= \frac{\rho v_2^2}{2}\cdot\frac{(S-s)^2}{S}.\end{align*}

Substituting $v_2^2=2gh$ from (1),

F = \frac{\rho(2gh)}{2}\cdot\frac{(S-s)^2}{S} = \rho gh\frac{(S-s)^2}{S}.

Substituting the given values in SI units (water’s density from the constants table, $\rho=1.00\times10^3$ kg/m $^3$ ),

F = (1.00\times10^3)(9.8)(4.6)\frac{(2.0\times10^{-4})^2}{3.0\times10^{-4}}\ \text{N}.

F = \rho gh\dfrac{(S-s)^2}{S} \approx 6.0\ \text{N}

(4)

Check

Dimensionally $[\rho][g][h] = \text{Pa}$ , and $(S-s)^2/S$ has units of area, so their product is a force, as required. Two limits confirm the shape: if $s\to S$ (no taper at all) the force vanishes, since a uniform tube exerts no net axial thrust; if $s\to0$ (the tube pinches almost shut) the boxed formula tends to $\rho gh S$ , exactly the static pressure $\rho gh$ acting on the full area $S$ that a nearly stagnant column of water would exert, which is the right limit once the outflow itself carries away almost no momentum.

Remark

The whole derivation only ever used the two end areas $S$ and $s$ and the height $h$ ; the taper’s actual profile in between never entered a single equation, because Bernoulli and the momentum theorem are both statements about the boundary of the control volume. Any nozzle joining the same two end areas over the same drop would be pulled on with exactly this same force, straight-sided or bell-mouthed, so long as its interior is a shape the flow follows without separating, which is what Bernoulli was assumed for. (A stepped bore is the instructive exception: real water separates at the step and loses head there, so the loss-free result stops applying, and it is only the problem’s own instruction to neglect viscosity that keeps such a case inside the formula.) It is a fact worth remembering whenever a “shape of the nozzle” seems relevant to a thrust calculation and turns out not to be.