Irodov 1.320: A Submerged Mouth Drives a Fountain – Amit & Deepali Sinha Foundation

Problem Irodov 1.320

A bent tube is lowered into a wide body of flowing water: its lower end is submerged, open and horizontal, facing directly into the oncoming current, while its upper end rises vertically out of the water, to a height $h_0 = 12$ cm above the free surface, where it is otherwise closed except for a small orifice. The water’s speed relative to the tube is $v = 2.5$ m/s. To what height $h$ above that orifice does the water jet spurt?

The submerged mouth faces upstream into the current v and reads the stagnation pressure. The tube carries it a height h_0 above the free surface to its closed top, where a small orifice sends the jet a further height h. — The submerged mouth faces upstream into the current $v$ and reads the stagnation pressure. The tube carries it a height $h_0$ above the free surface to its closed top, where a small orifice sends the jet a further height $h$ .

Given

$v = 2.5$ m/s
$h_0 = 12$ cm

Find

Find

h

, the height the jet rises above the orifice

Solution

Step 1Bernoulli from the free stream to the tube’s mouth

Let $d$ be the depth of the submerged mouth below the free surface. Far from the tube the water moves at speed $v$ relative to the tube at the local static pressure $p_d$ ; right at the mouth, facing directly into the current, the water is brought to rest relative to the tube and the pressure rises to the stagnation value $p_s$ . This is where the given “small orifice” earns its keep: because the outlet at the top is small compared with the tube’s bore, almost no water actually passes through the tube, so the mouth is a genuine stagnation point and not a working intake with the current still running past it. Bernoulli’s equation along this streamline gives

p_d + \frac{1}{2}\rho v^2 = p_s

(1)

Step 2The static pressure at the mouth’s own depth

Away from the tube the water is simply in hydrostatic equilibrium beneath the free surface, so the static pressure at depth $d$ is the atmospheric pressure $p_0$ plus the weight of the column above it:

p_d = p_0 + \rho g d.

Putting this into (1) gives the stagnation pressure at the mouth in terms of the depth and the stream speed alone:

p_s = p_0 + \rho g d + \frac{1}{2}\rho v^2

(2)

Step 3Bernoulli from the mouth up to the peak of the jet

The stagnation pressure $p_s$ then drives water up through the tube and out of the orifice, at height $h_0$ above the free surface; a particle leaving the orifice on the jet’s axis rises until it momentarily comes to rest again, at the jet’s peak, a further height $h$ above the orifice, back at atmospheric pressure $p_0$ . Applying Bernoulli’s equation (with the elevation term this time, since the two points are at different heights) between the mouth, at depth $d$ below the free surface, and the peak, at height $h_0+h$ above it,

p_s + \rho g(-d) = p_0 + \rho g\left(h_0+h\right).

Substituting (2) for $p_s$ , the depth $d$ cancels:

p_0 + \rho g d + \frac{1}{2}\rho v^2 - \rho g d = p_0 + \rho g\left(h_0+h\right),

and the atmospheric pressure cancels with it, leaving a pure statement about heads:

\frac{1}{2}\rho v^2 = \rho g\left(h_0+h\right)

(3)

Step 4Solve for

h

Dividing (3) through by $\rho g$ turns both sides into heights,

\frac{v^2}{2g} = h_0 + h,

and moving $h_0$ across isolates the rise above the orifice:

h = \frac{v^2}{2g} - h_0.

With $v=2.5$ m/s, $g=9.8$ m/s $^2$ and $h_0=0.12$ m,

h = \frac{(2.5)^2}{2(9.8)}\ \text{m} - 0.12\ \text{m} = 0.3189\ \text{m} - 0.12\ \text{m} = 0.199\ \text{m}.

h = \dfrac{v^2}{2g} - h_0 \approx 0.20\ \text{m} = 20\ \text{cm}

(4)

Check

The submersion depth $d$ dropped out completely on the way to (3), so the answer depends only on quantities visible above the surface, $v$ , $h_0$ and $g$ , exactly as the given data requires. Also, had $h_0$ exceeded $v^2/2g$ the formula would return a negative $h$ : physically that says the stagnation pressure cannot even lift water to the height of the orifice, so no jet forms at all, which is the correct degenerate limit.

Remark

The height $h_0+h=v^2/2g$ is precisely the velocity head of the stream, the height a free jet of that same initial speed would reach if launched straight up with no tube in the way at all; the bent tube’s only job is to intercept the stream’s kinetic energy at the mouth and carry the resulting stagnation pressure, undiminished, up past the free surface before releasing it, so the final rise above the orifice is simply that full velocity head with the climb to $h_0$ already spent.