Irodov 1.321: Radial Flow Under a Submerged Cap – Amit & Deepali Sinha Foundation

Problem Irodov 1.321

A wide open vessel is filled with an ideal (incompressible, inviscid) fluid of density $\rho$ to a height $h$ above its horizontal bottom. The bottom carries a round orifice of radius $R_1$ at its center. A closed round cylinder of larger radius $R_2$ ( $R_2 > R_1$ ) is mounted concentrically directly over this orifice, its flat underside separated from the vessel’s bottom by a clearance that is very small compared with $R_1$ and $R_2$ . The cylinder is closed above, so fluid can leave the vessel only by way of that clearance. The vessel is wide enough that the free surface itself sinks negligibly slowly. Find the static pressure $p(r)$ of the fluid inside the clearance, as a function of the radial distance $r$ from the common axis, for $R_1 < r < R_2$ .

Cross section through the common axis. Fluid of height h drains radially inward through the thin clearance under the closed cylinder of radius R_2 toward the orifice of radius R_1; the arrows mark the inward speed v(r). — Cross section through the common axis. Fluid of height $h$ drains radially inward through the thin clearance under the closed cylinder of radius $R_2$ toward the orifice of radius $R_1$ ; the arrows mark the inward speed $v(r)$ .

Given

$R_1$
$R_2 > R_1$
$\rho$
$h$
$p_0$ (atmospheric pressure)

Find

p(r)

for

R_1 < r < R_2

Solution

Step 1Continuity fixes how the radial speed varies with

r

The clearance is a thin gap of some fixed height, and every drop of fluid crossing the circle of radius $r$ must also cross every other circle in the gap, so the volume flow rate through each is the same.

Q = v(r)\cdot 2\pi r\,\delta = \text{const}, \tag{1}

with $\delta$ the (unknown, uniform) clearance height.

v(r)\,r = v(R_1)\,R_1

(2)

Step 2Bernoulli between the free surface and a point in the gap

Take the vessel’s bottom as the height reference. The free surface sits at height $h$ with pressure $p_0$ and negligible speed (the vessel is wide), and a point in the gap at radius $r$ sits at essentially that same reference height with pressure $p(r)$ and speed $v(r)$ . Along the streamline joining them,

p_0 + \rho g h + 0 = p(r) + 0 + \dfrac{1}{2}\rho v(r)^2.

p_0+\rho gh = p(r)+\dfrac{1}{2}\rho v(r)^2

(3)

Step 3The free-discharge condition at the orifice fixes

v(R_1)

Right at $r=R_1$ the fluid turns and falls directly into the open air below the vessel, so the pressure there equals the atmospheric pressure, $p(R_1)=p_0$ . Substituting into (3) at $r=R_1$ and solving for the speed there,

\begin{align*}p_0+\rho gh &= p_0 + \dfrac{1}{2}\rho v(R_1)^2 \\ \rho gh &= \dfrac{1}{2}\rho v(R_1)^2\end{align*}

v(R_1) = \sqrt{2gh}

(4)

exactly Torricelli’s speed for a hole at depth $h$ .

Step 4The speed profile across the clearance

Substituting (4) into the continuity relation (2) and solving for $v(r)$ ,

v(r)\,r = R_1\sqrt{2gh} \quad\Longrightarrow\quad v(r) = \frac{R_1}{r}\sqrt{2gh}. \tag{5}

Step 5The pressure field in the clearance

Squaring the speed of (5) and substituting into the Bernoulli relation (3), solved for $p(r)$ ,

\begin{align*}p(r) &= p_0+\rho gh-\dfrac{1}{2}\rho v(r)^2 \\ &= p_0+\rho gh-\dfrac{1}{2}\rho\cdot\frac{R_1^2}{r^2}\cdot 2gh\end{align*}

p(r) = p_0+\rho gh\left(1-\dfrac{R_1^2}{r^2}\right)

(6)

Check

Dimensionally $\rho gh(1-R_1^2/r^2)$ is $(\text{kg/m}^3)(\text{m/s}^2)(\text{m}) = \text{Pa}$ , matching $p_0$ , and the bracket is always between $0$ and $1$ on $R_1<r<R_2$ , so $p$ stays between $p_0$ and the static value $p_0+\rho gh$ , as it must for a pressure driven purely by flow. The result also delivers something no step assumed: the gap pressure falls short of that static value everywhere, by $\rho gh\,R_1^2/r^2$ , so starting the flow removes an upward push of $\int_{R_1}^{R_2}\rho gh\,(R_1^2/r^2)\,2\pi r\,dr = 2\pi\rho ghR_1^2\ln(R_2/R_1)$ from the cylinder’s underside. The draining fluid sucks the cap down onto the bottom rather than lifting it off, which is the striking consequence of the boxed formula and is nowhere reachable from the boundary condition that produced it.

Remark

The clearance height $\delta$ , the one dimension of the gap that is never actually given, never appears in the final result: it divides out of (1) on the spot, because the gap height is the same at every radius, leaving $p(r)$ set only by the two radii and $h$ . Physically this says the cylinder could be mounted at any sufficiently small clearance without changing the speed profile or the pressure field at all; what a smaller $\delta$ does change is the discharge, $Q = 2\pi R_1\delta\sqrt{2gh}$ , which here is an output of the problem rather than an input to it. The idealization that fails first as $\delta$ shrinks is the inviscid one: a genuinely thin gap turns lubrication dominated, and both $Q$ and $v(r)$ then drop far below these ideal values, so neglecting viscosity is what the small clearance eventually destroys, not what it needs.