Irodov 1.318: Two-Layer Torricelli – Amit & Deepali Sinha Foundation

Problem Irodov 1.318

A wide vessel contains a layer of water of thickness $h_1 = 30$ cm at the bottom, topped by a layer of kerosene of thickness $h_2 = 20$ cm, with a small hole pierced in the base. Neglecting viscosity, find the speed of the water jet leaving the hole.

Water below, kerosene floating above it, draining through a small hole in the base.

Given

$h_1 = 30$ cm
$h_2 = 20$ cm

Find

Find

v

, the velocity of the water jet leaving the hole in the bottom

Solution

Step 1Pressure at the hole from the two overlying layers

The vessel is wide, so the free surface sinks negligibly slowly and the liquid above the hole is, at every instant, very nearly in hydrostatic equilibrium. Building the pressure up from the free surface (atmospheric, $p_0$ ) through the kerosene layer and then the water layer down to the hole gives

p_{\text{hole}} = p_0 + \rho_2 g h_2 + \rho_1 g h_1. \tag{1}

Step 2Bernoulli in the water, from the hole to the open jet

Follow a streamline in the water from a point at the hole’s level, where the water is essentially at rest at pressure $p_{\text{hole}}$ , out to the free jet just outside, at atmospheric pressure $p_0$ and exit speed $v$ . With negligible height change at the hole itself, Bernoulli’s equation along this streamline gives

p_{\text{hole}} = p_0 + \frac{1}{2}\rho_1 v^2. \tag{2}

Step 3Combine and solve for

v

Substituting the pressure (1) into the Bernoulli relation (2) equates the two expressions for $p_{\text{hole}}$ ; dividing through by $\frac{1}{2}\rho_1$ then isolates $v^2$ .

\begin{align*}\rho_2 g h_2 + \rho_1 g h_1 &= \frac{1}{2}\rho_1 v^2 \\ v^2 &= \frac{2}{\rho_1}\left(\rho_2 g h_2 + \rho_1 g h_1\right) \\ &= 2g\left(h_1 + h_2\frac{\rho_2}{\rho_1}\right).\end{align*}

Taking the square root gives the exit speed.

v = \sqrt{2g\left(h_1+h_2\dfrac{\rho_2}{\rho_1}\right)}

(3)

Step 4Evaluate

With $\rho_1 = 1.00\times10^3$ kg/m $^3$ (water) and $\rho_2 = 0.80\times10^3$ kg/m $^3$ (kerosene),

v = \sqrt{2(9.8)\left(0.30 + 0.20\times0.8\right)} = \sqrt{19.6\times0.46} \text{ m/s}.

v = \sqrt{2g\left(h_1+h_2\dfrac{\rho_2}{\rho_1}\right)} \approx 3.0\ \text{m/s}

(4)

Check

With no kerosene at all ( $h_2=0$ ) the result collapses to $v=\sqrt{2gh_1}$ , the ordinary Torricelli speed for a single liquid, exactly as it should: the second layer only matters through the extra pressure its weight adds.

Remark

The kerosene layer enters only through $h_2\rho_2/\rho_1$ : it behaves exactly like an extra $h_2\rho_2/\rho_1$ of water stacked on top, an equivalent head that any immiscible layer above the hole can be converted to before summing. Stack a third, fourth liquid on top and the same trick just adds one more term, each layer’s own thickness rescaled by its density ratio to the flowing fluid.