The Shock Wave in Your Kitchen Sink — Froude Number and the Hydraulic Jump

Turn on the faucet and let water fall onto the flat bottom of your sink. A thin, fast sheet spreads out from the point of impact, and then at some radius the water suddenly piles up into a thick ring. That ring is a hydraulic jump. This everyday kitchen event, remarkably, obeys the same equations as a shock wave born in a supersonic nozzle. This post walks through the Froude number that governs open-channel flow, why the water depth can leap in an instant, and how shallow water mimics a compressible gas — using a little Python and two simulations. By the end you'll see why the ring in your sink and the shock in a de Laval nozzle are cousins.

Flow That Doesn't Fill a Pipe#

An open channel is a waterway with its top open to the atmosphere. Rivers, drainage ditches, and irrigation canals all qualify. Water filling a closed pipe is pushed by a pressure difference between inlet and outlet. Open-channel water is different. Bed slope and gravity drive the flow.

The key is that the free surface can rise and fall. Waves propagate along that surface. In shallow water, the speed of a small surface wave depends on depth alone.

Here $c$ is the shallow-water wave speed, $g$ is gravitational acceleration, and $h$ is the flow depth. Deeper water carries faster waves.

Specific Energy and Critical Depth#

The energy of the flow measured relative to the bed, expressed as a head (energy per unit weight, in length units), is the specific energy. In a rectangular channel of constant width $b$, writing the discharge per unit width as $q = Q/b$,

The first term $h$ is potential energy (depth); the second is kinetic energy. Fix $q$ and plot $E$ as a function of $h$, and an interesting curve emerges. Make the depth too shallow and velocity blows up the kinetic term; make it too deep and the potential term dominates. Somewhere in between, $E$ reaches a minimum.

Solving $dE/dh = 0$ gives the depth at that point.

This $h_c$ is the critical depth. There, $E_{\min} = \tfrac{3}{2} h_c$, and the velocity exactly equals the wave speed: $u_c = \sqrt{g h_c}$.

Play with the diagram below. Change $q$ and the depth $h$ to see which branch of the curve you ride.

Notice that one value of $E$ maps to two depths: the upper branch (deep, slow flow) and the lower branch (shallow, fast flow). Pass the nose at the critical depth $h_c$ and the two worlds split apart.

The Froude Number: Tranquil and Rapid#

The ratio of flow speed to wave speed sets the character of the flow. That dimensionless group is the Froude number (ratio of inertia to gravity-wave speed).

When $Fr < 1$, the flow is slower than its waves. A surface disturbance from a downstream obstacle can travel back upstream. This deep, tranquil flow is called subcritical. When $Fr > 1$, the flow outruns its waves. Downstream signals never reach upstream. This shallow, fast flow is supercritical. $Fr = 1$ is the critical state, and there the depth is exactly $h_c$.

You can feel where this is going. The Froude number plays the very same role as the Mach number (ratio of flow speed to sound speed) in gas dynamics. Just as sound carries information in a compressible gas, shallow-water waves carry it in an open channel. Supercritical is supersonic; subcritical is subsonic.

The Hydraulic Jump: When Water Thickens#

Trouble appears when supercritical flow must become subcritical. On the specific-energy curve there is no smooth path from the lower branch to the upper one. Nature resolves it with a discontinuity: shallow, fast water abruptly leaps to thick, slow water. That is the hydraulic jump.

The relation across the jump comes not from energy but from momentum conservation, because a turbulent roller inside the jump devours energy without mercy. Balancing momentum flux and hydrostatic pressure on both sides gives the conjugate-depth relation between the upstream depth $h_1$ and the downstream depth $h_2$.

$Fr_1$ is the Froude number just before the jump (supercritical). The specific energy lost in the jump falls out cleanly.

That loss is exactly why engineers deliberately create hydraulic jumps at dam spillways and stilling basins. They scatter the destructive kinetic energy of fast water into turbulence, protecting the downstream bed from erosion.

Try the simulation below. Vary the upstream $Fr_1$ and watch the thin supercritical sheet turn into thick subcritical flow, seen from the side.

Drop $Fr_1$ toward 1.5 and the jump weakens into a gentle swell. Push it past 5 and $h_2/h_1$ exceeds six, with a violent white foam roller.

Reproducing It in Code#

Let's compute the conjugate depth and the energy loss directly. Take a unit-width discharge $q = 0.8\ \mathrm{m^2/s}$ and an upstream depth $h_1 = 0.2\ \mathrm{m}$.

import numpy as np
 
g = 9.81  # m/s^2
 
def froude(u, h):
    """Froude number = flow speed / shallow-water wave speed."""
    return u / np.sqrt(g * h)
 
def critical_depth(q):
    """Critical depth for unit-width discharge q (minimum specific energy)."""
    return (q**2 / g) ** (1.0 / 3.0)
 
def specific_energy(h, q):
    """Specific energy E = h + q^2 / (2 g h^2)."""
    return h + q**2 / (2.0 * g * h**2)
 
def conjugate_depth(h1, q):
    """Downstream depth h2 for an upstream (supercritical) depth h1."""
    u1 = q / h1
    fr1 = froude(u1, h1)
    return 0.5 * h1 * (np.sqrt(1.0 + 8.0 * fr1**2) - 1.0)
 
def jump_dissipation(h1, h2):
    """Specific energy lost in the jump (turbulent dissipation)."""
    return (h2 - h1)**3 / (4.0 * h1 * h2)
 
q, h1 = 0.8, 0.2
u1 = q / h1
h2 = conjugate_depth(h1, q)
u2 = q / h2
 
print(f"upstream (super)  u1={u1:.2f} m/s  Fr1={froude(u1,h1):.2f}")
print(f"after jump        h2={h2:.3f} m  u2={u2:.2f} m/s  Fr2={froude(u2,h2):.2f}")
print(f"critical depth    hc={critical_depth(q):.3f} m")
print(f"energy loss       dE={jump_dissipation(h1,h2):.3f} m")

The output:

upstream (super)  u1=4.00 m/s  Fr1=2.86
after jump        h2=0.714 m  u2=1.12 m/s  Fr2=0.42
critical depth    hc=0.403 m
energy loss       dE=0.238 m

A supercritical flow at $Fr_1 = 2.86$ became a subcritical flow at $Fr_2 = 0.42$. The depth thickened by a factor of 3.6 and the velocity slowed by the same factor (continuity, $u h = \text{const}$). The price is 0.238 m of specific energy scattered into turbulence. That the downstream depth $h_2 = 0.714$ m exceeds the critical depth $h_c = 0.403$ m confirms the flow is now subcritical.

When Shallow Water Impersonates a Gas#

Here is the highlight. The steady continuity equation for shallow water in a channel of constant width is $u h = \text{const}$. The steady continuity equation for a compressible gas in a duct of constant area is $\rho v = \text{const}$. Depth $h$ maps exactly onto gas density $\rho$.

Differentiate the shallow-water energy equation $h + u^2/(2g) = \text{const}$ to get $u\,du + g\,dh = 0$, insert the wave speed $u_c = \sqrt{g h}$, and you find that shallow-water waves occupy the same seat as acoustic waves in a gas. Collecting it all yields this analogy.

This analogy corresponds to a fictitious gas with $\kappa = 2$ ($h \leftrightarrow T$, $h^2 \leftrightarrow p$). That is why early aeronautical researchers, instead of expensive supersonic wind tunnels, placed objects in a "shallow water table" and studied shock patterns qualitatively by watching the surface jumps. The ring in your sink resembles a bow shock ahead of a two-dimensional body for exactly this reason.

Key Takeaways#

• The Froude number $Fr = u/\sqrt{gh}$ is the Mach number of open-channel flow. $Fr<1$ is subcritical (tranquil), $Fr>1$ is supercritical (rapid), $Fr=1$ is critical.
• The hydraulic jump is the discontinuity where supercritical flow becomes subcritical. The conjugate depth $h_2/h_1 = \tfrac{1}{2}(\sqrt{1+8Fr_1^2}-1)$ follows from momentum, not energy, conservation, and the leftover energy dissipates into turbulence.
• The shallow-water/compressible analogy maps depth to density, shallow-water waves to sound, and the hydraulic jump to a shock. The water ring in your kitchen sink is a shock-wave experiment you run every day.