Why Putting Your Thumb Over a Hose Makes the Water Shoot Farther

A fluid at rest is described by pressure alone. Once it flows, speed enters the picture — and speed and pressure turn out to be linked.

That link explains a jet from a hose, water spurting from a hole in a tank, an aeroplane staying up, and a spinning ball curving through the air.

This part covers streamline and turbulent flow, the equation of continuity, Bernoulli's theorem, and its applications. Take $g=9.8$ m/s${}^2$ and water density $1000$ kg/m${}^3$.

In streamline flow every particle passing a point follows the same path with the same velocity, so the paths never cross; above a critical velocity the flow becomes turbulent, with irregular swirls and eddies.

Whether flow stays smooth is judged by the Reynolds number

$$R_e=\frac{\rho vD}{\eta }$$

where $D$ is the pipe diameter and $\eta$ the viscosity. Flow is usually streamline for $R_e$ below about $1000$ and turbulent above about $2000$.

Worked example. Water ($\eta =10^{-3}$ Pa s) flows in a $2$ cm pipe.

$$v_c=\frac{R_e\eta }{\rho D}=\frac{1000\times 10^{-3}}{1000\times 0.02}=0.05\text{ m/s}(\text{for }{R}_e=1000)$$

At $1$ m/s, $R_e=\frac{1000\times 1\times 0.02}{10^{-3}}=20000$ — clearly turbulent.

An everyday example. Smoke from an agarbatti rises in a smooth thread for a few centimetres, then breaks into curling swirls as it speeds up.

The substance. Two streamlines can never cross, since a particle at a crossing would need two velocities at once.

**For an incompressible fluid in steady flow, the volume entering a pipe per second must equal the volume leaving, so $A_1{v}_1=A_2{v}_2$ — a narrower section means faster flow.

Derivation.** In time $\Delta t$, a volume $A_1{v}_1\Delta t$ enters and $A_2{v}_2\Delta t$ leaves. With no fluid created, stored or compressed, these are equal. The volume flow rate $Q=Av$ is the same everywhere along the pipe.

Worked example 1 — a nozzle. A hose of radius $1$ cm carries water at $2$ m/s into a nozzle of radius $0.25$ cm.

$$v_2=v_1\frac{A_1}{A_2}=2\times {\left(\frac{1}{0.25}\right)}^2=32\text{ m/s}$$

Worked example 2 — filling a bucket.

$$Q=\pi (0.01{)}^2\times 2\approx 6.28\times 10^{-4}{\text{ m}}^3/\text{s}\approx 0.63\text{ L/s}$$

A $15$ L bucket fills in about $24$ s.

Worked example 3. If a blood vessel narrows to half its diameter, its area falls to a quarter, so the blood speeds up four times.

An everyday example. Pressing your thumb over the end of a garden hose shrinks the opening, so the same flow must leave much faster.

The substance. The nozzle speeds the water up but does not increase the flow rate — the bucket still fills in the same time.

**For an ideal fluid in steady flow, $P+\frac{1}{2}\rho v^2+\rho gh$ stays constant along a streamline, so where the fluid moves faster or higher, its pressure is lower.

The ideal fluid is incompressible and non-viscous, and the flow is steady.

Derivation outline.** Follow a small volume $\Delta V$ from section 1 to section 2. The net work done on it by pressure is $(P_1-P_2)\Delta V$. By the work-energy theorem, this equals its gain in kinetic and potential energy:

$$(P_1-P_2)\Delta V=\frac{1}{2}\rho \Delta V(v_2^2-v_1^2)+\rho \Delta Vg(h_2-h_1)$$

Rearranging gives $P_1+\frac{1}{2}\rho v_1^2+\rho g{h}_1=P_2+\frac{1}{2}\rho v_2^2+\rho g{h}_2$.

Worked example — a narrowing pipe. In a horizontal pipe, water at $2$ m/s has pressure $1.5\times 10^5$ Pa; further on it speeds up to $8$ m/s.

$$P_2=1.5\times 10^5-\frac{1}{2}(1000)(8^2-2^2)=1.5\times 10^5-30000=1.2\times 10^5\text{ Pa}$$

An everyday example. Blow across the top of a strip of paper held below your lips, and it rises — the moving air above has lower pressure.

The substance. Bernoulli's theorem is energy conservation per unit volume, and it applies along one streamline of one flow.

**Applying Bernoulli between the open surface and a hole gives the efflux speed $v=\sqrt{2gh}$; faster air over a wing or one side of a spinning ball lowers the pressure there, producing lift or a sideways force.

Torricelli's law.** Both the surface and the jet are at atmospheric pressure, and the surface is nearly still:

$$\rho gh=\frac{1}{2}\rho v^2\Rightarrow v=\sqrt{2gh}$$

Worked example 1. A hole $5$ m below the water surface of a tank:

$$v=\sqrt{2\times 9.8\times 5}\approx 9.9\text{ m/s}$$

Through a $1$ cm${}^2$ hole, about $9.9\times 10^{-4}$ m${}^3$ — nearly $1$ litre — leaves each second.

Worked example 2 — lift. Suppose air flows at $70$ m/s over a $20$ m${}^2$ wing and $60$ m/s under it, with $\rho =1.2$ kg/m${}^3$.

$$\Delta P=\frac{1}{2}(1.2)(70^2-60^2)=780\text{ Pa}\text{lift}=780\times 20=15600\text{ N}$$

Wings are also tilted to push air downward, which adds to the lift.

Magnus effect. A spinning ball drags air round with it, so air moves faster on one side than the other. The pressure difference pushes the ball sideways, curving its path.

An everyday example. A table tennis ball hit with heavy topspin dips sharply, pushed down by the pressure difference.

The substance. **The efflux speed is the same as for a stone falling through height $h$.**