My question is about the speed of a water jet from a garden hose without and with nozzle. I know the answer as experimental findings and ask for the physical reason.

I calculate with Bernoulli along a current thread, but see no need to apply a continuity equation. I also do not model frictional losses in the piping system of my house. If I am wrong here, please explain.

My house is supplied to water pressure $p_{\text{external}}$. I let water flow from a garden hose with cross-sectional area $A_1$ into the air with pressure $p_0=0$. According to Bernoulli, the water of density $\rho=1$ flows out with velocity $u=\sqrt{2p_{\text{external}}}$. It is independent of $A_1$. Now I use a slimmer hose with the cross section $A_2<A_1$, $u$ does not change thereby.

But if I put a nozzle with the cross-sectional area $A_2$ on the first tube, a laminar jet continues to flow out of it, but its velocity has increased. The water flies further into the garden until it reaches the ground after its parabolic flight.

What is the reason for this? Can I explain this with Bernoulli? And if not, why can't I exclusively Meiine question is about the velocity of a jet of water from a garden hose without and with nozzle.

Many thanks in advance and best regards!


1 Answer 1


There is pressure loss between the supply at $p_{external}$ and the end of the hose, from flow separation at increases of area at water fittings, and from viscous and turbulent transfer of energy from the fluid to the walls in each pipe

Bernoulli only applies in the absence of viscosity, which is why it's misleading here

In general, losses from a pipe or fitting are of the form $\Delta P_0=K_L\frac{1}{2}\rho V^2$, where $K_L$ is a loss coefficient which depends on the geometry (so is high for a sudden expansion, low for a smooth venturi)

A small pipe exiting into a big still tank has a loss coefficient $K_L=1$, as none of the kinetic energy of the water streaming out is recovered into pressure

So the open hose allows more flow, leading to higher velocities in the pipes, fittings, and hose, and leads to a lower stagnation pressure just before the end of the hose, so a lower velocity

The nozzle reduces the total flow rate, and with a very small nozzle and very low flow, the stagnation pressure just before the nozzle will be almost equal to the supply and your Bernoulli calculation will work

  • 2
    $\begingroup$ Excellent first answer! Welcome to PSE $\endgroup$
    – Dale
    Apr 30, 2022 at 15:15
  • $\begingroup$ Please let me restate my understanding to make sure I got it. Without a nozzle, thawing of the water in front of the nozzle does not occur. That is why the volume flow of water and its speed in the pipes of the house increases. Consequently, the viscous frictional losses also increase in them and the water flows out of the garden hose with a lower velocity (u0=\sqrt{2*(p_external-p_friction)}. With the nozzle at the end of the hose, p_friction=0 and the exiting velocity increases). Further, due to the mentioned thawing the volume rate decreases while the velocity is increased. Correct? $\endgroup$
    – Lucy
    May 1, 2022 at 7:36
  • $\begingroup$ @Lucy I'm not sure what you mean by "thawing" here. $p_{friction}$ isn't zero with a nozzle, but is much lower than without. As the nozzle hole gets smaller, the total water/second gets lower (similar speed, smaller hole, volumetric flowrate $\dot{Q}$ in $m^3/s$ = speed*area), so the speed in the pipe gets lower ($V_{pipe}=\dot{Q}/A_{pipe}$) and the losses get lower. $\endgroup$
    – sqek
    May 1, 2022 at 9:57
  • $\begingroup$ Thanks, that's exactly my understanding of your first (very good) explanation. Now I want to make sure I got the last step in your argumentation right: lower velocity in the hoses in front of rhe nozzle -> lower pressure losses -> higher pressure right before the nozzle -> higher velocity after the nozzle. Is this correct? $\endgroup$
    – Lucy
    May 2, 2022 at 8:11
  • $\begingroup$ @Lucy yes that's right $\endgroup$
    – sqek
    May 2, 2022 at 11:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.