NBPhO 2026

Overview

The problem stitches together three classical pieces of pipe-flow physics:

1. The Reynolds number as a competition between two drag mechanisms. Laminar wall friction scales as $\mu v/D$ while turbulent wall friction scales as $\rho v^2$; their ratio is dimensionless and is the Reynolds number $Re$.
2. Turbulent friction is quadratic in flow speed. Once $\delta\ll D$, the wall stress sits at the dynamic-pressure scale $\rho v^2$, so the pressure drop along a hose of fixed geometry obeys $\Delta p_f \propto v^2 \propto Q^2$. This is what allows us to calibrate the friction from a single operating point of the pump and then extrapolate it to a new flow rate.
3. A nozzle is just an accelerator. Squeezing the cross-section by a factor $f$ multiplies the exit speed by $1/f$ (continuity), the exit kinetic-energy density by $1/f^2$, and the stagnation pressure that the jet deposits on a wall by the same factor. The price is paid in flow rate, because pumping fluid through a high-KE jet is energetically expensive.

The physical small/large parameters in the numerical part are

$$\frac{\rho g h}{P/Q_0}\;\approx\;\tfrac{1}{3},\qquad \frac{\tfrac{1}{2}\rho v_0^2}{P/Q_0}\;\approx\;0.008,\qquad \frac{1}{f^2}\;\approx\;44,$$

which tells us up front that (a) friction in the hose dominates the no-nozzle pressure budget, (b) the kinetic-energy term is utterly negligible without the nozzle, and (c) the nozzle promotes the kinetic-energy term to the leading channel for energy loss, comparable in size to friction.

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Part (i) — Reynolds number from drag-force scales

Estimating the laminar drag

The laminar drag per unit area is $F_L = \mu\,du/dr$. Across the pipe radius the speed drops from $\sim v$ on the axis to $0$ at the wall over a length of order $D$, so

$$\frac{du}{dr}\;\sim\;\frac{v}{D}\qquad\Longrightarrow\qquad F_L\;\sim\;\frac{\mu v}{D}.$$

Estimating the turbulent drag

The problem hands us $F_T \sim \rho v^2$ directly: the dynamic pressure carried by the vortices that batter the wall.

The ratio

The Reynolds number is the ratio of these two scales:

$$R\;=\;\frac{F_T}{F_L}\;\sim\;\frac{\rho v^2}{\mu v/D}\;=\;\frac{\rho v D}{\mu}.$$

Reading off the powers,

$$\boxed{\;R\;=\;\rho^{1}\,v^{1}\,D^{1}\,\mu^{-1}\;}$$

Educational remark

The same answer emerges from pure dimensional analysis: $\rho$, $v$, $D$, $\mu$ have units that admit a single dimensionless monomial up to a power, and that monomial is $\rho v D/\mu$. The mechanistic argument above is what justifies naming this group “the ratio of turbulent to laminar drag” — the dimensional answer alone would not commit to a physical interpretation. Note also that $R$ is built only from bulk quantities; the boundary-layer thickness $\delta$ that appears inside the turbulent-drag picture has cancelled out, since $\mu v/\delta \sim \rho v^2$ implies $\delta \sim \mu/(\rho v) = D/R$, which is a consequence of $R$, not an independent input.

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Part (ii) — Flow type in the hose

Plugging in numbers

The mean speed in the hose is

$$v_0\;=\;\frac{Q}{A}\;=\;\frac{4Q}{\pi d^{2}}.$$

With $Q = 25\,\text{L}\cdot\text{min}^{-1} = \tfrac{25}{60}\times 10^{-3}\,\text{m}^{3}\cdot\text{s}^{-1} = 4.167\times 10^{-4}\,\text{m}^{3}\cdot\text{s}^{-1}$ and $d = 0.013\,\text{m}$,

$$A\;=\;\frac{\pi(0.013)^{2}}{4}\;=\;1.327\times 10^{-4}\,\text{m}^{2},\qquad v_0\;\approx\;3.14\,\text{m}\cdot\text{s}^{-1}.$$

The Reynolds number is

$$R\;=\;\frac{\rho_w v_0 d}{\mu_w}\;=\;\frac{1000\times 3.14\times 0.013}{1.1\times 10^{-3}}\;\approx\;3.7\times 10^{4}.$$

Conclusion

Since $R \approx 3.7\times 10^{4} \gg 2500$, the flow in the hose is

$$\boxed{\;\text{turbulent.}\;}$$

Physical remark

$R \sim 4\times 10^{4}$ is well into the developed-turbulence regime. The boundary-layer thickness implied by the heuristic $\delta\sim D/R$ is only

$$\delta\;\sim\;\frac{d}{R}\;\sim\;\frac{0.013}{3.7\times 10^{4}}\;\sim\;0.3\,\mu\text{m},$$

confirming $\delta \ll d$, so the assumption that velocity is essentially uniform across the cross-section (with a thin near-wall transition) is excellent. This is precisely why we may treat the kinetic-energy density of the flow as $\tfrac{1}{2}\rho v^2$ with $v = Q/A$ in part (iii) without a “kinetic-energy correction factor”.

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Part (iii) — Adding the nozzle

The energy budget of one operating point

Take the pump as a black box that delivers a fixed output power $P$ to the fluid, regardless of operating point. Per unit time, this power must pay for three things:

Term	Per unit volume	Why
Lifting water by $h$	$\rho g h$	Source is at depth $h$, exit is at the gardener
Wall friction in the hose	$\Delta p_f$	Turbulent dissipation along $s$
Kinetic energy of the exiting jet	$\tfrac{1}{2}\rho v_\text{exit}^{2}$	Water leaves with this KE; never recovered

A Bernoulli accounting from the still water in the well (atmospheric pressure, height $-h$, speed $\approx 0$) to the free jet (atmospheric pressure, height $0$, speed $v_\text{exit}$), with the pump pressure rise $\Delta p_\text{pump}$ and the friction loss $\Delta p_f$ both in the line, gives

$$\Delta p_\text{pump}\;=\;\rho g h\;+\;\Delta p_f\;+\;\tfrac{1}{2}\rho v_\text{exit}^{2},$$

and multiplying by $Q$ converts this to the power balance

$$P\;=\;Q\!\left[\rho g h\;+\;\Delta p_f\;+\;\tfrac{1}{2}\rho v_\text{exit}^{2}\right]. \tag{$\star$}$$

Modelling the friction

For turbulent flow, the wall stress at the dynamic-pressure scale $\rho v_h^2$ (with $v_h = Q/A$ the hose mean speed) gives

$$\Delta p_f\;=\;\frac{k}{2}\,\rho v_h^{2}, \qquad k\;\equiv\;\text{dimensionless friction coefficient (constant in $v$).}$$

The numerical constant $k$ depends on the hose’s length-to-diameter ratio and roughness in ways the problem has not handed us — but we do not need to derive it from first principles, because the no-nozzle operating point of the pump measures it.

Calibrating the friction from the baseline

In the no-nozzle case $v_\text{exit} = v_h = v_0$, so $(\star)$ becomes

$$\frac{P}{Q_0}\;=\;\rho g h\;+\;\frac{k+1}{2}\,\rho v_0^{2}.$$

Numerical evaluation with $P = 250\,\text{W}$ and $Q_0 = 4.167\times 10^{-4}\,\text{m}^{3}/\text{s}$:

$$\frac{P}{Q_0}\;=\;6.00\times 10^{5}\,\text{Pa},\qquad \rho g h\;=\;1.96\times 10^{5}\,\text{Pa},\qquad \tfrac{1}{2}\rho v_0^{2}\;=\;4.93\times 10^{3}\,\text{Pa}.$$

Solving for $k$:

$$(k+1)\,\tfrac{1}{2}\rho v_0^{2}\;=\;\frac{P}{Q_0}\,-\,\rho g h\;=\;4.04\times 10^{5}\,\text{Pa},$$ $$k+1\;=\;\frac{4.04\times 10^{5}}{4.93\times 10^{3}}\;\approx\;81.97, \qquad k\;\approx\;80.97.$$

So in the no-nozzle case roughly $33\,\%$ of the pump’s power goes to lifting the water, $67\,\%$ to wall friction, and well under $1\,\%$ to exit kinetic energy. The gardener’s hose is, energetically, a heater.

The same balance with a nozzle

The nozzle reduces the exit area to $fA$. In steady flow continuity demands $Q_1 = v_h A = v_\text{exit}\,fA$, so

$$v_\text{exit}\;=\;\frac{v_h}{f}.$$

The friction inside the hose is still set by $v_h$ (the hose geometry has not changed and the nozzle is short), but the kinetic-energy term now uses the exit speed. Equation $(\star)$ becomes

$$P\;=\;Q_1\!\left[\rho g h\;+\;\frac{k}{2}\,\rho v_h^{2}\;+\;\frac{1}{2 f^{2}}\,\rho v_h^{2}\right] \;=\;Q_1\!\left[\rho g h\;+\;\frac{k+1/f^{2}}{2}\,\rho v_h^{2}\right].$$

Defining the unknown ratio

$$\alpha\;\equiv\;\frac{Q_1}{Q_0}\;=\;\frac{v_{h,1}}{v_0},$$

and using $v_{h,1} = \alpha v_0$, the balance reads

$$\frac{P}{Q_0}\;=\;\rho g h\,\alpha\;+\;(k+1/f^{2})\,\tfrac{1}{2}\rho v_0^{2}\,\alpha^{3}. \tag{$\star\star$}$$

The cubic-in-$\alpha$ structure is the only complication that distinguishes this from the no-nozzle case. It comes from the kinetic-energy and friction terms scaling as $v_h^2$ once $Q_1 = \alpha Q_0$ is factored out.

Reducing to a numerical cubic

With $1/f^{2} = 1/0.15^{2} \approx 44.44$,

$$k\,+\,\tfrac{1}{f^{2}}\;\approx\;80.97 + 44.44\;\approx\;125.42,$$ $$(k+1/f^{2})\,\tfrac{1}{2}\rho v_0^{2}\;\approx\;125.42\times 4.928\times 10^{3}\,\text{Pa}\;\approx\;6.181\times 10^{5}\,\text{Pa}.$$

Plugging into $(\star\star)$, with $P/Q_0 = 6.000\times 10^{5}\,\text{Pa}$ and $\rho g h = 1.960\times 10^{5}\,\text{Pa}$:

$$6.181\times 10^{5}\,\alpha^{3}\;+\;1.960\times 10^{5}\,\alpha\;-\;6.000\times 10^{5}\;=\;0.$$

Solving the cubic

A natural starting estimate ignores gravity and friction’s adjustment, treating $\alpha\approx 1$ as the seed. Newton’s iteration with $f(\alpha) = 6.181\times 10^{5}\alpha^{3}+1.960\times 10^{5}\alpha-6.000\times 10^{5}$ and $f'(\alpha) = 1.854\times 10^{6}\alpha^{2}+1.960\times 10^{5}$ converges quickly:

• $f(1.000) = 6.181\times 10^{5}+1.960\times 10^{5}-6.000\times 10^{5} = 2.141\times 10^{5}$
• $f(0.900) = 4.508\times 10^{5}+1.764\times 10^{5}-6.000\times 10^{5} = 2.71\times 10^{4}$
• $f(0.884) \approx 2.5\times 10^{2}$
• $f(0.8838) \approx 6$

To within better than $1\,\%$,

$$\boxed{\;\alpha\;=\;\frac{Q_1}{Q_0}\;\approx\;0.884.\;}$$

So the volumetric flow rate falls by about $11.6\,\%$ once the nozzle is fitted: $Q_1 \approx 22.1\,\text{L}\cdot\text{min}^{-1}$.

Sanity check that the flow is still turbulent

The hose Reynolds number is now

$$R_1\;=\;\alpha R_0\;\approx\;0.884\times 3.7\times 10^{4}\;\approx\;3.3\times 10^{4},$$

still well above $2500$, so the $\Delta p_f \propto v^2$ scaling we used for the friction is consistent with the operating regime. (At the nozzle exit the Reynolds number based on the smaller diameter $d\sqrt{f}$ and larger speed $v_h/f$ is even higher, $\sim 9\times 10^{4}$, so the jet is also fully turbulent — but we never relied on that, since we ignored friction inside the short nozzle itself.)

Excess pressure on the dirty surface

When a free jet of speed $v_\text{jet}$ at atmospheric pressure $p_\text{atm}$ strikes a perpendicular surface, the streamline that passes through the stagnation point on the surface decelerates to rest. Bernoulli along that streamline gives

$$p_\text{atm}\;+\;\tfrac{1}{2}\rho v_\text{jet}^{2}\;=\;p_\text{stag},$$

so the excess pressure at the dirty surface above atmospheric is

$$\Delta p_\text{excess}\;=\;p_\text{stag}-p_\text{atm}\;=\;\tfrac{1}{2}\rho v_\text{jet}^{2}.$$

This is the pressure that does the cleaning. (Off the stagnation point the streamline pressure is lower, but the central stagnation pressure is the relevant scale for “force on the dirt directly under the jet”.)

Without a nozzle $v_\text{jet} = v_0$, with a nozzle $v_\text{jet} = v_{h,1}/f = \alpha v_0/f$. The ratio is

$$\frac{\Delta p_\text{excess,1}}{\Delta p_\text{excess,0}}\;=\;\frac{v_\text{jet,1}^{2}}{v_\text{jet,0}^{2}}\;=\;\frac{\alpha^{2}}{f^{2}}.$$

Plugging in $\alpha = 0.8838$, $f = 0.15$:

$$\frac{\alpha^{2}}{f^{2}}\;=\;\frac{0.7811}{0.02250}\;\approx\;34.7.$$

Hence

$$\boxed{\;\frac{\Delta p_\text{excess,1}}{\Delta p_\text{excess,0}}\;\approx\;34.7.\;}$$

In absolute terms the cleaning pressure rises from $\Delta p_\text{excess,0} = \tfrac{1}{2}\rho v_0^2 \approx 4.9\,\text{kPa}$ to $\Delta p_\text{excess,1} \approx 1.7\times 10^{5}\,\text{Pa} \approx 1.7\,\text{bar}$ — a couple of atmospheres of overpressure on the dirt, which is what makes nozzles useful.

Educational remarks

Why the gain in cleaning pressure ($\sim 35\times$) is less than the naive geometric factor ($1/f^{2}\approx 44\times$). Continuity alone would suggest $v_\text{exit}/v_0 = 1/f \approx 6.7$, hence $\Delta p_\text{excess} \propto v^2$ should rise by $1/f^2 \approx 44$. The actual gain is smaller because the flow rate itself drops by $\alpha \approx 0.884$, and the pressure factor goes as $\alpha^2/f^2$. The closure of the budget is: when you ask the pump to push water through a much higher kinetic-energy channel, the pump cannot maintain the same $Q$ at fixed $P$, so it backs off slightly and finds a new operating point.

The slow approach to the “blocked-pipe” limit. Suppose we shrank $f$ further. Equation $(\star\star)$ predicts $\alpha \propto f^{2/3}$ in the small-$f$ limit (because the kinetic-energy term then dominates), so $\Delta p_\text{excess} \propto \alpha^2/f^2 \propto f^{-2/3}$. The cleaning pressure does not diverge as $1/f^2$ would suggest — the pump’s finite power tames it. Practically, this means that there are diminishing returns to going to ever-narrower nozzles.

Why we may treat the pump as a constant-power source. The problem hands us “output power $P = 250\,\text{W}$” without a pump curve. In the language of pump engineering this is the simplest constant-power model: the pump moves the operating point along the hyperbola $\Delta p \cdot Q = P$. A real centrifugal pump has a more complex H–Q curve, but the constant-power idealisation is the only model that uses no information beyond what is given, and it is precisely consistent with the way $P$ enters energy balance $(\star)$.

Why we ignored friction inside the nozzle. The nozzle is short (its length is of order its diameter, both $\sim$ a centimetre at most), so its friction-loss contribution scales as $\Delta p_f^\text{nozzle}\sim 4\rho v_\text{exit}^2 \cdot L_n/d_n \sim 4\rho v_\text{exit}^2 \cdot O(1)$. Numerically that is at most a few times $\rho v_\text{exit}^2$ — comparable to the kinetic-energy term up to a numerical factor. Without a stated nozzle length, attempting to keep this term invents a free parameter; standard physics-Olympiad practice is to take the nozzle as a short, frictionless transition. The flat $1\,\%$ accuracy target is consistent with this approximation.

A consistency check on the budget at the new operating point. At $\alpha = 0.884$,

$$\frac{P}{Q_1}\;=\;\frac{P/Q_0}{\alpha}\;\approx\;6.79\times 10^{5}\,\text{Pa}.$$

Allocating: gravity $\rho g h = 1.96\times 10^5\,\text{Pa}$ ($29\%$), friction $\frac{k}{2}\rho v_{h,1}^2 = k\cdot \tfrac{1}{2}\rho v_0^2\cdot\alpha^2 \approx 80.97\times 4928 \times 0.781 \approx 3.12\times 10^5\,\text{Pa}$ ($46\%$), exit KE $\frac{1}{2f^2}\rho v_{h,1}^2 \approx 44.44\times 4928 \times 0.781 \approx 1.71\times 10^5\,\text{Pa}$ ($25\%$). Sum: $6.79\times 10^5\,\text{Pa}$. ✓ The KE term has gone from $<1\%$ of the budget to a quarter of it — consistent with our headline observation that the nozzle promotes KE to a leading channel.

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Summary of results

Part	Quantity	Symbolic answer	Numerical
(i)	Reynolds number from drag scales	$R = \rho v D/\mu$	(dimensionless)
(ii)	Flow type in the hose	$R \gg 2500$ → turbulent	$R \approx 3.7\times 10^{4}$
(iii)	Flow-rate ratio	$\alpha$ such that $\rho g h\,\alpha + (k+1/f^2)\tfrac{\rho v_0^2}{2}\alpha^3 = P/Q_0$	$\alpha \approx 0.884$
(iii)	Excess-pressure ratio	$\alpha^{2}/f^{2}$	$\approx 34.7$