NBPhO 2026

Difficulty medium

Prerequisites

• Continuity and Bernoulli for incompressible flow
• Wall-stress versus pressure-drop relation in a pipe
• Dimensional analysis of dynamic viscosity $\mu$ (units $\text{Pa}\cdot\text{s}$)
• Power = pressure rise × volumetric flow rate for a pump
• Solving a cubic equation numerically (Newton iteration / bisection)

Learning objectives

• Derive the Reynolds number $R = \rho v D/\mu$ as the ratio of turbulent to laminar drag scales
• Use the heuristic $\mu v/\delta \sim \rho v^2$ to identify the boundary-layer thickness $\delta \sim D/R$
• Apply $\Delta p_f \propto v^2$ for fully developed turbulent flow and calibrate the unknown coefficient from a single measured operating point
• Write the pump energy balance $P = Q[\rho g h + \Delta p_f + \tfrac{1}{2}\rho v_\text{exit}^2]$ and identify which channel dominates
• Recognise that adding a nozzle inflates the kinetic-energy term by $1/f^2$ and turns the balance into a cubic in $\alpha = Q_1/Q_0$
• Identify the excess pressure on a target wall as the Bernoulli stagnation pressure $\tfrac{1}{2}\rho v_\text{jet}^2$

Watch out for

• The friction inside the hose is set by the hose mean speed $v_h$, not by the exit speed $v_\text{exit} = v_h/f$. Using $v_\text{exit}$ for friction over-estimates the friction loss by $1/f^2 \approx 44$ and yields a much smaller flow rate.
• The kinetic-energy term in the power balance uses the exit speed, not the hose speed. Forgetting that the nozzle accelerates the jet drops the leading new physics introduced by attaching the nozzle.
• Treating $\Delta p_f$ as if its prefactor were $\rho v^2$ on the nose (with no calibration) gives a friction loss roughly $300\times$ too large, because the order-of-magnitude estimate $\mu v/\delta \sim \rho v^2$ does not fix the empirical numerical coefficient. Always calibrate from a measured operating point.
• Assuming the nozzle leaves $Q$ unchanged and only inflates the exit pressure by $1/f^2$ misses the back-reaction: at fixed pump power the flow rate must drop. The correct answer for the pressure ratio is $\alpha^2/f^2$, not $1/f^2$.
• The excess pressure on the dirty surface is the stagnation pressure $\tfrac{1}{2}\rho v_\text{jet}^2$, not the static pressure inside the jet (which is atmospheric in a free jet) and not the dynamic pressure inside the hose.