How hard does a plate have to push back on a jet of water it's deflecting? Drag the controls and watch the force change.

The idea

A moving stream of fluid carries momentum with it. When something in its path bends that stream — a turbine blade, a fire‑hose nozzle, the plate to the right — the fluid's momentum changes direction, and Newton's second law says that change has to show up somewhere as a force on whatever did the bending.

This setup

A jet of water in an inlet pipe splits into two pipes, and the flow distribution between the two outlet directions depends on the angle of the pipes. Force must be applied to the plate on which the two pipes rest to prevent the plate from moving. Drag the sliders to change the flow rate, the pipe diameters, and the tilt, and watch the split and the normal force Fn respond.

Sweep the outlet diameter end to end — does Fn move?

It doesn't. The normal force comes entirely from the inlet stream — the outlet flows run parallel to the plate, so they carry no momentum perpendicular to it, whatever their speed or pipe size. Fn only cares about the mass flow rate, the inlet diameter, and θ — the outlet diameter you're dragging changes the outlet velocities and the flow split, not the force.

Model assumptions Steady, incompressible flow of water (ρ = 1000 kg/m³); atmospheric pressure and equal elevation at the inlet and both outlets; a frictionless plate, so it exerts no force along its own surface; gravity and viscous losses neglected.

Inlet area and velocity, from continuity, with water at ρ = 1000 kg/m³:

\[ A_1=\frac{\pi}{4}D_{in}^{2} \qquad\qquad v_1=\frac{\dot m_1}{\rho A_1} \]

where $\dot m_1$ is the mass flow rate into the tube.

The continuity equation relates the outlet velocities to the inlet velocity,

\[ A_1 v_1 = A_A v_A + A_B v_B = A_A\left(v_A+v_B\right) \]

since $A_A = A_B$.

The momentum balances used in the simulation — normal to the plate, giving the force needed to hold it stationary, and along the frictionless plate, which applies no tangential force to the jets:

\[ F_n=\dot m_1 v_1\sin(\theta) \]
\[ \dot m_1 v_1\cos(\theta)+\dot m_B v_B=\dot m_A v_A \]

Solving the continuity equation and the tangential momentum balance together gives the outlet velocities and the flow split shown in the figure:

\[ v_A=\frac{v_1}{2}\left(\frac{A_1}{A_A}+\cos\theta\right) \qquad\qquad v_B=\frac{v_1}{2}\left(\frac{A_1}{A_A}-\cos\theta\right) \]
\[ \dot m_A=\dot m_1\left(\frac{1+(A_A/A_1)\cos\theta}{2}\right) \qquad\qquad \dot m_B=\dot m_1\left(\frac{1-(A_A/A_1)\cos\theta}{2}\right) \]

The outlet pipe diameter is limited to the inlet pipe diameter, so that $\dot m_B\ge 0$.

This simulation was created in Python in the Department of Chemical and Biological Engineering at University of Colorado Boulder by Ankur Gupta and modified for LearnChemE.com by John L. Falconer using Claude AI. It was prepared with financial support from the National Science Foundation (DUE 2336987 and 2336988). Address any questions or comments to LearnChemE@gmail.com.