Momentum

Moving fluid carries momentum; changing its speed or direction creates force. Learn the momentum equation for control volumes, jet forces on plates and vanes, pipe-bend reactions, thrust and water hammer basics.

A fire hose kicks back, a jet engine pushes an aircraft, water swirling through a pipe bend tries to tear it off its supports. All are momentum effects: force equals the rate of change of momentum of the fluid.

The control-volume momentum equation sizes thrust blocks, turbine blades, rockets and nozzles.

An analogy

Catching a ball: stopping it needs a force. Catching a steady stream of balls needs a steady force — equal to balls per second × speed change.

Force from changing momentum

Newton's second law applied to a flowing fluid: the force needed equals mass flow × change in velocity (a vector). Stopping a jet, turning it, or accelerating it all require force — and produce an equal reaction on the device.

Control volume approach

Draw a box around the region, include pressure forces at inlets/outlets, weight and support reactions, then apply ΣF = Σṁ V_out − Σṁ V_in in each direction. It's a free-body diagram for flowing fluid.

Bends, reducers and thrust blocks

Buried water mains need concrete thrust blocks at bends because pressure and momentum forces push the fitting outward — often tens of kN in large pipes.

Angular momentum and turbomachines

Euler's turbomachine equation, T = ṁ(r₂V_t2 − r₁V_t1), comes from angular momentum and is the foundation of pump, fan, compressor and turbine design. Sudden valve closure causes water hammer, a pressure surge Δp = ρ c ΔV (c ≈ 1200–1400 m/s in pipes).

Formulas

• Momentum equation (steady): ΣF = ṁ (V_out − V_in)
• Jet on a flat plate (normal): F = ρ Q V = ρ A V²
• Jet on a moving vane: F = ρ A (V − u)² (1 − cos θ)
• Rocket / jet thrust: F = ṁ(V_e − V₀) + (p_e − p₀)A_e

Worked example: Jet on a flat plate

Given: Ø50 mm water jet at 20 m/s; Plate normal to jet

Find: Force.

• Q = π/4 × 0.05² × 20 = 0.0393 m³/s
• F = 998 × 0.0393 × 20 = 784 N

Answer: ≈ 784 N

Worked example: Fire hose nozzle reaction

Given: Nozzle exit Ø25 mm at 30 m/s

Find: Reaction force on the firefighter (exit momentum).

• ṁ = 998 × π/4 × 0.025² × 30 = 14.7 kg/s
• F ≈ ṁV = 14.7 × 30 = 441 N

Answer: ≈ 440 N — why big hoses need two people

Common mistakes

• Wrong: Treating velocity as a scalar in the momentum equation. — Right: Resolve into components; direction changes matter.
• Wrong: Forgetting pressure forces at inlet/outlet sections. — Right: Include pA terms (gauge) at each boundary.

Applications

• Pipeline thrust blocks: Sized from pressure + momentum forces at bends, tees and dead ends.
• Propulsion: Jet engines, rockets, waterjets and propellers generate thrust by accelerating fluid.
• Impulse turbines: Pelton wheels turn high-speed jets almost 180° to capture their momentum.

Key takeaways

• ΣF = ṁ(V_out − V_in) as vectors.
• Jet on plate: F = ρAV².
• Include pressure forces at boundaries.
• Angular momentum → Euler's turbomachine equation.

Going further

• Propeller theory: Actuator-disk momentum theory gives ideal efficiency 2/(1 + V_e/V₀) — bigger, slower jets are more efficient (turbofans).
• Surge protection: Slow-closing valves, surge tanks and air vessels limit water-hammer pressures.
• Momentum in CFD: The Navier–Stokes momentum equations add viscous stress to this control-volume balance.