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Fluid Mechanics · Chapter 6 of 10 · Intermediate

Momentum Analysis of Flow Systems

A jet that changes speed or direction pushes back. The control-volume momentum equation turns that idea into the force on a pipe bend, a vane, or a rocket, without ever solving the flow inside.

Readiness check

This chapter applies Newton's second law to a control volume. Tick only what you can do closed-notes.

Recall momentum as mass times velocity.
Compute the mass flow rate ṁ = ρAV.
Resolve forces and velocities into x and y components.
Apply Newton's second law (force equals rate of change of momentum).
Recall continuity from Chapter 5.

0 or 1 weak itemsContinue with this chapter.

2 weak itemsReview mass flow and vectors in Chapter 5.

3 or more weak itemsRevisit Newton's laws in Physics: Newton's Laws first.

The core idea

The net force on a control volume equals the rate at which momentum leaves minus the rate at which it enters: ΣF = Σṁ_outV_out − Σṁ_inV_in.

ΣF = Σṁ_outV_out − Σṁ_inV_inṁ = ρAVjet on plate: F = ṁV = ρAV²

Newton's second law, applied to all the fluid inside a chosen control volume, says the net external force equals the net rate of momentum outflow. Because it is a vector equation, you solve it component by component. The beauty is that you only need conditions at the inlet and outlet, not the messy flow in between, so a jet hitting a plate, water turning through a bend, or exhaust leaving a rocket all yield to the same recipe. Pressure forces on the boundary join the applied and reaction forces in ΣF.

The skill works when: you draw a control volume, take momentum flux as ṁV at each opening, and balance components.

The skill breaks down when: momentum is treated as a scalar, or pressure and reaction forces are left out of ΣF.

The concept. Momentum flows into and out of a control volume. The net external force equals the difference in momentum flux, found from conditions at the openings alone.

The skills, taught in order

Momentum analysis is Newton's law in control-volume form. Five skills cover the equation, control-volume choice, bends, vanes, and the correction factor.

6.1 The linear momentum equation

For a control volume, ΣF = Σṁ_outV_out − Σṁ_inV_in in steady flow. ΣF includes pressure forces on the boundary, body forces (weight), and the reaction holding the device. It is a vector equation, applied separately in each direction.

6.2 Choosing a control volume and signs

Pick a control volume that cuts the inlets and outlets where conditions are known, and a sign convention for each axis. Momentum flux is ṁV directed along the flow; carry velocity signs carefully, since a reversed jet contributes a 2ṁV change.

6.3 Forces on bends and nozzles

Water turning through a bend changes its momentum direction, so the bend must supply a force to redirect it, plus resist the pressure. The anchoring force is found by balancing momentum flux and pressure at the two faces. Nozzles and reducers work the same way.

6.4 Jets on vanes

A free jet striking a surface is pure momentum (atmospheric pressure throughout). On a fixed flat plate the normal force is F = ṁV. On a curved vane the force depends on the turning angle. For a vane moving at speed u, the relevant rate uses the relative velocity (V − u), the foundation of turbine theory.

6.5 The momentum-flux correction factor

Real velocity profiles are non-uniform, so the true momentum flux differs from ṁV_avg by a factor β: about 1.33 for laminar pipe flow and near 1.02 to 1.05 for turbulent. It is often close enough to 1 to ignore, but matters for laminar or precise work.

Flow	β (momentum-flux factor)
Uniform	1.00
Laminar pipe	1.33
Turbulent pipe	1.02 to 1.05

Engineering connection: pipe-bend anchors, nozzle thrust, sprinkler reaction, rocket and jet propulsion, and the blade forces in turbomachinery all come from this one equation.

Worked example 1: jet on a flat plate

A water jet (ρ = 998 kg/m³) 5 cm in diameter strikes a stationary flat plate head-on at 20 m/s. Find the force on the plate.

Figure 1. The plate destroys the jet's axial momentum, so the force equals the incoming momentum flux ṁV. The water spreads sideways, carrying no net axial momentum away.

ProblemFind the force on the plate in Figure 1.
Given / findρ = 998 kg/m³, d = 0.05 m, V = 20 m/s, stationary plate. Find F.
AssumptionsSteady free jet (atmospheric pressure), water spreads with no axial velocity, friction negligible.
ModelMomentum equation in the jet direction: the plate removes all axial momentum, so F = ṁV.
Equationsṁ = ρAV F = ṁV = ρAV²
SolveA = π(0.025)² = 1.963×10⁻³ m². ṁ = 998 × 1.963×10⁻³ × 20 = 39.2 kg/s. F = ṁV = 39.2 × 20 = 784 N.
CheckEquivalently F = ρAV² = 998 × 1.963×10⁻³ × 20² = 784 N, matching. The force grows with the square of the jet speed, so doubling V quadruples the force.
ConclusionStopping a jet's axial momentum costs a force ṁV. Only the inlet and outlet states are needed, the power of the control-volume method.

Result. The jet pushes the plate with 784 N.

Worked example 2: jet turned by a vane

The same jet (5 cm, 20 m/s, ρ = 998 kg/m³) is now deflected through 90° by a stationary curved vane, leaving at the same speed. Find the magnitude and direction of the force on the vane.

Figure 2. Turning the jet 90° changes momentum in both directions, so the vane force has equal x and y components and a magnitude √2 times the single-component value.

ProblemFind the force on the 90° vane in Figure 2.
Given / findSame jet (ṁ = 39.2 kg/s), V = 20 m/s, turned 90°, speed unchanged. Find F and its direction.
AssumptionsFree jet at atmospheric pressure, no friction so speed is preserved, stationary vane.
ModelApply the momentum equation in x and y; the jet enters in +x and leaves in +y.
EquationsF_x = ṁ(0 − V) = −ṁV F_y = ṁ(V − 0) = ṁV F = √(F_x² + F_y²) = √2 ṁV
SolveEach component has magnitude ṁV = 39.2 × 20 = 784 N. So F = √2 × 784 = 1108 N, directed at 45° to the incoming jet.
CheckTurning the jet (1108 N) loads the vane harder than merely stopping it (784 N), because momentum changes in two directions, not one. A full 180° turn would give 2ṁV = 1568 N, the maximum.
ConclusionThe more the jet turns, the larger the force. Using the relative velocity for a moving vane extends this directly to turbine and pump blades.

Result. Force on the vane 1108 N, at 45° to the jet (components 784 N each).

Misconceptions and diagnostics

Mistake	Symptom	Diagnostic question	Correction
Treating momentum as a scalar	Wrong direction or magnitude	"Did I work in components?"	Momentum is a vector; balance each axis separately.
Omitting pressure forces	Bend anchor force too low	"Are there pressure forces on the faces?"	Include gauge-pressure forces at inlet and outlet in ΣF.
Using absolute velocity on a moving vane	Turbine force overestimated	"Is the surface moving?"	Use the relative velocity (V − u) for a moving vane.
Ignoring β for laminar flow	Momentum flux underestimated	"Is the profile uniform?"	Apply β (1.33 laminar) when the profile is non-uniform.

Practice ladder

Level 1 · Direct skill

A 3 cm water jet at 15 m/s hits a flat plate normally. Find the force (ρ = 998 kg/m³).

Show answer

A = π(0.015)² = 7.07×10⁻⁴ m²; ṁ = 998 × 7.07×10⁻⁴ × 15 = 10.6 kg/s; F = ṁV = 10.6 × 15 = 159 N. Or F = ρAV² = 159 N.

Level 2 · Mixed concept

The Worked Example 2 jet is turned through 180° (a full reversal) instead of 90°. Find the force.

Show answer

The axial momentum reverses, so ΔV = 2V and F = ṁ(2V) = 39.2 × 40 = 1568 N, all along the jet axis. A full reversal doubles the single-direction force, the maximum a vane can develop.

Level 3 · Independent problem

A nozzle reduces a pipe from 10 cm to 4 cm; water flows at 0.02 m³/s. Estimate the momentum-flux change that the nozzle anchor must react (ignore pressure).

Show answer

V₁ = Q/A₁ = 0.02/(π·0.05²) = 2.55 m/s; V₂ = 0.02/(π·0.02²) = 15.9 m/s. ṁ = ρQ = 998 × 0.02 = 19.96 kg/s. Momentum change = ṁ(V₂ − V₁) = 19.96 × 13.4 = 267 N (a real anchor must also carry the pressure force).

Level 4 · Transfer to real engineering

Find a real device that relies on momentum change (a fire-hose nozzle, a Pelton wheel, a rocket, a pipe elbow). Identify the inlet and outlet momentum and the reaction force.

What good work looks like

A clear control volume, momentum flux ṁV at each opening, pressure forces where relevant, and a reaction force with direction.

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that I balanced momentum in components and included pressure forces."

"Give me five flows; I will choose the control volume and label the momentum flux."

"Compute the bend force." Setting up ΣF = ṁΔV yourself is the skill.

"What is the thrust?" Identifying inlet and outlet momentum is the point.

Portfolio task

Analyse one momentum problem: choose a control volume, write the momentum equation in components, include pressure where needed, and find the anchoring or reaction force.

Must include: a labelled control volume, ṁV at each opening, and a reaction force with direction.

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. Write the control-volume momentum equation.

ΣF = Σṁ_outV_out − Σṁ_inV_in (steady), a vector equation.

2. What does ΣF include?

Pressure forces on the boundary, body forces (weight), and the reaction or anchoring force.

3. Force of a jet on a normal flat plate?

F = ṁV = ρAV².

4. How does a moving vane change the analysis?

Use the relative velocity (V − u) in the momentum flux.

5. What is the momentum-flux correction factor β?

It corrects ṁV_avg for a non-uniform profile: ~1.33 laminar, ~1.02 to 1.05 turbulent.

TodayFinish this quiz and Levels 1 and 2 of the ladder.

+1 dayRe-derive the plate and vane forces from a blank page.

+3 daysOne jet-force and one bend problem.

+7 daysUse dimensionless reasoning in Chapter 7.

+30 daysApply momentum to turbine blades in Chapter 10.

Textbook mapping

Item	Mapping
Primary source	Çengel and Cimbala, Fluid Mechanics: Fundamentals and Applications, Chapter 6 (Momentum Analysis of Flow Systems)
Cross-reference	White, Ch. 3 · Munson, Ch. 5
Core topics	6.1 Momentum equation · 6.2 Control volume and signs · 6.3 Bends and nozzles · 6.4 Jets on vanes · 6.5 Correction factor β
Engineering connection	Pipe-bend anchors, nozzle and rocket thrust, and turbine blade forces.
Read next	Chapter 7: Dimensional Analysis and Modeling.