---
title: "Momentum equation and its applications"
subject: "Fluid Mechanics"
chapter: 4
key: "semester_4/fluid_mechanics_mechanical:momentum-equation-control-volume-applications"
source: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/momentum-equation-control-volume-applications
site: BCT Engineering
---

# Momentum equation and its applications

Fluid Mechanics · Chapter 4: Basic Equations for Fluid Flow

- Asked in: year not given, 8 marks
- Video: [Fluid Mechanics: The Momentum Equation](https://www.youtube.com/watch?v=_fCv98W0L90) — Caspar Hewett
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/momentum-equation-control-volume-applications
- Whole chapter: https://www.bctengineer.com/md/semester_4/fluid_mechanics_mechanical/chapter-4

## Question

Derive the linear momentum equation for a control volume in steady flow. State the assumptions, and explain how it is applied to (i) a pipe bend, (ii) a jet striking a vane, and (iii) a jet-propelled body.

## Answer

#### Derivation
Newton's second law for a system: the resultant external force equals the time rate of change of linear momentum of the system.

$$
\sum \vec F = \frac{d(m\vec V)}{dt}\bigg|_{system}
$$

Using the Reynolds transport theorem to convert from the system to a control volume (CV):

$$
\sum \vec F = \frac{\partial}{\partial t}\int_{CV}\rho\vec V\,dV + \int_{CS}\rho\vec V(\vec V\cdot\hat n)\,dA
$$

For **steady flow** the first term is zero. For a CV with one inlet (1) and one outlet (2), with uniform velocity over each area:

$$
\sum \vec F = \dot m(\vec V_2 - \vec V_1) = \rho Q(\vec V_2 - \vec V_1)
$$

In components:

$$
\sum F_x = \rho Q(V_{2x} - V_{1x}), \qquad \sum F_y = \rho Q(V_{2y} - V_{1y})
$$

$\sum \vec F$ includes the pressure forces on the control surface, body forces (weight) and the force $\vec R$ exerted by the solid boundary on the fluid.

**Assumptions:** steady flow, uniform velocity at the inlet and outlet sections, a fixed control volume, and an inertial frame.

#### Applications
1. **Pipe bend:** take the CV as the fluid inside the bend. The forces are the end pressure forces $p_1A_1$ and $p_2A_2$ and the reaction $R$ from the bend. The change in the direction and speed of the flow gives the force on the bend, which is used to design anchor blocks and thrust blocks.

2. **Jet on a vane:** the vane changes the direction of the jet. For a smooth vane deflecting the jet through $\theta$, the force in the jet direction is $F = \rho a V^2(1-\cos\theta)$ for a fixed vane and $F = \rho a (V-u)^2(1-\cos\theta)$ for a single vane moving at speed $u$, found from the change of momentum using the relative velocities. This is used in impulse turbines (Pelton wheel).

3. **Jet propulsion:** a fluid is accelerated backward and the reaction thrust pushes the body forward. For ships with water jets and for aircraft engines, the thrust is $F = \dot m(V_{jet} - V_{body})$, where $V_{jet}$ is the jet velocity relative to the body. For a rocket, $F = \dot m V_e + (p_e - p_a)A_e$.

Other applications: force on a nozzle, thrust on a sluice gate, and force in a hydraulic jump.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For steady flow with one inlet and one outlet, the linear momentum equation is:
   A. Sum F = rho Q (V1 + V2)
   B. Sum F = rho Q (V2 - V1)
   C. Sum F = rho Q V2 only
   D. Sum F = Q (V2 - V1)/rho
   Answer: B. The first term (time rate of change inside the CV) vanishes in steady flow, leaving the net momentum outflow. (Key point: Steady-flow momentum equation)

2. Which forces make up the resultant force sum F in the control volume equation?
   A. Pressure forces on the control surface, body force (weight) and the reaction from the solid boundary
   B. Only pressure forces
   C. Only the weight of the fluid
   D. Only viscous forces inside the CV
   Answer: A. Sum F includes surface pressure forces, body forces and the force exerted by the solid boundary on the fluid. (Key point: Forces included in sum F)

3. For a single vane moving at speed u away from the jet, the force is rho a (V - u)^2 (1 - cos theta) because:
   A. the vane is heavier
   B. the mass striking the vane per second and the relative velocity both equal (V - u)-based values
   C. the jet loses pressure
   D. friction reduces the velocity
   Answer: B. The vane receives only rho a (V - u) of mass per second at a relative speed (V - u). (Key point: Jet on a moving vane)

4. The thrust on a body propelled by a water jet is:
   A. m_dot x V_jet
   B. m_dot (V_jet - V_body)
   C. m_dot (V_jet + V_body)
   D. rho g Q
   Answer: B. Thrust equals the rate of change of momentum of the water relative to the body: m_dot (V_jet - V_body). (Key point: Jet propulsion thrust)

## Related in Fluid Mechanics

- [Jet striking fixed and moving curved vanes](https://www.bctengineer.com/question.md?key=semester_4%2Ffluid_mechanics_mechanical%3Ajet-on-fixed-and-moving-vane) — Fluid Mechanics, chapter 4
- [Force on a reducing pipe bend](https://www.bctengineer.com/question.md?key=semester_4%2Ffluid_mechanics_mechanical%3Areducing-bend-force-numerical) — Fluid Mechanics, chapter 4
- [Laminar oil flow in a pipe](https://www.bctengineer.com/question.md?key=semester_4%2Ffluid_mechanics_mechanical%3Alaminar-oil-pipe-pressure-drop) — Fluid Mechanics, chapter 6
- [Thrust and efficiency of a water-jet ship](https://www.bctengineer.com/question.md?key=semester_4%2Ffluid_mechanics_mechanical%3Aship-water-jet-propulsion) — Fluid Mechanics, chapter 4
- [Hagen-Poiseuille flow in a circular tube](https://www.bctengineer.com/question.md?key=semester_4%2Ffluid_mechanics_mechanical%3Ahagen-poiseuille-derivation-circular-tube) — Fluid Mechanics, chapter 6
- [Types of flow, streamline, pathline, streakline](https://www.bctengineer.com/question.md?key=semester_4%2Ffluid_mechanics_mechanical%3Aflow-types-streamline-pathline-streakline) — Fluid Mechanics, chapter 3

## Related in other subjects

- [Linear momentum and impulse-momentum equation](https://www.bctengineer.com/question.md?key=semester_5%2Ffluid_machines%3Alinear-momentum-impulse-equation) — Fluid Machines, chapter 1
- [Derive continuity equation for unsteady pipe flow](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Acontinuity-equation-unsteady-pipe-flow) — Hydraulics, chapter 4
- [Water jet on fixed vane, 100 mm2](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Anozzle-fixed-vane-force-100mm2) — Applied Mechanics (Dynamics), chapter 3
- [Force by vane on steady stream](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Asteady-stream-force-vane-deduce) — Applied Mechanics (Dynamics), chapter 3