Momentum equation applied to control volumes
Linear momentum equation for control volumes: forces to include, one-dimensional form, moving CVs, and applications to jets on plates, pipe bends and jet-engine and rocket thrust.
Drafted with Aria, reviewed by the AiCanCode.org team. Spotted an error? Use Give Feedback at the bottom of the page.
Why it matters
The control-volume momentum equation gives forces without needing the detailed flow inside: the thrust of a jet engine or rocket, the load on a pipe bend or a thrust reverser, the force of a jet on a vane, and the drag of a body found from a wake survey in a wind tunnel. It is Newton's second law written for fluid that flows through a region.
Key ideas
From system to control volume. Newton's second law applies to a fixed mass: ΣF = d(mV)/dt. The Reynolds transport theorem turns it into a control-volume statement: the sum of forces on the fluid in the CV equals the rate of change of momentum stored inside plus the net outflow of momentum across the control surface.
It is a vector equation. Write it separately for x, y and z. Velocities carry signs in the chosen axes; the mass flux ρ(V·n)dA is positive for outflow and negative for inflow.
Forces to include.
- Surface forces: pressure on every part of the control surface (always acting inward, i.e. along −n), and shear stress where the CS cuts through fluid next to walls (usually negligible at inlets and outlets).
- Reaction forces from solid boundaries cut by the CS (bolts, supports, the wall of a bend). This is usually the unknown.
- Body forces: weight of the fluid and any structure inside, if vertical. Use gauge pressure when atmospheric pressure acts all round the outside; the atmospheric parts then cancel.
One-dimensional form. With uniform properties at each inlet and outlet and steady flow, ΣF = Σ(ṁV)_out − Σ(ṁV)_in. This is the form used in almost every problem. Non-uniform profiles need a momentum correction factor β = (1/(AV̄²))∫u² dA (β = 4/3 for laminar pipe flow, about 1.02 for turbulent).
Choosing the control volume. Cut the CS where you know velocities and pressures (or where p is atmospheric), and cut through the supports whose force you want. Make inlets and outlets normal to the flow.
Moving and accelerating control volumes. For a CV moving at constant velocity (a vane on a cart, an engine in steady flight), use velocities relative to the CV; the equation keeps the same form. An accelerating CV such as a rocket needs an extra inertial term, which leads to the rocket equation.
Typical results.
- Jet on a fixed flat plate normal to it: F = ρAV².
- Jet on a plate moving away at u: F = ρA(V − u)².
- Jet engine thrust: T = (ṁ_a + ṁ_f)V_e − ṁ_a·V∞ + (p_e − p_a)A_e.
- Rocket: T = ṁV_e + (p_e − p_a)A_e.
- Wake survey: the momentum deficit in the wake behind a body equals its drag.
Formulas
ΣF = ∂/∂t ∭ρ·V dV + ∯V·ρ·(V·n) dA
- ΣF: total force on the fluid in the CV (N); V: velocity vector (m/s); ρ: density (kg/m³); n: outward unit normal.
ΣF_x = Σ(ṁ·V_x)_out − Σ(ṁ·V_x)_in
- Steady, uniform inlets and outlets; ṁ in kg/s, V_x in m/s; similar for y and z.
F = ρ·A·V² (fixed plate, normal jet) and F = ρ·A·(V − u)² (single plate moving at u)
- A: jet area (m²); V: jet speed (m/s); u: plate speed (m/s).
T = (ṁ_a + ṁ_f)·V_e − ṁ_a·V∞ + (p_e − p_a)·A_e
- T: thrust (N); ṁ_a, ṁ_f: air and fuel mass flow (kg/s); V_e: exhaust speed (m/s); V∞: flight speed (m/s); p_e, p_a: exit and ambient pressure (Pa); A_e: exit area (m²).
β = (1/(A·V̄²))·∫u² dA
- Momentum correction factor (dimensionless).
Worked examples
Example 1 (standard): jet on a plate. Given: water jet, d = 50 mm, V = 20 m/s, hits a flat plate normally.
- A = (π/4)(0.05²) = 1.9635×10⁻³ m².
- Fixed plate:
F = ρ·A·V²= 1000 × 1.9635×10⁻³ × 20² = 785.4 N. - Plate moving away at u = 5 m/s:
F = ρ·A·(V − u)²= 1000 × 1.9635×10⁻³ × 15² = 441.8 N. Answer: 785 N fixed; 442 N moving.
Example 2 (GATE level): 90° pipe bend in a horizontal plane. Given: d = 0.2 m throughout, water Q = 0.1 m³/s, inlet flowing in +x with p₁ = 200 kPa (gauge), outlet flowing in +y with p₂ = 190 kPa (gauge). Find the force of the water on the bend.
- A = (π/4)(0.2²) = 0.031416 m², V = Q/A = 3.183 m/s, ṁ = 100 kg/s, ṁV = 318.3 N.
- Let R be the force of the bend on the fluid. x:
R_x + p₁·A = ṁ(0 − V), so R_x = −318.3 − 6283.2 = −6601.5 N. - y: outlet pressure pushes on the CV in −y:
R_y − p₂·A = ṁ(V − 0), so R_y = 318.3 + 5969.0 = 6287.3 N. - Force of water on the bend is −R: F_x = +6601.5 N, F_y = −6287.3 N.
- Magnitude = √(6601.5² + 6287.3²) = 9116 N. Answer: ≈ 9.12 kN, directed outward from the corner, about 44° below the +x axis. Note how the pressure terms dominate the momentum terms.
Example 3 (GATE level): turbojet thrust. Given: ṁ_a = 50 kg/s, ṁ_f = 1 kg/s, V∞ = 250 m/s, V_e = 600 m/s, A_e = 0.3 m², p_e − p_a = 20 kPa.
T = (ṁ_a + ṁ_f)·V_e − ṁ_a·V∞ + (p_e − p_a)·A_e.- = 51 × 600 − 50 × 250 + 20 000 × 0.3 = 30 600 − 12 500 + 6000. Answer: T = 24.1 kN.
Common mistakes
- Treating it as a scalar equation. Write each component with signs.
- Forgetting the pressure forces at inlets and outlets, which often dominate in pipe bends.
- Sign errors for pressure: pressure on the CS always pushes into the CV.
- Mixing the force on the fluid with the force on the structure; they are equal and opposite.
- Using absolute velocities on a moving vane instead of velocities relative to the vane.
- Using Q × (V_out − V_in) without ρ, or using ṁ = ρAV with the wrong area.
- Forgetting the ram drag ṁ_a·V∞ or the pressure thrust in engine problems.
For GATE AE
Expect: thrust of turbojets, turbofans and rockets including the pressure-thrust term; force on stationary and moving vanes and plates; forces on bends, nozzles and reducers; and drag from a wake velocity profile. Practise drawing the control volume first, listing every force, and writing x and y equations separately. Thrust questions are frequent in the aerospace paper, so learn the general thrust equation thoroughly.
Quick check
- A jet of area 0.001 m² and speed 30 m/s (water) hits a fixed plate normally. What is the force?
- If the velocity in a duct doubles at fixed area and density, by what factor does the momentum flux change?
- Write the rocket thrust equation.
- Which way does pressure act on an inlet face of a control volume?
Answers: 1. 900 N 2. Four times 3. T = ṁV_e + (p_e − p_a)A_e 4. Into the control volume, along the flow direction
Interview questions
All Fluid Mechanics interview questionsTry answering each one aloud before you open it.
1.What is the momentum equation in fluid mechanics, and how is it applied to control volumes?Concept
The momentum equation in fluid mechanics is derived from Newton's second law and is used to analyze the motion of fluid within a control volume. It states that the rate of change of momentum of a fluid is equal to the sum of the external forces acting on it. In control volumes, it is applied by considering the inflow and outflow of momentum across the control surfaces and the forces acting on the fluid within the volume.
2.Explain the difference between a control volume and a control surface in the context of fluid mechanics.Concept
A control volume is a defined region in space through which fluid flows, and it is used to analyze the behavior of the fluid within that region. A control surface, on the other hand, is the boundary of the control volume. It is the surface through which the fluid enters or exits the control volume. The control surface is crucial for applying the momentum equation as it helps in determining the flow of momentum across the boundaries.
3.Why is the momentum equation important in the design of aerospace vehicles?Application
The momentum equation is important in the design of aerospace vehicles because it helps engineers understand how forces and momentum are transferred within the fluid around the vehicle. This understanding is crucial for predicting the aerodynamic forces and moments acting on the vehicle, which in turn affects its stability, control, and performance. Accurate application of the momentum equation ensures that the vehicle can withstand the forces encountered during flight.
4.What happens if the control volume is not properly defined when applying the momentum equation?Application
If the control volume is not properly defined, it can lead to incorrect calculations of the momentum flux and external forces. This can result in errors in predicting the behavior of the fluid and the forces acting on the system. A poorly defined control volume may miss important inflows or outflows, leading to an incomplete analysis and potentially flawed design decisions.
5.How does the choice of control volume affect the complexity of solving the momentum equation?Application
The choice of control volume can significantly affect the complexity of solving the momentum equation. A well-chosen control volume simplifies the analysis by aligning with the natural boundaries of the problem, reducing the number of terms in the equation. Conversely, a poorly chosen control volume may introduce unnecessary complexity by requiring additional terms to account for artificial boundaries, making the problem more difficult to solve.
6.Explain how the momentum equation is used to analyze the thrust produced by a jet engine.Application
The momentum equation is used to analyze the thrust produced by a jet engine by considering the control volume around the engine. The equation accounts for the momentum of the air entering the engine and the momentum of the exhaust gases leaving the engine. The difference in momentum between the inflow and outflow, along with the pressure forces acting on the control surfaces, determines the thrust produced by the engine.
7.What role do external forces play in the momentum equation for a control volume?Concept
External forces play a crucial role in the momentum equation for a control volume as they contribute to the change in momentum of the fluid within the volume. These forces can include pressure forces, gravitational forces, and any other body forces acting on the fluid. The sum of these external forces, along with the momentum flux across the control surfaces, determines the overall change in momentum within the control volume.
8.Consider a steady flow through a fixed duct section. If the velocity is doubled with the same density and area, what happens to the momentum flux across that section?Application
Momentum flux is ṁV = ρAV², because the mass flow ρAV itself carries the velocity. Doubling V therefore doubles the mass flow and doubles the momentum carried by each kilogram, so the momentum flux becomes four times larger. This square law is why jet forces and thrust rise so steeply with jet speed.
9.Calculate the thrust produced by a jet engine if the mass flow rate of air is 100 kg/s, the velocity of air entering the engine is 200 m/s, and the velocity of exhaust gases is 600 m/s.Numerical
Neglecting fuel mass flow and assuming the nozzle is perfectly expanded (exit pressure equal to ambient), thrust T = ṁ(V_e − V_in) = 100 × (600 − 200) = 40 000 N = 40 kN. In a full analysis you would add the fuel flow to the exit mass flow and the pressure thrust (p_e − p_a)A_e, so it is worth stating these assumptions in an interview.
10.A fluid flows steadily through a pipe with a cross-sectional area of 0.5 m². If the velocity of the fluid is 3 m/s and the density is 1.2 kg/m³, calculate the momentum flux across the pipe's cross-section.Numerical
The momentum flux across the pipe's cross-section can be calculated using the formula: Momentum flux = ρ * A * V², where ρ = 1.2 kg/m³, A = 0.5 m², and V = 3 m/s. Momentum flux = 1.2 kg/m³ * 0.5 m² * (3 m/s)² = 1.2 kg/m³ * 0.5 m² * 9 m²/s² = 5.4 kg·m/s². Therefore, the momentum flux is 5.4 N.
Finished this topic? Mark it so your progress, study plan and readiness keep up.
Stuck on something here?