Control volume analysis: continuity and momentum equations

Why it matters

Control-volume analysis lets you find the force on a pipe bend, the reaction on a nozzle, the thrust of a jet or the torque on a pump impeller without knowing every detail of the flow inside. It is the method behind the anchor-block design of pipelines, jet-impact problems and the Euler turbomachine equation used later for pumps, turbines and turbochargers.

Key ideas

• System vs control volume. A system is a fixed mass of fluid that moves and deforms. A control volume (CV) is a region of space you choose, bounded by a control surface (CS) that fluid crosses. Conservation laws are written for systems; the Reynolds transport theorem converts them to CV form: rate of change inside the CV plus net outflow through the CS.
• Choosing the CV. Cut the CS perpendicular to the flow where velocity and pressure are known (inlets and outlets), and cut through supports or bolts where you want the reaction force.
• Conservation of mass (continuity). For steady flow the mass flow rate in equals the mass flow rate out. For incompressible flow this reduces to equal volume flow rates, so velocity rises where area falls. For unsteady flow, any imbalance changes the mass stored inside (for example a tank filling).
• Linear momentum (Newton's second law for a CV). For steady flow, the vector sum of all external forces on the fluid in the CV equals the net rate of momentum outflow. External forces are:
  • pressure forces on the control surface (use gauge pressure; atmospheric acts all round and cancels),
  • the force from walls, vanes or bends on the fluid (usually the unknown),
  • body forces such as the weight of fluid (often neglected in horizontal problems).
• Direction matters. Momentum is a vector — write separate x and y equations, and use the velocity component in each direction with its sign. The force of the fluid on the bend is equal and opposite to the force of the bend on the fluid.
• Momentum correction factor β. With non-uniform velocity, the true momentum flux is β·ṁ·V_avg, with β = 1 for uniform flow, about 1.33 for fully developed laminar pipe flow and about 1.01–1.04 for turbulent flow. Problems normally assume β = 1 unless told otherwise.
• Moment of momentum. Taking moments of the momentum equation gives the torque on a rotating CV (sprinklers, impellers): torque = ṁ·(r₂·V_w2 − r₁·V_w1). This becomes Euler's turbomachine equation in later topics.
• Assumptions to state: steady flow, uniform velocity across each inlet and outlet, incompressible fluid, and whether friction and weight are neglected.

Formulas

ρ₁·A₁·V₁ = ρ₂·A₂·V₂ = ṁ ṁ mass flow rate (kg/s), ρ density (kg/m³), A area (m²), V mean velocity (m/s). Steady flow.

Q = A₁·V₁ = A₂·V₂ Q volume flow rate (m³/s). Steady incompressible flow.

ΣF = ṁ·(V_out − V_in) Steady-flow momentum equation, one inlet and one outlet, in vector form. Write it as ΣF_x = ṁ·(V_x,out − V_x,in) and ΣF_y = ṁ·(V_y,out − V_y,in). ΣF in N, all forces acting on the fluid.

p₁·A₁ − p₂·A₂·cos θ + R_x = ρ·Q·(V₂·cos θ − V₁) x-momentum for a bend of angle θ in a horizontal plane; R_x is the x-force of the bend on the fluid (N); pressures in gauge Pa. The force of the fluid on the bend is −R_x.

F = ρ·A·V² Force (N) of a jet of area A and velocity V striking a fixed flat plate normally; the jet leaves tangentially, so it has zero normal velocity afterwards.

T = ṁ·(r₂·V_w2 − r₁·V_w1) Torque (N·m) on the fluid from a rotor; r radius (m), V_w tangential (whirl) velocity (m/s).

Worked examples

Example 1 (standard) — force on a nozzle. Given: water (ρ = 1000 kg/m³) flows at Q = 0.02 m³/s through a horizontal pipe of 100 mm diameter that ends in a nozzle of 40 mm exit diameter, discharging to atmosphere. Neglect losses. Find the force on the nozzle bolts.

1. Areas: A₁ = π × 0.1²/4 = 7.854 × 10⁻³ m²; A₂ = π × 0.04²/4 = 1.2566 × 10⁻³ m².
2. Velocities: V₁ = Q/A₁ = 2.546 m/s; V₂ = Q/A₂ = 15.915 m/s.
3. Bernoulli with p₂ = 0 (gauge): p₁ = ½ρ(V₂² − V₁²) = 500 × (253.29 − 6.48) = 123 409 Pa.
4. x-momentum on the fluid in the nozzle: p₁A₁ + R_x = ρQ(V₂ − V₁).
5. p₁A₁ = 123 409 × 7.854 × 10⁻³ = 969.3 N; ρQ(V₂ − V₁) = 1000 × 0.02 × 13.369 = 267.4 N.
6. R_x = 267.4 − 969.3 = −701.9 N (the nozzle pushes the fluid upstream).
7. The fluid therefore pushes the nozzle downstream with 701.9 N. Answer: the bolts must resist about 702 N, acting in the flow direction.

Example 2 (GATE level) — 90° pipe bend. Given: a horizontal 90° bend of constant diameter 0.3 m carries water at Q = 0.3 m³/s. The flow enters along +x and leaves along +y. Gauge pressure is 150 kPa at both ends (losses neglected). Find the resultant force of the water on the bend.

1. A = π × 0.3²/4 = 0.070686 m²; V = Q/A = 4.244 m/s; ρQ = 300 kg/s.
2. x-direction, forces on fluid: p₁A + R_x = ρQ(0 − V) → R_x = −(p₁A + ρQV).
3. p₁A = 150 000 × 0.070686 = 10 602.9 N; ρQV = 300 × 4.244 = 1273.2 N → R_x = −11 876 N.
4. y-direction: −p₂A + R_y = ρQ(V − 0) → R_y = p₂A + ρQV = +11 876 N.
5. Force of water on bend: F_x = +11 876 N, F_y = −11 876 N.
6. Resultant: F = √2 × 11 876 = 16 795 N, at 45° — pointing outward, away from the inside of the bend. Answer: F ≈ 16.8 kN at 45° to the inlet direction. An anchor block must resist it.

Common mistakes

• Writing pressure forces with absolute pressure, so atmospheric pressure appears on one side only.
• Confusing the force on the fluid with the force of the fluid on the solid; they are equal and opposite.
• Dropping the sign of a velocity component — a jet turned back through 180° changes momentum by 2V, not 0.
• Using Bernoulli across a sudden expansion or a hydraulic jump, where the energy loss is unknown; use momentum there.
• Forgetting the outlet pressure force at a bend, or applying the momentum equation with a non-uniform profile and β = 1 without saying so.

For GATE ME

Typical questions: force on a reducing bend or nozzle, the jet force on fixed or moving plates and vanes, the reaction of a tank discharging a jet, forces in a pipe with a sudden expansion, the torque and speed of a lawn sprinkler, and continuity in pipes and tanks with changing level. Practise drawing the CV, listing the forces on the fluid with directions, and then writing the x and y equations separately.

Quick check

1. Water enters a pipe at 3 m/s through 0.5 m² and leaves through 0.25 m². What is the exit velocity?
2. A water jet of area 0.02 m² at 10 m/s hits a fixed flat plate normally. What is the force?
3. Which law gives the continuity equation?
4. In the momentum equation, do we use gauge or absolute pressure for a pipe surrounded by atmosphere?
5. What is the momentum correction factor for fully developed laminar pipe flow? Answers: 1. 6 m/s. 2. ρAV² = 1000 × 0.02 × 100 = 2000 N. 3. Conservation of mass. 4. Gauge. 5. 4/3 (about 1.33).