Control volume analysis: continuity and momentum equations

Why it matters

Control-volume analysis is the workhorse of engineering fluid mechanics. With just mass and momentum balances drawn around a carefully chosen region, you can size pipe-bend anchors and nozzle bolts, find the thrust of a jet or rocket, compute the force on a turbine blade, and predict the reaction a hydraulic hose exerts when a valve opens — without knowing the detailed flow inside.

Key ideas

System versus control volume. A system is a fixed quantity of matter; the laws of mechanics (mass conservation, Newton's second law) are stated for systems. A control volume (CV) is a chosen region in space through which fluid flows; its boundary is the control surface (CS). The CV may be fixed, moving at constant velocity (a moving vane or a jet engine on an aircraft) or deforming (a cylinder with a moving piston).

Reynolds transport theorem (RTT). It converts a law for a system into one for a CV: the rate of change of a property for the system equals the rate of change inside the CV plus the net outflow of that property across the CS. Applying it to mass and to momentum gives the two equations of this topic.

Continuity. Mass inflow minus mass outflow equals the rate of mass accumulation inside the CV. For steady flow, mass in equals mass out. For steady incompressible flow with uniform velocity over each section, Q = A·V is the same at every section; where the area shrinks, velocity rises. The average velocity V = Q/A is used when the profile is not uniform.

Linear momentum. The sum of external forces on the fluid in the CV equals the rate of change of momentum inside the CV plus the net momentum outflow through the CS. For steady flow with uniform inlet and outlet sections, the force on the fluid equals ṁ(V_out − V_in) as vectors. Important points:

• The equation is a vector equation: write it separately in x, y (and z) and keep signs consistent with your axes.
• External forces are: pressure forces on the cut sections (use gauge pressure if the outside is at atmosphere), the force exerted by walls or a solid object on the fluid, and body forces (weight).
• The force by the fluid on the pipe, bend or vane is equal and opposite to the force by the wall on the fluid (Newton's third law). Most mistakes happen here.
• Pressure forces always act inward, normal to the cut section, on the fluid in the CV.
• For a non-uniform profile, momentum flux is β·ṁ·V with the momentum correction factor β (β = 4/3 for laminar pipe flow, about 1.02 to 1.05 for turbulent flow; β = 1 for uniform flow).

Moving control volumes. For a CV moving at constant velocity, use velocities relative to the CV in the flux terms; the mass flow rate entering is ρ·A·(V − u), not ρ·A·V.

Angular momentum. Taking moments of the momentum equation gives torque = ṁ(r₂·V_t2 − r₁·V_t1), which becomes Euler's turbomachinery equation (used in the turbine and pump topics).

Limits. These integral balances give net forces, not pressure or stress distributions. Viscous losses are not given by the momentum equation itself; when pressures are not stated, you often need the energy (Bernoulli) equation, with its own assumptions, to find them.

Formulas

ṁ = ρ·A·V = ρ·Q

• ṁ = mass flow rate (kg/s), ρ = density (kg/m³), A = flow area (m²), V = mean velocity (m/s), Q = volume flow rate (m³/s).

ρ₁·A₁·V₁ = ρ₂·A₂·V₂ (steady); A₁·V₁ = A₂·V₂ (steady, incompressible)

ΣF = d(M_CV)/dt + Σ(ṁ·V)_out − Σ(ṁ·V)_in

• ΣF = vector sum of external forces on the fluid in the CV (N), M_CV = momentum inside the CV (kg·m/s).

ΣF_x = ṁ·(V_2x − V_1x) (steady, one inlet, one outlet, uniform sections)

ΣF_x = p₁·A₁ − p₂·A₂·cosθ + R_x (typical bend, inlet along x, outlet at angle θ)

• p = gauge pressure (Pa), R_x = force of the bend on the fluid (N).

F_jet = ρ·A·V² (jet striking a fixed flat plate normally)

• F = force on the plate (N); for a plate moving away at speed u, F = ρ·A·(V − u)².

T = ṁ·(r₂·V_t2 − r₁·V_t1)

• T = torque on the fluid (N·m), r = radius (m), V_t = tangential velocity component (m/s).

Worked examples

Example 1 — fire-hose nozzle (standard). Water (ρ = 1000 kg/m³) flows at 0.02 m³/s through a horizontal nozzle from an 80 mm hose to a 25 mm outlet that discharges to atmosphere. Neglecting losses, find the force the water exerts on the nozzle.

1. Areas: A₁ = π × 0.04² = 5.027 × 10⁻³ m²; A₂ = π × 0.0125² = 4.909 × 10⁻⁴ m².
2. Velocities: V₁ = Q/A₁ = 3.979 m/s; V₂ = Q/A₂ = 40.74 m/s.
3. Inlet gauge pressure from Bernoulli (p₂ = 0): p₁ = ρ·(V₂² − V₁²)/2 = 500 × (1660.0 − 15.8) = 822.1 kPa.
4. Momentum in x on the fluid (ṁ = 20 kg/s): p₁·A₁ + R_x = ṁ·(V₂ − V₁), so R_x = 735.3 − 4132.4 = −3397 N (the nozzle pushes the fluid upstream).
5. The fluid pushes the nozzle the opposite way. Answer: about 3.40 kN on the nozzle, acting downstream — this is the load the coupling must hold.

Example 2 — horizontal reducing bend (GATE level). A horizontal 90° bend reduces from 300 mm to 200 mm diameter and carries 0.25 m³/s of water. The inlet (flow in +x) gauge pressure is 200 kPa; the outlet flow is in +y. Neglecting losses, find the resultant force of the water on the bend.

1. A₁ = 0.07069 m², A₂ = 0.03142 m²; V₁ = 3.537 m/s, V₂ = 7.958 m/s; ṁ = 250 kg/s.
2. Outlet pressure: p₂ = p₁ + ρ·(V₁² − V₂²)/2 = 200 000 + 500 × (12.51 − 63.33) = 174.6 kPa.
3. x on the fluid: p₁·A₁ + R_x = ṁ·(0 − V₁), so R_x = −(884.2 + 14 137.2) = −15 021 N.
4. y on the fluid: −p₂·A₂ + R_y = ṁ·(V₂ − 0), so R_y = 1989.4 + 5485.0 = 7474 N.
5. Force of water on bend = −R: F_x = +15 021 N, F_y = −7474 N.
6. Resultant = √(15 021² + 7474²) = 16 778 N, at tan⁻¹(7474/15 021) = 26.5° below the +x axis. Answer: about 16.8 kN at 26.5° to the inlet direction, pointing outward from the bend.

Common mistakes

• Mixing up the force on the fluid with the force on the bend — they are equal and opposite.
• Taking pressure forces as acting outward from the CV; on the fluid they always push inward on the cut sections.
• Using absolute pressure when atmosphere also acts on the outside of the pipe. Use gauge pressure.
• Adding speeds as scalars. Momentum flux is a vector; resolve each velocity into components.
• Forgetting that outlet pressure is lower than inlet pressure in a reducer, and assuming p₂ = p₁.
• For a moving vane, using ρAV for the mass flow that actually strikes it instead of ρA(V − u).

For GATE ME

Expect numericals on continuity in branching pipes, forces on bends, reducers and nozzles, jet forces on fixed and moving plates, thrust of a jet, and simple unsteady problems such as tank draining using mass balance. Conceptual MCQs test what forces enter the momentum equation and the values of the momentum correction factor. Practise drawing the CV, marking all forces on the fluid with signs, and then converting to the force on the solid at the end.

Quick check

1. Water enters a junction at 0.05 m³/s and leaves through two pipes; one carries 0.03 m³/s. What does the other carry?
2. What is the force of a 20 mm water jet at 30 m/s striking a fixed flat plate normally?
3. In which direction do pressure forces act on the fluid in a control volume?
4. What is the momentum correction factor for fully developed laminar pipe flow?
5. Is the force on the fluid equal to the force on the bend?

Answers: 1. 0.02 m³/s; 2. F = ρAV² = 1000 × 3.142 × 10⁻⁴ × 900 ≈ 283 N; 3. normal to and into the control surface; 4. 4/3; 5. equal in magnitude, opposite in direction.