Momentum Control Volumes and Turbomachinery

The Linear Momentum Equation

The FE tests a clean division of labor: use the energy equation for heads, velocities, and power, but use the momentum equation whenever a question asks for a force — the reaction on a pipe bend, the thrust of a jet, the force on a deflecting vane, the anchor force on a nozzle. For steady flow through a control volume:

ΣF = ṁ(V_out − V_in) = ṁ ΔV

where ṁ = ρQ = ρAV is the mass flow rate. The sum of forces on the control volume includes pressure forces on the inlet/outlet faces (p·A acting inward on each face), the support/anchor reaction, and gravity/weight if relevant. Because momentum is a vector, you must split into components and solve each axis independently:

• ΣF_x = ṁ(V_2x − V_1x)
• ΣF_y = ṁ(V_2y − V_1y)

Set a consistent sign convention (e.g., right and up positive), track the signed velocity components, then combine: F_R = √(F_x² + F_y²) with direction θ = arctan(F_y/F_x). The single most common error is using volume flow Q where ṁ = ρQ is required; the second is dropping the inlet pressure force when inlet pressure is gauge-nonzero.

Worked Example — Reducing Elbow

Water (ρ = 1000 kg/m³) flows through a 90° elbow. Inlet: V₁ = 4 m/s, A₁ = 0.02 m², gauge pressure p₁ = 200 kPa, flow in +x. Outlet: discharges to atmosphere (p₂ = 0 gauge) in +y, with V₂ = 8 m/s (smaller exit area).

Mass flow: ṁ = ρA₁V₁ = 1000 × 0.02 × 4 = 80 kg/s.

x-momentum: p₁A₁ + R_x = ṁ(0 − V₁). p₁A₁ = 200,000 × 0.02 = 4000 N. So R_x = ṁ(−V₁) − p₁A₁ = 80(−4) − 4000 = −4320 N (the support pushes in −x).

y-momentum: R_y = ṁ(V₂ − 0) = 80 × 8 = 640 N.

Resultant anchor force: F_R = √(4320² + 640²) ≈ 4367 N. Note the pressure term dominates here — never drop it when inlet pressure is gauge-nonzero.

Jets, Vanes, and Thrust

A free jet striking a stationary flat plate normal to the flow exerts F = ṁV = ρAV². For a jet turned through angle θ by a vane, the force components follow directly from the inlet/outlet velocity components. A jet reversed 180° by a curved vane delivers F = ṁ(V − (−V)) = 2ṁV — twice the force of a flat plate, a classic FE distinction.

For a moving vane (turbine blade), the relative velocity (V − u) replaces V, where u is the blade speed, and the effective mass flow is ρA(V − u). Power extracted = F × u, and it is maximized when u = V/2 for a single moving plate.

Rocket/jet thrust: Thrust = ṁ_exit V_exit + (p_exit − p_atm)A_exit. The momentum-flux term ṁV usually dominates; the pressure term matters only for under- or over-expanded nozzles.

Worked example — flat plate. A 25 mm-diameter water jet at 15 m/s hits a stationary plate normal to flow. A = π(0.025)²/4 = 4.91 × 10⁻⁴ m². ṁ = ρAV = 1000 × 4.91e-4 × 15 = 7.36 kg/s. F = ṁV = 7.36 × 15 = 110 N.

Turbomachinery and the Affinity Laws

For pumps, fans, and turbines, the FE rarely asks blade geometry — it asks head, flow, power, efficiency, and scaling. The affinity laws relate performance of a single pump at different speeds N (or geometrically similar pumps at diameter ratio D):

• Flow: Q₂/Q₁ = (N₂/N₁) — flow scales linearly with speed.
• Head: h₂/h₁ = (N₂/N₁)² — head scales with speed squared.
• Power: P₂/P₁ = (N₂/N₁)³ — power scales with speed cubed.

Worked example. A pump runs at 1750 rpm delivering Q = 0.10 m³/s, h = 25 m, P = 30 kW. Sped up to 2100 rpm (ratio 1.2): Q = 0.10 × 1.2 = 0.12 m³/s; h = 25 × 1.2² = 36 m; P = 30 × 1.2³ = 51.8 kW.

Specific speed N_s = N√Q/h^(3/4) classifies the impeller type independent of size. Low N_s indicates radial (centrifugal, high-head, low-flow) pumps; mid N_s indicates mixed-flow; high N_s indicates axial (propeller, low-head, high-flow) pumps. It tells a designer which geometry suits a given duty.

Worked example — turbine power. Water (Q = 0.5 m³/s) drops through a net head of 60 m to a turbine with overall efficiency 0.85. Available hydraulic power = γQh = 9810 × 0.5 × 60 = 294 kW. Shaft power out = 294 × 0.85 = 250 kW. Note efficiency reduces output for a turbine but increases required input for a pump — the direction flips with the device.

Sign convention: a pump/fan adds head and power to the fluid (h_p positive in the energy equation); a turbine extracts it (h_t positive on the downstream side). Confusing the direction flips the entire energy balance and the efficiency placement.