Topic 3 of 10 — free theory
Continuity, Energy & Momentum
The three governing equations of fluid mechanics: continuity, Bernoulli, and momentum.
Take the free 5-question mini-quiz ↓
Three equations govern nearly all exam fluid mechanics. Continuity says mass is conserved, the energy equation says energy is conserved (minus losses), and momentum says force equals the rate of momentum change. Learn when each one is the right tool.
1. Continuity — conservation of mass
Q = A1V1 = A2V2
| Q | volumetric flow rate (discharge) |
| A | cross-sectional area normal to the flow |
| V | mean velocity through the section |
ρ1A1V1 = ρ2A2V2
| Used when density changes (gases, compressible flow). For water, ρ cancels and you get Q = AV. |
Velocity and area trade off inversely: halve the diameter and the velocity quadruples (area goes as D²).
2. Energy equation — Bernoulli with losses
Bernoulli's equation is an energy balance per unit weight (each term is a “head” in feet or metres). Real flows lose head to friction, and pumps/turbines add or remove it:
p1/γ + V1²/2g + z1 + hA = p2/γ + V2²/2g + z2 + hT + hL
| p/γ | pressure head |
| V²/2g | velocity head |
| z | elevation head (same datum for both points!) |
| hA | head added by a pump |
| hT | head removed by a turbine |
| hL | total head loss between 1 and 2 |
Pick your two points strategically: free surfaces (p = 0 gauge, V ≈ 0 for large reservoirs) and pipe exits (p = 0 gauge) make terms vanish.
3. Momentum equation — forces from changing flow
When flow changes direction or speed — bends, reducers, nozzles, vanes — use momentum, not energy. It is a vector equation, so signs matter:
ΣF = ρQ(β2V2 − β1V1)
| ΣF | vector sum of forces on the fluid (pressure + weight + reaction), in the chosen direction |
| β | momentum correction factor, ≈ 1.0 for turbulent flow — the exam usually lets you drop it |
Solve for the force on the fluid, then reverse it: the force on the bend or reducer is equal and opposite.
PE depth: turn pressure difference into flow
A venturi meter is a deliberate contraction. Continuity relates the two velocities; Bernoulli converts the measured pressure-head difference into discharge. For a horizontal meter carrying an incompressible fluid:
Q = CdA2√[2gΔh / (1 − (A2/A1)²)]
| Δh | difference in piezometric head, (p1/γ + z1) − (p2/γ + z2) |
| Cd | discharge coefficient; use the supplied value rather than assuming ideal flow |
Decision sequence: write continuity, write energy, cancel equal elevations only if the meter is horizontal, then apply Cd. A differential manometer reading is not automatically Δh; first convert it to pressure head in the flowing fluid.
Worked example Force on a pipe reducer
Given:
- Horizontal reducer: 12-in. diameter → 6-in. diameter.
- Discharge Q = 3 cfs of water. Pressure at section 1: p1 = 20 psi (gauge).
- Neglect friction losses and the weight of water in the reducer.
Solution:
- Areas: A1 = π/4 × 1² = 0.785 ft²; A2 = π/4 × 0.5² = 0.196 ft².
- Velocities: V1 = 3/0.785 = 3.82 ft/s; V2 = 3/0.196 = 15.28 ft/s.
- Bernoulli (horizontal, no loss): p1/γ + V1²/2g = p2/γ + V2²/2g. p1/γ = 20×144/62.4 = 46.15 ft; V1²/2g = 0.23 ft; V2²/2g = 3.63 ft. Hence p2/γ = 42.75 ft and p2 = 18.53 psi.
- Momentum in x: p1A1 − p2A2 − Fx = ρQ(V2 − V1). p1A1 = 2,262 lb; p2A2 = 524 lb; ρQ(V2−V1) = 1.94×3×11.46 = 67 lb.
- Fx = 2,262 − 524 − 67 = 1,671 lb on the fluid (to the left), so the fluid pushes the reducer with ≈ 1.67 kips to the right.
Answer: Pressure drops to ≈ 18.5 psi; the reducer feels ≈ 1.67 kips in the flow direction.