Bernoulli, Energy Equation, Pipe Losses, and Pumps

Continuity and When Bernoulli Applies

Mass conservation for incompressible steady flow is continuity: ṁ = ρAV = constant, and for constant density Q = A₁V₁ = A₂V₂. Halving the cross-sectional area doubles the velocity. Always solve continuity first — it links the velocities that the energy equation needs. For a circular pipe, A = πD²/4, so V scales with 1/D²: cutting the diameter in half quadruples the velocity.

Bernoulli's equation (per unit weight, in head form) is:

p₁/γ + V₁²/2g + z₁ = p₂/γ + V₂²/2g + z₂

Each term is a head with units of length: p/γ is pressure head, V²/2g is velocity head, and z is elevation head. Bernoulli simply says these three trade off along a streamline at constant total head. It is valid only for steady, incompressible, frictionless flow along a streamline with no pump or turbine. The instant a problem mentions friction, a long pipe, a valve, a fitting, or a pump/turbine, you must upgrade to the mechanical energy equation. A classic Bernoulli application is the Pitot tube: stagnation minus static pressure gives velocity, V = √(2(p_stag − p_static)/ρ).

The Mechanical Energy Equation

The full handbook form adds shaft work and losses to Bernoulli:

p₁/γ + V₁²/2g + z₁ + h_p = p₂/γ + V₂²/2g + z₂ + h_t + h_L

where h_p is pump head added (energy in), h_t is turbine head extracted (energy out), and h_L is total head loss (friction + fittings). This single equation handles almost every FE pipe problem; Bernoulli is just the special case with h_p = h_t = h_L = 0.

Strategy for setting up points 1 and 2: choose the two points where you know the most. Reservoir/tank surfaces are ideal because there p = 0 gauge and V ≈ 0 (large area). Free jet exits to atmosphere have p = 0 gauge. This zeroes terms and shrinks the algebra.

Worked example. Water flows from a large open reservoir through a pipe to a lower open outlet 30 m below, with 12 m of total head loss. Both surfaces are at atmospheric pressure with negligible reservoir velocity. The energy equation reduces to z₁ = V₂²/2g + z₂ + h_L, so V₂²/2g = (z₁ − z₂) − h_L = 30 − 12 = 18 m, giving V₂ = √(2 × 9.81 × 18) = 18.8 m/s. Without losses it would be √(2 × 9.81 × 30) = 24.3 m/s — friction costs ~5.5 m/s here.

Pipe Friction and the Moody Chart

Major (friction) loss uses Darcy-Weisbach:

h_f = f (L/D)(V²/2g)

The Darcy friction factor f depends on Reynolds number Re = ρVD/μ and relative roughness ε/D.

• Laminar (Re < 2300): f = 64/Re exactly — no chart needed.
• Turbulent (Re > 4000): read f from the Moody chart or solve the Colebrook equation using ε/D.

Worked example. Water (ν = 1.0 × 10⁻⁶ m²/s) flows at 1.5 m/s through a 0.05 m pipe, 100 m long. Re = VD/ν = 1.5 × 0.05 / 1.0e-6 = 75,000 (turbulent). For a smooth pipe, f ≈ 0.019. h_f = 0.019 × (100/0.05) × (1.5²/(2×9.81)) = 0.019 × 2000 × 0.115 = 4.4 m.

Minor losses add fittings: h_m = K(V²/2g), where K is a loss coefficient from the handbook. Typical K values: sharp entrance 0.5, exit 1.0, fully open gate valve 0.2, fully open globe valve ~10, 90° elbow 0.9. In short piping systems with several valves and fittings, minor losses can exceed the straight-pipe friction loss, so never assume they are negligible just because the name says "minor." Total loss is h_L = h_f + Σh_m. An equivalent-length method also exists: each fitting is replaced by L_eq/D of straight pipe, then folded into one Darcy-Weisbach term — handy when a single friction factor covers the whole line.

Pump Power, Efficiency, and NPSH

The hydraulic (water) power delivered to the fluid is:

W_hydraulic = γ Q h_p = ρ g Q h_p

The brake power the motor must supply accounts for pump efficiency η_p: W_brake = W_hydraulic / η_p. Efficiency always increases the required input — if your brake power came out smaller than hydraulic power, you divided the wrong way.

Worked example. A pump delivers Q = 0.05 m³/s at h_p = 40 m with η_p = 0.75. Hydraulic power = 9810 × 0.05 × 40 = 19,620 W ≈ 19.6 kW. Brake power = 19.6 / 0.75 = 26.2 kW.

NPSH (Net Positive Suction Head) is a cavitation question, not a power question. Available NPSH must exceed required NPSH (NPSH_avail > NPSH_req) or the local pressure at the impeller eye drops to the fluid's vapor pressure, vapor bubbles form, and they collapse violently — eroding the impeller and choking flow. NPSH_avail = (p_atm − p_vapor)/γ − z_suction − h_L,suction. Raising the pump above the supply (large positive suction lift z_suction), high suction-line losses, or hot/volatile liquid (high p_vapor) all reduce NPSH_avail and invite cavitation. The fix is to lower the pump, shorten/enlarge the suction line, or cool the fluid.