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Fluid Mechanics · Chapter 5 of 10 · Intermediate

The Bernoulli and Energy Equations

Speed up a flow and its pressure drops. Bernoulli captures that trade between pressure, speed, and height, and the energy equation extends it to real systems with pumps, turbines, and losses.

Readiness check

This chapter applies conservation of mass and energy to flows. Tick only what you can do closed-notes.

Use the flow rate Q = AV and ṁ = ρAV.
Recall kinetic and potential energy per unit mass.
Rearrange an equation for one unknown.
Convert between pressure and head (P = ρgh).
Recall the convective acceleration idea from Chapter 4.

0 or 1 weak itemsContinue with this chapter.

2 weak itemsReview flow rate and pressure-head in Chapter 3.

3 or more weak itemsRevisit energy conservation in Physics: Fluids first.

The core idea

Mass is conserved (A₁V₁ = A₂V₂), and along a streamline of an ideal flow the sum of pressure, kinetic, and potential energy is constant: the Bernoulli equation.

A₁V₁ = A₂V₂P/ρ + V²/2 + gz = constanth_pump = Δz + ΔV²/2g + ΔP/ρg + h_L

Continuity says that for incompressible flow, narrowing the area speeds the fluid up. Bernoulli, which is energy conservation along a streamline for steady, incompressible, frictionless flow, then says that where the fluid is fast the pressure is low. Real systems add or remove energy and lose some to friction, so the energy equation generalises Bernoulli with pump head, turbine head, and a head-loss term. These two equations solve most everyday flow problems.

The skill works when: you apply continuity for the speeds and Bernoulli (or the energy equation) between two well-chosen points.

The skill breaks down when: Bernoulli is used across a pump, a turbine, or a lossy region where its assumptions fail.

The concept. Through a constriction, continuity speeds the fluid up and Bernoulli drops its pressure. The pressure tap at the narrow throat reads lower, the principle behind venturi meters and carburettors.

The skills, taught in order

Two conservation laws drive this chapter. Five skills cover mass, the Bernoulli equation, pressure types, the energy equation, and the grade lines.

5.1 Conservation of mass

Mass cannot accumulate in steady flow, so the mass flow rate ṁ = ρAV is the same at every section. For incompressible flow this reduces to A₁V₁ = A₂V₂: halve the area and the speed doubles. This is the first equation to write in almost every problem.

5.2 The Bernoulli equation

Along a streamline, for steady, incompressible, frictionless flow, P/ρ + V²/2 + gz is constant. Dividing by g gives the head form: pressure head + velocity head + elevation head. Its assumptions are strict, so know when they hold.

Bernoulli requires	Fails when
Steady flow	flow is accelerating in time
Incompressible	high-speed gas (Ma > 0.3)
Frictionless	long pipes, sharp fittings (losses)
Along a streamline, no shaft work	across a pump or turbine

5.3 Static, dynamic, and stagnation pressure

Bernoulli splits pressure into static (P), dynamic (½ρV²), and their sum, the stagnation pressure. A pitot-static tube measures the difference to find speed: V = √(2(P_stag − P)/ρ). This is how aircraft and many flow meters read velocity.

5.4 The energy equation

Generalising Bernoulli, the steady-flow energy equation adds shaft work and losses: between inlet and outlet, h_pump − h_turbine = Δ(P/ρg) + Δ(V²/2g) + Δz + h_L. This is what sizes real pumps and turbines once head loss (Chapter 8) is known.

5.5 Hydraulic and energy grade lines

The energy grade line (EGL) plots the total head; the hydraulic grade line (HGL) plots pressure plus elevation head, one velocity head below the EGL. They drop along lossy pipe, step up at a pump, and down at a turbine, giving a clear picture of where energy goes.

Engineering connection: venturi and pitot meters, siphons, draining tanks, spray nozzles, and pump and turbine sizing all come straight from these two equations.

Worked example 1: flow through a contraction

Water (ρ = 998 kg/m³) flows steadily through a horizontal pipe that contracts from 10 cm to 5 cm diameter. The upstream velocity is 2 m/s. Find the downstream velocity and the pressure drop (assume no losses).

Figure 1. Continuity quadruples the speed when the diameter halves (area drops fourfold), and Bernoulli converts that gain in kinetic energy into a drop in pressure.

ProblemFind V₂ and the pressure drop for the contraction in Figure 1.
Given / findD₁ = 0.10 m, D₂ = 0.05 m, V₁ = 2 m/s, ρ = 998 kg/m³, horizontal, no loss. Find V₂ and ΔP.
AssumptionsSteady, incompressible, frictionless, same elevation (z₁ = z₂).
ModelContinuity for V₂, then Bernoulli for the pressure change.
EquationsV₂ = V₁(D₁/D₂)² P₁ − P₂ = ½ρ(V₂² − V₁²)
SolveV₂ = 2 × (10/5)² = 2 × 4 = 8 m/s. ΔP = ½ × 998 × (8² − 2²) = ½ × 998 × 60 = 29.9 kPa.
CheckThe faster section has the lower pressure, as Bernoulli requires. The diameter ratio enters as the fourth power in the pressure drop (through V² and the squared area ratio), so small contractions cause large drops.
ConclusionContinuity sets the speeds and Bernoulli converts them to pressures. The same logic reads flow rate from a venturi's pressure drop.

Result. V₂ = 8 m/s; pressure drop 29.9 kPa across the contraction.

Worked example 2: sizing a pump

A pump delivers 0.05 m³/s of water (ρ = 998 kg/m³) from a lower to an upper reservoir 20 m higher, against a head loss of 4 m, with 75% efficiency. Find the required pump head and the electrical (brake) power.

Figure 2. The pump must supply the lift (20 m) plus the head lost to friction (4 m). Dividing the hydraulic power by efficiency gives the electrical power drawn.

ProblemFind the pump head and brake power for the system in Figure 2.
Given / findQ = 0.05 m³/s, z = 20 m, h_L = 4 m, η = 0.75, ρ = 998 kg/m³. Find h_pump and brake power.
AssumptionsBoth reservoirs open to atmosphere and large (V ≈ 0 at surfaces), so pressure and velocity head differences vanish.
ModelEnergy equation reduces to pump head = elevation gain + head loss; hydraulic power is ρgQh, brake power divides by efficiency.
Equationsh_pump = Δz + h_L Ẇ_water = ρgQh_pump Ẇ_brake = Ẇ_water/η
Solveh_pump = 20 + 4 = 24 m. Ẇ_water = 998 × 9.81 × 0.05 × 24 = 11.7 kW. Ẇ_brake = 11.7/0.75 = 15.7 kW.
CheckWithout losses the head would be 20 m and the power lower; the 4 m loss adds 20% to the head. Efficiency raises the electrical draw above the useful hydraulic power, as it must.
ConclusionThe energy equation turns a lift-and-loss requirement into a pump head and a motor size. Head loss, computed in Chapter 8, is the missing piece for any real pipeline.

Result. Pump head 24 m; hydraulic power 11.7 kW; brake power 15.7 kW.

Misconceptions and diagnostics

Mistake	Symptom	Diagnostic question	Correction
Bernoulli across a pump	Energy appears or vanishes	"Is there shaft work or loss between the points?"	Use the energy equation with h_pump, h_turbine, and h_L.
Bernoulli in a long pipe	Predicted pressure too high	"Is friction negligible here?"	Add head loss; Bernoulli alone is frictionless.
Forgetting continuity	Wrong downstream velocity	"Did I use A₁V₁ = A₂V₂?"	Find the speeds from continuity before Bernoulli.
Mixing pressure and head	Units inconsistent	"Is every term in the same form?"	Use all-pressure or all-head form, not a mix.

Practice ladder

Level 1 · Direct skill

A tank has a small hole 5 m below the surface. Find the jet velocity (Torricelli).

Show answer

V = √(2gh) = √(2 × 9.81 × 5) = √98.1 = 9.90 m/s. Bernoulli from the open surface to the open jet gives the free-fall speed.

Level 2 · Mixed concept

A pitot-static tube on an aircraft reads a stagnation-minus-static pressure of 2.5 kPa in air (ρ = 1.2 kg/m³). Find the airspeed.

Show answer

V = √(2ΔP/ρ) = √(2 × 2500/1.2) = √4167 = 64.5 m/s. The dynamic pressure ½ρV² equals the measured difference.

Level 3 · Independent problem

The pump of Worked Example 2 now has negligible head loss. What brake power is needed, and how much had the 4 m loss added?

Show answer

h_pump = 20 m, Ẇ_water = 998 × 9.81 × 0.05 × 20 = 9.79 kW, brake = 9.79/0.75 = 13.1 kW. The loss raised it from 13.1 to 15.7 kW, about 20% more, matching the 4/20 head ratio.

Level 4 · Transfer to real engineering

Find a real device that uses Bernoulli or the energy equation (a venturi meter, a siphon, a spray gun, a building water pump). Identify the two points and the equation that links them.

What good work looks like

Two well-chosen points, continuity applied for the speeds, the correct equation (Bernoulli or energy with pump/loss), and a verified result.

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that Bernoulli's assumptions actually hold between my two points."

"Give me five flows; I will say whether to use Bernoulli or the energy equation."

"Compute the pressure drop." Applying continuity then Bernoulli yourself is the skill.

"What pump power?" Setting up the energy equation is the point.

Portfolio task

Analyse one flow system: use continuity and Bernoulli for an ideal section, or the energy equation with pump or turbine head and a loss term, and verify the result.

Must include: a continuity step, the energy or Bernoulli equation between two points, and a head or power result.

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. Write continuity for incompressible flow.

A₁V₁ = A₂V₂ (and ṁ = ρAV is constant).

2. State the Bernoulli equation and its assumptions.

P/ρ + V²/2 + gz = constant; steady, incompressible, frictionless, along a streamline, no shaft work.

3. How does a pitot-static tube find speed?

V = √(2(P_stag − P)/ρ), from the dynamic pressure.

4. Write the energy equation with a pump.

h_pump = Δz + Δ(V²/2g) + Δ(P/ρg) + h_L.

5. What do the HGL and EGL show?

The EGL is total head; the HGL is one velocity head below it. They drop along lossy pipe and step at machines.

TodayFinish this quiz and Levels 1 and 2 of the ladder.

+1 dayRe-derive the contraction and pump results from a blank page.

+3 daysOne Bernoulli and one energy-equation problem.

+7 daysAdd momentum forces in Chapter 6.

+30 daysCombine the energy equation with head loss from Chapter 8.

Textbook mapping

Item	Mapping
Primary source	Çengel and Cimbala, Fluid Mechanics: Fundamentals and Applications, Chapter 5 (Bernoulli and Energy Equations)
Cross-reference	White, Ch. 3 · Munson, Ch. 3 and 5
Core topics	5.1 Conservation of mass · 5.2 Bernoulli · 5.3 Stagnation pressure · 5.4 Energy equation · 5.5 HGL and EGL
Engineering connection	Flow meters, siphons, nozzles, and pump and turbine sizing.
Read next	Chapter 6: Momentum Analysis of Flow Systems.