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Fluid Mechanics · Chapter 9 of 10 · Advanced

External Flow: Drag and Lift

Air resists a car, water slows a sinking grain, and the same physics lifts a wing. Drag and lift come from a thin boundary layer and where it separates, captured by a single coefficient.

01

Readiness check

This chapter applies drag and force balance to bodies in a stream. Tick only what you can do closed-notes.

  • Recall the dynamic pressure ½ρV².
  • Recall the drag coefficient idea from Chapter 7.
  • Balance forces at equilibrium.
  • Compute the Reynolds number.
  • Recall buoyancy from Chapter 3.
0 or 1 weak itemsContinue with this chapter.
2 weak itemsReview the drag coefficient in Chapter 7.
3 or more weak itemsRevisit dynamic pressure and force balance first.
02

The core idea

The force on a body in a stream is its dynamic pressure times area times a coefficient: FD = ½ρV²CDA. The coefficient hides the boundary-layer physics and depends on the Reynolds number and shape.

FD = ½ρV²CDAdrag = friction drag + pressure dragterminal V: FD + FB = W

A body moving through fluid feels drag (along the flow) and sometimes lift (across it). Drag has two parts: friction drag from shear in the thin boundary layer, and pressure drag from the low-pressure wake left when that layer separates. Streamlined shapes delay separation and cut pressure drag. The drag coefficient CD rolls all of this into one number, found from charts or tables, so the force follows from the dynamic pressure and frontal area. A falling body reaches terminal velocity when drag plus buoyancy balance its weight.

The skill works when: you use the right CD and area, and balance drag against weight and buoyancy for terminal velocity.
The skill breaks down when: Stokes' law is used outside its low-Reynolds range, or the wrong reference area is taken.
The concept. Flow over a bluff body: a thin boundary layer hugs the front, then separates, leaving a low-pressure wake. Friction drag comes from the layer, pressure drag from the wake; streamlining shrinks the wake.
03

The skills, taught in order

External flow is the drag coefficient plus boundary-layer reasoning. Five skills cover drag and lift, the coefficient, the boundary layer, separation, and terminal velocity.

9.1 Drag and lift

The force on a body splits into drag (parallel to the flow) and lift (perpendicular). Each is reported as a coefficient: CD = FD/(½ρV²A) and CL = FL/(½ρV²A). Drag itself is friction drag (shear) plus pressure (form) drag.

9.2 The drag coefficient

Once CD is known, the force is FD = ½ρV²CDA. The reference area A is the frontal area for bluff bodies (cars, spheres) and the planform area for wings. CD depends on shape and Reynolds number, read from tables or charts.

BodyTypical CD
Streamlined airfoil~0.04
Modern car0.25 to 0.35
Sphere (high Re)~0.4 to 0.5
Flat plate (normal)~1.1 to 1.2

9.3 The boundary layer

Because of no-slip, a thin boundary layer forms where velocity rises from zero at the wall to the free-stream value. It starts laminar and may transition to turbulent. Its shear is the friction drag, and its behaviour controls whether and where the flow separates.

9.4 Separation and wakes

Against an adverse pressure gradient (pressure rising downstream), the boundary layer can separate, leaving a wide low-pressure wake and large pressure drag. A turbulent boundary layer resists separation better, which is why a golf ball's dimples (tripping it turbulent) reduce drag, the drag crisis.

9.5 Terminal velocity and lift

A body falling through fluid accelerates until drag plus buoyancy equal its weight: terminal velocity. For very small, slow spheres (Re < 1), drag follows Stokes' law FD = 3πμDV, giving Vt = (ρs − ρf)gD²/(18μ). Lift on a wing comes from the pressure difference set up by circulation, reported as CL.

Engineering connection: vehicle aerodynamics, wind loads on buildings, settling of particles, sports-ball flight, and aircraft lift all rest on drag, lift, and boundary-layer behaviour.

04

Worked example 1: drag and power for a car

A car has frontal area 2.2 m² and drag coefficient 0.30, driving at 30 m/s (108 km/h) through air (ρ = 1.20 kg/m³). Find the aerodynamic drag and the power needed to overcome it.

Figure 1. Drag is the dynamic pressure times the frontal area and the drag coefficient. Because power is drag times speed, it grows with the cube of velocity, the reason highway driving burns so much fuel.
  1. ProblemFind the drag force and power for the car in Figure 1.
  2. Given / findA = 2.2 m², CD = 0.30, V = 30 m/s, ρ = 1.20 kg/m³. Find FD and power.
  3. AssumptionsSteady level driving, drag dominated by aerodynamics, frontal reference area.
  4. ModelDrag from the coefficient, power as drag times speed.
  5. EquationsFD = ½ρV²CDA P = FDV
  6. SolveFD = ½ × 1.20 × 30² × 0.30 × 2.2 = ½ × 1.20 × 900 × 0.66 = 356 N. P = 356 × 30 = 10,700 W = 10.7 kW.
  7. CheckSince P = ½ρV³CDA, power scales with V³: at 40 m/s it would jump to about 25 kW. This cubic growth is why fuel economy falls sharply at high speed.
  8. ConclusionOne coefficient and the frontal area give the drag, and the cube law dominates high-speed energy use. Lowering CD or area is the lever for efficiency.
Result. Drag 356 N; power to overcome it 10.7 kW (growing as V³).
05

Worked example 2: terminal velocity of a particle

A 50 μm silt particle (ρ = 2600 kg/m³) settles in still water at 20 °C (ρ = 998 kg/m³, μ = 1.002×10⁻³ Pa·s). Find its terminal velocity, and confirm Stokes' law applies.

Figure 2. At terminal velocity the upward drag and buoyancy balance the downward weight. For this tiny, slow particle the Reynolds number is well below 1, so Stokes' law holds.
  1. ProblemFind the terminal velocity of the particle in Figure 2 and check Stokes validity.
  2. Given / findD = 50 μm = 5×10⁻⁵ m, ρs = 2600, ρf = 998 kg/m³, μ = 1.002×10⁻³ Pa·s. Find Vt and Re.
  3. AssumptionsSmall slow sphere (creeping flow, Re < 1), so Stokes' drag FD = 3πμDV applies.
  4. ModelBalance weight against buoyancy plus Stokes drag, then check Re afterwards.
  5. EquationsVt = (ρs − ρf)gD²/(18μ) Re = ρfVtD/μ
  6. SolveVt = (2600 − 998)(9.81)(5×10⁻⁵)²/(18 × 1.002×10⁻³) = (1602)(9.81)(2.5×10⁻⁹)/0.018036 = 2.18×10⁻³ m/s (2.18 mm/s). Re = 998 × 2.18×10⁻³ × 5×10⁻⁵/1.002×10⁻³ = 0.11, well below 1.
  7. CheckRe = 0.11 confirms creeping flow, so Stokes' law was valid. A larger grain would settle faster and exceed Re = 1, where Stokes overpredicts the speed and a drag-coefficient approach is needed instead.
  8. ConclusionTerminal velocity is a force balance, and for fine particles Stokes' law gives it in closed form. Always check Re to confirm the law applies, the accuracy step that matters here.
Result. Terminal velocity 2.18 mm/s; Re = 0.11, so Stokes' law is valid.
06

Misconceptions and diagnostics

MistakeSymptomDiagnostic questionCorrection
Stokes' law outside its rangeTerminal velocity overpredicted"Is Re below 1?"Stokes holds only for creeping flow; otherwise use CD.
Wrong reference areaDrag off by a large factor"Frontal area or planform?"Frontal for bluff-body drag, planform for wing lift.
Turbulent always means more dragDimples seem pointless"Does turbulence delay separation?"A turbulent layer can shrink the wake and cut pressure drag.
Forgetting the cube law for powerHigh-speed power underestimated"Does power go as V or V³?"Drag goes as V², power as V³.
07

Practice ladder

Level 1 · Direct skill

A sign 1 m × 2 m (CD = 1.2) faces a 20 m/s wind (ρ = 1.20 kg/m³). Find the drag force.

Show answer

FD = ½ρV²CDA = ½ × 1.20 × 20² × 1.2 × 2 = 576 N. Flat plates have a high CD, so wind loads are large.

Level 2 · Mixed concept

The Worked Example 1 car slows to 20 m/s. By what factor does the drag power drop?

Show answer

Power ∝ V³, so the ratio is (20/30)³ = 0.296. Power falls to about 3.2 kW, less than a third, the strong reward of slowing down.

Level 3 · Independent problem

Confirm that the 50 μm particle's terminal velocity also follows from balancing Stokes drag 3πμDV against the net weight (ρs − ρf)g(πD³/6).

Show answer

Set 3πμDV = (ρs − ρf)g(πD³/6). Solving, V = (ρs − ρf)gD²/(18μ), the formula used, giving 2.18 mm/s. The two routes agree because Stokes' law is exactly that force balance.

Level 4 · Transfer to real engineering

Find a real drag or lift situation (a cyclist, a parachute, a wind-loaded sign, an aircraft wing). Estimate the coefficient and area, and compute the force or terminal velocity.

What good work looks like

The correct coefficient and reference area, F = ½ρV²C A computed (or a terminal-velocity balance), and a Reynolds-number check on any law used.

08

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that Stokes' law is valid here by confirming Re < 1."
"Give me five bodies; I will name the reference area and rough CD."
"Compute the drag." Choosing CD and the area yourself is the skill.
"What is the terminal velocity?" Setting up the force balance is the point.

Portfolio task

Analyse one external flow: compute drag (and power) from CD and area, or a terminal velocity from a force balance, with a Reynolds-number validity check.

Must include: the coefficient and reference area, the force or terminal velocity, and a Re check.
09

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. Write the drag force in terms of the coefficient.

FD = ½ρV²CDA.

2. Name the two contributions to drag.

Friction (shear) drag and pressure (form) drag from the wake.

3. What causes pressure drag?

Boundary-layer separation, which leaves a low-pressure wake.

4. When does Stokes' law apply, and what does it give?

For Re < 1; Vt = (ρs − ρf)gD²/(18μ).

5. How does drag power scale with speed?

As V³ (drag as V²), so high speed is costly.

TodayFinish this quiz and Levels 1 and 2 of the ladder.
+1 dayRe-derive the car drag and terminal velocity from a blank page.
+3 daysOne drag-force and one terminal-velocity problem.
+7 daysFinish with turbomachinery, Chapter 10.
+30 daysConnect CD = f(Re) back to the dimensional analysis of Chapter 7.
10

Textbook mapping

ItemMapping
Primary sourceÇengel and Cimbala, Fluid Mechanics: Fundamentals and Applications, Chapter 11 (Flow over Bodies: Drag and Lift)
Cross-referenceWhite, Ch. 7 · Munson, Ch. 9
Core topics9.1 Drag and lift · 9.2 Drag coefficient · 9.3 Boundary layer · 9.4 Separation and wakes · 9.5 Terminal velocity
Engineering connectionVehicle aerodynamics, wind loads, particle settling, and aircraft lift.
Read nextChapter 10: Turbomachinery.