The central idea of Prandtl's boundary-layer concept at high Reynolds number is that

viscous effects are confined to a thin layer near the wall, with inviscid flow outside it

viscosity matters equally everywhere in the flow, and the pressure field is the only thing it changes

viscous effects are confined to a thin layer near the wall, with inviscid flow outside it

the velocity at the wall equals the free-stream velocity, so that only the outer flow needs to be solved

the pressure varies strongly across the thin layer, so the outer pressure cannot be imposed on it

The displacement thickness δ* of a boundary layer represents

the outward shift of the outer streamlines caused by the layer

the distance from the wall at which the velocity first reaches 0.99 of U

the distance over which the wall shear stress acts on the passing fluid

the loss of momentum flux in the layer, divided by ρU² and by unit width

Boundary-layer separation from a smooth curved wall occurs

where pressure rises along the flow and the near-wall fluid stops

where the pressure gradient is favourable and the outer flow accelerates strongly

only when the layer is turbulent, because eddies carry momentum away from the wall

at the point where the wall shear stress is largest, which is near the leading edge

At a high Reynolds number, the total drag of a well-designed streamlined body is made up mainly of

friction drag, because the layer stays attached and the wake is thin

pressure (form) drag, because the wake is wide

wave drag, because the body is long

friction drag, because the layer stays attached and the wake is thin

induced drag, because the body is symmetrical

The drag coefficient of a smooth sphere falls sharply near Re ≈ 3 × 10⁵ (drag crisis). The main reason is that

the layer turns turbulent and separation moves downstream

the skin friction on the front half drops to zero once the layer is turbulent

the flow becomes compressible as the sphere approaches the speed of sound

the flow becomes inviscid because the viscosity of air falls at high Re

the layer turns turbulent and separation moves downstream

For a laminar boundary layer on a flat plate with dp/dx = 0, the plate-average drag coefficient C_D (one side) is compared with the local skin-friction coefficient C_f at the trailing edge. C_D is

twice C_f, since the shear is larger upstream

equal to C_f, since the plate is uniform

half of C_f, since the shear falls downstream

twice C_f, since the shear is larger upstream

four times C_f, since drag grows as the square of length

A thin aerofoil is held at a steadily increasing angle of attack in a fixed stream. Beyond the stall angle

the flow separates over much of the upper surface, lift falls and drag rises

the lift coefficient keeps rising along the same straight line, with the drag staying small

the flow separates over much of the upper surface, lift falls and drag rises

the flow becomes irrotational everywhere and the lift is then fixed by circulation alone

the drag falls because the frontal area of the section decreases

Which of the following statements about the displacement and momentum thicknesses are correct?

δ* = ∫₀^∞ (1 − u/U) dy measures the outward displacement of the outer streamlines

For a flat plate with dp/dx = 0, the drag per unit width on one side equals ρU²θ at the trailing edge

δ* = ∫₀^∞ (1 − u/U) dy measures the outward displacement of the outer streamlines

For a flat plate with dp/dx = 0, the drag per unit width on one side equals ρU²θ at the trailing edge

The displacement thickness is always larger than δ

The momentum thickness is always larger than the displacement thickness

Which of the following statements about boundary-layer separation are correct?

It occurs in a region of adverse pressure gradient

A turbulent layer resists separation better than a laminar layer

At the separation point the wall shear stress is zero

It occurs in a region of adverse pressure gradient

A laminar layer on a flat plate with dp/dx = 0 separates once the plate is long enough

A turbulent layer resists separation better than a laminar layer

At the separation point the wall shear stress is zero

Which of the following statements about drag are correct?

For a bluff body at high Reynolds number, pressure drag usually exceeds friction drag

Streamlining reduces drag mainly by delaying separation and shrinking the wake

C_D = F_D/(½ρU²A) is dimensionless, with A the reference area used in its definition

For a bluff body at high Reynolds number, pressure drag usually exceeds friction drag

Streamlining reduces drag mainly by delaying separation and shrinking the wake

C_D = F_D/(½ρU²A) is dimensionless, with A the reference area used in its definition

The drag coefficient of a sphere is 0.47 at all Reynolds numbers

External Flows: Boundary Layers, Separation, Drag and Lift

This chapter answers Section XE1.9 of XE1 Fluid Mechanics in the Engineering Sciences (XE) paper, one of the nine optional sections, any two of which a candidate chooses in the exam. The heading reads: “Concept of Prandtl boundary layer, Boundary layer thickness, Displacement thickness and momentum thickness. Qualitative idea of boundary layer separation, Streamlined and bluff bodies, Drag and lift forces.” It teaches the boundary-layer idea and its thicknesses, the laminar flat-plate results (the numerical constants are stated in the stems), separation and the adverse pressure gradient, streamlined and bluff bodies, drag and drag coefficients for plates, spheres and cylinders, the drag crisis, and lift and stall. The numericals are boundary-layer thickness, plate drag, thickness integrals, ball and sphere drag, and stall speed.

1. The Prandtl boundary-layer concept and its thickness

At high Reynolds number the effect of viscosity is felt only in a thin layer next to the wall, the boundary layer, where the speed rises from zero at the wall (no slip) to nearly the free-stream value U. Outside it the flow behaves as inviscid and can be treated by potential flow and Bernoulli. Prandtl's idea splits the problem in two: the outer flow gives the pressure, which the thin layer simply receives, and the layer then supplies the wall shear and the friction drag. The thickness δ is small compared with the body length L, so δ/L ∼ 1/√Re_L.

The boundary-layer thickness δ is the distance from the wall at which u reaches 99% of U. It is a convention, since the layer merges smoothly into the outer flow. Across the thin layer the pressure is almost constant in the direction normal to the wall (∂p/∂y ≈ 0), so the outer-flow pressure acts through the layer unchanged. On a flat plate at zero incidence the outer flow is uniform, so dp/dx = 0 and U is constant.

Boundary layer on a flat plate
Stage	Where	Features
Laminar	from the leading edge to Re_x ≈ 5 × 10⁵	smooth layers, thickness ∝ √x
Transition	around Re_x ≈ 5 × 10⁵ (varies with roughness and free-stream turbulence)	layer thickens quickly
Turbulent	downstream	thicker, fuller profile, larger wall shear

ℹ️ Re_x versus Re_L

Re_x = Ux/ν is measured from the leading edge to the station considered and decides local properties; Re_L uses the full plate length and decides the plate-average drag. Both use ν = μ/ρ, so stating air or water changes Re by a factor of about 15.

2. Displacement thickness and momentum thickness

Two integral thicknesses measure what the layer takes from the outer flow. The displacement thickness δ* = ∫₀^∞ (1 − u/U) dy is the distance by which the wall would have to be moved outward in an inviscid flow to give the same mass flow, so it measures the loss of volume flow: the outer streamlines are pushed away from the wall by δ*. The momentum thickness θ = ∫₀^∞ (u/U)(1 − u/U) dy measures the loss of momentum flux: the momentum lost in the layer equals that of a slab of fluid of thickness θ moving at U.

Why they matter. For a flat plate with dp/dx = 0 the momentum integral equation gives the wall shear as τ_w = ρU² dθ/dx, so the drag on one side of width b and length L is D = ρU² θ_L b, where θ_L is the momentum thickness at the trailing edge. The ratio H = δ*/θ is the shape factor, about 2.6 for a laminar flat-plate layer and about 1.3 to 1.4 for a turbulent one; H rises as a layer approaches separation, so a large H signals a layer close to separating.

Worked example. Take the profile u/U = 2η − η² for η = y/δ ≤ 1 (and u = U beyond). Then δ* = δ∫(1 − 2η + η²)dη = δ(1 − 1 + 1/3) = δ/3. For θ, (2η − η²)(1 − η)² = 2η − 5η² + 4η³ − η⁴, whose integral is 1 − 5/3 + 1 − 1/5 = 2/15, so θ = 2δ/15. For δ = 15 mm this gives δ* = 5.0 mm and θ = 2.0 mm, and H = 2.5. Always θ < δ* < δ.

3. The laminar flat-plate results (Blasius)

For a laminar boundary layer on a flat plate at zero incidence the exact (Blasius) solution gives these results, in which the constants are numerical solutions that exam questions state in the stem:

Laminar flat-plate results with Re_x = Ux/ν
Quantity	Result	Scaling
Thickness	δ = 5.0 x/√Re_x	δ ∝ √x, δ ∝ 1/√U
Displacement and momentum	δ* = 1.72 x/√Re_x, θ = 0.664 x/√Re_x	θ ≈ δ/7.5
Local skin friction	C_f = τ_w/(½ρU²) = 0.664/√Re_x	τ_w ∝ x^(−1/2), U^(3/2)
Average drag coefficient, one side, length L	C_D = 1.328/√Re_L	D ∝ √L, U^(3/2)

The wall shear is largest at the leading edge, where the layer is thinnest, and falls downstream, but the total drag keeps growing as √L. The average coefficient is twice the local value at the trailing edge: 1.328 = 2 × 0.664. The drag on one side is D = ½ρU² C_D L b. For a turbulent layer (about 5 × 10⁵ < Re_L < 10⁷) the common estimates are δ ≈ 0.37 x/Re_x^(1/5) and C_D ≈ 0.074/Re_L^(1/5), which show faster growth (δ ∝ x^(0.8)) and more drag.

Worked example. Air (ν = 1.5 × 10⁻⁵ m²/s, ρ = 1.2 kg/m³) at U = 10 m/s over a plate: at x = 0.50 m, Re_x = 10 × 0.5/1.5 × 10⁻⁵ = 3.33 × 10⁵ (laminar), so δ = 5.0 × 0.5/577 = 4.33 mm. For U = 5 m/s over a plate L = 1.0 m and b = 2.0 m, Re_L = 3.33 × 10⁵, C_D = 1.328/577 = 0.0023 and D = ½ × 1.2 × 25 × 0.0023 × 2.0 = 0.069 N on one side.

⚠️ Only below transition and zero pressure gradient

The numbers 5.0 and 0.664 hold only for a laminar layer on a flat plate with dp/dx = 0; check Re_L < about 5 × 10⁵ first. Double the speed and δ falls only by 1/√2, while the drag rises by 2^(3/2) = 2.83, not 4.

4. Boundary-layer separation and the adverse pressure gradient

Where the outer flow speeds up (dp/dx < 0, a favourable gradient, as on the front of a cylinder) pressure pushes the near-wall fluid along. Where the outer flow slows (dp/dx > 0, an adverse gradient, as on the rear of a cylinder, in a diffuser or on the upper surface behind an aerofoil's suction peak), the near-wall fluid, already slowed by friction, loses momentum to the pressure rise. It is brought to rest and then reversed, and the layer separates from the wall.

At the separation point (∂u/∂y) at the wall is zero, so the wall shear stress is zero; downstream of it there is reversed flow and a wake of eddies.
A flat plate with dp/dx = 0 never separates in this way; separation needs an adverse gradient, so it happens on curved or diverging surfaces and at sharp edges.
A turbulent layer has a fuller profile and more momentum near the wall, so it resists the adverse gradient longer than a laminar one.

Separation turns the rear pressure into a low, nearly uniform wake pressure instead of the recovery that ideal flow predicts. This is why pressure drag appears and why d'Alembert's paradox fails for real bodies. Boundary-layer wall shear alone cannot cause separation; the pressure field does.

5. Streamlined and bluff bodies, drag and the drag coefficient

Drag is the force component along the flow, F_D = ½ρU² C_D A, with A the frontal (projected) area for blunt bodies or the plan area for a plate. It has two parts: friction (skin-friction) drag from wall shear, and pressure (form) drag from the pressure difference between front and rear. A streamlined body (aerofoil, fairing, a slender ellipse) is shaped so that the layer stays attached over almost the whole surface, the wake is thin, and most of the drag is friction. A bluff body (cylinder, sphere, plate normal to the flow) separates early, has a wide wake and gets most of its drag as pressure drag.

Typical drag behaviour
Case	Drag coefficient	Dominant part
Flat plate along the flow, laminar	C_D = 1.328/√Re_L (plate area)	friction
Sphere, Re about 10³ to 2 × 10⁵	C_D ≈ 0.47 (frontal area)	pressure
Smooth cylinder, Re about 10³ to 2 × 10⁵	C_D ≈ 1.2 (projected area)	pressure
Sphere, Re < 1 (creeping flow)	F = 3πμVD, C_D = 24/Re	viscous

Drag crisis. C_D for a sphere or cylinder is nearly constant over a wide range of Re, then falls sharply near Re ≈ 3 × 10⁵ (for a sphere, from about 0.47 to roughly 0.1 to 0.2). The cause is that the layer turns turbulent before separating; its fuller profile resists the adverse gradient, so separation moves downstream, the wake narrows, and pressure drag drops by more than the friction rises. The exact Re depends on surface roughness and free-stream turbulence, which is why dimples on a golf ball make its drag lower at the same speed.

Worked examples. A ball of D = 0.070 m at 30 m/s in air (ρ = 1.2 kg/m³) with C_D = 0.47: A = π × 0.07²/4 = 3.85 × 10⁻³ m², so F = ½ × 1.2 × 900 × 0.47 × 3.85 × 10⁻³ = 0.98 N. A steel bead of D = 2 mm (ρ = 7800 kg/m³) falls in oil (ρ = 900 kg/m³, μ = 0.50 Pa·s). At terminal speed weight minus buoyancy equals the Stokes drag 3πμVD, so V = (ρ_s − ρ)gD²/(18μ) = 6900 × 9.81 × 4 × 10⁻⁶/9 = 0.0301 m/s, and Re = 0.11 confirms the creeping-flow assumption.

6. Lift, the lift coefficient, aerofoils and stall

Lift is the force component perpendicular to the free stream, F_L = ½ρU² C_L A, with A the plan area of the wing. An aerofoil at a small angle of attack α produces lift because its shape and incidence make the flow faster over the upper (suction) surface and slower on the lower, so the pressure is lower above and higher below; equivalently the flow carries a net circulation. For a thin aerofoil in ideal flow, C_L ≈ 2πα (α in radians), so C_L rises linearly with α at small angles, and the drag stays small and mostly frictional.

Stall. As α increases, the suction peak near the leading edge grows and the adverse gradient behind it steepens. At a critical angle (of order 15° for many sections, depending on the shape and Re) the layer separates over much of the upper surface. The lift then drops, the drag rises sharply, and the aerofoil is stalled. The maximum lift coefficient C_L,max sets the minimum flying speed: in level flight lift equals weight W, so the stall speed is V_s = √(2W/(ρ A C_L,max)).

Worked example. W = 20,000 N, A = 16 m², C_L,max = 1.6, ρ = 1.2 kg/m³. V_s = √(2 × 20000/(1.2 × 16 × 1.6)) = √1302 = 36.1 m/s. In ideal flow a lifting body has zero drag; real drag comes from friction, separation and, for a finite wing, the wing-tip vortices (induced drag).

⚠️ Stall is not a speed limit of the engine

Stall is a loss of lift through separation at a high angle of attack, not low engine power. A wing can stall at any speed if α is too large, and the stall speed rises with the weight and falls with the density. The area A in F_L and F_D must be the one the coefficient was defined with: plan area for a wing, frontal area for a sphere.

Key takeaways

The boundary layer is a thin viscous layer where u rises from 0 to 0.99U at thickness δ; outside it the flow is inviscid and sets the pressure.
δ* = ∫(1 − u/U)dy is the outward displacement of streamlines; θ = ∫(u/U)(1 − u/U)dy is the momentum loss, and for dp/dx = 0 the one-side drag is ρU²θ_L b. Always θ < δ* < δ.
Laminar plate: δ = 5.0x/√Re_x, C_f = 0.664/√Re_x, C_D = 1.328/√Re_L; δ ∝ √x, drag ∝ U^(3/2) and ∝ √L, valid below about Re ≈ 5 × 10⁵.
Separation needs an adverse pressure gradient; at the separation point the wall shear is zero. A turbulent layer resists separation longer than a laminar one.
Streamlined bodies have mostly friction drag; bluff bodies mostly pressure drag. F_D = ½ρU²C_D A; the sphere drag crisis near Re ≈ 3 × 10⁵ comes from a turbulent layer delaying separation.
Lift F_L = ½ρU²C_L A grows with angle of attack until stall, when separation on the upper surface cuts lift and raises drag; stall speed is √(2W/(ρAC_L,max)).

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

The central idea of Prandtl's boundary-layer concept at high Reynolds number is that
1. viscosity matters equally everywhere in the flow, and the pressure field is the only thing it changes
2. viscous effects are confined to a thin layer near the wall, with inviscid flow outside it
3. the velocity at the wall equals the free-stream velocity, so that only the outer flow needs to be solved
4. the pressure varies strongly across the thin layer, so the outer pressure cannot be imposed on it
Show answer
Answer: B — viscous effects are confined to a thin layer near the wall, with inviscid flow outside it
The thin layer carries all the viscous shear and the no-slip condition, and the outer inviscid flow supplies the pressure. At the wall the velocity is zero, not U, and across the thin layer the pressure is nearly constant (∂p/∂y ≈ 0), which are the traps in the other statements.
The conventional boundary-layer thickness δ is the distance from the wall at which the local velocity reaches
1. 50% of the free-stream velocity
2. 90% of the free-stream velocity, where the shear stress vanishes
3. exactly the free-stream velocity
4. 99% of the free-stream velocity
Show answer
Answer: D — 99% of the free-stream velocity
By convention δ is where u = 0.99U, because the layer merges asymptotically into the outer flow and never reaches U exactly. The 50% and 90% values are not the definition, and the shear stress becomes small but is not exactly zero there.
The displacement thickness δ* of a boundary layer represents
1. the outward shift of the outer streamlines caused by the layer
2. the distance from the wall at which the velocity first reaches 0.99 of U
3. the distance over which the wall shear stress acts on the passing fluid
4. the loss of momentum flux in the layer, divided by ρU² and by unit width
Show answer
Answer: A — the outward shift of the outer streamlines caused by the layer
δ* = ∫(1 − u/U)dy measures the deficit in volume flow, equivalent to shifting the wall outward by δ*. The 99% definition is δ, and the momentum loss divided by ρU² is the momentum thickness θ, which is the tempting confusion.
A laminar boundary layer grows on a flat plate at zero incidence (Blasius solution, δ = 5.0x/√Re_x). At a fixed station x, the free-stream speed is doubled while the layer stays laminar. The thickness δ becomes
1. 0.5 times the original
2. 1.41 times the original
3. 0.71 times the original
4. 2 times the original
Show answer
Answer: C — 0.71 times the original
δ ∝ 1/√Re_x ∝ 1/√U, so doubling U multiplies δ by 1/√2 = 0.71. A factor of 0.5 would need δ ∝ 1/U, and the factors above 1 would mean the layer thickens with speed, which is the opposite of what happens.
Boundary-layer separation from a smooth curved wall occurs
1. where pressure rises along the flow and the near-wall fluid stops
2. where the pressure gradient is favourable and the outer flow accelerates strongly
3. only when the layer is turbulent, because eddies carry momentum away from the wall
4. at the point where the wall shear stress is largest, which is near the leading edge
Show answer
Answer: A — where pressure rises along the flow and the near-wall fluid stops
Pressure rising along the flow robs the slow near-wall fluid of momentum until it stops (∂u/∂y = 0 at the wall) and reverses. A favourable gradient keeps the layer attached, and turbulence delays separation rather than causing it. The wall shear is zero, not largest, at separation.
At a high Reynolds number, the total drag of a well-designed streamlined body is made up mainly of
1. pressure (form) drag, because the wake is wide
2. wave drag, because the body is long
3. friction drag, because the layer stays attached and the wake is thin
4. induced drag, because the body is symmetrical
Show answer
Answer: C — friction drag, because the layer stays attached and the wake is thin
Streamlining delays separation, so the wake and the pressure drag are small, leaving skin friction as the main part. Wide wakes and pressure-dominated drag belong to bluff bodies. Wave and induced drag are different mechanisms, not set by length or symmetry.
The drag coefficient of a smooth sphere falls sharply near Re ≈ 3 × 10⁵ (drag crisis). The main reason is that
1. the skin friction on the front half drops to zero once the layer is turbulent
2. the flow becomes compressible as the sphere approaches the speed of sound
3. the flow becomes inviscid because the viscosity of air falls at high Re
4. the layer turns turbulent and separation moves downstream
Show answer
Answer: D — the layer turns turbulent and separation moves downstream
A turbulent layer has more near-wall momentum and stays attached longer, so the wake shrinks and the pressure drag falls by more than the friction rises. The friction does not vanish, the speeds involved are low so compressibility is irrelevant, and the flow is certainly not inviscid.
For a laminar boundary layer on a flat plate with dp/dx = 0, the plate-average drag coefficient C_D (one side) is compared with the local skin-friction coefficient C_f at the trailing edge. C_D is
1. equal to C_f, since the plate is uniform
2. half of C_f, since the shear falls downstream
3. twice C_f, since the shear is larger upstream
4. four times C_f, since drag grows as the square of length
Show answer
Answer: C — twice C_f, since the shear is larger upstream
C_f = 0.664/√Re_L at the trailing edge and the average is C_D = 1.328/√Re_L, exactly twice, because the shear is highest near the leading edge and averaging ∫x^(−1/2)dx/L gives 2L^(1/2)/L. The drag grows as √L, not as L², and the shear is not uniform.
An aeroplane in level flight weighs W = 20,000 N, with wing plan area 16 m² and a maximum lift coefficient of 1.6. Take air density 1.2 kg/m³. The stall speed in m/s is approximately
1. 36.1
2. 25.5
3. 51.0
4. 72.2
Show answer
Answer: A — 36.1
At stall L = W = ½ρV²AC_L,max, so V = √(2W/(ρAC_L,max)) = √(40000/30.72) = 36.1 m/s. Dropping the factor 2 gives 25.5 m/s, and the values 51.0 and 72.2 are the two results above multiplied by 2, which no consistent step gives.
A thin aerofoil is held at a steadily increasing angle of attack in a fixed stream. Beyond the stall angle
1. the lift coefficient keeps rising along the same straight line, with the drag staying small
2. the flow separates over much of the upper surface, lift falls and drag rises
3. the flow becomes irrotational everywhere and the lift is then fixed by circulation alone
4. the drag falls because the frontal area of the section decreases
Show answer
Answer: B — the flow separates over much of the upper surface, lift falls and drag rises
Stall is the loss of lift from large-scale separation on the upper surface. Lift does not follow the linear trend any further, and the drag rises, since the wake is now wide and pressure drag dominates. Flow does not become ideal, and the frontal area in fact grows with α.
A ball of diameter 0.070 m moves at 30 m/s through air of density 1.2 kg/m³ with C_D = 0.47 based on frontal area. The drag force is approximately
1. 0.49 N
2. 1.95 N
3. 3.9 N
4. 0.98 N
Show answer
Answer: D — 0.98 N
F = ½ρV²C_D A with A = π × 0.07²/4 = 3.85 × 10⁻³ m²: F = 0.5 × 1.2 × 900 × 0.47 × 3.85 × 10⁻³ = 0.98 N. Using the diameter squared (0.0049 m²) for A gives 1.24 N, which is not an option, and dropping the factor ½ gives 1.95 N.
Which of the following statements about the displacement and momentum thicknesses are correct?
1. δ* = ∫₀^∞ (1 − u/U) dy measures the outward displacement of the outer streamlines
2. For a flat plate with dp/dx = 0, the drag per unit width on one side equals ρU²θ at the trailing edge
3. The displacement thickness is always larger than δ
4. The momentum thickness is always larger than the displacement thickness
Show answer
Answer: A — δ* = ∫₀^∞ (1 − u/U) dy measures the outward displacement of the outer streamlines; B — For a flat plate with dp/dx = 0, the drag per unit width on one side equals ρU²θ at the trailing edge
δ* is the streamline displacement, and the momentum integral with dp/dx = 0 gives D = ρU²θb per side. Since u/U ≤ 1, (u/U)(1 − u/U) is below (1 − u/U), so θ < δ, and δ < δ since the integrand vanishes beyond δ. Both of the 'always larger' statements are reversed.
Which of the following statements about boundary-layer separation are correct?
1. It occurs in a region of adverse pressure gradient
2. A laminar layer on a flat plate with dp/dx = 0 separates once the plate is long enough
3. A turbulent layer resists separation better than a laminar layer
4. At the separation point the wall shear stress is zero
Show answer
Answer: A — It occurs in a region of adverse pressure gradient; C — A turbulent layer resists separation better than a laminar layer; D — At the separation point the wall shear stress is zero
Separation needs dp/dx > 0, comes with ∂u/∂y = 0 at the wall, and is delayed by a turbulent layer's fuller profile. A flat plate with dp/dx = 0 has no adverse gradient, so the layer only becomes turbulent and thickens, never separating by itself.
Which of the following statements about drag are correct?
1. For a bluff body at high Reynolds number, pressure drag usually exceeds friction drag
2. Streamlining reduces drag mainly by delaying separation and shrinking the wake
3. C_D = F_D/(½ρU²A) is dimensionless, with A the reference area used in its definition
4. The drag coefficient of a sphere is 0.47 at all Reynolds numbers
Show answer
Answer: A — For a bluff body at high Reynolds number, pressure drag usually exceeds friction drag; B — Streamlining reduces drag mainly by delaying separation and shrinking the wake; C — C_D = F_D/(½ρU²A) is dimensionless, with A the reference area used in its definition
A bluff body separates early, so the wake pressure deficit dominates; streamlining attaches the flow; and C_D is a ratio of forces. The 0.47 value holds only over a range of Re (about 10³ to 2 × 10⁵) and falls in the drag crisis and rises at low Re, where C_D = 24/Re for Re < 1.
Air (kinematic viscosity 1.5 × 10⁻⁵ m²/s) flows at 10 m/s over a flat plate at zero incidence. Take the laminar result δ = 5.0 x/√Re_x. The boundary-layer thickness at x = 0.50 m, in millimetres to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 4.33
Re_x = Ux/ν = 10 × 0.50/1.5 × 10⁻⁵ = 3.333 × 10⁵ < 5 × 10⁵, so the layer is laminar. √Re_x = 577.35, so δ = 5.0 × 0.50/577.35 = 4.330 × 10⁻³ m = 4.33 mm. Forgetting the square root gives 7.5 μm, an absurdly thin layer.
A flat plate of length 1.0 m (flow direction) and width 2.0 m lies at zero incidence in air (ρ = 1.2 kg/m³, ν = 1.5 × 10⁻⁵ m²/s) moving at 5.0 m/s. The layer is laminar, with plate-average C_D = 1.328/√Re_L based on plate area. The friction drag on one side, in millinewtons to one decimal place, is ____.
Numerical answer — type the value.
Show answer
Answer: 69.0
Re_L = 5.0 × 1.0/1.5 × 10⁻⁵ = 3.333 × 10⁵ (laminar), so C_D = 1.328/577.35 = 2.300 × 10⁻³. D = ½ρU²C_D(Lb) = 0.5 × 1.2 × 25 × 2.300 × 10⁻³ × 2.0 = 0.0690 N = 69.0 mN. For both sides the force would be 138.0 mN.
A boundary layer of thickness δ = 15 mm has the velocity profile u/U = 2η − η², with η = y/δ for y ≤ δ and u = U for y > δ. The momentum thickness θ = ∫₀^δ (u/U)(1 − u/U) dy, in millimetres to one decimal place, is ____.
Numerical answer — type the value.
Show answer
Answer: 2.0
θ = δ∫₀¹(2η − η²)(1 − η)² dη. The integrand expands to 2η − 5η² + 4η³ − η⁴, with integral 1 − 5/3 + 1 − 1/5 = 2/15. So θ = 15 mm × 2/15 = 2.0 mm. Taking only (1 − u/U) would give δ* = δ/3 = 5.0 mm.
A steel bead (diameter 2.0 mm, density 7800 kg/m³) falls steadily in oil (density 900 kg/m³, dynamic viscosity 0.50 Pa·s). Assume creeping flow with Stokes drag F = 3πμVD and g = 9.81 m/s². The terminal speed, in mm/s to one decimal place, is ____.
Numerical answer — type the value.
Show answer
Answer: 30.1
At terminal speed (ρ_s − ρ)g(πD³/6) = 3πμVD, so V = (ρ_s − ρ)gD²/(18μ) = 6900 × 9.81 × (0.002)²/(18 × 0.50) = 0.030084 m/s = 30.1 mm/s. Check: Re = 900 × 0.0301 × 0.002/0.5 = 0.11 < 1, so Stokes drag is valid. Omitting buoyancy (using 7800) would give 34.0 mm/s.