A velocity potential function φ with V = ∇φ can be defined for a flow only when the flow is

irrotational, since then ∇ × V = 0

incompressible, since then ∇·V = 0

steady, since then ∂V/∂t = 0

irrotational, since then ∇ × V = 0

two-dimensional, since then only u and v exist

In a two-dimensional, incompressible, irrotational flow, which statement about φ and ψ is correct?

Both satisfy Laplace's equation, and lines of constant φ cross lines of constant ψ at right angles

Only φ satisfies Laplace's equation, while ψ never does, because ψ is fixed by continuity alone

Lines of constant φ coincide with lines of constant ψ, so the flow runs along equipotentials

ψ exists only for irrotational flow, and φ only for rotational flow, since φ needs vorticity

In a free vortex in an incompressible fluid, with gravity effects ignored, the pressure

rises as r increases, since speed falls as 1/r

is uniform, since the flow is irrotational

falls as r increases, since speed rises with r

rises as r increases, since speed falls as 1/r

is greatest at the centre, where speed is greatest

For potential flow of speed U past a circular cylinder, the largest speed on the cylinder surface is

2U, at θ = 90° from the front stagnation point

U, at the front stagnation point

3U, at θ = 90° from the front stagnation point

The potential-flow solution for a steady flow past a non-rotating circular cylinder predicts zero drag. The real drag arises mainly because

viscosity makes the boundary layer separate, leaving a low-pressure wake

the fluid is compressible, so the pressure field loses its fore-and-aft symmetry

viscosity makes the boundary layer separate, leaving a low-pressure wake

the stagnation pressure is lower than p_∞ + ½ρU², so the front face carries less load than ideal

the stream function is not single valued around the cylinder, so the surface integral fails

For flow past a rotating cylinder of radius R, the surface speed is q = |2U sinθ − Γ/(2πR)|. When Γ = 4πUR, the stagnation points

merge into one point on the surface, at 90° to the stream direction

remain at the front and rear of the cylinder, exactly as without circulation

move off the surface and into the fluid away from the cylinder

disappear because the lift vanishes at this value of circulation

Which of the following statements about the velocity potential and stream function are correct?

A velocity potential exists whenever the flow is irrotational

A stream function exists for any two-dimensional incompressible flow, rotational or not

A velocity potential exists whenever the flow is irrotational

A stream function exists for any two-dimensional incompressible flow, rotational or not

The stream function satisfies Laplace's equation for every two-dimensional incompressible flow

In uniform flow the lines of constant φ are streamlines

Which of the following are correct for the superposition of potential flows?

The velocity potentials of the component flows add, because Laplace's equation is linear

The velocity vectors of the component flows add at each point

The velocity potentials of the component flows add, because Laplace's equation is linear

The pressure fields of the component flows add, because pressure is a scalar

The velocity vectors of the component flows add at each point

The stream functions multiply to give the stream function of the combined flow

For potential flow of speed U past a circular cylinder, which of the following statements are correct?

The pressure coefficient at a stagnation point is +1

The minimum pressure coefficient on the surface is −3

The surface pressure at θ = 30° equals the free-stream pressure

The pressure coefficient at a stagnation point is +1

The minimum pressure coefficient on the surface is −3

The surface speed at θ = 30° from the front stagnation point is 2U

The surface pressure at θ = 30° equals the free-stream pressure

Potential Flows: Velocity Potential, Source, Sink and Vortex, Superposition and Flow past a Cylinder

This chapter answers Section XE1.8 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: “Velocity potential function, Uniform flow, Source, Sink, and Vortex.” It teaches the velocity potential and stream function and Laplace's equation, the four elementary flows, superposition (source-sink pair, half-body, doublet and the flow past a circular cylinder), the pressure coefficient on the cylinder, the d'Alembert paradox, and the lift on a rotating cylinder with the Kutta-Joukowski result stated. The numericals are stagnation-point location, speeds and pressure differences in vortex and cylinder flow, and lift per unit span.

1. Velocity potential, stream function and Laplace's equation

A flow is irrotational when its vorticity ∇ × V is zero everywhere. Then the velocity is the gradient of a scalar, the velocity potential φ, defined here by V = ∇φ, so u = ∂φ/∂x, v = ∂φ/∂y and, in polar form, v_r = ∂φ/∂r, v_θ = (1/r)∂φ/∂θ (some books put a minus sign in the definition; the physics does not change). If the flow is also incompressible, ∇·V = 0 gives Laplace's equation ∇²φ = ∂²φ/∂x² + ∂²φ/∂y² = 0. Lines of constant φ are equipotential lines.

In two-dimensional incompressible flow the stream function ψ satisfies continuity automatically: u = ∂ψ/∂y, v = −∂ψ/∂x, with v_r = (1/r)∂ψ/∂θ and v_θ = −∂ψ/∂r. Lines of constant ψ are streamlines, and the difference of ψ between two streamlines is the volume flow per unit depth between them. Vorticity is ω_z = −∇²ψ, so ψ obeys Laplace's equation only when the flow is also irrotational. For such a flow φ and ψ are conjugate: ∂φ/∂x = ∂ψ/∂y and ∂φ/∂y = −∂ψ/∂x, so equipotential lines and streamlines cross at right angles.

What exists when
Function	Needs	Satisfies Laplace's equation when
Velocity potential φ	irrotational flow (any dimension)	also incompressible
Stream function ψ	2D incompressible flow, rotational or not	also irrotational

⚠️ Check ψ before calling a flow irrotational

For ψ = 2xy, ∇²ψ = 0, so the flow is irrotational: u = 2x, v = −2y. For ψ = x² + y², ∇²ψ = 4 ≠ 0, so the flow is rotational (it is solid-body rotation). Having a stream function says nothing about irrotationality; only ∇²ψ = 0 does.

2. Uniform flow, source, sink and free vortex

Four building blocks solve Laplace's equation in the plane. Let U be the stream speed along +x, m the source strength (volume flow per unit depth, taken positive for a source, negative for a sink) and Γ the circulation of a vortex (counter-clockwise positive).

The elementary plane flows
Flow	Velocity	φ	ψ
Uniform	u = U, v = 0	Ux	Uy
Source / sink	v_r = m/(2πr), v_θ = 0	(m/2π) ln r	(m/2π) θ
Free vortex	v_θ = Γ/(2πr), v_r = 0	Γθ/2π	−(Γ/2π) ln r

The source speed v_r falls as 1/r because the same flow m crosses a circle of circumference 2πr. The free vortex is irrotational everywhere except the singular centre, though the fluid circulates: the circulation round any loop that encloses the centre is Γ, while the circulation round a loop that does not enclose it is zero. Velocity is infinite at the centre of both, so these flows describe only the region away from the origin.

Pressure in a free vortex. Bernoulli applies between any two points of an irrotational flow. With v_θ = Γ/(2πr), p + ½ρΓ²/(4π²r²) is constant, so pressure rises outward: p(r) = p_∞ − ρΓ²/(8π²r²). For water, a vortex with 8 m/s at r = 0.5 m has 2 m/s at r = 2 m, so p₂ − p₁ = ½ × 1000 × (8² − 2²) = 30 kPa.

⚠️ Free vortex versus forced vortex

In a free vortex speed is proportional to 1/r and the flow is irrotational. In a forced (solid-body) vortex speed is proportional to r and the vorticity is 2ω. Only the first has a velocity potential. Real vortices (bathtub, tornado) have a forced-vortex core inside a free-vortex outer region.

3. Superposition: source-sink pair and the half-body

Laplace's equation is linear, so the sum of any solutions is a solution: φ = Σφᵢ and ψ = Σψᵢ, and velocity vectors add. Pressure does not add, because Bernoulli is quadratic in velocity; compute the total velocity first and then find p. Superposition lets simple flows build flows past shapes: the streamline ψ = 0 (or another chosen value) that separates fluid from the source is taken as a solid surface, because no fluid crosses a streamline.

Source in a uniform stream (Rankine half-body). ψ = Uy + (m/2π)θ = U r sinθ + (m/2π)θ. The source sends fluid upstream and the stream pushes it back, so they balance at a stagnation point on the axis upstream of the source at distance a = m/(2πU). The streamline through it is a body shape that is open downstream. Far downstream its total width is m/U (half-width m/2U), since the source's whole flow m must pass through a strip of width w moving at U.

Worked example. U = 10 m/s and m = 12 m²/s. The stagnation point lies a = 12/(2π × 10) = 0.191 m upstream of the source, and the body is 12/10 = 1.2 m wide far downstream. Changing m to 24 doubles both the distance and the width.

A source-sink pair of equal strength, separated by 2b, has streamlines that leave the source and enter the sink (the flow is that of a dipole as b → 0). Placed in a uniform stream, the source-sink pair gives a closed oval, the Rankine oval, which is a closed body; the half-body above is its open-ended cousin.

4. Doublet and flow past a circular cylinder

A doublet is the limit of a source-sink pair brought together (b → 0) while the product m·2b = κ is held fixed. Its stream function is ψ = −(κ/2π) sinθ/r. Add it to a uniform stream: ψ = U sinθ (r − R²/r), with R² = κ/(2πU). On r = R, ψ = 0, so the circle r = R is a streamline and can be a solid cylinder of radius R. The velocity at the cylinder surface is purely tangential:

v_r = U cosθ (1 − R²/r²) = 0 on r = R, and v_θ = −U sinθ (1 + R²/r²) = −2U sinθ there.
The two stagnation points are at θ = 0 and θ = 180° (front and rear), where the speed is zero.
The surface speed is largest at θ = 90° and 270°, where it is 2U, twice the free stream.

The pressure coefficient is C_p = (p − p_∞)/(½ρU²) = 1 − (V/U)², by Bernoulli between the surface point and far upstream. With V = 2U sinθ on the surface, C_p = 1 − 4 sin²θ.

Pressure coefficient on the cylinder, potential flow
θ	V/U	C_p	Meaning
0°, 180°	0	+1	stagnation, maximum pressure
30°	1	0	pressure equals p_∞
60°	1.73	−2	suction
90°	2	−3	maximum speed, minimum pressure

Worked example. Air (ρ = 1.2 kg/m³) at U = 40 m/s meets a cylinder. At the shoulder (θ = 90°), p − p_∞ = −3 × ½ × 1.2 × 40² = −2880 Pa, and at the front stagnation point p − p_∞ = +960 Pa. The pressure difference between the two is 4 × 960 = 3840 Pa.

5. Pressure distribution and the d'Alembert paradox

The potential-flow pressure distribution C_p = 1 − 4 sin²θ is symmetric about the flow axis (front to back) and about the stream direction (top to bottom). Integrating pressure over the surface, the forces along and across the stream both vanish. A body in steady, inviscid, irrotational, unseparated flow of unbounded extent therefore feels no drag. This is d'Alembert's paradox, since real bodies certainly experience drag.

The paradox is resolved by the viscous boundary layer (Section XE1.9). A real boundary layer cannot follow the pressure recovery on the rear half, where the pressure rises, so the flow separates and leaves a low-pressure wake. The pressure on the rear is then not restored, front and rear differ, and pressure (form) drag arises, along with a small skin-friction drag. Potential flow is therefore a good model of the outer flow and of the front part of the body, and a poor model of the wake.

⚠️ Same potential flow, no net force

Potential flow gives a surface pressure close to the measured one on the front part of a real cylinder (at high Reynolds number), but the zero-drag result does not carry over. A common wrong answer says the paradox arises because the fluid is compressible; it arises because viscosity is absent and so separation cannot occur.

6. Rotating cylinder, circulation and the Kutta-Joukowski lift

Add a free vortex of circulation Γ at the centre of the cylinder: ψ = U sinθ (r − R²/r) − (Γ/2π) ln(r/R). Still ψ = 0 on r = R, so the cylinder remains a streamline. On the surface the speed is q = |2U sinθ − Γ/(2πR)|. The stagnation points satisfy sinθ = Γ/(4πUR) (with Γ taken counter-clockwise): they move toward each other along the surface as Γ grows, meet at the top (or bottom) of the cylinder when Γ = 4πUR, and leave the surface for a larger Γ.

With circulation the pressure distribution is no longer symmetric top to bottom: the surface speed is higher on the side where the rotation adds to the stream, so the pressure is lower there, and a net transverse force results. The Kutta-Joukowski theorem gives the lift per unit span on any cylinder of arbitrary cross-section in steady, ideal, irrotational flow: L′ = ρUΓ. The force is perpendicular to the free stream, directed toward the side where the surface moves with the stream, and the drag is still zero.

Worked example. A spinning cylinder of R = 0.10 m has surface speed ωR = 15 m/s. Assuming the circulation equals that of the surface, Γ = 2πR(ωR) = 9.42 m²/s. In air (ρ = 1.2 kg/m³) at U = 20 m/s the lift per metre is L′ = 1.2 × 20 × 9.42 = 226 N/m. This is the Magnus effect; in real flow, viscosity and separation make the lift smaller and some drag remains.

⚠️ Where the lift comes from

Without circulation a cylinder in potential flow has zero lift whatever the speed. The lift is tied to Γ, not to the rotation of the cylinder as such; the theorem needs Γ to be known or fixed by a trailing-edge (Kutta) condition. Use L′ = ρUΓ with Γ in m²/s, and note that L′ is a force per metre of span (N/m).

Key takeaways

A velocity potential exists for irrotational flow; a stream function exists for any 2D incompressible flow. Both satisfy Laplace's equation only for 2D incompressible irrotational flow, where the two families of lines cross at right angles.
Source: v_r = m/(2πr); free vortex: v_θ = Γ/(2πr); both fall as 1/r. Pressure in a free vortex rises outward, p = p_∞ − ρΓ²/(8π²r²).
Velocities add under superposition; pressures do not. A source in a uniform stream gives a half-body with its stagnation point at m/(2πU) upstream and far-downstream width m/U.
Doublet plus uniform stream is flow past a cylinder: surface speed 2U sinθ, C_p = 1 − 4 sin²θ, maximum speed 2U and minimum C_p = −3 at θ = 90°.
The symmetric potential-flow pressure gives zero drag (d'Alembert's paradox); real drag comes from separation and the wake, which need viscosity.
Adding circulation Γ gives lift L′ = ρUΓ per unit span (Kutta-Joukowski), perpendicular to the stream, with zero drag in ideal flow.

Practice questions (17)

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.

A velocity potential function φ with V = ∇φ can be defined for a flow only when the flow is
1. incompressible, since then ∇·V = 0
2. steady, since then ∂V/∂t = 0
3. irrotational, since then ∇ × V = 0
4. two-dimensional, since then only u and v exist
Show answer
Answer: C — irrotational, since then ∇ × V = 0
A curl-free field is the gradient of a scalar, so irrotationality is the condition. Incompressibility is the condition for a stream function (2D) and decides whether φ satisfies Laplace's equation; steadiness and dimension do not decide the existence of φ.
In a two-dimensional, incompressible, irrotational flow, which statement about φ and ψ is correct?
1. Both satisfy Laplace's equation, and lines of constant φ cross lines of constant ψ at right angles
2. Only φ satisfies Laplace's equation, while ψ never does, because ψ is fixed by continuity alone
3. Lines of constant φ coincide with lines of constant ψ, so the flow runs along equipotentials
4. ψ exists only for irrotational flow, and φ only for rotational flow, since φ needs vorticity
Show answer
Answer: A — Both satisfy Laplace's equation, and lines of constant φ cross lines of constant ψ at right angles
Irrotationality gives ∇²ψ = 0 and incompressibility gives ∇²φ = 0, and ∇φ·∇ψ = 0 follows from u = ∂φ/∂x = ∂ψ/∂y and v = ∂φ/∂y = −∂ψ/∂x. The statement that ψ never satisfies Laplace's equation is wrong for exactly this flow, and the claim that ψ is only for irrotational and φ only for rotational flow reverses the truth: φ needs irrotationality.
Which of the following stream functions, with u = ∂ψ/∂y and v = −∂ψ/∂x, represents an irrotational flow?
1. ψ = x² + y²
2. ψ = x²y
3. ψ = y³
4. ψ = 2xy
Show answer
Answer: D — ψ = 2xy
Vorticity is ω_z = −∇²ψ. For 2xy, ∇²ψ = 0, so the flow is irrotational. For x² + y² it is 4, for x²y it is 2y and for y³ it is 6y, none of which vanishes everywhere. x² + y² is the tempting one because of its symmetry; it is a solid-body rotation, not a potential flow.
A two-dimensional source of strength m = 4π m²/s (volume flow per unit depth) is placed in still water. The radial speed at r = 2.0 m is
1. 0.5 m/s
2. 2.0 m/s
3. 4.0 m/s
4. 1.0 m/s
Show answer
Answer: D — 1.0 m/s
v_r = m/(2πr) = 4π/(2π × 2.0) = 1.0 m/s. Using m/r (2π m/s ≈ 6.3) or m/(4πr) = 0.5 m/s mistakes the geometry: the flow crosses a circle of circumference 2πr, not a sphere of area 4πr².
In a free vortex in an incompressible fluid, with gravity effects ignored, the pressure
1. is uniform, since the flow is irrotational
2. falls as r increases, since speed rises with r
3. rises as r increases, since speed falls as 1/r
4. is greatest at the centre, where speed is greatest
Show answer
Answer: C — rises as r increases, since speed falls as 1/r
Bernoulli holds across the whole irrotational flow: p + ½ρv² is constant with v = Γ/(2πr), so p rises outward and is lowest near the core. Uniform pressure would mean no centripetal force. The options with speed rising with r or the greatest pressure at the centre reverse the Bernoulli relation: higher speed goes with lower pressure.
A source placed in a uniform stream gives the flow past a semi-infinite body that is open downstream. This body is called
1. a circular cylinder
2. a Rankine half-body
3. a Rankine oval
4. a Joukowski aerofoil
Show answer
Answer: B — a Rankine half-body
A source plus a uniform stream gives the half-body. The oval needs a source and a sink of equal strength, the circular cylinder comes from a doublet, and an aerofoil needs a transformation with circulation.
For potential flow of speed U past a circular cylinder, the largest speed on the cylinder surface is
1. 2U, at θ = 90° from the front stagnation point
2. U, at the front stagnation point
3. 3U, at θ = 90° from the front stagnation point
4. 2U, at both stagnation points
Show answer
Answer: A — 2U, at θ = 90° from the front stagnation point
The surface speed is 2U|sinθ|, which is largest at θ = 90°. The number 3 is the magnitude of the minimum pressure coefficient, not a speed ratio, and at the stagnation points the speed is zero.
The potential-flow solution for a steady flow past a non-rotating circular cylinder predicts zero drag. The real drag arises mainly because
1. the fluid is compressible, so the pressure field loses its fore-and-aft symmetry
2. viscosity makes the boundary layer separate, leaving a low-pressure wake
3. the stagnation pressure is lower than p_∞ + ½ρU², so the front face carries less load than ideal
4. the stream function is not single valued around the cylinder, so the surface integral fails
Show answer
Answer: B — viscosity makes the boundary layer separate, leaving a low-pressure wake
The potential-flow pressure is symmetric fore and aft, so the force cancels. Real boundary layers separate in the adverse pressure gradient on the rear half, so the rear pressure is not recovered. Compressibility is not needed (the paradox holds at low speed), and the stagnation pressure is correctly p_∞ + ½ρU².
For flow past a rotating cylinder of radius R, the surface speed is q = |2U sinθ − Γ/(2πR)|. When Γ = 4πUR, the stagnation points
1. merge into one point on the surface, at 90° to the stream direction
2. remain at the front and rear of the cylinder, exactly as without circulation
3. move off the surface and into the fluid away from the cylinder
4. disappear because the lift vanishes at this value of circulation
Show answer
Answer: A — merge into one point on the surface, at 90° to the stream direction
q = 0 needs sinθ = Γ/(4πUR) = 1, so θ = 90°, a single point. For Γ below 4πUR there are two stagnation points on the surface and for Γ above it they move off into the fluid. The lift ρUΓ is not zero; it is proportional to Γ.
Which of the following statements about the velocity potential and stream function are correct?
1. A velocity potential exists whenever the flow is irrotational
2. A stream function exists for any two-dimensional incompressible flow, rotational or not
3. The stream function satisfies Laplace's equation for every two-dimensional incompressible flow
4. In uniform flow the lines of constant φ are streamlines
Show answer
Answer: A — A velocity potential exists whenever the flow is irrotational; B — A stream function exists for any two-dimensional incompressible flow, rotational or not
Irrotationality gives φ, and incompressibility in 2D gives ψ. But ψ satisfies Laplace's equation only if the flow is also irrotational, since ω_z = −∇²ψ. In uniform flow along x, φ = Ux, so constant-φ lines are vertical and cross the horizontal streamlines at right angles.
Which of the following are correct for the superposition of potential flows?
1. The velocity potentials of the component flows add, because Laplace's equation is linear
2. The pressure fields of the component flows add, because pressure is a scalar
3. The velocity vectors of the component flows add at each point
4. The stream functions multiply to give the stream function of the combined flow
Show answer
Answer: A — The velocity potentials of the component flows add, because Laplace's equation is linear; C — The velocity vectors of the component flows add at each point
Linearity means φ and ψ add, and so do the velocities. Pressure follows Bernoulli, p = p₀ − ½ρ|V|², which is quadratic in the total velocity, so it must be computed from the summed velocity. Stream functions add, they do not multiply.
For potential flow of speed U past a circular cylinder, which of the following statements are correct?
1. The pressure coefficient at a stagnation point is +1
2. The minimum pressure coefficient on the surface is −3
3. The surface speed at θ = 30° from the front stagnation point is 2U
4. The surface pressure at θ = 30° equals the free-stream pressure
Show answer
Answer: A — The pressure coefficient at a stagnation point is +1; B — The minimum pressure coefficient on the surface is −3; D — The surface pressure at θ = 30° equals the free-stream pressure
C_p = 1 − 4 sin²θ gives +1 at θ = 0 and −3 at θ = 90°. At θ = 30°, sinθ = 0.5, so the speed is 2U × 0.5 = U (not 2U), C_p = 0 and the pressure equals p_∞.
A two-dimensional source of strength m = 12 m²/s (volume flow per unit depth) is placed in a uniform stream of speed U = 10 m/s. The stagnation point lies on the axis upstream of the source. The distance of this stagnation point from the source, in metres to three decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 0.191
At the stagnation point the source speed m/(2πa) points upstream and cancels U, so U − m/(2πa) = 0. So a = m/(2πU) = 12/(2π × 10) = 0.19099 m, which is 0.191 m. Forgetting the 2π gives 1.2 m, which is the far-downstream width m/U.
Air of density 1.2 kg/m³ flows steadily past a long circular cylinder with an undisturbed speed of 30 m/s. Using ideal (potential) flow, find the difference in pressure between the front stagnation point and the point on the surface at 90° from it, in pascals, as a whole number.
Numerical answer — type the value.
Show answer
Answer: 2160
With ½ρU² = 0.5 × 1.2 × 900 = 540 Pa, C_p = +1 at the front and −3 at 90°. The difference is (1 − (−3)) × 540 = 4 × 540 = 2160 Pa. Equivalently, at 90° the speed is 60 m/s, so the difference is ½ × 1.2 × 60² = 2160 Pa.
A free vortex in water (ρ = 1000 kg/m³) has a tangential speed of 8.0 m/s at r = 0.50 m. Ignoring gravity, the pressure at r = 2.0 m exceeds the pressure at r = 0.50 m by ____ kPa (answer as a whole number).
Numerical answer — type the value.
Show answer
Answer: 30
v ∝ 1/r gives v₂ = 8 × 0.5/2.0 = 2.0 m/s. Bernoulli: p₂ − p₁ = ½ρ(v₁² − v₂²) = 0.5 × 1000 × (64 − 4) = 30,000 Pa = 30 kPa. Taking v₂ = 4 m/s (treating it as a forced vortex) would give 24 kPa.
A two-dimensional flow has the stream function ψ = 3xy (in m²/s, with x and y in metres), and u = ∂ψ/∂y, v = −∂ψ/∂x. The speed at the point (x, y) = (2, 1) m, in m/s to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 6.71
u = ∂ψ/∂y = 3x = 6 m/s and v = −∂ψ/∂x = −3y = −3 m/s at (2, 1). Speed = √(36 + 9) = √45 = 6.708, which is 6.71 m/s. Swapping the formulas for u and v gives the same speed here, but not the direction.
A long cylinder of radius 0.10 m rotates in a uniform air stream (ρ = 1.2 kg/m³, U = 20 m/s). Treat the circulation as Γ = 2πR(ωR) for a surface speed ωR = 15 m/s in ideal flow. Using the Kutta-Joukowski result L′ = ρUΓ, the lift per metre of span is approximately
1. 226 N/m
2. 360 N/m
3. 452 N/m
4. 113 N/m
Show answer
Answer: A — 226 N/m
Γ = 2π × 0.10 × 15 = 9.42 m²/s, and L′ = 1.2 × 20 × 9.42 = 226 N/m. Using Γ = ωR (15) gives 360 N/m, using 2ρUΓ gives 452 N/m and using ½ρUΓ gives 113 N/m; the theorem has no such factor.