Fluid Mechanics and Marine Hydrodynamics II: Potential Flow, Vortices, Lifting Surfaces, Added Mass, Model Testing and Surface Waves

The second chapter of NM Section 3 is the ideal fluid and what marine hydrodynamics builds on it. Vorticity, circulation, Stokes’s integral theorem and Kelvin’s theorem; potential flow — sources, sinks, doublets (dipoles), the line vortex and their superposition, flow with circulation, potential flow with rotational symmetry, the generalised Bernoulli equation, hydrodynamic forces, D’Alembert’s paradox and the Kutta–Joukowski theorem of hydrodynamic lift; vortex motion — the analogy with the Biot–Savart law, straight parallel vortex filaments and vortex sheets; aerofoils (hydrofoils) — lift, drag, circulation and pressure distribution, thin-aerofoil theory, wings of infinite and finite span, the circulation distribution, the linearised theory of lifting surfaces, and cavitation; added mass and slender-body theory; hydrodynamic model testing and scaling laws; and potential theory applied to surface waves — dispersion, energy transport and wave–body forces. A rudder, a propeller blade and a stabiliser fin are all lifting surfaces, and every seakeeping calculation starts from the linear wave.

1. Vorticity, circulation, Kelvin’s theorem and the velocity potential

Vorticity ω = ∇ × V is twice the local angular velocity of a fluid element. Circulation Γ = ∮V·dl around a closed curve equals, by Stokes’s theorem, the flux of vorticity through any surface spanning it, Γ = ∫∫ω·dA. Kelvin’s circulation theorem: in an inviscid, barotropic fluid under conservative body forces, the circulation around a closed curve moving with the fluid is constant in time. So a flow that starts from rest is irrotational and stays so — which is why the flow outside the boundary layer can be treated as potential, and why, when a foil starts moving and sheds a starting vortex, an equal and opposite circulation appears round the foil.

Irrotational flow has a velocity potential φ with V = ∇φ; if it is also incompressible, ∇·V = 0 gives Laplace’s equation ∇²φ = 0, which is linear, so solutions can be superposed. In two dimensions the stream function ψ (u = ∂ψ/∂y, v = −∂ψ/∂x) also satisfies Laplace’s equation; lines of constant ψ are streamlines, the difference of ψ between two streamlines is the flow rate between them, and streamlines cross equipotentials at right angles. For unsteady potential flow Euler’s equation integrates to the generalised Bernoulli equation ∂φ/∂t + p/ρ + ½|∇φ|² + gz = F(t), valid throughout the fluid and not just along a streamline — the form used for waves and for accelerating bodies.

2. Elementary flows, superposition, D’Alembert’s paradox and Kutta–Joukowski lift

The elementary two-dimensional flows
FlowPotential φVelocity
Uniform stream UUxu = U
Source (sink if m < 0) of strength m(m/2π) ln rRadial, m/(2πr)
Line vortex of circulation Γ(Γ/2π)θTangential, Γ/(2πr)
Doublet (dipole) of strength κκ cos θ/(2πr)Falls as 1/r²

Superposition builds bodies. A uniform stream plus a source gives the Rankine half-body, with a stagnation point at a distance m/(2πU) upstream of the source and a width tending to m/U far downstream. A stream plus a source and an equal sink gives a closed Rankine oval; letting the pair merge into a doublet gives the circular cylinder of radius a = √[κ/(2πU)], on whose surface the speed is 2U sin θ, the pressure coefficient Cp = 1 − 4 sin²θ, and the minimum Cp is −3 at the shoulders. In three dimensions, potential flow with rotational symmetry gives the sphere (a 3-D doublet in a stream), with surface speed 1.5U sin θ.

The pressure on the cylinder is symmetric front to back, so steady potential flow exerts no drag on any closed body — D’Alembert’s paradox, resolved by viscosity, which creates the boundary layer, separation and the wake. Adding a vortex of circulation Γ to the cylinder breaks the top–bottom symmetry: the stagnation points move to sin θ = −Γ/(4πUa) and the body feels a lift per unit span L′ = ρUΓ, perpendicular to the stream — the Kutta–Joukowski theorem, true for any two-dimensional body with circulation Γ. It explains the Magnus effect on a spinning cylinder (the Flettner rotor) and, with the Kutta condition, the lift of a foil.

3. Vortex motion: Biot–Savart, filaments and sheets

Vorticity induces velocity exactly as current induces a magnetic field, so the Biot–Savart law carries over: an element dl of a vortex filament of strength Γ induces dV = (Γ/4π)(dl × r)/r³. An infinite straight filament induces a purely tangential speed Γ/(2πr) at distance r; a semi-infinite one, Γ/(4πr) at its end. Helmholtz’s theorems: the strength of a vortex filament is constant along it, a filament cannot end in the fluid (it closes on itself or ends on a boundary), and in an inviscid fluid it moves with the fluid.

Straight parallel vortex filaments move one another: two of equal and opposite strength a distance d apart translate together at Γ/(2πd), perpendicular to the line joining them; two of the same sign rotate about their centroid. A vortex sheet is a continuous distribution of vorticity of strength γ per unit length, across which the tangential velocity jumps by γ. Thin-aerofoil theory replaces the foil by a vortex sheet on its camber line, and the horseshoe vortex — a bound vortex along the span and two trailing tip vortices — is the model of a finite wing.

4. Foils, finite wings, lifting surfaces and cavitation

A foil in a stream generates lift because the Kutta condition — the flow leaves the sharp trailing edge smoothly — fixes the circulation. Thin-aerofoil theory gives the section lift coefficient C_l = 2π(α − αL0) for small angles of attack α (in radians), with αL0 = 0 for a symmetric section and negative for positive camber; the aerodynamic centre is at the quarter chord, where the pitching moment does not change with α, and for a symmetric section the centre of pressure is there too. The pressure distribution shows a suction peak near the leading edge on the upper (suction) side, and it is this peak that cavitates first.

On a wing of finite span the pressure difference drives flow round the tips, the circulation falls to zero at the tips, and the vorticity shed along the span forms a trailing vortex sheet that rolls up into tip vortices. The downwash it induces tilts the local flow, reducing the effective angle and producing induced drag. Prandtl’s lifting-line theory shows that an elliptic circulation distribution gives uniform downwash and the least induced drag, C_Di = C_L²/(πAR) (with a span-efficiency factor e < 1 otherwise, C_L²/(πeAR)), and reduces the lift slope to a = a₀/[1 + a₀/(πAR)], AR being the aspect ratio span²/area. The linearised theory of lifting surfaces extends this to low-aspect-ratio planforms such as rudders and propeller blades by distributing vortices over the whole surface.

Cavitation is the formation of vapour cavities where the local pressure falls to the vapour pressure p_v. It is governed by the cavitation number σ = (p∞ − p_v)/(½ρV²); a foil cavitates when −Cp,min exceeds σ, so high speed, shallow submergence (low p∞) and a sharp suction peak all promote it. Collapsing cavities cause erosion, noise and vibration, and extensive cavitation causes lift or thrust breakdown. Its forms on propellers — sheet, bubble, cloud, tip-vortex and hub-vortex — are taken up with propulsion.

5. Added mass, slender bodies, model testing and scaling laws

A body accelerating through a fluid must accelerate some of the fluid with it, so it behaves as if its mass were increased by an added (hydrodynamic) mass; the extra force is −m_a (dU/dt), even in potential flow and even though a steadily moving body feels no drag. For a sphere the added mass is half the displaced mass, ½ρ(4/3)πa³; for a circular cylinder moving across its axis it is the displaced mass, ρπa² per unit length. Added mass depends on the direction of motion and on the free surface and depth, and it is the term that makes a ship’s heave and roll periods longer than its dry mass would suggest.

Slender-body theory exploits a body whose length greatly exceeds its beam: the flow at each cross-section is treated as two-dimensional, the section’s added mass is computed as if it were an infinite cylinder, and the forces follow from the rate at which the fluid momentum of the passing sections changes. A closed slender body in steady inviscid flow feels no net force but a destabilising (Munk) moment; the same cross-flow idea underlies strip theory in seakeeping and the prediction of manoeuvring forces.

Hydrodynamic model testing — in towing tanks, cavitation tunnels, manoeuvring and seakeeping basins — relies on scaling laws. Under Froude scaling with geometric scale λ = L_s/L_m and the same g, speeds and times scale as √λ, accelerations as 1, forces as λ³ (times the density ratio) and power as λ3.5. Viscous effects scale with Re and cannot be matched at the same time, so friction is corrected separately; cavitation tests must also match σ, which is why cavitation tunnels reduce the ambient pressure.

6. Surface waves: linear theory, energy transport and wave–body forces

Applying potential theory to the free surface with small amplitude gives linear (Airy) wave theory: a wave of height H, period T, wavenumber k = 2π/L and frequency ω = 2π/T obeys the dispersion relation ω² = gk tanh kh in depth h. In deep water (h > L/2, tanh kh → 1) ω² = gk, so the celerity is c = gT/(2π) and the length L = gT²/(2π) — a 10 s wave is about 156 m long and an 8 s wave travels at 12.5 m/s. In shallow water (h < L/20) c = √(gh), the same for all periods. Water particles move in circles in deep water, with radius decaying as ekz below the surface, and in flattening ellipses in shallow water.

The wave energy per unit surface area is E = ρgH²/8, half kinetic and half potential, and it is carried not at c but at the group velocity c_g, which is c/2 in deep water and equal to c in shallow water; the energy flux (wave power per unit crest length) is E·c_g. Wave–body forces are split, in linear theory, into the Froude–Krylov force from the pressure of the undisturbed wave, the diffraction force from the body’s disturbance of it, and the radiation forces — added mass and wave damping — from the body’s own motion. For slender members of offshore structures, Morison’s equation gives the in-line force per unit length f = ½ρC_D D u|u| + ρC_M(πD²/4)(du/dt), a drag term plus an inertia term, with C_M = 1 + C_a.

⚠️ The phase moves at c, the energy at c/2
In deep water a wave group travels at half the speed of the crests within it; crests appear at the back of a group, run through it and vanish at the front. Swell from a distant storm therefore arrives at c_g, and wave power is E·c_g, not E·c.

Key takeaways

  • Kelvin: circulation round a material curve is constant in inviscid flow, so flow from rest stays irrotational; V = ∇φ, ∇²φ = 0, and solutions superpose.
  • Cylinder: surface speed 2U sin θ, Cp,min = −3, zero drag (D’Alembert); with circulation L′ = ρUΓ (Kutta–Joukowski).
  • Thin foil C_l = 2π(α − α_L0), aerodynamic centre at c/4; finite wing C_Di = C_L²/(πeAR), least for elliptic loading.
  • Added mass: half the displaced mass for a sphere, the displaced mass for a cylinder; Froude scaling: V ∝ √λ, F ∝ λ³, P ∝ λ3.5.
  • Deep water c = gT/2π, L = gT²/2π, c_g = c/2; E = ρgH²/8; Morison f = ½ρC_D D u|u| + ρC_M(πD²/4)u̇.

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.

  1. Kelvin’s circulation theorem states that, in an inviscid barotropic fluid under conservative body forces, the circulation round a closed curve moving with the fluid

    1. remains constant in time
    2. decays exponentially
    3. is always zero
    4. grows with the vorticity
    Show answer

    Answer: A — remains constant in time

    DΓ/Dt = 0 under those conditions, so a flow that starts irrotational stays irrotational. It is not always zero — a foil carries a bound circulation, balanced by the starting vortex it shed — and it decays only when viscosity acts.
  2. A two-dimensional source of strength 4 m²/s is placed in a uniform stream of 2 m/s, forming a Rankine half-body. The distance of the stagnation point upstream of the source (in m), to three decimal places, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 0.318

    At the stagnation point the source’s outflow m/(2πr) cancels U, so r = m/(2πU) = 4/(4π) = 0.318 m. The half-body’s full width far downstream is m/U = 2 m; m/U mistaken for the stagnation distance gives 2.
  3. In steady potential flow past a circular cylinder without circulation, the minimum pressure coefficient on its surface is

    1. −3
    2. −1
    3. −4
    4. 0
    Show answer

    Answer: A — −3

    The surface speed is 2U sin θ, so Cp = 1 − (2 sin θ)² = 1 − 4 sin²θ, which is −3 at θ = 90°. −4 forgets the 1; −1 would be the value if the speed were only √2 U. For a sphere the peak speed is 1.5U and Cp,min = −1.25.
  4. A hydrofoil section in sea water of density 1025 kg/m³ moves at 5 m/s and carries a circulation of 2 m²/s. By the Kutta–Joukowski theorem, the lift per unit span (in N/m) is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 10250

    L′ = ρUΓ = 1025 × 5 × 2 = 10250 N/m, perpendicular to the stream. Inserting ½ρ as if it were a dynamic pressure gives 5125 N/m.
  5. Which statements about steady two-dimensional potential flow are correct?

    1. a closed body experiences no drag
    2. streamlines and equipotential lines cross at right angles
    3. a body with circulation experiences a force perpendicular to the stream
    4. the stream function satisfies Laplace’s equation only if the flow is rotational
    Show answer

    Answer: A — a closed body experiences no drag; B — streamlines and equipotential lines cross at right angles; C — a body with circulation experiences a force perpendicular to the stream

    D’Alembert’s paradox gives zero drag; ∇φ and ∇ψ are orthogonal; Kutta–Joukowski gives a lift ρUΓ normal to U. ∇²ψ = −ω, so ψ satisfies Laplace’s equation exactly when the flow is IRROTATIONAL — the reverse of the fourth statement.
  6. A straight line vortex of circulation 10 m²/s is in an otherwise still fluid. The speed it induces at 2 m from its axis (in m/s), to two decimal places, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 0.80

    v = Γ/(2πr) = 10/(4π) = 0.796 ≈ 0.80 m/s, tangential. Using the Biot–Savart result for a semi-infinite filament, Γ/(4πr), gives 0.40 m/s; dropping the 2π gives 5 m/s.
  7. Two straight parallel vortex filaments of equal strength but opposite sense, a fixed distance apart, in an otherwise still inviscid fluid

    1. translate together perpendicular to the line joining them
    2. rotate about the midpoint between them
    3. remain at rest
    4. move towards each other and annihilate
    Show answer

    Answer: A — translate together perpendicular to the line joining them

    Each induces on the other a speed Γ/(2πd) in the same direction, perpendicular to the joining line, so the pair moves as a unit at that speed — the model of a tip-vortex pair behind a wing. Filaments of the same sense would rotate about their centroid instead; in an inviscid fluid they do not decay.
  8. By thin-aerofoil theory, the lift coefficient of a symmetric section at an angle of attack of 4°, to two decimal places, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 0.44

    C_l = 2πα with α in radians: 2π × 4π/180 = 2π × 0.0698 = 0.439 ≈ 0.44. Putting α = 4 in degrees gives 25.1, which is not a lift coefficient; forgetting the 2 gives 0.22.
  9. A wing of aspect ratio 6 with an elliptic lift distribution operates at a lift coefficient of 0.5. Its induced-drag coefficient, to four decimal places, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 0.0133

    C_Di = C_L²/(πAR) = 0.25/(6π) = 0.01326 ≈ 0.0133, with e = 1 for elliptic loading. Forgetting to square C_L gives 0.0265; doubling the aspect ratio would halve the induced drag.
  10. A foil in sea water of density 1025 kg/m³ moves at 10 m/s where the ambient pressure is 120 kPa and the vapour pressure 2 kPa. The cavitation number, to two decimal places, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 2.30

    σ = (p∞ − p_v)/(½ρV²) = 118 000/(0.5 × 1025 × 100) = 118 000/51 250 = 2.30. The foil cavitates where −Cp exceeds 2.30. Omitting the vapour pressure gives 2.34; omitting the ½ gives 1.15.
  11. Which changes make a hydrofoil more likely to cavitate?

    1. increasing its speed
    2. reducing its depth of submergence
    3. increasing its angle of attack so the leading-edge suction peak sharpens
    4. raising the ambient pressure as in a pressurised tunnel
    Show answer

    Answer: A — increasing its speed; B — reducing its depth of submergence; C — increasing its angle of attack so the leading-edge suction peak sharpens

    Cavitation starts when −Cp,min > σ = (p∞ − p_v)/(½ρV²). Higher speed and shallower depth both lower σ; a sharper suction peak raises −Cp,min. Raising the ambient pressure raises σ and suppresses cavitation — a cavitation tunnel LOWERS its pressure to model the ship’s σ.
  12. A sphere of radius 1 m accelerates through deep sea water of density 1025 kg/m³. Its added mass (in kg), to the nearest integer, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 2147

    m_a = ½ρ(4/3)πa³ = 0.5 × 1025 × 4.1888 = 2147 kg, half the displaced mass of 4294 kg. Taking the whole displaced mass, as for a cylinder in cross-flow, doubles it.
  13. Under Froude scaling with geometric scale ratio λ and the same fluid, the power of the full-scale ship relative to the model scales as

    1. λ3.5
    2. λ³
    3. λ2.5
    4. λ⁴
    Show answer

    Answer: A — λ^{3.5}

    Forces scale as λ³ and speeds as √λ, so power = force × speed scales as λ3.5. λ³ is the force (or displacement) ratio; λ2.5 forgets that force carries the cube of length.
  14. A regular wave of period 8 s travels in deep water. Taking g = 9.81 m/s², its celerity (in m/s), to two decimal places, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 12.49

    Deep water: c = gT/(2π) = 9.81 × 8/6.2832 = 12.49 m/s, and L = cT = 99.9 m. The group velocity is half, 6.24 m/s; the shallow-water formula √(gh) needs a depth and does not apply.
  15. A regular wave of height 2 m travels in sea water of density 1025 kg/m³. Taking g = 9.81 m/s², its energy per unit surface area (in J/m²), to the nearest integer, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 5028

    E = ρgH²/8 = 1025 × 9.81 × 4/8 = 5027.6 ≈ 5028 J/m². Using the amplitude 1 m in place of the height, with the same 1/8, gives 1257; the correct amplitude form is ρga²/2.
  16. A vertical cylinder of diameter 1 m stands in sea water of density 1025 kg/m³. At an instant the water-particle velocity is 2 m/s and its acceleration 1 m/s². With C_D = 1.0 and C_M = 2.0, the Morison force per unit length (in N/m), to the nearest integer, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 3660

    Drag ½ρC_D D u|u| = 0.5 × 1025 × 1 × 1 × 4 = 2050 N/m; inertia ρC_M(πD²/4)u̇ = 1025 × 2 × 0.7854 × 1 = 1610 N/m; total 3660 N/m (the two are summed because both are given at the same instant). Using C_M − 1 = 1 in the inertia term gives 2855 N/m.
  17. Which statements about linear surface waves are correct?

    1. in deep water the group velocity is half the phase velocity
    2. in shallow water all wave periods travel at the same speed
    3. in deep water, longer-period waves travel faster
    4. the wave energy per unit area is proportional to the wave height
    Show answer

    Answer: A — in deep water the group velocity is half the phase velocity; B — in shallow water all wave periods travel at the same speed; C — in deep water, longer-period waves travel faster

    c_g = c/2 in deep water; c = √(gh) in shallow water is independent of period (non-dispersive); c = gT/(2π) in deep water rises with T, which is why swell outruns the storm. E = ρgH²/8 is proportional to the SQUARE of the height.