Fluid Mechanics and Marine Hydrodynamics I: Fluid Statics, Conservation Laws, Similarity, Viscous Flow and the Boundary Layer

Section 3 of the Naval Architecture and Marine Engineering (NM) paper is named for two subjects, Fluid Mechanics and Marine Hydrodynamics, and it is split between two chapters along the line where viscosity stops mattering. This first chapter is the real fluid: fluid properties; fluid statics and the stability of floating bodies; the conservation laws of mass, momentum and energy in integral and differential form, with the continuity, Euler and Bernoulli equations; dimensional analysis and dynamic similarity; viscous flow of incompressible fluids — the Navier–Stokes equations, Couette and plane Poiseuille flow, elementary turbulent flow and flow through pipes; and boundary-layer theory — Prandtl’s equations, the criterion for separation, the Blasius solution, skin friction, displacement and momentum thickness, the turbulent boundary layer and boundary-layer control. The second chapter takes the ideal fluid: potential flow, vortices, lifting surfaces, added mass, model testing and surface waves. The skin friction computed here is the frictional resistance of a ship, and the Froude and Reynolds numbers met here decide how a model test is scaled.

1. Fluid properties, fluid statics and the stability of floating bodies

A fluid deforms continuously under shear. Its density ρ (fresh water about 1000 kg/m³, sea water about 1025 kg/m³), its dynamic viscosity μ, defined by Newton’s law of viscosity τ = μ du/dy, and the kinematic viscosity ν = μ/ρ are the properties that matter most here; surface tension and vapour pressure matter for small scales and for cavitation. In a fluid at rest the pressure is the same in every direction and increases with depth as dp/dz = −ρg, so p = p₀ + ρgh in a liquid of constant density.

The hydrostatic force on a plane surface of area A whose centroid is at depth h_c is F = ρg h_c A, acting at the centre of pressure, which lies below the centroid at depth h_cp = h_c + I_G/(h_c A), I_G being the second moment of the area about its own horizontal centroidal axis. For a vertical rectangle of depth d with its top at the surface, h_c = d/2 and the centre of pressure is at 2d/3. On a curved surface the horizontal component equals the force on the vertical projection and the vertical component equals the weight of fluid above the surface.

A floating body displaces its own weight of fluid (Archimedes), and its buoyancy acts at the centre of buoyancy B, the centroid of the displaced volume. It is stable in small heel when the metacentre M lies above the centre of gravity G: GM = KB + BM − KG > 0, with BM = I/∇, I the second moment of the waterplane about the heel axis and ∇ the displaced volume. A fully submerged body has no waterplane, so BM = 0 and it is stable only if G is below B. The naval architecture chapter develops this into the whole of ship stability.

2. Conservation laws: continuity, momentum, energy, Euler and Bernoulli

For a control volume, conservation of mass says that the rate of increase of mass inside equals the net inflow; in steady flow through a duct this is ρ₁A₁V₁ = ρ₂A₂V₂, and for an incompressible fluid A₁V₁ = A₂V₂. In differential form it is ∂ρ/∂t + ∇·(ρV) = 0, reducing to ∇·V = 0 for incompressible flow. The momentum equation for a steady control volume, ΣF = Σ(ṁV)_out − Σ(ṁV)_in, gives the force on a pipe bend, a nozzle or a vane and the thrust of a jet or a propeller; the energy equation adds heat and shaft work to the balance of pressure, kinetic and potential energy, and its mechanical-energy form includes the head lost to friction.

For an inviscid fluid, Newton’s law per unit volume is Euler’s equation ρ DV/Dt = −∇p + ρg. Integrated along a streamline in steady flow it becomes Bernoulli’s equation, p/ρ + V²/2 + gz = constant, which holds along each streamline; if the flow is also irrotational the constant is the same everywhere. Its four conditions — steady, inviscid, incompressible, along a streamline — are what an examiner tests when the equation is misapplied. The generalised, unsteady form for potential flow is taken up in the next chapter.

🧠 Continuity first, Bernoulli second
Water (1000 kg/m³) flows from a 0.2 m pipe at 2 m/s into a horizontal 0.1 m pipe. Continuity: V₂ = 2 × (0.2/0.1)² = 8 m/s. Bernoulli: p₁ − p₂ = ½ × 1000 × (64 − 4) = 30 kPa. Halving the diameter quadruples the velocity; doubling the velocity does not.

3. Dimensional analysis and dynamic similarity

The Buckingham π theorem: a relation among n variables involving k fundamental dimensions can be written as a relation among n − k dimensionless groups. Ship resistance R depends on ρ, V, L, μ and g — six variables and three dimensions (M, L, T) — so R/(½ρV²L²) = f(Re, Fn), a relation among three groups. The groups that matter in marine hydrodynamics are the Reynolds number Re = VL/ν (inertia to viscous forces), the Froude number Fn = V/√(gL) (inertia to gravity, which governs the wave pattern), the Weber number (inertia to surface tension), the Euler number (pressure to inertia) and the cavitation number σ = (p − p_v)/(½ρV²).

Dynamic similarity between model and prototype needs geometric similarity and equal values of every relevant dimensionless group. For a ship model this is impossible: equal Fn requires V_m = V_s/√λ, while equal Re in water of similar viscosity requires V_m = λV_s. The way out is Froude’s hypothesis, which treats the resistance as a frictional part depending on Re and a residuary part depending on Fn; the model is run at the corresponding speed (equal Fn) and the frictional part is corrected between the two Reynolds numbers, as the resistance section of the naval architecture chapter shows.

4. Viscous flow: Navier–Stokes, Couette, Poiseuille, turbulence and pipes

The Navier–Stokes equations add the viscous term to Euler’s: ρ DV/Dt = −∇p + ρg + μ∇²V for an incompressible Newtonian fluid. They have exact solutions for parallel flows. Couette flow between a fixed plate and one moving at U, gap h, with no pressure gradient, has the linear profile u = Uy/h and shear stress τ = μU/h everywhere. Plane Poiseuille flow between fixed plates driven by a pressure gradient has the parabolic profile u = (1/2μ)(−dp/dx)y(h − y), with u_max = 1.5 ū. In a circular pipe (Hagen–Poiseuille) u_max = 2ū, Q = πR⁴(−dp/dx)/(8μ), and the Darcy friction factor is f = 64/Re.

Pipe flow is laminar below a Reynolds number of about 2000 and becomes turbulent above a few thousand, with random fluctuations that transport momentum far more effectively than viscosity — modelled as a Reynolds stress −ρu′v′ and an eddy viscosity. The turbulent profile is much flatter (roughly a 1/7-power law), with a thin viscous sublayer at the wall. In either regime the head lost to friction is given by the Darcy–Weisbach equation h_f = f(L/D)(V²/2g); in turbulent flow f depends on Re and the relative roughness ε/D, read from the Moody chart. Minor losses at bends, valves and fittings are K V²/2g each.

Exact viscous flows
FlowProfileUseful result
Couette (moving plate, no pressure gradient)Linear, u = Uy/hτ = μU/h, uniform across the gap
Plane Poiseuille (fixed plates)Parabolicu_max = 1.5 ū
Hagen–Poiseuille (circular pipe)Paraboloidalu_max = 2ū, f = 64/Re

5. Boundary-layer theory: Blasius, thicknesses, separation, turbulence and control

At high Reynolds number the viscous effects are confined to a thin boundary layer next to the surface. Prandtl’s order-of-magnitude argument reduces Navier–Stokes to the boundary-layer equations u∂u/∂x + v∂u/∂y = −(1/ρ)dp/dx + ν∂²u/∂y², with ∂p/∂y = 0 across the layer, so the pressure is impressed by the outer potential flow. For a flat plate with zero pressure gradient, Blasius’s similarity solution gives the thickness δ ≈ 5.0x/√Reₓ, the displacement thickness δ* = 1.72x/√Reₓ, the momentum thickness θ = 0.664x/√Reₓ, the local skin-friction coefficient cf = 0.664/√Reₓ and the plate drag coefficient C_D = 1.328/√Re_L. The shape factor H = δ*/θ is about 2.59 for the laminar layer.

The displacement thickness δ* = ∫(1 − u/U)dy is the distance the outer flow is pushed away by the velocity deficit; the momentum thickness θ = ∫(u/U)(1 − u/U)dy measures the momentum lost, and the drag per unit width of a plate is ρU²θ at its trailing edge. On a flat plate the layer becomes turbulent at Reₓ of roughly 5 × 10⁵, after which it grows faster, δ ≈ 0.37x/Reₓ1/5, with C_D ≈ 0.074/Re_L1/5 and H ≈ 1.3. A ship at full scale has a turbulent boundary layer over nearly its whole length, which is why its frictional resistance is computed from turbulent friction lines.

Separation occurs in an adverse pressure gradient (dp/dx > 0), where the slow fluid near the wall is decelerated until it reverses; the criterion is (∂u/∂y) at the wall = 0, that is zero wall shear. Downstream the flow is a recirculating wake and a large pressure drag appears. A turbulent layer, with more momentum near the wall, resists separation longer than a laminar one — which is why the drag of a sphere falls sharply when its boundary layer turns turbulent. Boundary-layer control delays separation by suction (removing the slow fluid), blowing or tangential injection (re-energising it), slots and slats, moving surfaces, and vortex generators that mix high-momentum fluid down to the wall.

⚠️ Zero wall shear marks separation; zero pressure gradient does not
A favourable pressure gradient (dp/dx < 0) never separates a boundary layer, and a flat plate at zero gradient never separates either. Separation needs dp/dx > 0, and its location is where (∂u/∂y) at the wall first reaches zero.

Key takeaways

  • F = ρg h_c A on a plane surface, acting 2d/3 down a vertical rectangle from the surface; floating stability needs GM = KB + BM − KG > 0 with BM = I/∇.
  • Bernoulli holds along a streamline in steady, inviscid, incompressible flow, and everywhere if the flow is also irrotational.
  • Ship resistance: R/(½ρV²L²) = f(Re, Fn); equal Fn and equal Re cannot both be met, so Froude’s hypothesis splits the resistance.
  • Couette τ = μU/h; plane Poiseuille u_max = 1.5ū; pipe u_max = 2ū and f = 64/Re; h_f = f(L/D)V²/2g.
  • Blasius: δ = 5x/√Re, C_D = 1.328/√Re, H = 2.59; separation where wall shear is zero in an adverse gradient.

Practice questions (14)

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. A vertical rectangular gate 2 m wide and 3 m deep has its top edge at the free surface of sea water of density 1025 kg/m³. Taking g = 9.81 m/s², the hydrostatic force on the gate (in kN), to one decimal place, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 90.5

    F = ρg h_c A = 1025 × 9.81 × 1.5 × 6 = 90 497 N = 90.5 kN, acting at 2 m depth (2/3 of the height). Using the full depth 3 m in place of the centroid depth doubles it to 181.0 kN.
  2. The centre of pressure on a submerged vertical plane surface lies

    1. below its centroid
    2. at its centroid
    3. above its centroid
    4. at the free surface
    Show answer

    Answer: A — below its centroid

    h_cp = h_c + I_G/(h_c A), and the added term is always positive, because pressure grows with depth and the lower part of the surface carries more of the force. The gap shrinks as the surface is submerged deeper.
  3. Water of density 1000 kg/m³ flows at 2 m/s in a horizontal pipe of 0.2 m diameter that reduces to 0.1 m. Neglecting losses, the pressure drop between the two sections (in kPa) is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 30

    Continuity gives V₂ = 2 × (0.2/0.1)² = 8 m/s; Bernoulli gives Δp = ½ × 1000 × (8² − 2²) = 30 000 Pa = 30 kPa. Scaling velocity with the diameter ratio instead of its square gives V₂ = 4 m/s and 6 kPa.
  4. Which conditions are required to apply Bernoulli’s equation between two points on the same streamline?

    1. steady flow
    2. negligible viscous losses
    3. incompressible flow
    4. irrotational flow
    Show answer

    Answer: A — steady flow; B — negligible viscous losses; C — incompressible flow

    Along a streamline Bernoulli needs steady, inviscid and incompressible flow. Irrotationality is needed only to apply it between points on DIFFERENT streamlines, where it makes the constant the same everywhere.
  5. The resistance of a ship is taken to depend on density, speed, length, viscosity and gravitational acceleration. The number of independent dimensionless groups in the relation, counting the resistance coefficient, is ____.

    Numerical answer — type the value.

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    Answer: 3

    Six variables (R, ρ, V, L, μ, g) minus three dimensions (M, L, T) gives three groups: the resistance coefficient, the Reynolds number and the Froude number. Forgetting to count R itself gives 2.
  6. Complete dynamic similarity of a ship model in water is not achievable because

    1. equal Froude number needs a lower model speed while equal Reynolds number needs a higher one
    2. the model cannot be made geometrically similar
    3. the Froude number depends on viscosity
    4. model tanks cannot measure resistance accurately
    Show answer

    Answer: A — equal Froude number needs a lower model speed while equal Reynolds number needs a higher one

    Equal Fn gives V_m = V_s/√λ, but equal Re with nearly the same ν gives V_m = λV_s — a speed that is λ1.5 times higher. Both cannot hold, so the model runs at equal Fn and friction is corrected separately. The Froude number contains no viscosity.
  7. Oil of viscosity 0.001 Pa·s fills a 2 mm gap between a fixed plate and a plate moving at 2 m/s. The shear stress on the moving plate (in Pa) is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 1

    Couette flow has a linear profile, so τ = μU/h = 0.001 × 2/0.002 = 1 Pa, the same across the whole gap. Leaving h in millimetres gives 0.001 Pa.
  8. Water flows at 2 m/s through a pipe 100 m long and 0.1 m in diameter with a Darcy friction factor of 0.02. Taking g = 9.81 m/s², the head lost to friction (in m), to two decimal places, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 4.08

    h_f = f(L/D)(V²/2g) = 0.02 × 1000 × 4/19.62 = 4.08 m. Using the Fanning factor convention with 4f would give 16.31 m; omitting the 2 in 2g gives 8.15 m.
  9. For fully developed laminar flow between two fixed parallel plates, the ratio of the maximum to the mean velocity is

    1. 1.5
    2. 2
    3. 1
    4. 1.33
    Show answer

    Answer: A — 1.5

    Plane Poiseuille flow is parabolic across the gap and its mean is two-thirds of its peak, so u_max/ū = 1.5. The value 2 belongs to the circular pipe, whose paraboloid has a mean of half its peak.
  10. Laminar flow in a pipe has a Reynolds number of 1600. The Darcy friction factor, to two decimal places, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 0.04

    f = 64/Re = 64/1600 = 0.04. The Fanning friction factor, a quarter of the Darcy factor, is 16/Re = 0.01, a different convention.
  11. Water of kinematic viscosity 1.0 × 10⁻⁶ m²/s flows at 0.5 m/s over a smooth flat plate. Using the Blasius result δ = 5.0x/√Reₓ, the boundary-layer thickness 0.5 m from the leading edge (in mm) is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 5

    Reₓ = Ux/ν = 0.5 × 0.5/10⁻⁶ = 2.5 × 10⁵, laminar; √Reₓ = 500; δ = 5 × 0.5/500 = 0.005 m = 5 mm. The displacement thickness at the same point would be 1.72 × 0.5/500 = 1.72 mm.
  12. The criterion for separation of a steady two-dimensional boundary layer is

    1. the velocity gradient normal to the wall, at the wall, becomes zero
    2. the pressure gradient along the wall becomes zero
    3. the boundary layer becomes turbulent
    4. the free-stream velocity reaches its maximum
    Show answer

    Answer: A — the velocity gradient normal to the wall, at the wall, becomes zero

    Separation is where the near-wall flow is brought to rest and begins to reverse, which is where (∂u/∂y) at the wall — the wall shear — falls to zero; it needs an adverse pressure gradient to get there. Transition to turbulence delays separation rather than causing it.
  13. Which statements about boundary layers are correct?

    1. the pressure is approximately constant across the thickness of the layer
    2. the shape factor of a laminar flat-plate layer is larger than that of a turbulent one
    3. suction at the wall helps delay separation
    4. the displacement thickness is smaller than the momentum thickness
    Show answer

    Answer: A — the pressure is approximately constant across the thickness of the layer; B — the shape factor of a laminar flat-plate layer is larger than that of a turbulent one; C — suction at the wall helps delay separation

    ∂p/∂y ≈ 0 is Prandtl’s key result; H is about 2.59 laminar and 1.3 turbulent; suction removes the retarded fluid. Since the integrand of δ*, (1 − u/U), is never less than that of θ, (u/U)(1 − u/U), the displacement thickness is always the LARGER, which is why H exceeds 1.
  14. A smooth flat plate is in laminar flow at a length Reynolds number of 2.5 × 10⁵. Its skin-friction drag coefficient from the Blasius solution, to four decimal places, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 0.0027

    C_D = 1.328/√Re_L = 1.328/500 = 0.002656, which is 0.0027 to four decimal places. The local coefficient at the trailing edge, 0.664/500 = 0.0013, is half the plate average.