Classification of Flows: Fluid, Viscosity, Newtonian and Non-Newtonian Behaviour, and Flow Regimes

This chapter answers Section XE1.1 of XE1 Fluid Mechanics, one of the nine optional sections of the Engineering Sciences (XE) paper, any two of which a candidate chooses in the exam: “Concept of a fluid, Viscous versus inviscid flows, Concept of viscosity, Newtonian versus non-Newtonian fluid, Incompressible versus compressible flows, Internal versus external flows, Steady versus unsteady flows, Laminar versus turbulent flows.” It teaches what a fluid and the continuum are, Newton’s law of viscosity with its units and temperature dependence, the power-law and Bingham models, and then the pairs of flow labels, with the Mach number and the Reynolds number as the criteria that decide two of them. The numericals are shear stress and force in a Couette gap and a journal bearing, apparent viscosity of power-law and Bingham fluids, Mach number, and Reynolds number and flow regime.

1. What a fluid is, and the continuum

A fluid is a substance that deforms continuously for as long as any shear stress, however small, is applied to it. A solid meets a shear stress with a fixed deformation and then rests; a fluid keeps flowing. Liquids and gases are both fluids: a liquid has a free surface and is nearly incompressible, a gas fills its container and is easily compressed. At rest a fluid can carry a normal (compressive) stress, the pressure, but no shear stress, which is why the pressure at a point in a static fluid is the same in every direction.

Fluid mechanics treats the fluid as a continuum: density, velocity and pressure are smooth functions of position, each defined by averaging over a small volume that still holds very many molecules. The test is the Knudsen number Kn = λ/L, the mean free path λ divided by the smallest length of interest L. For Kn ≪ 1 (conventionally below about 0.01) the continuum model and the no-slip wall condition hold. Air at atmospheric pressure has λ of tens of nanometres, so the continuum fails only for very rarefied gases or flow in nano-channels. Three properties are used throughout: density ρ (kg/m³), specific weight γ = ρg (N/m³) and relative density SG = ρ/ρ_water.

2. Viscosity and Newton’s law, with Couette-gap numericals

Viscosity is the internal friction of a fluid, its resistance to relative motion of neighbouring layers. At a solid wall the fluid sticks to the wall (the no-slip condition), so a moving plate drags the fluid next to it and that layer drags the next. For simple shear with velocity u(y), Newton’s law of viscosity is τ = μ du/dy: τ is the shear stress (Pa), du/dy the velocity gradient (equal to the rate of shear strain, s⁻¹) and μ the dynamic viscosity. The kinematic viscosity ν = μ/ρ has units m²/s and measures how fast momentum diffuses through the fluid. Units: μ in Pa·s = N·s/m² = kg/(m·s); 1 poise = 0.1 Pa·s and 1 cP = 1 mPa·s; 1 stokes = 10⁻⁴ m²/s. At 20 °C water has μ ≈ 1.0 × 10⁻³ Pa·s and ν ≈ 1.0 × 10⁻⁶ m²/s, air has μ ≈ 1.8 × 10⁻⁵ Pa·s and ν ≈ 1.5 × 10⁻⁵ m²/s.

Temperature. In a liquid the molecules are close together and the resistance to shear comes mainly from cohesive forces; heating loosens them and μ falls sharply (roughly as exp(B/T)). In a gas the molecules are far apart and shear stress arises from molecules crossing between layers and carrying momentum with them; heating makes them faster, the exchange grows and μ rises. Gas viscosity is almost independent of pressure at ordinary pressures.

Worked example 1 (Couette gap). Two large plates are h = 2 mm apart with oil of μ = 0.08 Pa·s between them. The upper plate, of area 0.4 m², moves at U = 1.5 m/s and the lower is fixed. With no pressure gradient the profile is linear, u = Uy/h, so du/dy = U/h = 1.5/0.002 = 750 s⁻¹. Then τ = μU/h = 0.08 × 750 = 60 Pa, the same across the whole gap, and F = τA = 60 × 0.4 = 24 N.

Worked example 2 (thin film on a rotating shaft). When the clearance c is much smaller than the diameter D, the film can be unrolled into a Couette gap: wall speed U = πDN/60, τ = μU/c on the area πDL, torque T = τ(πDL)(D/2) and power = Tω. For D = 50 mm, N = 600 rpm, c = 0.25 mm, L = 80 mm and μ = 0.1 Pa·s: U = π × 0.05 × 10 = 1.571 m/s; τ = 0.1 × 1.571/0.00025 = 628 Pa; F = 628 × π × 0.05 × 0.08 = 7.90 N; T = 7.90 × 0.025 = 0.197 N·m; ω = 2π × 10 = 62.83 rad/s, so power = 0.197 × 62.83 = 12.4 W, all dissipated as heat in the oil.

⚠️ Convert before you substitute
Gaps are given in mm, speeds in rpm and viscosities in cP. Convert to m, m/s (or rad/s) and Pa·s first: 100 cP = 0.1 Pa·s. Also remember that air has a lower μ than water (by a factor of about 55) but a higher ν (by a factor of about 15), because its density is so much smaller; a question that asks which fluid has the larger kinematic viscosity has the opposite answer to one about dynamic viscosity.

3. Newtonian and non-Newtonian fluids

A Newtonian fluid obeys τ = μ du/dy with one constant μ at a given temperature and pressure, so the plot of τ against shear rate γ̇ = du/dy is a straight line through the origin. Water, air, most gases, thin oils and simple liquids are Newtonian. In a non-Newtonian fluid the ratio τ/γ̇, called the apparent viscosity μ_app, depends on γ̇ (and sometimes on time).

Constitutive models for time-independent fluids
ModelLawApparent viscosityBehaviour and example
Newtonianτ = μγ̇μ, constantwater, air, thin oils
Shear-thinning (pseudoplastic)τ = Kγ̇ⁿ, n < 1Kγ̇ⁿ⁻¹, falls as γ̇ risespolymer solutions, blood, many paints
Shear-thickening (dilatant)τ = Kγ̇ⁿ, n > 1Kγ̇ⁿ⁻¹, rises as γ̇ risesdense cornstarch–water suspension
Bingham plasticτ = τ_y + μ_p γ̇ for τ > τ_y; γ̇ = 0 for τ ≤ τ_y(τ_y + μ_p γ̇)/γ̇, falls towards μ_ptoothpaste, many pastes and muds

In the power-law (Ostwald–de Waele) model K is the consistency index (units Pa·sⁿ) and n the flow-behaviour index; n = 1 recovers a Newtonian fluid. A Bingham plastic is a rigid solid until the stress exceeds the yield stress τ_y, then flows with constant plastic viscosity μ_p. Some fluids also vary with time: a thixotropic fluid thins with time under constant shear and recovers at rest, and a viscoelastic fluid recovers part of its deformation like a solid.

Worked example 3 (power law). A polymer solution has K = 0.9 Pa·sⁿ and n = 0.5. At γ̇ = 25 s⁻¹: τ = 0.9 × 25^0.5 = 4.5 Pa and μ_app = τ/γ̇ = 0.18 Pa·s. At γ̇ = 100 s⁻¹: τ = 9 Pa and μ_app = 0.09 Pa·s. Quadrupling γ̇ halves μ_app, since 4^(n−1) = 4^(−0.5) = 0.5, while the shear stress doubles.

Worked example 4 (Bingham plastic in a Couette gap). τ_y = 12 Pa, μ_p = 0.3 Pa·s, gap 4 mm, plate speed 0.2 m/s. In a Couette gap the shear stress is uniform, so the plate speed forces the whole gap to yield and γ̇ = U/h = 0.2/0.004 = 50 s⁻¹. Then τ = 12 + 0.3 × 50 = 27 Pa and μ_app = 27/50 = 0.54 Pa·s, larger than μ_p. Had the applied stress been only 8 Pa (below τ_y), nothing would flow.

⚠️ Two easy confusions
Shear-thinning describes the fall of apparent viscosity with shear rate; it is not the fall of viscosity with temperature, which every liquid shows. And a power-law fluid with n < 1 still has a shear stress that rises with shear rate; only the ratio τ/γ̇ falls. For a Bingham plastic, μ_p is the slope of the flow curve, not the apparent viscosity.

4. Viscous versus inviscid, incompressible versus compressible

In a viscous flow the shear stresses due to viscosity matter; an inviscid flow sets μ = 0 as a model, since no real fluid is inviscid. The model works where inertia dominates, at large Reynolds number and away from solid surfaces. At a wall viscosity cannot be dropped: it enforces no-slip and builds a thin boundary layer. For steady, incompressible, irrotational flow past a body the inviscid model predicts zero drag (d’Alembert’s paradox), so drag has to come from the viscous layer. An ideal fluid is one that is both inviscid and incompressible.

A flow is incompressible when the density of each fluid particle stays constant as it moves, Dρ/Dt = 0, which is the same as ∇·V = 0. This describes the flow, not the fluid: air at low speed is incompressible, while water is compressible in water hammer and sound waves. For gases the criterion is the Mach number M = V/c, where V is the flow speed and c the local speed of sound: c = √(γRT) for an ideal gas and c = √(K/ρ) for a liquid of bulk modulus K. The isentropic stagnation relation for γ = 1.4 is ρ₀/ρ = (1 + 0.2M²)^2.5, which is 1.046 at M = 0.3, a change of about 4.6 %, and for small M the change is about M²/2. By convention a flow with M < 0.3 is treated as incompressible. For air at 20 °C, c ≈ 343 m/s, so 0.3 corresponds to about 103 m/s.

Worked example 5 (Mach number). Air (γ = 1.4, R = 287 J/kg·K) moves at 250 m/s where T = 230 K. Then c = √(1.4 × 287 × 230) = √92,414 = 304.0 m/s and M = 250/304.0 = 0.82. This is subsonic but well above 0.3, so density changes are large and the flow must be treated as compressible. A car at 30 m/s in air at 20 °C has M = 30/343 = 0.087 and a density change of about M²/2 = 0.4 %: incompressible.

5. Internal versus external, steady versus unsteady

An internal flow is bounded by walls all round: pipes, ducts, channels and the passages of turbomachines. The walls and the cross-section decide the flow, and viscous effects eventually fill the whole section. An external flow develops round a body in an essentially unbounded stream (an aerofoil, a car, a building, a sphere): viscosity is confined to a boundary layer and wake, and the quantities of interest are drag and lift.

A flow is steady when the properties at every fixed point do not change with time, ∂/∂t = 0; otherwise it is unsteady. Steady does not mean uniform, which means the same velocity at all points at a given instant: a fluid particle in a steady flow still accelerates as it passes through a nozzle. Steadiness also depends on the observer. A body moving at constant velocity through still fluid creates an unsteady flow for a ground observer but a steady flow in the frame of the body, which is why wind tunnels hold the model fixed. Turbulent flow is unsteady in its instantaneous values but is called steady when the time-averaged values are constant.

⚠️ Steady, uniform and laminar are three different things
Water leaving a tap at a constant rate is steady but not uniform (the stream narrows and speeds up). Flow in a long constant-diameter pipe whose discharge is being raised is uniform along the pipe but unsteady. A turbulent flow can be steady in the mean. Each pair of labels answers a different question.

6. Laminar and turbulent flow: the Reynolds number

In laminar flow the fluid moves in smooth layers and mixing between layers is molecular only. In turbulent flow the velocity and pressure fluctuate irregularly and eddies of many sizes carry momentum, heat and mass across the flow far faster than molecular diffusion. At the same mean velocity a turbulent pipe flow has a flatter core profile, a steeper gradient at the wall and so a larger wall shear stress and friction than a laminar flow, and it mixes and transfers heat better.

The Reynolds number Re = ρVL/μ = VL/ν is the ratio of inertial to viscous forces. For pipe flow L = D and V is the mean velocity Q/A. By convention flow is laminar below Re ≈ 2300 (some texts use 2000), in transition up to about 4000, and turbulent above that. These are conventions, not constants of nature: the critical value depends on inlet disturbances, roughness and vibration, and in very quiet apparatus laminar flow has been kept up to far larger Re. For a boundary layer on a flat plate transition is conventionally taken near Re_x ≈ 5 × 10⁵. For a non-circular duct use the hydraulic diameter D_h = 4A/P, with A the flow area and P the wetted perimeter.

Worked example 6 (largest laminar flow). Water (ρ = 1000 kg/m³, μ = 1.0 × 10⁻³ Pa·s) in a 20 mm pipe stays laminar up to Re = 2300. V = Reμ/(ρD) = 2300 × 10⁻³/(1000 × 0.02) = 0.115 m/s and Q = V × πD²/4 = 0.115 × 3.142 × 10⁻⁴ = 3.61 × 10⁻⁵ m³/s ≈ 36 mL/s, a tiny flow; ordinary water pipes carry turbulent flow.

Worked example 7 (Re at fixed discharge). V = 4Q/(πD²), so Re = ρVD/μ = 4ρQ/(πμD): at fixed Q, Re ∝ 1/D, and doubling the diameter halves Re. Worked example 8 (duct). Air (ν = 1.5 × 10⁻⁵ m²/s) at 6 m/s in a 0.3 m × 0.1 m duct: A = 0.03 m², P = 0.8 m, D_h = 4 × 0.03/0.8 = 0.15 m and Re = 6 × 0.15/(1.5 × 10⁻⁵) = 6.0 × 10⁴, which is turbulent.

Key takeaways

  • A fluid deforms continuously under any shear stress and carries no shear at rest; it is modelled as a continuum when Kn ≪ 1, with no-slip at walls.
  • Newton’s law is τ = μ du/dy with μ in Pa·s and ν = μ/ρ in m²/s; liquid viscosity falls and gas viscosity rises with temperature; a Couette gap gives τ = μU/h and F = τA.
  • A power-law fluid has τ = Kγ̇ⁿ and μ_app = Kγ̇ⁿ⁻¹: shear-thinning for n < 1, shear-thickening for n > 1. A Bingham plastic is rigid below τ_y and has τ = τ_y + μ_p γ̇ above it.
  • Incompressible flow means Dρ/Dt = 0 (∇·V = 0), a property of the flow; the gas criterion is M = V/c < 0.3 with c = √(γRT), about 4.6 % density change at M = 0.3.
  • Inviscid flow is a model valid outside boundary layers at large Re. Steady means ∂/∂t = 0 at a fixed point and depends on the observer; steady is not uniform; internal flows are wall-bounded and external flows surround a body.
  • Re = ρVD/μ is inertial over viscous force; pipe flow is laminar below about 2300 and turbulent above about 4000 by convention; use D_h = 4A/P for ducts, and at fixed discharge Re ∝ 1/D.

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.

  1. Which statement correctly distinguishes a fluid from a solid?

    1. A fluid has a constant density, whereas the density of a solid changes with the load applied to it
    2. A fluid at rest cannot sustain a stress of any kind, normal or tangential
    3. A fluid goes on deforming for as long as a shear stress, however small, is applied to it
    4. A fluid has zero viscosity, whereas a solid resists shear through viscous friction
    Show answer

    Answer: C — A fluid goes on deforming for as long as a shear stress, however small, is applied to it

    The defining property of a fluid is continuous deformation under any sustained shear stress. A fluid at rest can carry a normal stress, the pressure, so the claim that it carries no stress of any kind is the tempting but wrong one: it is only the shear stress that must vanish at rest. Fluids can be compressible and real fluids are viscous, so constant density and zero viscosity are not defining properties.

  2. The temperature of both liquid water and air is raised from 20 °C to 60 °C at atmospheric pressure. How do their dynamic viscosities change?

    1. Both rise, because faster molecules transfer more momentum between layers
    2. The viscosity of water falls, while that of air rises
    3. The viscosity of water rises, while that of air falls
    4. Both fall, because the fluid expands and becomes less dense on heating
    Show answer

    Answer: B — The viscosity of water falls, while that of air rises

    In a liquid the shear resistance is mainly cohesion between close molecules, which heating weakens, so μ falls. In a gas it comes from molecules crossing between layers and carrying momentum, which speeds up on heating, so μ rises. Applying the gas mechanism to the liquid is the tempting error: it explains the air but not the water.

  3. Two large parallel plates 2 mm apart contain oil of dynamic viscosity 0.08 Pa·s. The upper plate has a wetted area of 0.4 m² and moves at 1.5 m/s relative to the fixed lower plate. Assume a linear velocity profile and no pressure gradient. The force needed to move the upper plate is ____ N.

    Numerical answer — type the value.

    Show answer

    Answer: 24

    Velocity gradient du/dy = U/h = 1.5/0.002 = 750 s⁻¹. Shear stress τ = μ du/dy = 0.08 × 750 = 60 Pa. Force F = τA = 60 × 0.4 = 24 N. Using the gap in mm without converting to m is the usual slip.

  4. A shaft of diameter 50 mm turns at 600 rpm inside a concentric stationary sleeve of length 80 mm. The radial clearance of 0.25 mm is filled with oil of dynamic viscosity 0.1 Pa·s. Treat the film as a Couette gap with a linear velocity profile and neglect end effects. The power dissipated in the oil is ____ W (to one decimal place).

    Numerical answer — type the value.

    Show answer

    Answer: 12.4

    Surface speed U = πDN/60 = π × 0.05 × 600/60 = 1.5708 m/s. τ = μU/c = 0.1 × 1.5708/0.00025 = 628.3 Pa. Area = πDL = π × 0.05 × 0.08 = 0.012566 m², so F = 7.896 N. Torque T = F × D/2 = 7.896 × 0.025 = 0.1974 N·m. ω = 2π × 10 = 62.83 rad/s, so power = Tω = 12.40 W. Check: P = Fv = 7.896 × 1.5708 = 12.40 W.

  5. A fluid obeys the power law τ = Kγ̇ⁿ with n = 0.5. The shear rate is increased by a factor of 4 at constant temperature. What happens to its apparent viscosity?

    1. It doubles, because the shear stress doubles when the shear rate is quadrupled
    2. It is unchanged, because K and n are constants of the fluid
    3. It falls to one quarter, because the stress is proportional to the square root of the shear rate
    4. It halves, because μ_app = Kγ̇ⁿ⁻¹ and 4⁻⁰·⁵ = 0.5
    Show answer

    Answer: D — It halves, because μ_app = Kγ̇ⁿ⁻¹ and 4⁻⁰·⁵ = 0.5

    Apparent viscosity is τ/γ̇ = Kγ̇ⁿ⁻¹. For n = 0.5 it scales as γ̇^(−0.5), so quadrupling γ̇ multiplies it by 1/2. The tempting answer confuses stress with viscosity: the stress does double (4^0.5 = 2), but the ratio stress over rate halves. K and n are constants, but μ_app is not.

  6. A Bingham plastic has yield stress 12 Pa and plastic viscosity 0.3 Pa·s. It fills a gap of 4 mm between a fixed plate and a plate moving at 0.2 m/s, and the whole gap is sheared with a linear velocity profile. The apparent viscosity, defined as shear stress divided by shear rate, is ____ Pa·s (to two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 0.54

    Shear rate γ̇ = U/h = 0.2/0.004 = 50 s⁻¹. Bingham law: τ = τ_y + μ_p γ̇ = 12 + 0.3 × 50 = 27 Pa. Apparent viscosity = τ/γ̇ = 27/50 = 0.54 Pa·s. Answering 0.3 (the plastic viscosity) forgets the yield-stress contribution τ_y/γ̇ = 0.24 Pa·s.

  7. Which description fits a shear-thickening (dilatant) fluid?

    1. Its apparent viscosity increases as the shear rate increases, as in a power-law fluid with n > 1
    2. Its apparent viscosity decreases as the shear rate increases, as in a power-law fluid with n < 1
    3. It does not flow until the shear stress exceeds a yield stress and then flows with a constant viscosity
    4. Its viscosity is independent of the shear rate and falls when the fluid is heated
    Show answer

    Answer: A — Its apparent viscosity increases as the shear rate increases, as in a power-law fluid with n > 1

    A shear-thickening fluid has μ_app = Kγ̇ⁿ⁻¹ with n > 1, so μ_app grows with shear rate (a dense cornstarch suspension is the standard example). The fall of μ_app with γ̇ is shear-thinning, the yield-stress description is a Bingham plastic, and a rate-independent viscosity is Newtonian.

  8. A Bingham plastic of yield stress 10 Pa and plastic viscosity μ_p lies in a thin film in which the shear stress is 6 Pa everywhere. What does the material do?

    1. It flows with shear rate (6 − 10)/μ_p, which is negative, so the layers move backwards
    2. It flows as a Newtonian fluid of viscosity μ_p, because the yield stress only matters for solids
    3. It does not flow: it behaves as a rigid solid with zero shear rate
    4. It flows with shear rate 6/μ_p, because the stress is positive
    Show answer

    Answer: C — It does not flow: it behaves as a rigid solid with zero shear rate

    The Bingham law gives γ̇ = 0 for τ ≤ τ_y. Since 6 Pa is below the 10 Pa yield stress, no shearing occurs and the film moves as a rigid body. The formula γ̇ = (τ − τ_y)/μ_p applies only above yield; applying it below yield gives a negative, meaningless rate, which is the error in the tempting distractors.

  9. Which of the following statements about viscosity are correct?

    1. The kinematic viscosity ν = μ/ρ has the SI unit m²/s
    2. 1 poise is equal to 0.1 Pa·s
    3. The viscosity of a gas decreases when it is heated at constant pressure
    4. The viscosity of a liquid arises mainly from momentum carried by molecules that cross between layers, so it rises with temperature
    Show answer

    Answer: A — The kinematic viscosity ν = μ/ρ has the SI unit m²/s; B — 1 poise is equal to 0.1 Pa·s

    ν = μ/ρ has units (kg/m·s)/(kg/m³) = m²/s, and 1 poise = 1 g/(cm·s) = 0.1 Pa·s. Gas viscosity rises with temperature, and liquid viscosity comes mainly from cohesion and falls with temperature, so the other two statements are wrong.

  10. A rigid body moves at a constant velocity through a large volume of fluid at rest. Idealise the flow as smooth, with no turbulent fluctuations. How should the flow be classified?

    1. Steady for an observer fixed to the ground, because the body’s velocity does not change
    2. Unsteady for an observer fixed to the ground, but steady for an observer moving with the body
    3. Unsteady in both frames, because the velocity at a point near the body changes as the body approaches
    4. Uniform in both frames, because the body’s velocity is the same everywhere along its path
    Show answer

    Answer: B — Unsteady for an observer fixed to the ground, but steady for an observer moving with the body

    A ground observer sees the velocity at a fixed point change as the body passes, so ∂/∂t ≠ 0 and the flow is unsteady. In the body’s frame the pattern is fixed and the incoming stream is steady. The body’s constant velocity does not make the ground-frame flow steady; steadiness is about the flow field at fixed points of the chosen frame. The flow is also not uniform, since velocity varies from point to point.

  11. Water of density 1000 kg/m³ and dynamic viscosity 1.0 × 10⁻³ Pa·s flows in a circular pipe of diameter 20 mm. Take the flow to be laminar for Re up to 2300, where Re is based on the mean velocity and the diameter. The largest flow rate for which the flow remains laminar is ____ mL/s (to the nearest whole number).

    Numerical answer — type the value.

    Show answer

    Answer: 36

    Re = ρVD/μ = 2300 gives V = 2300 × 10⁻³/(1000 × 0.02) = 0.115 m/s. Area = π(0.02)²/4 = 3.1416 × 10⁻⁴ m². Q = VA = 3.613 × 10⁻⁵ m³/s = 36.1 mL/s, so 36. Cross-check with Q = Re π D μ/(4ρ) = 2300 × π × 0.02 × 10⁻³/4000 = 3.613 × 10⁻⁵ m³/s.

  12. Oil is pumped through a circular pipe at a fixed volumetric flow rate Q. The pipe is replaced by one of twice the diameter and the oil temperature is unchanged. By what factor does the Reynolds number based on the pipe diameter change?

    1. It halves, because Re = 4ρQ/(πμD) is inversely proportional to D at fixed Q
    2. It doubles, because D appears in the numerator of ρVD/μ
    3. It is unchanged, because the same oil flows at the same rate
    4. It becomes one quarter, because the area increases four times
    Show answer

    Answer: A — It halves, because Re = 4ρQ/(πμD) is inversely proportional to D at fixed Q

    At fixed Q the mean velocity is V = 4Q/(πD²), so Re = ρVD/μ = 4ρQ/(πμD). Doubling D halves Re. The tempting answer counts only the explicit D and forgets that the velocity falls to a quarter; the quarter answer does the opposite and forgets the explicit D.

  13. Which of the following statements about the classification of flows are correct?

    1. Water flowing through a pipe is an internal flow, while air flowing past a building is an external flow
    2. Flow in a long pipe of constant diameter, whose discharge is being steadily increased by opening a valve, is unsteady
    3. In a steady flow every fluid particle must have zero acceleration
    4. A flow in which the velocity has the same value at every point at a given instant must be steady
    Show answer

    Answer: A — Water flowing through a pipe is an internal flow, while air flowing past a building is an external flow; B — Flow in a long pipe of constant diameter, whose discharge is being steadily increased by opening a valve, is unsteady

    Pipe flow is wall-bounded (internal) and flow round a building is unbounded (external). A rising discharge means the velocity at a fixed point changes with time, so the flow is unsteady. Steady flow can still have convective acceleration, as in a nozzle, and a uniform flow can be unsteady, as in the rising-discharge pipe, so the statements about zero particle acceleration and about a uniform flow being steady both fail.

  14. Air at 20 °C, in which the speed of sound is 343 m/s, flows past a car at 60 m/s. Treat a flow as incompressible when M < 0.3, the usual convention. Which statement is correct?

    1. The flow must be treated as compressible, because air is a gas and every gas flow is compressible
    2. The flow must be treated as compressible, because the Mach number 0.175 exceeds 0.1
    3. The flow is incompressible, and the density changes by about 17 %, because the Mach number equals the fractional density change
    4. Incompressible is a good model: M = 60/343 = 0.175, below 0.3, and the density change is about M²/2 ≈ 1.5 %
    Show answer

    Answer: D — Incompressible is a good model: M = 60/343 = 0.175, below 0.3, and the density change is about M²/2 ≈ 1.5 %

    M = 60/343 = 0.175 is well below 0.3, so the flow is treated as incompressible. The density change is of order M²/2 = 0.0153, about 1.5 %, and not M itself: the Mach number is not the fractional density change, which is the error in the option that equates the Mach number with the density change. Being a gas does not make a flow compressible, and 0.1 is not the convention stated.

  15. Which of the following statements about compressibility are correct?

    1. A flow is incompressible when ∇·V = 0, which can happen for a gas as well as for a liquid
    2. Water can be treated as incompressible in every flow, including pressure waves in a pipe after a sudden valve closure
    3. The limit M < 0.3 corresponds to a density change of roughly 5 % or less, from the isentropic relation
    4. Air in a duct at M = 0.8 can be treated as having a constant density to within 5 %
    Show answer

    Answer: A — A flow is incompressible when ∇·V = 0, which can happen for a gas as well as for a liquid; C — The limit M < 0.3 corresponds to a density change of roughly 5 % or less, from the isentropic relation

    Incompressibility is ∇·V = 0, a statement about the flow, and for γ = 1.4 the stagnation density change is 4.6 % at M = 0.3. Water is compressible in water hammer, where the finite sound speed (about 1.5 km/s) controls the pressure wave. At M = 0.8 the isentropic ratio ρ₀/ρ = (1 + 0.2 × 0.64)^2.5 = 1.35, a change of about 35 %, nowhere near 5 %.

  16. Air, treated as an ideal gas with γ = 1.4 and R = 287 J/kg·K, flows at 250 m/s at a point where its static temperature is 230 K. The local Mach number is ____ (to two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 0.82

    Speed of sound c = √(γRT) = √(1.4 × 287 × 230) = √92,414 = 304.0 m/s. Mach number M = V/c = 250/304.0 = 0.822, so 0.82. Using 340 m/s or the sea-level temperature instead of 230 K is the usual slip.

  17. Which statement about laminar and turbulent flow in a circular pipe is correct?

    1. Pipe flow always turns turbulent at exactly Re = 2300, whatever the inlet disturbance and wall roughness
    2. At the same mean velocity the wall shear stress is lower in turbulent flow than in laminar flow, because eddies reduce friction
    3. The limits of about 2300 and 4000 are conventions: transition depends on inlet disturbances and roughness
    4. The Reynolds number is the ratio of viscous to inertial forces, so a small Re means that inertia dominates
    Show answer

    Answer: C — The limits of about 2300 and 4000 are conventions: transition depends on inlet disturbances and roughness

    The 2300 and 4000 limits are conventions for typical commercial conditions; with very quiet inlets laminar flow has been held to much larger Re, and strong disturbances can trigger turbulence earlier. Turbulent flow has a steeper wall gradient and higher wall shear stress at the same mean velocity. Re is inertial over viscous force, so small Re means viscosity dominates, which reverses the statement that small Re means inertia dominates.

  18. Air flows at a very high Reynolds number over a streamlined body. In which region is the inviscid-flow model an acceptable approximation?

    1. In the whole flow including the layer at the wall, because at high Re viscous forces vanish
    2. In the region outside the thin boundary layer and wake, where velocity gradients are small
    3. Only inside the boundary layer, where the velocity gradient is large and viscosity is negligible
    4. Nowhere, because an inviscid model cannot give the pressure distribution on any real body
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    Answer: B — In the region outside the thin boundary layer and wake, where velocity gradients are small

    Viscous stress is the product of μ and the velocity gradient. Outside the thin boundary layer and wake the gradients are small, so the viscous stress is negligible and the inviscid model gives good pressure distributions. Inside the boundary layer the gradient is large, so viscosity is important there; and no-slip at the wall cannot be satisfied by an inviscid model. High Re makes the viscous region thin, it does not remove it.