Physics · Book 4 · Bachelor Year 2

University Physics — Year 2

University Physics — Year 2 · Bachelor Year 2

4Viscous Flows

Tip a jar of honey and it pours in a slow, thick ribbon; tip a jar of water and it is gone. Drag a spoon through each: the honey resists, and keeps resisting at any speed; the water barely notices until you move fast, and then it swirls. The difference is viscosity — the internal friction of a fluid — and the fight between viscosity and inertia, summed up in one number, Reynolds’, decides whether a flow creeps in orderly layers or tumbles into turbulence. This chapter adds the viscous force to Euler’s equation, solves the two laminar flows every engineer uses, and explains why, at high Reynolds number, viscosity hides in a thin layer at the wall and yet still costs every car, ship and aircraft most of its fuel.

Honey pours in a thin, steady ribbon that coils as it lands: a viscosity ten thousand times that of water makes inertia irrelevant — a creeping flow.
Honey pours in a thin, steady ribbon that coils as it lands: a viscosity ten thousand times that of water makes inertia irrelevant — a creeping flow.

4.1 Viscosity and the Navier–Stokes equation

Definition 4.1 (Newtonian viscosity)

In a plane shear flow v→=vx(y) e→x\vect v = v_x(y)\,\vect e_x, the fluid above a surface y=y = const exerts on the fluid below it (through a surface element  ⁣dS\dd S) the tangential force

 ⁣dF→=η  ⁣dvx ⁣dy  ⁣dS  e→x,\dd\vect F = \eta\,\frac{\dd v_x}{\dd y}\,\dd S\;\vect e_x ,

dragging the slower layer forward and being dragged back in return. The shear stress is τ=η  ⁣dvx/ ⁣dy\tau = \eta\,\dd v_x/\dd y; the coefficient η\eta (unit Pa s\mathrm{Pa}\,\mathrm{s}) is the dynamic viscosity of the fluid, and ν=η/ρ\nu = \eta/\rho (m2/s\mathrm{m}^{2}/\mathrm{s}) its kinematic viscosity. A fluid for which η\eta does not depend on the shear rate is Newtonian: water, air, oils, alcohol; blood, paint, ketchup and polymer melts are not. Orders of magnitude at 20∘C20{}^{\circ}\mathrm{C}: air η=1.8×10−5 Pa s\eta = 1.8 \times 10^{-5}\,\mathrm{Pa}\,\mathrm{s} (ν=1.5×10−5 m2/s\nu = 1.5 \times 10^{-5}\,\mathrm{m}^{2}/\mathrm{s}), water 1.0×10−3 Pa s1.0 \times 10^{-3}\,\mathrm{Pa}\,\mathrm{s} (ν=1.0×10−6 m2/s\nu = 1.0 \times 10^{-6}\,\mathrm{m}^{2}/\mathrm{s}), olive oil 0.08 Pa s0.08\,\mathrm{Pa}\,\mathrm{s}, glycerol 1.5 Pa s1.5\,\mathrm{Pa}\,\mathrm{s}, honey ∼10 Pa s\sim10\,\mathrm{Pa}\,\mathrm{s}. Liquids get thinner when heated, gases thicker.

Proposition 4.2 (Viscous force density; no-slip condition)

In the plane shear flow, the net viscous force on a slab of fluid between yy and y+ ⁣dyy + \dd y is, per unit volume, η  ⁣d2vx/ ⁣dy2\eta\,\dd^2v_x/\dd y^2: viscosity acts as a volume force density η Δv→\eta\,\Delta\vect v (the Laplacian taken component by component), a result that holds for any incompressible flow. At a solid wall the fluid sticks: its velocity equals the wall’s (the no-slip condition), which is why dust stays on a fan blade and why a wall feels a viscous drag.

Proof. The slab is pulled by η vx′(y+ ⁣dy) ⁣dS\eta\,v_x'(y + \dd y)\dd S from above and by −η vx′(y) ⁣dS-\eta\,v_x'(y)\dd S from below: net ηvx′′  ⁣dy  ⁣dS\eta v_x''\,\dd y\,\dd S. The general form is admitted (it follows from writing the stress as a symmetric linear function of the velocity gradients). The no-slip condition is experimental. ∎

Theorem 4.3 (Navier–Stokes equation)

An incompressible Newtonian fluid obeys

ρ(∂v→∂t+(v→⋅grad→⁡)v→)=−grad→⁡P+ρg→+η Δv→,div⁡v→=0,\rho\Bigl(\frac{\partial\vect v}{\partial t} + (\vect v\cdot\operatorname{\vect{grad}}) \vect v\Bigr) = -\operatorname{\vect{grad}}P + \rho\vect g + \eta\,\Delta\vect v , \qquad \operatorname{div}\vect v = 0 ,

with the no-slip condition on every wall. Euler’s equation is the case η=0\eta = 0.

Proof. Admitted at this level. ∎

Left: plane Couette flow between a fixed plate and a plate moving at U — a linear profile, the same shear stress on every layer. Right: the viscous forces on a slab of fluid; they cancel unless the velocity profile is curved.
Left: plane Couette flow between a fixed plate and a plate moving at UU — a linear profile, the same shear stress on every layer. Right: the viscous forces on a slab of fluid; they cancel unless the velocity profile is curved.

Example 4.4 (Shear stress on a table)

A 0.1 mm0.1\,\mathrm{mm} film of oil (η=0.1 Pa s\eta = 0.1\,\mathrm{Pa}\,\mathrm{s}) between a plate and a table, the plate sliding at 1 m/s1\,\mathrm{m}/\mathrm{s}: τ=ηU/h=1 kPa\tau = \eta U/h = 1\,\mathrm{kPa}, a force of 100 N100\,\mathrm{N} on a tenth of a square metre — the principle of the lubricated bearing, where the viscous drag replaces dry friction a hundred times larger. Honey on a spoon, 10 Pa s10\,\mathrm{Pa}\,\mathrm{s}, 1 mm1\,\mathrm{mm} thick, draining at 1 mm/s1\,\mathrm{mm}/\mathrm{s}: τ=10 Pa\tau = 10\,\mathrm{Pa}, exactly the weight of a millimetre of honey per unit area — which is why it drains at that speed.

4.2 The Reynolds number

Definition 4.5 (Reynolds number)

For a flow of characteristic speed UU and length LL, the Reynolds number

Re=ρULη=ULν\mathrm{Re} = \frac{\rho UL}{\eta} = \frac{UL}{\nu}

is the ratio of the orders of magnitude of the inertial term ρ(v→⋅grad→⁡)v→∼ρU2/L\rho(\vect v\cdot\operatorname{\vect{grad}})\vect v \sim \rho U^2/L and the viscous term ηΔv→∼ηU/L2\eta\Delta\vect v \sim \eta U/L^2 in the Navier–Stokes equation. At Re≪1\mathrm{Re} \ll 1 inertia is negligible and the flow is creeping (viscous, reversible, smooth); at Re≫1\mathrm{Re} \gg 1 viscosity is negligible except near walls and the flow is that of a perfect fluid — until, beyond a threshold of order 10310^3 (about 20002000 to 30003000 in a pipe, based on the diameter), it becomes turbulent: unsteady, chaotic, full of eddies at every scale. Between the two, a flow is laminar: steady and layered.

Example 4.6 (Reynolds numbers of everyday flows)

A bacterium (1 µm1\,\text{µ}\mathrm{m}, 30 µm/s30\,\text{µ}\mathrm{m}/\mathrm{s}, water): Re=3×10−5\mathrm{Re} = 3 \times 10^{-5} — it lives in a world without inertia, where stopping its flagellum stops it within an atomic diameter. Blood in a capillary: 10−310^{-3}; a falling dust grain: 10−210^{-2}; a goldfish: 10310^3; a swimmer: 10610^6; a car at 30 m/s30\,\mathrm{m}/\mathrm{s} (4 m4\,\mathrm{m}): 8×1068 \times 10^6; an airliner: 10810^8; the Gulf Stream: 101110^{11}. Two flows with the same geometry and the same Reynolds number are similar: this is what allows a 1 m1\,\mathrm{m} model of a ship to be tested in a towing tank, or an aircraft in a wind tunnel.

4.3 Laminar flows: Couette, Poiseuille, Stokes

Proposition 4.7 (Poiseuille flow in a pipe)

In a horizontal cylindrical pipe of radius RR and length LL, driven by the pressure difference ΔP=P1−P2\Delta P = P_1 - P_2 between its ends, the steady laminar flow is

v(r)=ΔP4ηL (R2−r2),DV=πR48ηL ΔP:v(r) = \frac{\Delta P}{4\eta L}\,(R^2 - r^2) , \qquad D_V = \frac{\pi R^4}{8\eta L}\,\Delta P :

a parabolic profile, maximal on the axis, with a mean speed ⟨v⟩=vmax⁡/2\langle v\rangle = v_{\max}/2. The flow rate is proportional to the pressure drop (Poiseuille’s law) with the hydraulic resistance Rh=ΔP/DV=8ηL/πR4R_h = \Delta P/D_V = 8\eta L/\pi R^4 — and to the fourth power of the radius. The wall shear stress is τw=ΔP R/2L=4η⟨v⟩/R\tau_w = \Delta P\,R/2L = 4\eta\langle v \rangle/R. Valid while Re=2R⟨v⟩/ν≲2000\mathrm{Re} = 2R\langle v\rangle/\nu \lesssim 2000.

Proof. Steady, v→=v(r)e→z\vect v = v(r)\vect e_z (so the convective term vanishes: v→⋅grad→⁡\vect v \cdot\operatorname{\vect{grad}} acts along zz, along which nothing varies). Navier–Stokes along zz: 0=−∂zP+ηΔv0 = -\partial_zP + \eta\Delta v, and radially ∂rP=0\partial_rP = 0: PP depends on zz only, Δv\Delta v on rr only, so both sides equal a constant, ∂zP=−ΔP/L\partial_zP = -\Delta P/L. In cylindrical coordinates Δv=1r ⁣d ⁣dr(r ⁣dv ⁣dr)\Delta v = \frac1r\frac{\dd}{\dd r}\bigl(r\frac{\dd v}{\dd r}\bigr) (stated; it is the balance of viscous forces on a cylindrical shell, 2πrL ηv′2\pi rL\,\eta v' at rr and at r+ ⁣drr + \dd r). Integrating twice with vv finite on the axis and v(R)=0v(R) = 0 gives the parabola; DV=∫0Rv 2πr  ⁣drD_V = \int_0^Rv\,2\pi r\,\dd r. The wall stress is −ηv′(R)-\eta v'(R); equivalently, the pressure force πR2ΔP\pi R^2\Delta P balances the wall friction 2πRLτw2\pi RL\tau_w. ∎

Left: Poiseuille flow — the parabolic profile in a pipe, driven by the pressure drop P_1 - P_2. Right: the force balance on a cylindrical shell of fluid: the pressure difference on its ends is balanced by the viscous stresses on its inner and outer surfaces.
Left: Poiseuille flow — the parabolic profile in a pipe, driven by the pressure drop P1−P2P_1 - P_2. Right: the force balance on a cylindrical shell of fluid: the pressure difference on its ends is balanced by the viscous stresses on its inner and outer surfaces.

Example 4.8 (The fourth power)

A syringe needle of inner diameter 0.4 mm0.4\,\mathrm{mm} and length 3 cm3\,\mathrm{cm}, water, thumb pressure 20 kPa20\,\mathrm{kPa}: DV=π(2×10−4)4×2×104/(8×10−3×0.03)=0.42 mL/sD_V = \pi(2 \times 10^{-4})^4 \times 2 \times 10^4/(8 \times 10^{-3} \times 0.03) = 0.42\,\mathrm{mL}/\mathrm{s}; a needle twice as wide gives sixteen times more. An artery narrowed by a plaque to 70%70\% of its radius offers (1/0.7)4=4.2(1/0.7)^4 = 4.2 times its resistance: the body compensates by raising the pressure upstream, which is part of why atherosclerosis and hypertension go together.

Proposition 4.9 (Stokes’ drag)

A sphere of radius RR moving at speed vv through a fluid at Re=2ρvR/η≪1\mathrm{Re} = 2\rho vR/\eta \ll 1 feels the drag

F=6πηRv,F = 6\pi\eta Rv ,

opposite to its velocity. A sphere of density ρs\rho_s falling in a fluid of density ρ\rho reaches the terminal velocity v∞=2R2(ρs−ρ)g9ηv_\infty = \dfrac{2R^2(\rho_s - \rho)g}{9\eta} — the law of sedimentation, centrifuges, and Millikan’s oil drops.

Proof. The drag is admitted (it is the exact solution of the creeping-flow equations around a sphere, two thirds of it viscous friction, one third pressure). Terminal speed: weight minus buoyancy, 43πR3(ρs−ρ)g\tfrac43\pi R^3(\rho_s - \rho)g, equals the drag; always check Re≪1\mathrm{Re} \ll 1 afterwards. ∎

Example 4.10 (Fog, rain and dust)

A fog droplet of 10 µm10\,\text{µ}\mathrm{m} diameter in air falls at 2×(5×10−6)2×104/(9×1.8×10−5)=3 mm/s2 \times (5 \times 10^{-6})^2 \times 10^4/(9 \times 1.8 \times 10^{-5}) = 3\,\mathrm{mm}/\mathrm{s} (Re=2×10−3\mathrm{Re} = 2 \times 10^{-3}): the slightest updraft keeps it aloft, which is why clouds float. A 1 mm1\,\mathrm{mm} raindrop would give 30 m/s30\,\mathrm{m}/\mathrm{s} and Re≈2000\mathrm{Re} \approx 2000: Stokes no longer applies, and the quadratic drag of the next section gives the observed 4 m/s4\,\mathrm{m}/\mathrm{s}. A 10 µm10\,\text{µ}\mathrm{m} dust grain (ρs=2500 kg/m3\rho_s = 2500\,\mathrm{kg}/\mathrm{m}^{3}) settles at 8 mm/s8\,\mathrm{mm}/\mathrm{s}, a metre in two minutes — finer dust hangs for hours.

4.4 High Reynolds number: boundary layers and drag

Proposition 4.11 (The boundary layer)

At high Reynolds number the flow around a body is that of a perfect fluid everywhere except in a thin boundary layer along the wall, in which the velocity falls from its outer value UU to zero (no slip). Along a flat plate its thickness grows as

δ(x)∼νxU=xRex,Rex=Uxν,\delta(x) \sim \sqrt{\frac{\nu x}{U}} = \frac{x}{\sqrt{\mathrm{Re}_x}} , \qquad \mathrm{Re}_x = \frac{Ux}{\nu} ,

with xx the distance from the leading edge; the layer itself turns turbulent beyond Rex∼5×105\mathrm{Re}_x \sim 5 \times 10^5. The viscous friction on the wall, and the separation of the layer behind a blunt body (which leaves a turbulent wake of low pressure), are the two sources of drag.

Proof. Order of magnitude: a fluid particle spends the time x/Ux/U travelling along the plate, during which the wall’s influence diffuses into the fluid — the viscous term ν ∂2v/∂y2\nu\,\partial^2v/\partial y^2 is a diffusion of momentum with diffusivity ν\nu, reaching the distance νt\sqrt{\nu t} in the time tt (Chapter 24). Hence δ∼νx/U\delta \sim \sqrt{\nu x /U}. The full (Prandtl) theory is admitted. ∎

Proposition 4.12 (Drag at high Reynolds number)

The drag on a body of frontal area SS moving at vv through a fluid is written

F=12ρv2 S Cx,F = \tfrac12\rho v^2\,S\,C_x ,

where the drag coefficient CxC_x depends on the shape and on Re\mathrm{Re}. For a sphere, Cx=24/ReC_x = 24/\mathrm{Re} at Re≪1\mathrm{Re} \ll 1 (Stokes, with Re\mathrm{Re} based on the diameter), then about 0.40.4–0.50.5, nearly constant, from Re≈103\mathrm{Re} \approx 10^3 to 2×1052 \times 10^5, where it drops suddenly to about 0.10.1 (the drag crisis: the boundary layer turns turbulent, sticks longer to the surface and the wake shrinks — the dimples on a golf ball trigger it early). Streamlined bodies have Cx≈0.05C_x \approx 0.05; a car 0.30.3; a cyclist 0.90.9; a flat plate facing the flow 1.21.2. The power needed to move at vv is Fv∝v3Fv \propto v^3.

Proof. Admitted at this level. ∎

Left: the drag coefficient of a sphere against the Reynolds number (log scales): Stokes’ law at low Re, a plateau near 0.45, and the drag crisis near Re = 2 × 105. Right: the boundary layer thickening along a flat plate; outside it the flow is that of a perfect fluid. Left: the drag coefficient of a sphere against the Reynolds number (log scales): Stokes’ law at low Re, a plateau near 0.45, and the drag crisis near Re = 2 × 105. Right: the boundary layer thickening along a flat plate; outside it the flow is that of a perfect fluid.
Left: the drag coefficient of a sphere against the Reynolds number (log scales): Stokes’ law at low Re\mathrm{Re}, a plateau near 0.450.45, and the drag crisis near Re=2×105\mathrm{Re} = 2 \times 10^5. Right: the boundary layer thickening along a flat plate; outside it the flow is that of a perfect fluid.

Example 4.13 (A cyclist, a car)

A cyclist (SCx=0.35 m2SC_x = 0.35\,\mathrm{m}^{2}) at 10 m/s10\,\mathrm{m}/\mathrm{s}: F=0.5×1.2×100×0.35=21 NF = 0.5 \times 1.2 \times 100 \times 0.35 = 21\,\mathrm{N}, P=210 WP = 210\,\mathrm{W} — a sustained effort; at 15 m/s15\,\mathrm{m}/\mathrm{s} (54 km/h54\,\mathrm{km}/\mathrm{h}) it would take 710 W710\,\mathrm{W}, which is why sprinters draft. A car (SCx=0.7 m2SC_x = 0.7\,\mathrm{m}^{2}) at 130 km/h130\,\mathrm{km}/\mathrm{h}: 550 N550\,\mathrm{N}, 20 kW20\,\mathrm{kW} against air alone — the fuel consumption at motorway speed is mostly drag.

Remark 4.14 (Surface tension)

A second force lives at the boundaries of a liquid: its free surface behaves like a stretched membrane of tension γ\gamma (N/m\mathrm{N}/\mathrm{m}; water 0.072 N/m0.072\,\mathrm{N}/\mathrm{m}), because surface molecules lack half their neighbours. The pressure inside a drop of radius RR exceeds the outside pressure by the Laplace pressure 2γ/R2\gamma/R (1.4 kPa1.4\,\mathrm{kPa} in a 0.1 mm0.1\,\mathrm{mm} drop, 14 bar14\,\mathrm{bar} in a 10 nm10\,\mathrm{nm} one); surface tension beats gravity below the capillary length γ/ρg=2.7 mm\sqrt{\gamma/\rho g} = 2.7\,\mathrm{mm} for water — the size of dew drops, of the meniscus in a glass, of the rise in a capillary tube (h=2γ/ρgrh = 2\gamma/\rho gr, Jurin). It is why small insects walk on water, why a paintbrush’s hairs cling together when wet, and why a liquid jet breaks into drops.

Method 4.15 (Which regime?)

Before any viscous-flow calculation, compute Re\mathrm{Re}. Re≪1\mathrm{Re} \ll 1: Stokes and creeping flow, drag ∝v\propto v, pressure drops ∝\propto flow rate. Re\mathrm{Re} up to ∼ ⁣2000\sim\!2000 in a pipe: laminar, Poiseuille. Above: turbulent — Poiseuille underestimates the loss badly; use the empirical law ΔP=λ(L/D) 12ρ⟨v⟩2\Delta P = \lambda(L/D)\,\tfrac12\rho\langle v\rangle^2 with λ≈0.3/Re1/4\lambda \approx 0.3/\mathrm{Re}^{1/4} for smooth pipes, and F=12ρv2SCxF = \tfrac12\rho v^2SC_x for bodies. In every case, check afterwards that the Re\mathrm{Re} computed from the answer agrees with the regime assumed.

4.5 Exercises

Exercise 4.1 ★

A 0.50 m20.50\,\mathrm{m}^{2} plate slides at 2.0 m/s2.0\,\mathrm{m}/\mathrm{s} on a 0.20 mm0.20\,\mathrm{mm} film of oil (η=0.25 Pa s\eta = 0.25\,\mathrm{Pa}\,\mathrm{s}). Shear stress, force on the plate, power dissipated; compare with dry friction (f=0.3f = 0.3) if the plate weighs 10 kg10\,\mathrm{kg}.

Solution

Solution of Exercise 4.1.

τ=ηU/h=0.25×2/2×10−4=2.5 kPa\tau = \eta U/h = 0.25 \times 2/2 \times 10^{-4} = 2.5\,\mathrm{kPa}; F=1.25 kNF = 1.25\,\mathrm{kN}; P=Fv=2.5 kWP = Fv = 2.5\,\mathrm{kW}. Dry friction: 0.3×98=29 N0.3 \times 98 = 29\,\mathrm{N} — here the thin, fast-sheared film resists far more; the viscous drag grows as U/hU/h, so lubrication pays at lower speeds and with thicker films (and it never seizes).

Exercise 4.2 ★

Reynolds numbers of: (a) a 2 µm2\,\text{µ}\mathrm{m} bacterium at 30 µm/s30\,\text{µ}\mathrm{m}/\mathrm{s} in water; (b) a swimmer (1.8 m1.8\,\mathrm{m}, 1.5 m/s1.5\,\mathrm{m}/\mathrm{s}); (c) a car (4 m4\,\mathrm{m}, 30 m/s30\,\mathrm{m}/\mathrm{s}) in air; (d) blood in the aorta (2.5 cm2.5\,\mathrm{cm}, 0.33 m/s0.33\,\mathrm{m}/\mathrm{s}, ν=3.3×10−6 m2/s\nu = 3.3 \times 10^{-6}\,\mathrm{m}^{2}/\mathrm{s}); (e) oil (η=0.1 Pa s\eta = 0.1\,\mathrm{Pa}\,\mathrm{s}, ρ=900 kg/m3\rho = 900\,\mathrm{kg}/\mathrm{m}^{3}) at 0.5 m/s0.5\,\mathrm{m}/\mathrm{s} in a 1 cm1\,\mathrm{cm} pipe. Regime of each.

Solution

Solution of Exercise 4.2.

(a) 2×10−6×3×10−5/10−6=6×10−52 \times 10^{-6} \times 3 \times 10^{-5}/10^{-6} = 6 \times 10^{-5}, creeping. (b) 1.8×1.5/10−6=2.7×1061.8 \times 1.5/10^{-6} = 2.7 \times 10^6, turbulent. (c) 4×30/1.5×10−5=8×1064 \times 30/1.5 \times 10^{-5} = 8 \times 10^6, turbulent. (d) 0.025×0.33/3.3×10−6=25000.025 \times 0.33/3.3 \times 10^{-6} = 2500, borderline laminar. (e) 900×0.5×0.01/0.1=45900 \times 0.5 \times 0.01/0.1 = 45, laminar.

Exercise 4.3 ★

Water is pushed through a 0.40 mm0.40\,\mathrm{mm} (inner diameter), 3.0 cm3.0\,\mathrm{cm} needle by 20 kPa20\,\mathrm{kPa}. Flow rate; time to inject 5 mL5\,\mathrm{mL}; maximum speed and Reynolds number; wall shear stress. With a 0.80 mm0.80\,\mathrm{mm} needle?

Solution

Solution of Exercise 4.3.

DV=πR4ΔP/8ηL=π(2×10−4)4×2×104/(8×10−3×0.03)=0.42 mL/sD_V = \pi R^4\Delta P/8\eta L = \pi(2 \times 10^{-4})^4 \times 2 \times 10^4/(8 \times 10^{-3} \times 0.03) = 0.42\,\mathrm{mL}/\mathrm{s}; 12 s12\,\mathrm{s} for 5 mL5\,\mathrm{mL}. vmax⁡=ΔPR2/4ηL=6.7 m/sv_{\max} = \Delta PR^2/ 4\eta L = 6.7\,\mathrm{m}/\mathrm{s}, ⟨v⟩=3.3 m/s\langle v\rangle = 3.3\,\mathrm{m}/\mathrm{s}, Re=2R⟨v⟩/ν=1300\mathrm{Re} = 2R\langle v \rangle/\nu = 1300: laminar. τw=ΔPR/2L=67 Pa\tau_w = \Delta PR/2L = 67\,\mathrm{Pa}. At 0.8 mm0.8\,\mathrm{mm}: sixteen times the flow, 6.7 mL/s6.7\,\mathrm{mL}/\mathrm{s} — but Re≈104\mathrm{Re} \approx 10^4: the flow turns turbulent and the gain is less.

Exercise 4.4 ★

Terminal speeds in air (η=1.8×10−5 Pa s\eta = 1.8 \times 10^{-5}\,\mathrm{Pa}\,\mathrm{s}) of (a) a 20 µm20\,\text{µ}\mathrm{m} pollen grain (ρs=1100 kg/m3\rho_s = 1100\,\mathrm{kg}/\mathrm{m}^{3}), (b) a 2 µm2\,\text{µ}\mathrm{m} smoke particle (ρs=2000 kg/m3\rho_s = 2000\,\mathrm{kg}/\mathrm{m}^{3}), (c) a 0.2 mm0.2\,\mathrm{mm} drizzle drop. Check Re\mathrm{Re} in each case; time to fall 1 m1\,\mathrm{m}.

Solution

Solution of Exercise 4.4.

v∞=2R2Δρ g/9ηv_\infty = 2R^2\Delta\rho\,g/9\eta. (a) 1.3 cm/s1.3\,\mathrm{cm}/\mathrm{s}, Re=0.02\mathrm{Re} = 0.02, 75 s75\,\mathrm{s} per metre. (b) 0.24 mm/s0.24\,\mathrm{mm}/\mathrm{s}, Re=3×10−5\mathrm{Re} = 3 \times 10^{-5}, 70 min70\,\mathrm{min} per metre. (c) 1.2 m/s1.2\,\mathrm{m}/\mathrm{s} but Re=16\mathrm{Re} = 16: Stokes overestimates (real speed about 0.7 m/s0.7\,\mathrm{m}/\mathrm{s}), about a second per metre.

Exercise 4.5 ★★

A water main of diameter 30 cm30\,\mathrm{cm} carries 0.10 m3/s0.10\,\mathrm{m}^{3}/\mathrm{s} over 1.0 km1.0\,\mathrm{km}. (a) Mean speed and Reynolds number. (b) Pressure drop if Poiseuille’s law applied. (c) Actual drop with λ=0.3/Re1/4\lambda = 0.3/\mathrm{Re}^{1/4}; ratio to (b); pumping power. (d) The same flow in two parallel 21 cm21\,\mathrm{cm} mains (same total section): drop and power.

Solution

Solution of Exercise 4.5.

(a) ⟨v⟩=0.1/0.0707=1.4 m/s\langle v\rangle = 0.1/0.0707 = 1.4\,\mathrm{m}/\mathrm{s}, Re=4.2×105\mathrm{Re} = 4.2 \times 10^5. (b) 8ηLDV/πR4=500 Pa8\eta LD_V/\pi R^4 = 500\,\mathrm{Pa}. (c) λ=0.3/25.5=0.0118\lambda = 0.3/25.5 = 0.0118, ΔP=0.0118×3333×1000=39 kPa\Delta P = 0.0118 \times 3333 \times 1000 = 39\,\mathrm{kPa}: 8080 times more; P=ΔPDV=3.9 kWP = \Delta PD_V = 3.9\,\mathrm{kW}. (d) Same speed, Re=3×105\mathrm{Re} = 3 \times 10^5, λ=0.0128\lambda = 0.0128, ΔP=0.0128×4717×1000=60 kPa\Delta P = 0.0128 \times 4717 \times 1000 = 60\,\mathrm{kPa}, 6 kW6\,\mathrm{kW}: more wall per unit flow, more loss.

Exercise 4.6 ★★

Cells of diameter 10 µm10\,\text{µ}\mathrm{m} and density 1050 kg/m31050\,\mathrm{kg}/\mathrm{m}^{3} in water. (a) Sedimentation speed under gravity; time to settle 5 cm5\,\mathrm{cm}. (b) In a centrifuge at 1000 g1000\,g. (c) Minimum rotation rate for 1000 g1000\,g at 10 cm10\,\mathrm{cm} radius. (d) Why can the largest particles be separated first?

Solution

Solution of Exercise 4.6.

(a) v=2×25×10−12×50×9.81/(9×10−3)=2.7 µm/sv = 2 \times 25 \times 10^{-12} \times 50 \times 9.81/(9 \times 10^{-3}) = 2.7\,\text{µ}\mathrm{m}/\mathrm{s}; 5 cm5\,\mathrm{cm} in 1.8×1041.8 \times 10^4 s, five hours. (b) 2.7 mm/s2.7\,\mathrm{mm}/\mathrm{s}, 18 s18\,\mathrm{s}. (c) Ω=1000g/r=313 rad/s=3000 rpm\Omega = \sqrt{1000g/r} = 313\,\mathrm{rad}/\mathrm{s} = 3000\,\mathrm{rpm}. (d) v∝R2v \propto R^2: large particles pellet first; successive spins at increasing speed separate sizes.

Exercise 4.7 ★★

A cyclist (SCx=0.35 m2SC_x = 0.35\,\mathrm{m}^{2}, rolling resistance 4 N4\,\mathrm{N}) rides at 12 m/s12\,\mathrm{m}/\mathrm{s}. (a) Drag and total power. (b) At 15 m/s15\,\mathrm{m}/\mathrm{s}. (c) Into a 5 m/s5\,\mathrm{m}/\mathrm{s} headwind at 12 m/s12\,\mathrm{m}/\mathrm{s} ground speed. (d) Drafting reduces CxC_x by 30%30\%: power saved. (e) Descending a 5%5\% slope with no pedalling (mass 80 kg80\,\mathrm{kg}): terminal speed.

Solution

Solution of Exercise 4.7.

(a) F=0.6×144×0.35=30 NF = 0.6 \times 144 \times 0.35 = 30\,\mathrm{N}, +4 N+4\,\mathrm{N}: P=34×12=410 WP = 34 \times 12 = 410\,\mathrm{W}. (b) 47+4=51 N47 + 4 = 51\,\mathrm{N}, 770 W770\,\mathrm{W}. (c) Relative speed 17 m/s17\,\mathrm{m}/\mathrm{s}: 61+4=65 N61 + 4 = 65\,\mathrm{N}, times the ground speed: 780 W780\,\mathrm{W}. (d) Drag 21 N21\,\mathrm{N}: 300 W300\,\mathrm{W}, saving 110 W110\,\mathrm{W}. (e) mgsin⁡α=39 N=4+0.21v2mg\sin\alpha = 39\,\mathrm{N} = 4 + 0.21v^2: v=13 m/sv = 13\,\mathrm{m}/\mathrm{s}, 47 km/h47\,\mathrm{km}/\mathrm{h}.

Exercise 4.8 ★★

Quadratic drag. (a) Terminal speed of a golf ball (46 g46\,\mathrm{g}, 43 mm43\,\mathrm{mm}, Cx=0.25C_x = 0.25) and of a skydiver (80 kg80\,\mathrm{kg}, SCx=0.7 m2SC_x = 0.7\,\mathrm{m}^{2}, then 25 m225\,\mathrm{m}^{2} with the parachute open). (b) Equation of motion for a fall from rest with quadratic drag; show v=v∞tanh⁡(gt/v∞)v = v_\infty\tanh(gt/v_\infty). (c) Time and distance for the skydiver to reach 90%90\% of v∞v_\infty.

Solution

Solution of Exercise 4.8.

(a) Golf: 12ρCxS=2.2×10−4 kg/m\tfrac12\rho C_xS = 2.2 \times 10^{-4}\,\mathrm{kg}/\mathrm{m}, v∞=0.45/2.2×10−4=45 m/sv_\infty = \sqrt{0.45/2.2 \times 10^{-4}} = 45\,\mathrm{m}/\mathrm{s}; skydiver 785/0.42=43 m/s\sqrt{785/0.42} = 43\,\mathrm{m}/\mathrm{s}; parachute 785/15=7.2 m/s\sqrt{785/15} = 7.2\,\mathrm{m}/\mathrm{s}. (b) mv˙=mg−kv2m\dot v = mg - kv^2, i.e. v˙=g(1−v2/v∞2)\dot v = g(1 - v^2/v_\infty^2), whose solution from rest is v∞tanh⁡(gt/v∞)v_\infty\tanh(gt/v_\infty). (c) tanh⁡x=0.9\tanh x = 0.9 at x=1.47x = 1.47: t=1.47×43/9.81=6.5 st = 1.47 \times 43/9.81 = 6.5\,\mathrm{s}; z=(v∞2/g)ln⁡cosh⁡x=190×0.83=160 mz = (v_\infty^2/g)\ln\cosh x = 190 \times 0.83 = 160\,\mathrm{m}.

Exercise 4.9 ★★

Couette viscometer. A cylinder of radius 5.0 cm5.0\,\mathrm{cm} and height 10 cm10\,\mathrm{cm} rotates at 10 rad/s10\,\mathrm{rad}/\mathrm{s} inside a fixed cylinder, the gap 1.0 mm1.0\,\mathrm{mm} being filled with the liquid to test; the torque needed is 2.0×10−2 N m2.0 \times 10^{-2}\,\mathrm{N}\,\mathrm{m}. (a) Treating the gap as plane Couette flow, shear rate and shear stress. (b) Viscosity. (c) Reynolds number of the gap flow; is the flow laminar? (d) Why is the gap made thin?

Solution

Solution of Exercise 4.9.

(a) Shear rate ΩR/e=0.5/10−3=500 s−1\Omega R/e = 0.5/10^{-3} = 500\,\mathrm{s}^{-1}; Γ=τ⋅2πRh⋅R\Gamma = \tau\cdot2\pi Rh \cdot R: τ=0.02/(2π×2.5×10−3×0.1)=12.7 Pa\tau = 0.02/(2\pi \times 2.5 \times 10^{-3} \times 0.1) = 12.7\,\mathrm{Pa}. (b) η=12.7/500=2.5×10−2 Pa s\eta = 12.7/500 = 2.5 \times 10^{-2}\,\mathrm{Pa}\,\mathrm{s}. (c) Re=ρUe/η≈1000×0.5×10−3/0.025=20\mathrm{Re} = \rho Ue/\eta \approx 1000 \times 0.5 \times 10^{-3}/0.025 = 20: laminar. (d) A thin gap makes the shear rate uniform (the plane approximation) and keeps the flow laminar.

Exercise 4.10 ★★★

A vascular network. Blood (η=3.5 mPa s\eta = 3.5\,\mathrm{mPa}\,\mathrm{s}) flows at 5.0 L/min5.0\,\mathrm{L}/\mathrm{min}. (a) Hydraulic resistance of the aorta (radius 1.25 cm1.25\,\mathrm{cm}, length 40 cm40\,\mathrm{cm}) and pressure drop along it. (b) The 101010^{10} capillaries (radius 4 µm4\,\text{µ}\mathrm{m}, length 1 mm1\,\mathrm{mm}) are in parallel: resistance of one, of all, and the pressure drop across the capillary bed. (c) The arterioles (10810^8, radius 10 µm10\,\text{µ}\mathrm{m}, length 2 mm2\,\mathrm{mm}): same questions; where is the resistance of the circulation? (d) The total drop is 13 kPa13\,\mathrm{kPa}: power of the heart’s left ventricle, and what fraction of the body’s 100 W100\,\mathrm{W}. (e) A drug dilates the arterioles by 10%10\%: effect on the total resistance.

Solution

Solution of Exercise 4.10.

DV=8.3×10−5 m3/sD_V = 8.3 \times 10^{-5}\,\mathrm{m}^{3}/\mathrm{s}. (a) Rh=8ηL/πR4=8×3.5×10−3×0.4/(π×2.4×10−8)=1.5×105 Pa s/m3R_h = 8\eta L/\pi R^4 = 8 \times 3.5 \times 10^{-3} \times 0.4/(\pi \times 2.4 \times 10^{-8}) = 1.5 \times 10^{5}\,\mathrm{Pa}\,\mathrm{s}/\mathrm{m}^{3}, ΔP=12 Pa\Delta P = 12\,\mathrm{Pa}. (b) One capillary 3.5×1016 Pa s/m33.5 \times 10^{16}\,\mathrm{Pa}\,\mathrm{s}/\mathrm{m}^{3}; 101010^{10} in parallel: 3.5×106 Pa s/m33.5 \times 10^{6}\,\mathrm{Pa}\,\mathrm{s}/\mathrm{m}^{3}; ΔP=290 Pa\Delta P = 290\,\mathrm{Pa} (2 mmHg2\,\mathrm{mmHg}). (c) One arteriole 8×3.5×10−3×2×10−3/(π×10−20)=1.8×1015 Pa s/m38 \times 3.5 \times 10^{-3} \times 2 \times 10^{-3}/(\pi \times 10^{-20}) = 1.8 \times 10^{15}\,\mathrm{Pa}\,\mathrm{s}/\mathrm{m}^{3}; the set: 1.8×107 Pa s/m31.8 \times 10^{7}\,\mathrm{Pa}\,\mathrm{s}/\mathrm{m}^{3}; ΔP=1.5 kPa\Delta P = 1.5\,\mathrm{kPa}: the arterioles dominate the two (the small vessels, not the large ones, are where the pressure falls). (d) P=13000×8.3×10−5=1.1 WP = 13000 \times 8.3 \times 10^{-5} = 1.1\,\mathrm{W}, about 1%1\%. (e) (1/1.1)4=0.68(1/1.1)^4 = 0.68: a third of the arteriolar resistance removed — vasodilators lower the blood pressure this way.

Exercise 4.11 ★★★

Boundary layer and friction drag. A flat plate of length LL and width bb moves edge-on at UU. In the laminar regime the wall shear stress is τw=0.332 ρU2/Rex\tau_w = 0.332\,\rho U^2/\sqrt{\mathrm{Re}_x} (Blasius). (a) Show that the friction force on one face is F=0.664 ρU2bL/ReLF = 0.664\,\rho U^2bL/\sqrt{\mathrm{Re}_L}. (b) Boundary-layer thickness at the trailing edge of a 1 m1\,\mathrm{m} plate in water at 0.3 m/s0.3\,\mathrm{m}/\mathrm{s}, and the force on a 1 m21\,\mathrm{m}^{2} plate (both faces). (c) At 10 m/s10\,\mathrm{m}/\mathrm{s}: ReL\mathrm{Re}_L, and why the formula no longer applies. (d) A ship’s hull (100 m100\,\mathrm{m}, 10 m/s10\,\mathrm{m}/\mathrm{s}): boundary-layer thickness estimate at the stern assuming the turbulent law δ≈0.37x/Rex1/5\delta \approx 0.37x/\mathrm{Re}_x^{1/5}.

Solution

Solution of Exercise 4.11.

(a) F=b∫0Lτw  ⁣dx=0.332ρU2bν/U∫0L ⁣dx/x=0.664ρU2bνL/U=0.664ρU2bL/ReLF = b\int_0^L\tau_w\,\dd x = 0.332\rho U^2b\sqrt{\nu/U}\int_0^L\dd x/\sqrt x = 0.664\rho U^2b\sqrt{\nu L/U} = 0.664\rho U^2bL/\sqrt{\mathrm{Re}_L}. (b) ReL=3×105\mathrm{Re}_L = 3 \times 10^5 (laminar); δ≈5νL/U=9 mm\delta \approx 5\sqrt{\nu L/U} = 9\,\mathrm{mm}; F=0.664×1000×0.09×1/548=0.11 NF = 0.664 \times 1000 \times 0.09 \times 1/548 = 0.11\,\mathrm{N} per face, 0.22 N0.22\,\mathrm{N}. (c) ReL=107\mathrm{Re}_L = 10^7: the layer is turbulent beyond the first few centimetres, the laminar formula fails (the real drag is several times larger). (d) Rex=109\mathrm{Re}_x = 10^9: δ≈0.37×100/63=0.6 m\delta \approx 0.37 \times 100/63 = 0.6\,\mathrm{m}.

Exercise 4.12 ★★★

A falling film. A liquid film of thickness hh flows down a vertical wall under gravity, steadily, the air exerting no stress on its free surface. (a) Write Navier–Stokes for v→=v(y)e→z\vect v = v(y)\vect e_z (yy the distance from the wall, zz downward) and the boundary conditions. (b) Show v(y)=ρg2η(2hy−y2)v(y) = \dfrac{\rho g}{2\eta}(2hy - y^2) and compute the flow rate per unit width q=ρgh3/3ηq = \rho gh^3/3\eta. (c) A fresh coat of paint (η=0.5 Pa s\eta = 0.5\,\mathrm{Pa}\,\mathrm{s}, ρ=1200 kg/m3\rho = 1200\,\mathrm{kg}/\mathrm{m}^{3}) 0.2 mm0.2\,\mathrm{mm} thick: surface speed and flow rate; how far does the surface move in 10 min10\,\mathrm{min}? Why does paint sag when applied too thick? (d) Honey (η=10 Pa s\eta = 10\,\mathrm{Pa}\,\mathrm{s}) 2 mm2\,\mathrm{mm} thick on a vertical spoon: surface speed.

Solution

Solution of Exercise 4.12.

(a) 0=ρg+ηv′′(y)0 = \rho g + \eta v''(y), v(0)=0v(0) = 0 (no slip), v′(h)=0v'(h) = 0 (no stress at the free surface). (b) Integrate twice: v=(ρg/2η)(2hy−y2)v = (\rho g/2\eta)(2hy - y^2); q=∫0hv  ⁣dy=ρgh3/3ηq = \int_0^hv\,\dd y = \rho gh^3/3\eta. (c) vs=ρgh2/2η=1200×9.81×4×10−8/1=0.47 mm/sv_s = \rho gh^2/2\eta = 1200 \times 9.81 \times 4 \times 10^{-8}/1 = 0.47\,\mathrm{mm}/\mathrm{s}; q=6.3×10−8 m2/sq = 6.3 \times 10^{-8}\,\mathrm{m}^{2}/\mathrm{s}; 28 cm28\,\mathrm{cm} in ten minutes — a Newtonian paint of that thickness would run off; real paints are shear-thinning and thixotropic, and sag anyway when too thick because vs∝h2v_s \propto h^2. (d) 1400×9.81×4×10−6/20=2.7 mm/s1400 \times 9.81 \times 4 \times 10^{-6}/20 = 2.7\,\mathrm{mm}/\mathrm{s}.

A von Kármán vortex street in the clouds downstream of an island, seen from space: the wake of a cylinder at a Reynolds number of a few hundred, drawn on the scale of a hundred kilometres (NASA).
A von Kármán vortex street in the clouds downstream of an island, seen from space: the wake of a cylinder at a Reynolds number of a few hundred, drawn on the scale of a hundred kilometres (NASA).

4.6 Problem: Blood, crude oil and rain

Problem 4.1

Weekend problem — the same viscous laws in an artery, in a thousand-kilometre pipeline, and in a falling drop

Part I — Blood. Blood: ρ=1060 kg/m3\rho = 1060\,\mathrm{kg}/\mathrm{m}^{3}, η=3.5×10−3 Pa s\eta = 3.5 \times 10^{-3}\,\mathrm{Pa}\,\mathrm{s} (treated as Newtonian); cardiac output DV=5.0 L/minD_V = 5.0\,\mathrm{L}/\mathrm{min}.

  1. Aorta, radius 1.25 cm1.25\,\mathrm{cm}: mean speed, Reynolds number, regime.
  2. If the flow in the aorta were Poiseuille’s, what would be the maximum speed and the wall shear stress? (The endothelium senses stresses of order 1 Pa1\,\mathrm{Pa}.)
  3. A capillary, radius 4.0 µm4.0\,\text{µ}\mathrm{m}, carries blood at 0.30 mm/s0.30\,\mathrm{mm}/\mathrm{s}: Reynolds number; pressure drop over its 1.0 mm1.0\,\mathrm{mm} length, in pascals and in millimetres of mercury (1 mmHg1\,\mathrm{mmHg} = 133 Pa133\,\mathrm{Pa}).
  4. Time for a red cell to cross the capillary; why is this the right order for gas exchange?
  5. Arterioles (radius 10 µm10\,\text{µ}\mathrm{m}, length 2.0 mm2.0\,\mathrm{mm}, 10810^8 in parallel): resistance of the set and the pressure drop across it at full cardiac output.
  6. The total arterial-to-venous drop is 13 kPa13\,\mathrm{kPa}: mechanical power delivered by the left ventricle. Compare with a 100 W100\,\mathrm{W} body.
  7. A stenosis reduces the radius of a coronary artery by 40%40\% over a short length: factor by which its resistance rises; why the heart muscle downstream may still be supplied at rest but not during effort.
  8. During a hard effort the cardiac output reaches 25 L/min25\,\mathrm{L}/\mathrm{min}: Reynolds number in the aorta, and regime.
  9. Blood is not Newtonian: its apparent viscosity falls at high shear rate and rises at low. Which of the vessels above is most affected, and in which direction?

Part II — The pipeline. A crude-oil pipeline: diameter D=1.2 mD = 1.2\,\mathrm{m}, length L=1300 kmL = 1300\,\mathrm{km}, flow DV=3.0 m3/sD_V = 3.0\,\mathrm{m}^{3}/\mathrm{s}; oil ρ=850 kg/m3\rho = 850\,\mathrm{kg}/\mathrm{m}^{3}, η=2.0×10−2 Pa s\eta = 2.0 \times 10^{-2}\,\mathrm{Pa}\,\mathrm{s} when kept warm. Turbulent friction: ΔP=λ(L/D)12ρ⟨v⟩2\Delta P = \lambda (L/D)\tfrac12\rho\langle v\rangle^2, λ=0.316/Re1/4\lambda = 0.316/\mathrm{Re}^{1/4}.

  1. Mean speed and Reynolds number; regime.
  2. Pressure drop per kilometre and over the whole line.
  3. The pipe tolerates 80 bar80\,\mathrm{bar}: minimum number of pump stations along the line.
  4. Total pumping power; energy per cubic metre of oil delivered, compared with the energy content of the oil (36 GJ/m336\,\mathrm{GJ}/\mathrm{m}^{3}).
  5. What drop would Poiseuille’s law predict? Comment on the cost of turbulence (and on why polymer additives that delay turbulence are injected in some pipelines).
  6. In winter the unheated oil would reach η=2.0 Pa s\eta = 2.0\,\mathrm{Pa}\,\mathrm{s}: new Reynolds number and regime; pressure drop over the line with the appropriate law. Conclusion for the design (the real line is insulated and the oil kept warm).
  7. Transit time of the oil from one end to the other.
  8. Wall shear stress in the warm, turbulent case, from the pressure drop; total friction force on the pipe wall.
  9. The same flow in a 1.0 m1.0\,\mathrm{m} pipe: by what factor does the turbulent pressure drop change?
  10. A 0.10 mm0.10\,\mathrm{mm} grain of sand (ρs=2650 kg/m3\rho_s = 2650\,\mathrm{kg}/\mathrm{m}^{3}) is carried in the warm oil: Stokes settling speed, Reynolds number of the grain, time to settle across the pipe. Compare with the turbulent velocity fluctuations, a few percent of ⟨v⟩\langle v \rangle: does sand settle in the running pipeline? In a stopped one?

Part III — Drops. Air: ρ=1.2 kg/m3\rho = 1.2\,\mathrm{kg}/\mathrm{m}^{3}, η=1.8×10−5 Pa s\eta = 1.8 \times 10^{-5}\,\mathrm{Pa}\,\mathrm{s}; water ρw=1000 kg/m3\rho_w = 1000\,\mathrm{kg}/\mathrm{m}^{3}, γ=0.072 N/m\gamma = 0.072\,\mathrm{N}/\mathrm{m}.

  1. Cloud droplet, diameter 10 µm10\,\text{µ}\mathrm{m}: Stokes terminal speed, Reynolds number, time to fall 1 km1\,\mathrm{km}; why the cloud stays.
  2. Drizzle, 0.20 mm0.20\,\mathrm{mm}: Stokes speed and Reynolds number; is Stokes still reliable?
  3. Raindrop, 2.0 mm2.0\,\mathrm{mm}: show Stokes fails, and find the terminal speed with Cx=0.5C_x = 0.5. Time to fall from 1 km1\,\mathrm{km}.
  4. In an updraft of 1 cm/s1\,\mathrm{cm}/\mathrm{s}, does the cloud droplet rise or fall? What updraft would hold the drizzle drop?
  5. Laplace pressure inside the three drops; compare with the dynamic pressure 12ρv∞2\tfrac12\rho v_\infty^2 for each, and explain why large raindrops flatten and break up above about 5 mm5\,\mathrm{mm}.
  6. In one table, give for the eight flows of this problem the Reynolds number and the law that governs it.
Solution

Solution of Problem 4.1.

1. S=4.9 cm2S = 4.9\,\mathrm{cm}^{2}, ⟨v⟩=83/4.9=0.17 m/s\langle v\rangle = 83/4.9 = 0.17\,\mathrm{m}/\mathrm{s}; Re=1060×0.17×0.025/3.5×10−3=1300\mathrm{Re} = 1060 \times 0.17 \times 0.025/3.5 \times 10^{-3} = 1300: laminar (pulsatile peaks reach ∼5000\sim 5000).

2. vmax⁡=0.34 m/sv_{\max} = 0.34\,\mathrm{m}/\mathrm{s}; τw=4η⟨v⟩/R=0.19 Pa\tau_w = 4\eta\langle v\rangle/R = 0.19\,\mathrm{Pa}.

3. Re=1060×3×10−4×8×10−6/3.5×10−3=7×10−4\mathrm{Re} = 1060 \times 3 \times 10^{-4} \times 8 \times 10^{-6}/3.5 \times 10^{-3} = 7 \times 10^{-4}; ΔP=8ηL⟨v⟩/R2=8×3.5×10−3×10−3×3×10−4/1.6×10−11=525 Pa=3.9 mmHg\Delta P = 8\eta L\langle v\rangle/R^2 = 8 \times 3.5 \times 10^{-3} \times 10^{-3} \times 3 \times 10^{-4}/1.6 \times 10^{-11} = 525\,\mathrm{Pa} = 3.9\,\mathrm{mmHg}.

4. 1/0.3=3.3 s1/0.3 = 3.3\,\mathrm{s}: long enough for oxygen to diffuse the few micrometres to the tissue (Chapter 24).

5. One arteriole 8ηL/πR4=1.8×1015 Pa s/m38\eta L/\pi R^4 = 1.8 \times 10^{15}\,\mathrm{Pa}\,\mathrm{s}/\mathrm{m}^{3}; 10810^8 in parallel: 1.8×107 Pa s/m31.8 \times 10^{7}\,\mathrm{Pa}\,\mathrm{s}/\mathrm{m}^{3}; ΔP=1.8×107×8.3×10−5=1.5 kPa=11 mmHg\Delta P = 1.8 \times 10^7 \times 8.3 \times 10^{-5} = 1.5\,\mathrm{kPa} = 11\,\mathrm{mmHg}.

6. P=ΔP DV=13000×8.3×10−5=1.1 WP = \Delta P\,D_V = 13000 \times 8.3 \times 10^{-5} = 1.1\,\mathrm{W}, about 1%1\% of the body’s power.

7. (1/0.6)4=7.7(1/0.6)^4 = 7.7. At rest the vessels downstream dilate and compensate; during effort the demand rises four- or fivefold and the narrowed artery cannot deliver it: angina.

8. Five times the speed: Re≈6500\mathrm{Re} \approx 6500, turbulent — the murmurs a stethoscope hears during exercise.

9. In the slow, low-shear flow of small veins and capillaries the red cells aggregate and the apparent viscosity rises (the opposite in the fast aorta); all the pressure-drop estimates for small vessels are lower bounds.

10. ⟨v⟩=3/1.13=2.65 m/s\langle v\rangle = 3/1.13 = 2.65\,\mathrm{m}/\mathrm{s}; Re=850×2.65×1.2/0.02=1.35×105\mathrm{Re} = 850 \times 2.65 \times 1.2/0.02 = 1.35 \times 10^5: turbulent.

11. λ=0.316/19.2=0.0165\lambda = 0.316/19.2 = 0.0165; ΔP/L=(0.0165/1.2)×12×850×7.0=41 Pa/m=41 kPa/km\Delta P/L = (0.0165/1.2) \times \tfrac12 \times 850 \times 7.0 = 41\,\mathrm{Pa}/\mathrm{m} = 41\,\mathrm{kPa}/\mathrm{km}; 534 bar534\,\mathrm{bar} over the line.

12. 534/80=6.7534/80 = 6.7: at least seven stations.

13. P=ΔP DV=5.34×107×3=160 MWP = \Delta P\,D_V = 5.34 \times 10^7 \times 3 = 160\,\mathrm{MW}; 53 MJ/m353\,\mathrm{MJ}/\mathrm{m}^{3}, 0.15%0.15\% of the oil’s energy.

14. ΔP=32ηL⟨v⟩/D2=32×0.02×1.3×106×2.65/1.44=15 bar\Delta P = 32\eta L\langle v\rangle/D^2 = 32 \times 0.02 \times 1.3 \times 10^6 \times 2.65/1.44 = 15\,\mathrm{bar}: thirty-five times less. Turbulence is expensive; a few parts per million of long polymers delay it and cut the drop by tens of percent.

15. Re=1350\mathrm{Re} = 1350: laminar; Poiseuille: ΔP=32×2×1.3×106×2.65/1.44=1500 bar\Delta P = 32 \times 2 \times 1.3 \times 10^6 \times 2.65/1.44 = 1500\,\mathrm{bar} — impossible; the oil must be kept warm, hence the insulation and heating.

16. 1.3×106/2.65=4.9×105 s=5.7 days1.3 \times 10^6/2.65 = 4.9 \times 10^{5}\,\mathrm{s} = 5.7\,\mathrm{days}.

17. τw=(ΔP/L)D/4=41×0.3=12 Pa\tau_w = (\Delta P/L)D/4 = 41 \times 0.3 = 12\,\mathrm{Pa}; F=τwπDL=6×107 NF = \tau_w\pi DL = 6 \times 10^{7}\,\mathrm{N}.

18. ΔP∝λ⟨v⟩2/D∝D−4.75\Delta P \propto \lambda\langle v\rangle^2/D \propto D^{-4.75}: 1.24.75=2.41.2^{4.75} = 2.4.

19. v=2R2Δρ g/9η=2×2.5×10−9×1800×9.81/0.18=0.49 mm/sv = 2R^2\Delta\rho\,g/9\eta = 2 \times 2.5 \times 10^{-9} \times 1800 \times 9.81/ 0.18 = 0.49\,\mathrm{mm}/\mathrm{s}; Re=2×10−3\mathrm{Re} = 2 \times 10^{-3}; 41 min41\,\mathrm{min} to cross 1.2 m1.2\,\mathrm{m}. Turbulent fluctuations of a few centimetres per second keep it suspended while the oil runs; in a stopped line it settles in under an hour.

20. v=2×2.5×10−11×1000×9.81/1.62×10−4=3.0 mm/sv = 2 \times 2.5 \times 10^{-11} \times 1000 \times 9.81/1.62 \times 10^{-4} = 3.0\,\mathrm{mm}/\mathrm{s}; Re=2×10−3\mathrm{Re} = 2 \times 10^{-3}; 1 km1\,\mathrm{km} in 3.3×1053.3 \times 10^5 s, four days: any updraft above a few millimetres per second holds it.

21. 1.2 m/s1.2\,\mathrm{m}/\mathrm{s}, Re=16\mathrm{Re} = 16: Stokes overestimates by about a factor two (real: 0.7 m/s0.7\,\mathrm{m}/\mathrm{s}).

22. Stokes would give 120 m/s120\,\mathrm{m}/\mathrm{s} and Re∼104\mathrm{Re} \sim 10^4: absurd. Quadratic drag, 43πR3ρwg=12ρv2πR2Cx\tfrac43\pi R^3\rho_wg = \tfrac12\rho v^2\pi R^2C_x: v=8Rρwg/3ρCx=6.6 m/sv = \sqrt{8R \rho_wg/3\rho C_x} = 6.6\,\mathrm{m}/\mathrm{s} (Re=900\mathrm{Re} = 900, consistent); 150 s150\,\mathrm{s} from 1 km1\,\mathrm{km}.

23. 1 cm/s1\,\mathrm{cm}/\mathrm{s} >> 3 mm/s3\,\mathrm{mm}/\mathrm{s}: the droplet rises; the drizzle drop needs about 1 m/s1\,\mathrm{m}/\mathrm{s}, which convective clouds provide.

24. 2γ/R2\gamma/R: 29 kPa29\,\mathrm{kPa}, 1.4 kPa1.4\,\mathrm{kPa}, 144 Pa144\,\mathrm{Pa}; dynamic pressures 12ρv2\tfrac12\rho v^2: 5×10−65 \times 10^{-6}, 0.30.3 and 26 Pa26\,\mathrm{Pa}. The ratio reaches 0.20.2 at 2 mm2\,\mathrm{mm}; at 5 mm5\,\mathrm{mm} (v≈9 m/sv \approx 9\,\mathrm{m}/\mathrm{s}, 12ρv2=49 Pa\tfrac12\rho v^2 = 49\,\mathrm{Pa} against 58 Pa58\,\mathrm{Pa}) the aerodynamic pressure flattens the drop and breaks it: no raindrop exceeds about 5 mm5\,\mathrm{mm}.

25. Aorta 10310^3 (laminar, near transition); capillary 10−310^{-3} (Poiseuille); arteriole 10−210^{-2} (Poiseuille); warm pipeline 10510^5 (turbulent, λ(Re)\lambda(\mathrm{Re})); cold pipeline 10310^3 (Poiseuille); sand in oil 10−310^{-3} (Stokes); cloud droplet 10−310^{-3} (Stokes); raindrop 10310^3 (quadratic drag, Cx≈0.5C_x \approx 0.5).

Terms defined in this chapter

See all 393 terms in the glossary