---
title: "Viscous Flows"
book: "University Physics — Year 2"
subject: physics
language: en
chapter: 4
exercises: 12
source: https://one-course.com/books/physics/4/en/chapter/4-viscous-flows
---

# Chapter 4 — Viscous 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](#def-b2-viscous-flows-viscosity)* — the internal friction of a fluid — and the fight between [viscosity](#def-b2-viscous-flows-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](https://one-course.com/books/physics/4/en/chapter/3-perfect-fluids-euler-and-bernoulli#thm-b2-euler-bernoulli-euler), solves the two [laminar flows](#def-b2-viscous-flows-reynolds) every engineer uses, and explains why, at high [Reynolds number](#def-b2-viscous-flows-reynolds), [viscosity](#def-b2-viscous-flows-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.](https://one-course.com/images/onecourse/chapters/physics-4/b2-viscous-flows/img-529efba96899.jpg)

*Honey pours in a thin, steady ribbon that coils as it lands: a [viscosity](#def-b2-viscous-flows-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 $\vect v = v_x(y)\,\vect e_x$, the fluid above a surface $y =$ const exerts on the fluid below it (through a surface element $\dd S$) the tangential force

$$
\dd\vect F = \eta\,\frac{\dd v_x}{\dd y}\,\dd S\;\vect e_x ,
$$

dragging the slower layer forward and being dragged back in return. The *shear stress* is $\tau = \eta\,\dd v_x/\dd y$; the coefficient $\eta$ (unit $\mathrm{Pa}\,\mathrm{s}$) is the *dynamic viscosity* of the fluid, and $\nu = \eta/\rho$ ($\mathrm{m}^{2}/\mathrm{s}$) its *kinematic viscosity*. A fluid for which $\eta$ 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 $20{}^{\circ}\mathrm{C}$: air $\eta =
1.8 \times 10^{-5}\,\mathrm{Pa}\,\mathrm{s}$ ($\nu = 1.5 \times 10^{-5}\,\mathrm{m}^{2}/\mathrm{s}$), water $1.0 \times 10^{-3}\,\mathrm{Pa}\,\mathrm{s}$ ($\nu = 1.0 \times 10^{-6}\,\mathrm{m}^{2}/\mathrm{s}$), olive oil $0.08\,\mathrm{Pa}\,\mathrm{s}$, glycerol $1.5\,\mathrm{Pa}\,\mathrm{s}$, honey $\sim10\,\mathrm{Pa}\,\mathrm{s}$. 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 $y$ and $y + \dd y$ is, per unit volume, $\eta\,\dd^2v_x/\dd y^2$: [viscosity](#def-b2-viscous-flows-viscosity) acts as a volume force density $\eta\,\Delta\vect v$ (the Laplacian taken component by component), a result that holds for any [incompressible flow](https://one-course.com/books/physics/4/en/chapter/2-fluid-kinematics#cor-b2-fluid-kinematics-incompressible). At a solid wall the fluid *sticks*: its velocity equals the wall’s (the *[no-slip condition](#prop-b2-viscous-flows-density)*), which is why dust stays on a fan blade and why a wall feels a viscous drag.

**Proof.** The slab is pulled by $\eta\,v_x'(y + \dd y)\dd S$ from above and by $-\eta\,v_x'(y)\dd S$ from below: net $\eta v_x''\,\dd y\,\dd S$. The general form is admitted (it follows from writing the stress as a symmetric linear function of the velocity gradients). The [no-slip condition](#prop-b2-viscous-flows-density) is experimental. ∎

**Theorem 4.3 (Navier–Stokes equation).**

An incompressible [Newtonian fluid](#def-b2-viscous-flows-viscosity) obeys

$$
\rho\Bigl(\frac{\partial\vect v}{\partial t} + (\vect v\cdot\operatorname{\vect{grad}})
\vect v\Bigr) = -\operatorname{\vect{grad}}P + \rho\vect g + \eta\,\Delta\vect v ,
\qquad \operatorname{div}\vect v = 0 ,
$$

with the [no-slip condition](#prop-b2-viscous-flows-density) on every wall. [Euler’s equation](https://one-course.com/books/physics/4/en/chapter/3-perfect-fluids-euler-and-bernoulli#thm-b2-euler-bernoulli-euler) is the case $\eta = 0$.

**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.](https://one-course.com/images/onecourse/chapters/physics-4/b2-viscous-flows/fig-d036ed5a82a8.svg)

*Left: plane Couette flow between a fixed plate and a plate moving at $U$ — a linear profile, the same [shear stress](#def-b2-viscous-flows-viscosity) 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 $0.1\,\mathrm{mm}$ film of oil ($\eta = 0.1\,\mathrm{Pa}\,\mathrm{s}$) between a plate and a table, the plate sliding at $1\,\mathrm{m}/\mathrm{s}$: $\tau = \eta U/h = 1\,\mathrm{kPa}$, a force of $100\,\mathrm{N}$ on a tenth of a square metre — the principle of the lubricated [bearing](https://one-course.com/books/physics/4/en/chapter/1-rigid-body-mechanics#def-b2-rigid-body-mechanics-pivot), where the viscous drag replaces dry friction a hundred times larger. Honey on a spoon, $10\,\mathrm{Pa}\,\mathrm{s}$, $1\,\mathrm{mm}$ thick, draining at $1\,\mathrm{mm}/\mathrm{s}$: $\tau = 10\,\mathrm{Pa}$, 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 $U$ and length $L$, the *Reynolds number*

$$
\mathrm{Re} = \frac{\rho UL}{\eta} = \frac{UL}{\nu}
$$

is the ratio of the orders of magnitude of the inertial term $\rho(\vect v\cdot\operatorname{\vect{grad}})\vect v \sim \rho U^2/L$ and the viscous term $\eta\Delta\vect v \sim \eta U/L^2$ in the [Navier–Stokes equation](#thm-b2-viscous-flows-ns). At $\mathrm{Re} \ll 1$ inertia is negligible and the flow is *creeping* (viscous, reversible, smooth); at $\mathrm{Re} \gg 1$ [viscosity](#def-b2-viscous-flows-viscosity) is negligible except near walls and the flow is that of a [perfect fluid](https://one-course.com/books/physics/4/en/chapter/3-perfect-fluids-euler-and-bernoulli#def-b2-euler-bernoulli-perfect) — until, beyond a threshold of order $10^3$ (about $2000$ to $3000$ 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 ($1\,\text{µ}\mathrm{m}$, $30\,\text{µ}\mathrm{m}/\mathrm{s}$, water): $\mathrm{Re} =
3 \times 10^{-5}$ — it lives in a world without inertia, where stopping its flagellum stops it within an atomic diameter. Blood in a capillary: $10^{-3}$; a falling dust grain: $10^{-2}$; a goldfish: $10^3$; a swimmer: $10^6$; a car at $30\,\mathrm{m}/\mathrm{s}$ ($4\,\mathrm{m}$): $8 \times 10^6$; an airliner: $10^8$; the Gulf Stream: $10^{11}$. Two flows with the same geometry and the same [Reynolds number](#def-b2-viscous-flows-reynolds) are *similar*: this is what allows a $1\,\mathrm{m}$ 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 $R$ and length $L$, driven by the pressure difference $\Delta P = P_1 - P_2$ between its ends, the steady [laminar flow](#def-b2-viscous-flows-reynolds) is

$$
v(r) = \frac{\Delta P}{4\eta L}\,(R^2 - r^2) , \qquad
D_V = \frac{\pi R^4}{8\eta L}\,\Delta P :
$$

a parabolic profile, maximal on the axis, with a mean speed $\langle v\rangle = v_{\max}/2$. The flow rate is proportional to the pressure drop (Poiseuille’s law) with the *[hydraulic resistance](#prop-b2-viscous-flows-poiseuille)* $R_h = \Delta P/D_V = 8\eta L/\pi R^4$ — and to the *fourth power* of the radius. The wall [shear stress](#def-b2-viscous-flows-viscosity) is $\tau_w = \Delta P\,R/2L = 4\eta\langle v
\rangle/R$. Valid while $\mathrm{Re} = 2R\langle v\rangle/\nu \lesssim 2000$.

**Proof.** Steady, $\vect v = v(r)\vect e_z$ (so the convective term vanishes: $\vect v
\cdot\operatorname{\vect{grad}}$ acts along $z$, along which nothing varies). Navier–Stokes along $z$: $0 = -\partial_zP + \eta\Delta v$, and radially $\partial_rP = 0$: $P$ depends on $z$ only, $\Delta v$ on $r$ only, so both sides equal a constant, $\partial_zP = -\Delta P/L$. In cylindrical coordinates $\Delta v = \frac1r\frac{\dd}{\dd r}\bigl(r\frac{\dd v}{\dd r}\bigr)$ (stated; it is the balance of viscous forces on a cylindrical shell, $2\pi rL\,\eta v'$ at $r$ and at $r + \dd r$). Integrating twice with $v$ finite on the axis and $v(R) = 0$ gives the parabola; $D_V = \int_0^Rv\,2\pi
r\,\dd r$. The wall stress is $-\eta v'(R)$; equivalently, the pressure force $\pi R^2\Delta P$ balances the wall friction $2\pi RL\tau_w$. ∎

![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.](https://one-course.com/images/onecourse/chapters/physics-4/b2-viscous-flows/fig-50301880892a.svg)

*Left: [Poiseuille flow](#prop-b2-viscous-flows-poiseuille) — 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.*

**Example 4.8 (The fourth power).**

A syringe needle of inner diameter $0.4\,\mathrm{mm}$ and length $3\,\mathrm{cm}$, water, thumb pressure $20\,\mathrm{kPa}$: $D_V = \pi(2 \times 10^{-4})^4 \times 2 \times
10^4/(8 \times 10^{-3} \times 0.03) = 0.42\,\mathrm{mL}/\mathrm{s}$; a needle twice as wide gives sixteen times more. An artery narrowed by a plaque to $70\%$ of its radius offers $(1/0.7)^4 = 4.2$ 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 $R$ moving at speed $v$ through a fluid at $\mathrm{Re} = 2\rho vR/\eta \ll 1$ feels the drag

$$
F = 6\pi\eta Rv ,
$$

opposite to its velocity. A sphere of density $\rho_s$ falling in a fluid of density $\rho$ reaches the *[terminal velocity](#prop-b2-viscous-flows-stokes)* $v_\infty =
\dfrac{2R^2(\rho_s - \rho)g}{9\eta}$ — 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, $\tfrac43\pi R^3(\rho_s -
\rho)g$, equals the drag; always check $\mathrm{Re} \ll 1$ afterwards. ∎

**Example 4.10 (Fog, rain and dust).**

A fog droplet of $10\,\text{µ}\mathrm{m}$ diameter in air falls at $2 \times (5 \times
10^{-6})^2 \times 10^4/(9 \times 1.8 \times 10^{-5}) = 3\,\mathrm{mm}/\mathrm{s}$ ($\mathrm{Re} = 2 \times
10^{-3}$): the slightest updraft keeps it aloft, which is why clouds float. A $1\,\mathrm{mm}$ raindrop would give $30\,\mathrm{m}/\mathrm{s}$ and $\mathrm{Re} \approx 2000$: Stokes no longer applies, and the quadratic drag of the next section gives the observed $4\,\mathrm{m}/\mathrm{s}$. A $10\,\text{µ}\mathrm{m}$ dust grain ($\rho_s =
2500\,\mathrm{kg}/\mathrm{m}^{3}$) settles at $8\,\mathrm{mm}/\mathrm{s}$, 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](#def-b2-viscous-flows-reynolds) the flow around a body is that of a [perfect fluid](https://one-course.com/books/physics/4/en/chapter/3-perfect-fluids-euler-and-bernoulli#def-b2-euler-bernoulli-perfect) everywhere except in a thin *[boundary layer](#prop-b2-viscous-flows-bl)* along the wall, in which the velocity falls from its outer value $U$ to zero (no slip). Along a flat plate its thickness grows as

$$
\delta(x) \sim \sqrt{\frac{\nu x}{U}} = \frac{x}{\sqrt{\mathrm{Re}_x}} ,
\qquad \mathrm{Re}_x = \frac{Ux}{\nu} ,
$$

with $x$ the distance from the leading edge; the layer itself turns turbulent beyond $\mathrm{Re}_x \sim 5 \times 10^5$. 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](https://one-course.com/books/physics/4/en/chapter/2-fluid-kinematics#def-b2-fluid-kinematics-particle) spends the time $x/U$ travelling along the plate, during which the wall’s influence diffuses into the fluid — the viscous term $\nu\,\partial^2v/\partial y^2$ is a diffusion of momentum with diffusivity $\nu$, reaching the distance $\sqrt{\nu t}$ in the time $t$ ([Chapter 24](https://one-course.com/books/physics/4/en/chapter/24-particle-diffusion#ch-b2-particle-diffusion)). Hence $\delta \sim \sqrt{\nu x
/U}$. The full (Prandtl) theory is admitted. ∎

**Proposition 4.12 (Drag at high Reynolds number).**

The drag on a body of frontal area $S$ moving at $v$ through a fluid is written

$$
F = \tfrac12\rho v^2\,S\,C_x ,
$$

where the *[drag coefficient](#prop-b2-viscous-flows-drag)* $C_x$ depends on the shape and on $\mathrm{Re}$. For a sphere, $C_x = 24/\mathrm{Re}$ at $\mathrm{Re} \ll 1$ (Stokes, with $\mathrm{Re}$ based on the diameter), then about $0.4$–$0.5$, nearly constant, from $\mathrm{Re} \approx 10^3$ to $2 \times 10^5$, where it drops suddenly to about $0.1$ (the *[drag crisis](#prop-b2-viscous-flows-drag)*: the [boundary layer](#prop-b2-viscous-flows-bl) turns turbulent, sticks longer to the surface and the wake shrinks — the dimples on a golf ball trigger it early). Streamlined bodies have $C_x \approx 0.05$; a car $0.3$; a cyclist $0.9$; a flat plate facing the flow $1.2$. The power needed to move at $v$ is $Fv \propto v^3$.

**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.](https://one-course.com/images/onecourse/chapters/physics-4/b2-viscous-flows/fig-0e2bfe2eeddf.svg)

![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.](https://one-course.com/images/onecourse/chapters/physics-4/b2-viscous-flows/fig-f0a48b550772.svg)

*Left: the [drag coefficient](#prop-b2-viscous-flows-drag) of a sphere against the [Reynolds number](#def-b2-viscous-flows-reynolds) (log scales): [Stokes’ law](#prop-b2-viscous-flows-stokes) at low $\mathrm{Re}$, a plateau near $0.45$, and the [drag crisis](#prop-b2-viscous-flows-drag) near $\mathrm{Re} = 2 \times 10^5$. Right: the [boundary layer](#prop-b2-viscous-flows-bl) thickening along a flat plate; outside it the flow is that of a [perfect fluid](https://one-course.com/books/physics/4/en/chapter/3-perfect-fluids-euler-and-bernoulli#def-b2-euler-bernoulli-perfect).*

**Example 4.13 (A cyclist, a car).**

A cyclist ($SC_x = 0.35\,\mathrm{m}^{2}$) at $10\,\mathrm{m}/\mathrm{s}$: $F = 0.5 \times 1.2 \times 100
\times 0.35 = 21\,\mathrm{N}$, $P = 210\,\mathrm{W}$ — a sustained effort; at $15\,\mathrm{m}/\mathrm{s}$ ($54\,\mathrm{km}/\mathrm{h}$) it would take $710\,\mathrm{W}$, which is why sprinters draft. A car ($SC_x = 0.7\,\mathrm{m}^{2}$) at $130\,\mathrm{km}/\mathrm{h}$: $550\,\mathrm{N}$, $20\,\mathrm{kW}$ 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 $\gamma$ ($\mathrm{N}/\mathrm{m}$; water $0.072\,\mathrm{N}/\mathrm{m}$), because surface molecules lack half their neighbours. The pressure inside a drop of radius $R$ exceeds the outside pressure by the *[Laplace pressure](#rem-b2-viscous-flows-capillarity)* $2\gamma/R$ ($1.4\,\mathrm{kPa}$ in a $0.1\,\mathrm{mm}$ drop, $14\,\mathrm{bar}$ in a $10\,\mathrm{nm}$ one); [surface tension](#rem-b2-viscous-flows-capillarity) beats gravity below the *[capillary length](#rem-b2-viscous-flows-capillarity)* $\sqrt{\gamma/\rho g} = 2.7\,\mathrm{mm}$ for water — the size of dew drops, of the meniscus in a glass, of the rise in a capillary tube ($h = 2\gamma/\rho gr$, 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 $\mathrm{Re}$. $\mathrm{Re} \ll 1$: Stokes and creeping flow, drag $\propto v$, pressure drops $\propto$ flow rate. $\mathrm{Re}$ up to $\sim\!2000$ in a pipe: laminar, Poiseuille. Above: turbulent — Poiseuille underestimates the loss badly; use the empirical law $\Delta P = \lambda(L/D)\,\tfrac12\rho\langle v\rangle^2$ with $\lambda
\approx 0.3/\mathrm{Re}^{1/4}$ for smooth pipes, and $F = \tfrac12\rho v^2SC_x$ for bodies. In every case, check afterwards that the $\mathrm{Re}$ computed from the answer agrees with the regime assumed.

## 4.5 Exercises

**Exercise 4.1 ★.**

A $0.50\,\mathrm{m}^{2}$ plate slides at $2.0\,\mathrm{m}/\mathrm{s}$ on a $0.20\,\mathrm{mm}$ film of oil ($\eta = 0.25\,\mathrm{Pa}\,\mathrm{s}$). [Shear stress](#def-b2-viscous-flows-viscosity), force on the plate, power dissipated; compare with dry friction ($f = 0.3$) if the plate weighs $10\,\mathrm{kg}$.

**Solution of Exercise 4.1.**

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

**Exercise 4.2 ★.**

[Reynolds numbers](#def-b2-viscous-flows-reynolds) of: (a) a $2\,\text{µ}\mathrm{m}$ bacterium at $30\,\text{µ}\mathrm{m}/\mathrm{s}$ in water; (b) a swimmer ($1.8\,\mathrm{m}$, $1.5\,\mathrm{m}/\mathrm{s}$); (c) a car ($4\,\mathrm{m}$, $30\,\mathrm{m}/\mathrm{s}$) in air; (d) blood in the aorta ($2.5\,\mathrm{cm}$, $0.33\,\mathrm{m}/\mathrm{s}$, $\nu = 3.3 \times 10^{-6}\,\mathrm{m}^{2}/\mathrm{s}$); (e) oil ($\eta = 0.1\,\mathrm{Pa}\,\mathrm{s}$, $\rho =
900\,\mathrm{kg}/\mathrm{m}^{3}$) at $0.5\,\mathrm{m}/\mathrm{s}$ in a $1\,\mathrm{cm}$ pipe. Regime of each.

**Solution of Exercise 4.2.**

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

**Exercise 4.3 ★.**

Water is pushed through a $0.40\,\mathrm{mm}$ (inner diameter), $3.0\,\mathrm{cm}$ needle by $20\,\mathrm{kPa}$. Flow rate; time to inject $5\,\mathrm{mL}$; maximum speed and [Reynolds number](#def-b2-viscous-flows-reynolds); wall [shear stress](#def-b2-viscous-flows-viscosity). With a $0.80\,\mathrm{mm}$ needle?

**Solution of Exercise 4.3.**

$D_V = \pi R^4\Delta P/8\eta L = \pi(2 \times 10^{-4})^4 \times 2 \times 10^4/(8 \times 10^{-3}
\times 0.03) = 0.42\,\mathrm{mL}/\mathrm{s}$; $12\,\mathrm{s}$ for $5\,\mathrm{mL}$. $v_{\max} = \Delta PR^2/
4\eta L = 6.7\,\mathrm{m}/\mathrm{s}$, $\langle v\rangle = 3.3\,\mathrm{m}/\mathrm{s}$, $\mathrm{Re} = 2R\langle v
\rangle/\nu = 1300$: laminar. $\tau_w = \Delta PR/2L = 67\,\mathrm{Pa}$. At $0.8\,\mathrm{mm}$: sixteen times the flow, $6.7\,\mathrm{mL}/\mathrm{s}$ — but $\mathrm{Re} \approx 10^4$: the flow turns turbulent and the gain is less.

**Exercise 4.4 ★.**

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

**Solution of Exercise 4.4.**

$v_\infty = 2R^2\Delta\rho\,g/9\eta$. (a) $1.3\,\mathrm{cm}/\mathrm{s}$, $\mathrm{Re} = 0.02$, $75\,\mathrm{s}$ per metre. (b) $0.24\,\mathrm{mm}/\mathrm{s}$, $\mathrm{Re} = 3 \times 10^{-5}$, $70\,\mathrm{min}$ per metre. (c) $1.2\,\mathrm{m}/\mathrm{s}$ but $\mathrm{Re} = 16$: Stokes overestimates (real speed about $0.7\,\mathrm{m}/\mathrm{s}$), about a second per metre.

**Exercise 4.5 ★★.**

A water main of diameter $30\,\mathrm{cm}$ carries $0.10\,\mathrm{m}^{3}/\mathrm{s}$ over $1.0\,\mathrm{km}$. (a) Mean speed and [Reynolds number](#def-b2-viscous-flows-reynolds). (b) Pressure drop if Poiseuille’s law applied. (c) Actual drop with $\lambda = 0.3/\mathrm{Re}^{1/4}$; ratio to (b); pumping power. (d) The same flow in two parallel $21\,\mathrm{cm}$ mains (same total section): drop and power.

**Solution of Exercise 4.5.**

(a) $\langle v\rangle = 0.1/0.0707 = 1.4\,\mathrm{m}/\mathrm{s}$, $\mathrm{Re} = 4.2 \times 10^5$. (b) $8\eta LD_V/\pi R^4 = 500\,\mathrm{Pa}$. (c) $\lambda = 0.3/25.5 = 0.0118$, $\Delta P = 0.0118
\times 3333 \times 1000 = 39\,\mathrm{kPa}$: $80$ times more; $P = \Delta PD_V = 3.9\,\mathrm{kW}$. (d) Same speed, $\mathrm{Re} = 3 \times 10^5$, $\lambda = 0.0128$, $\Delta P = 0.0128 \times 4717
\times 1000 = 60\,\mathrm{kPa}$, $6\,\mathrm{kW}$: more wall per unit flow, more loss.

**Exercise 4.6 ★★.**

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

**Solution of Exercise 4.6.**

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

**Exercise 4.7 ★★.**

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

**Solution of Exercise 4.7.**

(a) $F = 0.6 \times 144 \times 0.35 = 30\,\mathrm{N}$, $+4\,\mathrm{N}$: $P = 34 \times 12 =
410\,\mathrm{W}$. (b) $47 + 4 = 51\,\mathrm{N}$, $770\,\mathrm{W}$. (c) Relative speed $17\,\mathrm{m}/\mathrm{s}$: $61 + 4 = 65\,\mathrm{N}$, times the ground speed: $780\,\mathrm{W}$. (d) Drag $21\,\mathrm{N}$: $300\,\mathrm{W}$, saving $110\,\mathrm{W}$. (e) $mg\sin\alpha = 39\,\mathrm{N} =
4 + 0.21v^2$: $v = 13\,\mathrm{m}/\mathrm{s}$, $47\,\mathrm{km}/\mathrm{h}$.

**Exercise 4.8 ★★.**

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

**Solution of Exercise 4.8.**

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

**Exercise 4.9 ★★.**

*Couette viscometer.* A cylinder of radius $5.0\,\mathrm{cm}$ and height $10\,\mathrm{cm}$ rotates at $10\,\mathrm{rad}/\mathrm{s}$ inside a fixed cylinder, the gap $1.0\,\mathrm{mm}$ being filled with the liquid to test; the torque needed is $2.0 \times 10^{-2}\,\mathrm{N}\,\mathrm{m}$. (a) Treating the gap as plane Couette flow, shear rate and [shear stress](#def-b2-viscous-flows-viscosity). (b) [Viscosity](#def-b2-viscous-flows-viscosity). (c) [Reynolds number](#def-b2-viscous-flows-reynolds) of the gap flow; is the flow laminar? (d) Why is the gap made thin?

**Solution of Exercise 4.9.**

(a) Shear rate $\Omega R/e = 0.5/10^{-3} = 500\,\mathrm{s}^{-1}$; $\Gamma = \tau\cdot2\pi Rh
\cdot R$: $\tau = 0.02/(2\pi \times 2.5 \times 10^{-3} \times 0.1) = 12.7\,\mathrm{Pa}$. (b) $\eta
= 12.7/500 = 2.5 \times 10^{-2}\,\mathrm{Pa}\,\mathrm{s}$. (c) $\mathrm{Re} = \rho Ue/\eta \approx 1000 \times 0.5 \times
10^{-3}/0.025 = 20$: 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 ($\eta = 3.5\,\mathrm{mPa}\,\mathrm{s}$) flows at $5.0\,\mathrm{L}/\mathrm{min}$. (a) [Hydraulic resistance](#prop-b2-viscous-flows-poiseuille) of the aorta (radius $1.25\,\mathrm{cm}$, length $40\,\mathrm{cm}$) and pressure drop along it. (b) The $10^{10}$ capillaries (radius $4\,\text{µ}\mathrm{m}$, length $1\,\mathrm{mm}$) are in parallel: resistance of one, of all, and the pressure drop across the capillary bed. (c) The arterioles ($10^8$, radius $10\,\text{µ}\mathrm{m}$, length $2\,\mathrm{mm}$): same questions; where is the resistance of the circulation? (d) The total drop is $13\,\mathrm{kPa}$: power of the heart’s left ventricle, and what fraction of the body’s $100\,\mathrm{W}$. (e) A drug dilates the arterioles by $10\%$: effect on the total resistance.

**Solution of Exercise 4.10.**

$D_V = 8.3 \times 10^{-5}\,\mathrm{m}^{3}/\mathrm{s}$. (a) $R_h = 8\eta L/\pi R^4 = 8 \times 3.5 \times 10^{-3} \times
0.4/(\pi \times 2.4 \times 10^{-8}) = 1.5 \times 10^{5}\,\mathrm{Pa}\,\mathrm{s}/\mathrm{m}^{3}$, $\Delta P = 12\,\mathrm{Pa}$. (b) One capillary $3.5 \times 10^{16}\,\mathrm{Pa}\,\mathrm{s}/\mathrm{m}^{3}$; $10^{10}$ in parallel: $3.5 \times 10^{6}\,\mathrm{Pa}\,\mathrm{s}/\mathrm{m}^{3}$; $\Delta P = 290\,\mathrm{Pa}$ ($2\,\mathrm{mmHg}$). (c) One arteriole $8 \times 3.5 \times 10^{-3}
\times 2 \times 10^{-3}/(\pi \times 10^{-20}) = 1.8 \times 10^{15}\,\mathrm{Pa}\,\mathrm{s}/\mathrm{m}^{3}$; the set: $1.8 \times 10^{7}\,\mathrm{Pa}\,\mathrm{s}/\mathrm{m}^{3}$; $\Delta P = 1.5\,\mathrm{kPa}$: the arterioles dominate the two (the small vessels, not the large ones, are where the pressure falls). (d) $P = 13000 \times 8.3 \times 10^{-5} = 1.1\,\mathrm{W}$, about $1\%$. (e) $(1/1.1)^4 = 0.68$: a third of the arteriolar resistance removed — vasodilators lower the blood pressure this way.

**Exercise 4.11 ★★★.**

*[Boundary layer](#prop-b2-viscous-flows-bl) and friction drag.* A flat plate of length $L$ and width $b$ moves edge-on at $U$. In the laminar regime the wall [shear stress](#def-b2-viscous-flows-viscosity) is $\tau_w = 0.332\,\rho U^2/\sqrt{\mathrm{Re}_x}$ (Blasius). (a) Show that the [friction force](https://one-course.com/books/physics/4/en/chapter/1-rigid-body-mechanics#def-b2-rigid-body-mechanics-contact) on one face is $F = 0.664\,\rho U^2bL/\sqrt{\mathrm{Re}_L}$. (b) Boundary-layer thickness at the trailing edge of a $1\,\mathrm{m}$ plate in water at $0.3\,\mathrm{m}/\mathrm{s}$, and the force on a $1\,\mathrm{m}^{2}$ plate (both faces). (c) At $10\,\mathrm{m}/\mathrm{s}$: $\mathrm{Re}_L$, and why the formula no longer applies. (d) A ship’s hull ($100\,\mathrm{m}$, $10\,\mathrm{m}/\mathrm{s}$): boundary-layer thickness estimate at the stern assuming the turbulent law $\delta \approx
0.37x/\mathrm{Re}_x^{1/5}$.

**Solution of Exercise 4.11.**

(a) $F = b\int_0^L\tau_w\,\dd x = 0.332\rho U^2b\sqrt{\nu/U}\int_0^L\dd x/\sqrt x =
0.664\rho U^2b\sqrt{\nu L/U} = 0.664\rho U^2bL/\sqrt{\mathrm{Re}_L}$. (b) $\mathrm{Re}_L =
3 \times 10^5$ (laminar); $\delta \approx 5\sqrt{\nu L/U} = 9\,\mathrm{mm}$; $F = 0.664 \times 1000
\times 0.09 \times 1/548 = 0.11\,\mathrm{N}$ per face, $0.22\,\mathrm{N}$. (c) $\mathrm{Re}_L = 10^7$: the layer is turbulent beyond the first few centimetres, the laminar formula fails (the real drag is several times larger). (d) $\mathrm{Re}_x =
10^9$: $\delta \approx 0.37 \times 100/63 = 0.6\,\mathrm{m}$.

**Exercise 4.12 ★★★.**

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

**Solution of Exercise 4.12.**

(a) $0 = \rho g + \eta v''(y)$, $v(0) = 0$ (no slip), $v'(h) = 0$ (no stress at the free surface). (b) Integrate twice: $v = (\rho g/2\eta)(2hy - y^2)$; $q =
\int_0^hv\,\dd y = \rho gh^3/3\eta$. (c) $v_s = \rho gh^2/2\eta = 1200 \times 9.81 \times
4 \times 10^{-8}/1 = 0.47\,\mathrm{mm}/\mathrm{s}$; $q = 6.3 \times 10^{-8}\,\mathrm{m}^{2}/\mathrm{s}$; $28\,\mathrm{cm}$ 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 $v_s \propto h^2$. (d) $1400 \times 9.81 \times 4 \times 10^{-6}/20 = 2.7\,\mathrm{mm}/\mathrm{s}$.

![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).](https://one-course.com/images/onecourse/chapters/physics-4/b2-viscous-flows/img-e5c057789fa9.jpg)

*A von Kármán vortex street in the clouds downstream of an island, seen from space: the wake of a cylinder at a [Reynolds number](#def-b2-viscous-flows-reynolds) 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: $\rho = 1060\,\mathrm{kg}/\mathrm{m}^{3}$, $\eta = 3.5 \times 10^{-3}\,\mathrm{Pa}\,\mathrm{s}$ (treated as Newtonian); cardiac output $D_V = 5.0\,\mathrm{L}/\mathrm{min}$.

1. Aorta, radius $1.25\,\mathrm{cm}$ : mean speed, [Reynolds number](#def-b2-viscous-flows-reynolds) , regime.
2. If the flow in the aorta were Poiseuille’s, what would be the maximum speed and the wall [shear stress](#def-b2-viscous-flows-viscosity) ? (The endothelium senses stresses of order $1\,\mathrm{Pa}$ .)
3. A capillary, radius $4.0\,\text{µ}\mathrm{m}$ , carries blood at $0.30\,\mathrm{mm}/\mathrm{s}$ : [Reynolds number](#def-b2-viscous-flows-reynolds) ; pressure drop over its $1.0\,\mathrm{mm}$ length, in pascals and in millimetres of mercury ( $1\,\mathrm{mmHg}$ = $133\,\mathrm{Pa}$ ).
4. Time for a red cell to cross the capillary; why is this the right order for gas exchange?
5. Arterioles (radius $10\,\text{µ}\mathrm{m}$ , length $2.0\,\mathrm{mm}$ , $10^8$ in parallel): resistance of the set and the pressure drop across it at full cardiac output.
6. The total arterial-to-venous drop is $13\,\mathrm{kPa}$ : mechanical power delivered by the left ventricle. Compare with a $100\,\mathrm{W}$ body.
7. A stenosis reduces the radius of a coronary artery by $40\%$ 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 $25\,\mathrm{L}/\mathrm{min}$ : [Reynolds number](#def-b2-viscous-flows-reynolds) in the aorta, and regime.
9. Blood is not Newtonian: its apparent [viscosity](#def-b2-viscous-flows-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 $D = 1.2\,\mathrm{m}$, length $L = 1300\,\mathrm{km}$, flow $D_V = 3.0\,\mathrm{m}^{3}/\mathrm{s}$; oil $\rho = 850\,\mathrm{kg}/\mathrm{m}^{3}$, $\eta =
2.0 \times 10^{-2}\,\mathrm{Pa}\,\mathrm{s}$ when kept warm. Turbulent friction: $\Delta P = \lambda
(L/D)\tfrac12\rho\langle v\rangle^2$, $\lambda = 0.316/\mathrm{Re}^{1/4}$.

10. Mean speed and [Reynolds number](#def-b2-viscous-flows-reynolds) ; regime.
11. Pressure drop per kilometre and over the whole line.
12. The pipe tolerates $80\,\mathrm{bar}$ : minimum number of pump stations along the line.
13. Total pumping power; energy per cubic metre of oil delivered, compared with the energy content of the oil ( $36\,\mathrm{GJ}/\mathrm{m}^{3}$ ).
14. 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).
15. In winter the unheated oil would reach $\eta = 2.0\,\mathrm{Pa}\,\mathrm{s}$ : new [Reynolds number](#def-b2-viscous-flows-reynolds) 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).
16. Transit time of the oil from one end to the other.
17. Wall [shear stress](#def-b2-viscous-flows-viscosity) in the warm, turbulent case, from the pressure drop; total [friction force](https://one-course.com/books/physics/4/en/chapter/1-rigid-body-mechanics#def-b2-rigid-body-mechanics-contact) on the pipe wall.
18. The same flow in a $1.0\,\mathrm{m}$ pipe: by what factor does the turbulent pressure drop change?
19. A $0.10\,\mathrm{mm}$ grain of sand ( $\rho_s = 2650\,\mathrm{kg}/\mathrm{m}^{3}$ ) is carried in the warm oil: Stokes settling speed, [Reynolds number](#def-b2-viscous-flows-reynolds) of the grain, time to settle across the pipe. Compare with the turbulent velocity fluctuations, a few percent of $\langle v  \rangle$ : does sand settle in the running pipeline? In a stopped one?

**Part III — Drops.** Air: $\rho = 1.2\,\mathrm{kg}/\mathrm{m}^{3}$, $\eta = 1.8 \times 10^{-5}\,\mathrm{Pa}\,\mathrm{s}$; water $\rho_w =
1000\,\mathrm{kg}/\mathrm{m}^{3}$, $\gamma = 0.072\,\mathrm{N}/\mathrm{m}$.

20. Cloud droplet, diameter $10\,\text{µ}\mathrm{m}$ : Stokes terminal speed, [Reynolds number](#def-b2-viscous-flows-reynolds) , time to fall $1\,\mathrm{km}$ ; why the cloud stays.
21. Drizzle, $0.20\,\mathrm{mm}$ : Stokes speed and [Reynolds number](#def-b2-viscous-flows-reynolds) ; is Stokes still reliable?
22. Raindrop, $2.0\,\mathrm{mm}$ : show Stokes fails, and find the terminal speed with $C_x = 0.5$ . Time to fall from $1\,\mathrm{km}$ .
23. In an updraft of $1\,\mathrm{cm}/\mathrm{s}$ , does the cloud droplet rise or fall? What updraft would hold the drizzle drop?
24. [Laplace pressure](#rem-b2-viscous-flows-capillarity) inside the three drops; compare with the [dynamic pressure](https://one-course.com/books/physics/4/en/chapter/3-perfect-fluids-euler-and-bernoulli#thm-b2-euler-bernoulli-bernoulli) $\tfrac12\rho v_\infty^2$ for each, and explain why large raindrops flatten and break up above about $5\,\mathrm{mm}$ .
25. In one table, give for the eight flows of this problem the [Reynolds number](#def-b2-viscous-flows-reynolds) and the law that governs it.

**Solution of Problem 4.1.**

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

**2.** $v_{\max} = 0.34\,\mathrm{m}/\mathrm{s}$; $\tau_w = 4\eta\langle v\rangle/R = 0.19\,\mathrm{Pa}$.

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

**4.** $1/0.3 = 3.3\,\mathrm{s}$: long enough for oxygen to diffuse the few micrometres to the tissue ([Chapter 24](https://one-course.com/books/physics/4/en/chapter/24-particle-diffusion#ch-b2-particle-diffusion)).

**5.** One arteriole $8\eta L/\pi R^4 = 1.8 \times 10^{15}\,\mathrm{Pa}\,\mathrm{s}/\mathrm{m}^{3}$; $10^8$ in parallel: $1.8 \times 10^{7}\,\mathrm{Pa}\,\mathrm{s}/\mathrm{m}^{3}$; $\Delta P = 1.8 \times 10^7 \times 8.3 \times 10^{-5} =
1.5\,\mathrm{kPa} = 11\,\mathrm{mmHg}$.

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

**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: $\mathrm{Re} \approx 6500$, 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](#def-b2-viscous-flows-viscosity) rises (the opposite in the fast aorta); all the pressure-drop estimates for small vessels are lower bounds.

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

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

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

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

**14.** $\Delta P = 32\eta L\langle v\rangle/D^2 = 32 \times 0.02 \times 1.3 \times 10^6
\times 2.65/1.44 = 15\,\mathrm{bar}$: 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.** $\mathrm{Re} = 1350$: laminar; Poiseuille: $\Delta P = 32 \times 2 \times 1.3 \times
10^6 \times 2.65/1.44 = 1500\,\mathrm{bar}$ — impossible; the oil must be kept warm, hence the insulation and heating.

**16.** $1.3 \times 10^6/2.65 = 4.9 \times 10^{5}\,\mathrm{s} = 5.7\,\mathrm{days}$.

**17.** $\tau_w = (\Delta P/L)D/4 = 41 \times 0.3 = 12\,\mathrm{Pa}$; $F = \tau_w\pi DL =
6 \times 10^{7}\,\mathrm{N}$.

**18.** $\Delta P \propto \lambda\langle v\rangle^2/D \propto D^{-4.75}$: $1.2^{4.75} = 2.4$.

**19.** $v = 2R^2\Delta\rho\,g/9\eta = 2 \times 2.5 \times 10^{-9} \times 1800 \times 9.81/
0.18 = 0.49\,\mathrm{mm}/\mathrm{s}$; $\mathrm{Re} = 2 \times 10^{-3}$; $41\,\mathrm{min}$ to cross $1.2\,\mathrm{m}$. 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.** $v = 2 \times 2.5 \times 10^{-11} \times 1000 \times 9.81/1.62 \times 10^{-4} =
3.0\,\mathrm{mm}/\mathrm{s}$; $\mathrm{Re} = 2 \times 10^{-3}$; $1\,\mathrm{km}$ in $3.3 \times 10^5$ s, four days: any updraft above a few millimetres per second holds it.

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

**22.** Stokes would give $120\,\mathrm{m}/\mathrm{s}$ and $\mathrm{Re} \sim 10^4$: absurd. Quadratic drag, $\tfrac43\pi R^3\rho_wg = \tfrac12\rho v^2\pi R^2C_x$: $v = \sqrt{8R
\rho_wg/3\rho C_x} = 6.6\,\mathrm{m}/\mathrm{s}$ ($\mathrm{Re} = 900$, consistent); $150\,\mathrm{s}$ from $1\,\mathrm{km}$.

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

**24.** $2\gamma/R$: $29\,\mathrm{kPa}$, $1.4\,\mathrm{kPa}$, $144\,\mathrm{Pa}$; [dynamic pressures](https://one-course.com/books/physics/4/en/chapter/3-perfect-fluids-euler-and-bernoulli#thm-b2-euler-bernoulli-bernoulli) $\tfrac12\rho v^2$: $5 \times 10^{-6}$, $0.3$ and $26\,\mathrm{Pa}$. The ratio reaches $0.2$ at $2\,\mathrm{mm}$; at $5\,\mathrm{mm}$ ($v \approx 9\,\mathrm{m}/\mathrm{s}$, $\tfrac12\rho v^2 =
49\,\mathrm{Pa}$ against $58\,\mathrm{Pa}$) the aerodynamic pressure flattens the drop and breaks it: no raindrop exceeds about $5\,\mathrm{mm}$.

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