Fluids in Motion

Fluids in Motion

Part 1 of 2 — Fluid Dynamics: Continuity & Bernoulli

Where this fits. The previous lesson, Fluids at Rest, covered density, pressure, buoyancy and surface tension. This lesson sets the fluid moving: continuity and Bernoulli for ideal flow, then viscosity, Poiseuille's law and blood flow. The next lesson, Waves & Interference, turns from fluids to oscillations and waves.

Throughout this part we assume an ideal fluid, with three properties (Part 2 drops "non-viscous" to handle real blood flow):

• incompressible: its density stays constant;
• non-viscous: no internal friction between layers of fluid;
• laminar (and steady): it flows in smooth layers that do not mix, with the pattern unchanging in time.

Flow Rate

The volume flow rate $Q$ is the volume of fluid passing a cross-section per unit time:

$Q = Av$

where $A$ is the cross-sectional area and $v$ is the fluid's average speed. Why $Av$: in a time $t$, a slug of fluid of length $vt$ passes the cross-section, which is a volume $A(vt)$; divide by $t$.

• Units: $\text{m}^3/\text{s}$ in SI; mL/min or L/min in medicine (resting cardiac output is about 5 L/min).
• Conversions: $1\;\text{mL} = 1\;\text{cm}^3$; $1\;\text{L} = 1000\;\text{cm}^3 = 10^{-3}\;\text{m}^3$. Convert per minute to per second before mixing with speeds in m/s or cm/s.
• Mass flow rate is $\rho Q$, in kg/s.
• Flow RATE is not SPEED. $Q$ is volume per time; $v$ is distance per time. A wide, slow river carries far more water per second than a narrow, fast stream.

Continuity: Mass In = Mass Out

Fluid cannot pile up inside a rigid, leak-free pipe, so the mass entering each second equals the mass leaving. For an incompressible fluid the density is the same everywhere, so the volume flow rate is the same everywhere too:

$A_1 v_1 = A_2 v_2$

• Narrower means faster: $v \propto 1/A$.
• Circular pipes: $A = \pi r^2$, so $v \propto 1/r^2$. Halve the radius (or the diameter) and the area falls to one-quarter, so the speed rises fourfold.
• Branching: when a pipe splits, the branch flows add up to the parent flow: $A_0 v_0 = A_1 v_1 + A_2 v_2 + \dots$ The speed in identical branches is set by their TOTAL area. If the branches together have more area than the parent, the fluid slows down, even though each branch is narrower.
• Continuity links geometry to speed only. It says nothing about pressure; that is Bernoulli's job.
• Gases are compressible, so for them the conserved quantity is $\rho A v$. On the MCAT, liquids and blood are treated as incompressible.

Partly covering the end of a garden hose with a thumb shrinks the exit area, and the water leaves much faster: continuity in action.

Practice: Flow Rate & Continuity 🎯

Worked Example — Why Capillary Blood Crawls

Problem. The aorta (cross-section about $3\;\text{cm}^2$) carries blood at about $30\;\text{cm/s}$. The capillary bed's TOTAL cross-section is about $3000\;\text{cm}^2$.

• (a) What flow rate does the heart send into the aorta, and is it a sensible cardiac output?
• (b) How fast does blood move through the capillaries?
• (c) Each capillary is far narrower than the aorta. Why is the blood slower there, not faster?

(a) Flow rate, set by the heart. $Q = Av = (3)(30) = 90\;\text{cm}^3/\text{s}$. Sanity check: $90 \times 60 = 5400\;\text{cm}^3/\text{min} \approx 5.4\;\text{L/min}$, a normal cardiac output.

(b) Continuity across the whole bed. With $Q = 90\;\text{cm}^3/\text{s}$ from (a), $v_{\text{cap}} = \dfrac{Q}{A_{\text{total}}} = \dfrac{90}{3000} = 0.03\;\text{cm/s}$. That is 1000 times slower than in the aorta, in exactly the place where slow flow helps exchange.

(c) Interpret. Any single capillary IS far narrower than the aorta, but continuity applies to the summed area of all the parallel paths. "Narrow vessel, so fast blood" is single-tube thinking applied to a branching network.

Bernoulli: Energy Conservation for Ideal Fluids

Along a streamline (the path followed by one parcel of fluid) in an ideal fluid,

$P + \tfrac{1}{2}\rho v^2 + \rho g h = \text{constant}$

Every term is an energy per unit volume ($\text{J/m}^3$, which is the same unit as Pa):

Here $h$ is HEIGHT above a chosen reference level, so pressure falls as the fluid rises. (In the Fluids at Rest lesson $h$ was depth.) Bernoulli is simply conservation of energy, which is why it needs the ideal-fluid assumptions: with viscosity, energy leaks away as heat.

• Fluid at rest: $v = 0$, so $P + \rho g h$ is constant, which is the hydrostatic result from the Fluids at Rest lesson.
• Horizontal flow: the $\rho g h$ terms cancel, leaving $P + \tfrac{1}{2}\rho v^2 = \text{constant}$. Where the fluid moves faster, its pressure is lower.
• Why it feels backwards. Fast fluid hits hard when something STOPS it, but the pressure it exerts on the walls alongside it is lower. In fact, the fluid speeds up into a narrowing BECAUSE the pressure behind it is higher: the pressure difference is what accelerates it.
• The standard chain: area down, so speed up (continuity), so pressure down (Bernoulli).

Venturi effect and the Venturi meter. Pressure drops in a constriction. A Venturi meter is a horizontal pipe with a narrow throat; open vertical tubes above the wide section and the throat show the pressure difference as a difference in liquid heights ($\Delta P = \rho g \Delta h$, from the Fluids at Rest lesson). From $\Delta P$ and the two areas, the flow speed follows.

The same effect lets an atomizer or aspirator pull liquid up a tube: a fast air stream across the tube's top lowers the pressure there.

Torricelli's theorem. Liquid leaks from a small hole a depth $h$ below the open surface of a wide tank. The surface and the jet are both at atmospheric pressure, and the surface barely moves, so Bernoulli reduces to

$v = \sqrt{2gh}$

the speed of an object dropped from height $h$: each parcel cashes in the potential energy of its fall. Twice the depth gives $\sqrt{2}$ times the speed, not twice.

Pitot tube. A pitot tube points straight into a flow. Fluid entering its open tip is brought to rest there, so the pressure at the tip (the stagnation pressure) is $P + \tfrac{1}{2}\rho v^2$, where $P$ is the ordinary (static) pressure of the moving fluid. Measuring both gives $\Delta P = \tfrac{1}{2}\rho v^2$, so $v = \sqrt{2\Delta P/\rho}$.

Lift. A wing is shaped and tilted so that air flows faster over its top than beneath it. Faster flow on top means lower pressure on top, and the pressure difference times the wing's area is an upward force.

• Aircraft airspeed indicators are pitot-static probes, and hospital "Venturi masks" use the Venturi effect to draw in room air and deliver a fixed oxygen concentration.

Worked Example — A Venturi Meter (Continuity, Then Bernoulli)

Problem. Water ($\rho = 1000\;\text{kg/m}^3$) flows at $1\;\text{m/s}$ through a horizontal pipe of cross-section $10\;\text{cm}^2$ that narrows to $2\;\text{cm}^2$. Use $g = 10\;\text{m/s}^2$.

• (a) How fast does the water move in the narrow section?
• (b) How much lower is the pressure in the narrow section?
• (c) By how much do the water levels in the meter's two open tubes differ?
• (d) What is the volume flow rate through the pipe?

(a) Continuity for the new speed. $v_2 = v_1\dfrac{A_1}{A_2} = (1)\dfrac{10}{2} = 5\;\text{m/s}$

(b) Bernoulli for the pressure change. The pipe is horizontal, so the $\rho g h$ terms cancel. With $v_2 = 5\;\text{m/s}$ from (a):

$P_1 - P_2 = \tfrac{1}{2}\rho(v_2^2 - v_1^2) = \tfrac{1}{2}(1000)(25 - 1) = 12{,}000\;\text{Pa}$

The narrow, fast section is LOWER in pressure by $12\;\text{kPa}$. Squaring the difference instead, $(5 - 1)^2 = 16$, would give a wrong $8\;\text{kPa}$.

(c) Read it on the meter. With $\Delta P = 12{,}000\;\text{Pa}$ from (b), $\Delta h = \dfrac{\Delta P}{\rho g} = \dfrac{12{,}000}{(1000)(10)} = 1.2\;\text{m}$, lower over the throat.

(d) Flow rate. $Q = A_1 v_1 = (10 \times 10^{-4}\;\text{m}^2)(1\;\text{m/s}) = 1.0 \times 10^{-3}\;\text{m}^3/\text{s}$, or 1 L/s, the same in both sections. Order matters: continuity first (geometry to speed), Bernoulli second (speed to pressure).

Worked Example — Torricelli's Tank

Problem. A wide open drum of water has a small hole of area $1.0\;\text{cm}^2$, $5\;\text{m}$ below the water surface. Use $g = 10\;\text{m/s}^2$.

• (a) How fast does the water leave the hole?
• (b) How much water flows out each second?
• (c) As the tank drains, how fast does the water leave when the depth above the hole is $1.25\;\text{m}$?

(a) Exit speed. $v = \sqrt{2gh} = \sqrt{2(10)(5)} = \sqrt{100} = 10\;\text{m/s}$, the same speed a stone reaches after falling 5 m.

(b) Flow out. With $v = 10\;\text{m/s}$ from (a), $Q = Av = (1.0 \times 10^{-4}\;\text{m}^2)(10\;\text{m/s}) = 1.0 \times 10^{-3}\;\text{m}^3/\text{s}$, about 1 L/s.

(c) As the tank drains. $v = \sqrt{2(10)(1.25)} = \sqrt{25} = 5\;\text{m/s}$: a quarter of the depth gives half of (a)'s speed. The exit speed and the flow both fall as the depth above the hole shrinks.

Practice: Bernoulli, Venturi & Torricelli 🎯

Key Takeaways

• Flow rate is not speed. $Q = Av$ is volume per time; $1\;\text{mL} = 1\;\text{cm}^3$, and convert minutes to seconds.
• Continuity. $A_1v_1 = A_2v_2$: half the radius means four times the speed, and branches share the flow by TOTAL area.
• Bernoulli. $P + \tfrac{1}{2}\rho v^2 + \rho g h$ is constant, so at one height faster fluid has lower pressure (Venturi, lift).
• Torricelli and pitot. A tank leaks at $v = \sqrt{2gh}$; a pitot tube reads stagnation minus static pressure $= \tfrac{1}{2}\rho v^2$.

Common traps

• Fast means high pressure. Backwards: at the same height, faster fluid has LOWER pressure on the walls beside it.
• Unsquared radius. Speed goes as $1/r^2$, so half the radius gives four times the speed, not twice.
• Squaring the difference. Bernoulli uses $v_2^2 - v_1^2$, never $(v_2 - v_1)^2$.
• Missing square root. In $v = \sqrt{2gh}$, twice the depth gives $\sqrt{2}$ times the speed, not twice.

Low-yield extras

• aircraft pitot-static airspeed indicators; Venturi oxygen masks

Fluids in Motion

Part 2 of 2 — Viscosity, Poiseuille & Blood Flow

Part 1 of this lesson assumed a fluid with no internal friction. Real fluids, blood included, have friction, and that friction is why the heart has to keep pumping.

Viscosity: When Fluids Are Real

Viscosity $\eta$ (Greek "eta") is a fluid's internal friction: its resistance to layers of fluid sliding past one another. Its SI unit is the pascal-second (Pa·s). Honey is far more viscous than water.

• Blood is about 3 to 4 times as viscous as water. Its viscosity rises with hematocrit, the fraction of blood volume occupied by red cells: polycythemia (too many red cells) thickens blood, and anemia (too few) thins it.
• A liquid's viscosity falls as it warms: warm syrup pours easily.
• Viscosity costs energy. Friction between layers turns mechanical energy into heat, so pressure falls steadily along even a uniform, horizontal tube. Bernoulli's equation cannot predict that drop, because it assumes no energy is lost. A pressure difference is needed just to keep a viscous fluid moving at constant speed.
• Velocity profile. In smooth flow through a tube, the fluid touching the wall is at rest and the fluid at the centre moves fastest, giving a bullet-shaped (parabolic) profile. The "speed" in $Q = Av$ is the average across the tube.

• Turbulence becomes more likely with higher speed, larger diameter, higher density and lower viscosity. These four factors combine into one dimensionless number, the Reynolds number $Re = \rho v D/\eta$, where $D$ is the tube diameter: the larger $Re$, the more likely turbulence, and flow turns turbulent above a critical value. Once it does, neither Bernoulli's equation nor Poiseuille's law holds.
• Blood flow is mostly laminar. Turbulence appears where blood moves fast: past a narrowed valve or vessel, in the aorta during heavy exercise, and in severe anemia (thin blood plus a high cardiac output). It is heard with a stethoscope as a murmur over the heart or a bruit (a whooshing sound) over an artery.
• A blood-pressure cuff uses this. When the cuff pressure lies between systolic and diastolic pressure, the squeezed artery snaps open briefly with each beat and the jet is turbulent, making tapping sounds. As the cuff deflates, the sounds begin when cuff pressure falls to systolic and disappear when it falls below diastolic, when flow becomes smooth again.

• In a straight tube, flow usually turns turbulent above $Re \approx 2000$. The cuff sounds are called Korotkoff sounds.

Poiseuille's Law

For laminar flow of a viscous fluid through a long, rigid, cylindrical tube of radius $r$ and length $L$:

$Q = \frac{\pi \, \Delta P \, r^4}{8 \eta L}$

where $\Delta P$ is the pressure difference between the tube's ends (the driving pressure). Written like Ohm's law, $Q = \Delta P / R$, with the tube's resistance

$R = \frac{8\eta L}{\pi r^4}$

• The fourth power of the radius is the tested fact. A radius cut of only about 16% halves the flow, because $0.84^4 \approx 0.5$.
• If the flow must stay the same (the heart maintaining output), the driving pressure must rise in proportion to the resistance.
• Do not confuse Poiseuille with continuity. Continuity relates speed and area when $Q$ is FIXED; Poiseuille tells you what $Q$ will BE for a given pressure difference and tube.
• Assumptions: laminar, steady flow of a fluid with constant viscosity through a rigid tube. Vessels flex and blood is a suspension, but the $r^4$ conclusion holds well enough to be the MCAT's favourite.

Practice: Viscosity & Poiseuille's Law 🎯

The Circulation as a Fluids Problem

The MCAT rarely asks about pipes; it asks about blood vessels. The physiology is in Part 2 of the Cardiovascular System lesson; here is the physics behind it.

• $Q = \Delta P/R$ for the whole circuit. For the systemic circulation, $Q$ is the cardiac output (CO, about 5 L/min at rest), $\Delta P$ is the mean arterial pressure (MAP, the average pressure in the aorta, about 93 mmHg) minus the near-zero pressure in the right atrium, and $R$ is the total peripheral resistance (TPR) of all the systemic vessels. So $MAP \approx CO \times TPR$.
• Series and parallel. Blood passes through vessel types in SERIES (artery, arteriole, capillary, venule, vein): their resistances add, and the same flow $Q$ passes through each level. The organs branch off the aorta in PARALLEL: $1/R_{\text{total}} = 1/R_1 + 1/R_2 + \dots$, so the total is lower than any single organ's resistance. Opening more beds lowers TPR; closing or removing one raises it.
• Total cross-sectional area sets speed. Continuity applies to the SUMMED area at each level (worked in Part 1 of this lesson). Each capillary is microscopic, but there are billions of them, so the capillaries have by far the largest total area and the slowest blood (giving time for exchange). The aorta has the smallest total area and the fastest blood, and blood speeds up again as veins merge.
• Where pressure falls. The largest pressure drop is across the arterioles (from about 90 to about 35 mmHg). Each capillary is narrower than an arteriole, but so many capillaries sit in parallel that their combined resistance is lower.
• Why radius control is so powerful. Arteriole walls contain smooth muscle, and because $R \propto 1/r^4$, a small change in radius makes a large change in resistance. Constricting one organ's arterioles also diverts flow to the other parallel beds.
• Viscosity counts too. A higher hematocrit raises $\eta$, raising resistance and the heart's workload.
• Gravity still applies. Standing adds $\rho g h$ to pressures below the heart, as in the Fluids at Rest lesson.
• Air in the lungs. Airflow obeys the same $Q = \Delta P/R$ logic: the pressure difference between the alveoli and the atmosphere drives air through the airways, and airway radius dominates resistance through $r^4$. Narrowing of the airways in asthma therefore sharply cuts airflow. Passages supply the details; reason with $r^4$.

• In the lungs, the medium-sized bronchi (not the tiniest airways) have the highest total resistance, for the same parallel-numbers reason. Blood's apparent viscosity also falls in very narrow vessels (the Fåhræus–Lindqvist effect).

Grading a Stenosis: Which Law Applies?

A stenosis is an abnormal narrowing of a vessel or valve. Three laws answer three different questions about it.

• Continuity applies whenever the flow is known or assumed unchanged, as in a mild stenosis where the body maintains flow.
• Bernoulli fits a SHORT narrowing, where the speed changes quickly and friction in the brief segment is small. It explains the pressure dip in the jet. Downstream, the jet breaks into turbulence, which causes the bruit.
• Poiseuille fits a LONG narrowing, or any question about whether flow can be maintained at a fixed driving pressure. A severe or long stenosis raises resistance steeply, and flow falls.
• The reverse case: an aneurysm (an abnormal widening of an artery). Continuity says blood slows in the wider segment; Bernoulli says slower blood means higher pressure on the wall there; that pressure stretches the wall further, a vicious cycle that lets an aneurysm keep growing.
• They do not contradict each other. Continuity assumes $Q$ stays fixed; Poiseuille tells you what it costs (a larger $\Delta P$) to keep $Q$ fixed, or how far $Q$ falls if $\Delta P$ cannot rise.

Clinicians measure the jet speed with Doppler ultrasound; the wave physics behind that is in the Sound, Doppler & Ultrasound lesson.

Passage Strategy: Pick the Regime First

Then ask three follow-up questions: a single tube or a branching network (use total area)? Is the flow fixed, or the driving pressure? Laminar, or turbulent (where neither equation applies)?

Worked Example — Grading a Stenosis

Problem. A carotid ultrasound shows peak blood speed rising from $60\;\text{cm/s}$ upstream to $240\;\text{cm/s}$ inside a short stenosis. Take blood's density as $1000\;\text{kg/m}^3$ and $1\;\text{mmHg} \approx 133\;\text{Pa}$.

• (a) By how much has the vessel's radius (and diameter) been reduced?
• (b) How much lower is the pressure in the jet, in mmHg?
• (c) If the narrowing were long rather than short, how would each centimetre of it compare in resistance with a normal centimetre of the vessel?
• (d) What happens to the blood just beyond the stenosis, and how is it detected?

(a) Continuity, run backwards. At roughly constant flow, a 4-fold rise in speed means the area has fallen to one-quarter ($A \propto 1/v$). Since $A \propto r^2$, the radius is HALVED: a 50% reduction in diameter.

(b) Bernoulli for the pressure dip. In SI units the speeds are $0.6$ and $2.4\;\text{m/s}$:

$\Delta P = \tfrac{1}{2}\rho(v_2^2 - v_1^2) = \tfrac{1}{2}(1000)(5.76 - 0.36) = 500 \times 5.4 = 2700\;\text{Pa}$

That is $2700/133 \approx 20\;\text{mmHg}$ lower in the jet.

(c) Poiseuille's warning. With the radius halved, from (a), and $R \propto 1/r^4$, each centimetre would have $2^4 = 16$ times the resistance of a normal centimetre of the vessel. "Half the diameter" sounds mild; to a viscous fluid it is severe.

(d) Downstream. The fast jet becomes turbulent beyond the stenosis, and the turbulence is heard as a bruit.

Practice: The Circulation & Passage Reasoning 🎯

Key Takeaways

• Viscosity costs pressure. Internal friction makes pressure fall along even a uniform tube, a drop Bernoulli cannot see.
• Poiseuille's fourth power. $Q = \pi\Delta P r^4/(8\eta L)$ and $R = 8\eta L/(\pi r^4)$: double the radius, 16 times the flow.
• Circulation as a circuit. $MAP \approx CO \times TPR$; parallel beds keep TPR low, and the arterioles take the largest pressure drop.
• Grading a stenosis. Faster (continuity), lower jet pressure (Bernoulli), higher resistance (Poiseuille), then a turbulent bruit.

Common traps

• Fast capillary blood. Capillary blood is the SLOWEST in the circulation, because the capillaries' TOTAL area is the largest.
• Squared versus fourth power. Use $r^2$ for speed at fixed flow (continuity) and $r^4$ for flow at fixed pressure (Poiseuille).
• Bernoulli in long vessels. A long, viscous vessel loses pressure to friction, so Poiseuille applies there, not Bernoulli.
• Parallel as series. Parallel beds total less than any one bed, so removing a bed RAISES TPR rather than lowering it.

Low-yield extras

• the critical Reynolds number (about 2000); the Korotkoff name; airway resistance peaking in the medium bronchi; the Fåhræus–Lindqvist effect.