06 - Ideal Fluids

TARGET DECK: MED::I::Medical Physics::06 - Ideal Fluids

Fluid Mechanics – Ideal Fluids

Prof. Enrico Giampieri — DIMES

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Fluids — Static

Ideal Fluids

Definition:

Fluid Fluids are systems in the liquid or gas phase. They have no proper shape but a proper volume — they acquire the shape of their container, opposing no resistance to shear stress.

• Ideal fluid: incompressible, with null viscosity (no internal friction)
• Real fluid: partly compressible and viscous
• Fluid characteristics arise from short-range intermolecular forces, capable of transmitting pressure in all directions
• Gases have negligible intermolecular interactions (compressible); momentum is transmitted only via particle collisions

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An ideal fluid is {1:incompressible} and has {1:null viscosity}.

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Fluids transmit pressure in {1:all directions} due to {1:short-range intermolecular forces}.

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Intensive and Extensive Quantities

When describing fluids, we prefer intensive quantities because fluid systems often lack clear fixed boundaries.

Type	Definition	Examples
Extensive	Additive for subsystems	Mass, volume, linear momentum, energy
Intensive	Independent of system size	Density, pressure, temperature

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An {1:extensive} property is additive for subsystems (e.g., mass, volume), while an {1:intensive} property does not depend on system size (e.g., density, pressure).

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Density

Fluid mass is distributed continuously. Density is defined as:

$\rho =\frac{m}{V}$

$[\rho ]=\text{M}\cdot {\text{L}}^{-3}\Rightarrow \text{SI unit: }kg/{\mathrm{m}}^3$

Material	$\rho (kg/{\mathrm{m}}^3)$
Ether	736
Ethylic alcohol	791
Acetone	792
Benzene	809
Methylic alcohol	810
Water	1000
Blood	1050
Mercury	13600

of water is {1:1000} $kg/{\mathrm{m}}^3$, while mercury is {1:13600} $kg/{\mathrm{m}}^3$.

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Pressure

Definition

$P=\frac{F_n}{S}$

where $F_n$ is the force orthogonal to the surface and $S$ is the surface area.

$[P]=N/{\mathrm{m}}^2\Rightarrow 1N/{\mathrm{m}}^2=1\mathrm{Pa}(\mathrm{Pascal})$

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efined as $P=\frac{F_n}{S}$, measured in {1:Pascal (Pa)}, equivalent to {1:$N/{\mathrm{m}}^2$}.

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Pascal’s Principle

Pascal's Principle When pressure is applied to a static fluid, it propagates uniformly in every direction at every point of the fluid.

• If a pressure difference existed, there would be a net force causing fluid movement until pressure equalized.
• Clinical example: snorkelling decompression — increased lung pressure (vertical) is transmitted to the ear canal (horizontal).

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Pascal’s principle states that pressure applied to a static fluid is transmitted {1:uniformly in every direction}.

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Equilibrium Between Pressures

Given $|F_{P_M}|=12,000\mathrm{N}$, $A_2=0.1{\mathrm{m}}^2$, find $F_1$ for $A_1=0.002{\mathrm{m}}^2$:

$P_1=P_2\Rightarrow \frac{F_1}{A_1}=\frac{F_2}{A_2}\Rightarrow F_1=F_2\cdot \frac{A_1}{A_2}$

Energy Conservation?

The work done is conserved: a smaller force over a larger displacement equals a larger force over a smaller displacement.
$\Delta h_1\cdot A_1=\Delta h_2\cdot A_2\Rightarrow \text{Work is the same!}$

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Pressure Dependence on Fluid Weight (Hydrostatic Pressure)

$F_P=m_L\cdot g\Rightarrow P=\rho \cdot g\cdot h$

where $m_L={\rho }_L{V}_L$, and $V_L=S\cdot h$.

Important

• In a static fluid subject to gravity, pressure is equal at all points on a horizontal plane.
• The pressure difference between two points at different heights equals the weight of a fluid column of unit cross-section with height equal to the height difference.

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ressure is given by $P=\rho \cdot g\cdot h$, where $\rho$ is {1:fluid density}, $g$ is {1:gravitational acceleration}, and $h$ is {1:depth}.

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Torricelli’s Experiment

$P_A={\rho }_{Hg}\cdot g\cdot h$

$P_{atm}=13600\frac{\mathrm{kg}}{{\mathrm{m}}^3}\times 9.81\frac{\mathrm{m}}{{\mathrm{s}}^2}\times 760\times 10^{-3}\mathrm{m}\approx 101,080\mathrm{Pa}$

$1\mathrm{atm}=760\mathrm{mmHg}$

Torricelli's device was the first instrument to measure pressure. Key insight: pressure measurement = height measurement.

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Pressure Measurement Units

Unit	Conversion
Pascal (Pa)	SI base unit
Torr (mmHg)	133.3 Pa
Baria (CGS, dyn/cm²)	$10^{-1}$ Pa
Bar	$10^5$ Pa
Atmosphere (atm)	$1.013\times 10^5$ Pa
mmH₂O	9.81 Pa

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tm} = {1:1.013 \times 10^5}\ \mathrm{Pa} = {1:760}\ \mathrm{mmHg}$.

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Blood Pressure

Clinical Context

• Systolic pressure: maximum at peak of the sphygmotic wave → ~$120\mathrm{mmHg}$
• Diastolic pressure: minimum between peaks → ~$80\mathrm{mmHg}$
• Mean pressure at heart level: $\approx 100\mathrm{mmHg}$

In a supine position: average arterial pressure $\approx$ cardiac pressure $=100\mathrm{mmHg}$.

In an upright position, hydrostatic effects shift pressure:

$P_{\text{head}}=P_{\text{heart}}-{\rho }_{\text{blood}}\cdot g\cdot h=100\mathrm{mmHg}-39\mathrm{torr}=61\mathrm{mmHg}$

(head ~50 cm above heart)

$P_{\text{foot}}=P_{\text{heart}}+{\rho }_{\text{blood}}\cdot g\cdot h=100\mathrm{mmHg}+100\mathrm{mmHg}=200\mathrm{mmHg}$

(feet ~130 cm below heart)

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person, blood pressure at the feet (~130 cm below the heart) is approximately {1:200} mmHg, due to the added {1:hydrostatic} pressure of the blood column.

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Systolic pressure is {1:120} mmHg and diastolic pressure is {1:80} mmHg.

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Ideal Fluid Motion

Flow Rate

$Q=\frac{\Delta V}{\Delta t}$

Volume in a conduit: $V=S\cdot \Delta l$, therefore:

$Q=\frac{\Delta V}{\Delta t}=S\cdot \frac{\Delta l}{\Delta t}=S\cdot v$

For an ideal fluid, velocity is uniform across any cross-section of the conduit.

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defined as $Q=\frac{\Delta V}{\Delta t}=S\cdot v$, where $S$ is {1:cross-sectional area} and $v$ is {1:fluid velocity}.

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Conservation of Flow Rate (Continuity Equation)

An ideal fluid is incompressible → mass and volume are conserved during motion.

$V_1=V_2\Rightarrow S_1\cdot v_1=S_2\cdot v_2$

Continuity Equation

$S_1{v}_1=S_2{v}_2$
In narrower tube sections, speed increases(neglecting friction).

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y equation states $S_1{v}_1=S_2{v}_2$: in a narrower section of a tube, fluid velocity {1:increases}.

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Work of a Fluid

$\Delta L=F\cdot \Delta l=P\cdot S\cdot \Delta l=P\cdot V$

Fluid work is better expressed in terms of pressure and volume rather than force and displacement.

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Power of a Pump

$W=\frac{\Delta L}{\Delta t}=P\cdot \frac{\Delta V}{\Delta t}=P\cdot Q$

Power depends on pressure on the fluid and on the flow rate.

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a pump is $W=P\cdot Q$, where $P$ is {1:pressure} and $Q$ is {1:flow rate}.

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Heart as a Pump

Clinical Application

• Cardiac output: $\approx 6L/min=10^{-4}{\mathrm{m}}^3/s$
• Mean pressure: $100\mathrm{mmHg}=1.3\times 10^4\mathrm{Pa}$
• Average power: $W=P\cdot Q\approx 1.3\mathrm{W}$

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conditions, the heart pumps approximately {1:6 litres per minute}, with an average mechanical power of {1:1.3 W}.

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Bernoulli’s Theorem

Derivation

In an ideal fluid with no friction, only conservative forces act → total mechanical energy is conserved.

Energy contributions:

• Kinetic energy: $E_C=\frac{1}{2}m{v}^2$
• Gravitational potential energy: $U_G=mgh$
• Work by pressure forces: $\Delta L=P\cdot V$

$L=p_1{V}_1-p_2{V}_2+mg{h}_1-mg{h}_2=\frac{1}{2}m{v}_1^2-\frac{1}{2}m{v}_2^2$

Rearranging (conservation of total mechanical energy):

$p_1{V}_1+mg{h}_1+\frac{1}{2}m{v}_1^2=p_2{V}_2+mg{h}_2+\frac{1}{2}m{v}_2^2$

Dividing by volume $V$:

$\boxed{P+\rho gh+\frac{1}{2}\rho v^2=\text{const}}$

Bernoulli's Theorem

$P+\rho gh+\frac{1}{2}\rho v^2=\text{const}$
This is a direct application of conservation of mechanical energy for ideal fluids.

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heorem states $P+\rho gh+\frac{1}{2}\rho v^2=\text{const}$, which is a consequence of conservation of {1:mechanical energy} in ideal fluids.

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s equation, as fluid velocity increases, pressure {1:decreases} (assuming constant height).

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Key Consequences

• Flow rate is conserved ($S_1{v}_1=S_2{v}_2$)
• Energy is conserved along a streamline
• Speed depends only on the tube cross-section
• Pressure $P$ varies with height and tube size

Virtual Pipes

Bernoulli’s theorem applies not only to physical pipes but also to free fluid flows: ocean currents, air streams, etc. Each streamline can be treated as its own virtual pipe.

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Applications

Torricelli’s Theorem

Applying Bernoulli between the surface and the hole at the base of a tank:

• Upper surface: $v\approx 0$, $U=mgh$
• Lower hole: $v=v_{max}$, $U=0$

$\rho gh=\frac{\rho v^2}{2}\Rightarrow v=\sqrt{2gh}$

Mnemonic

“Falling fluid” — the exit velocity equals that of a freely falling object dropped from height $h$: same formula as $v=\sqrt{2gh}$ from kinematics!

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theorem gives the efflux velocity as $v=\sqrt{2gh}$, which is identical to the velocity of {1:a freely falling object} dropped from height $h$.

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Venturi Effect

When the conduit section is reduced (at constant height, $h_1=h_2$):

$P_1+\frac{1}{2}\rho v_1^2=P_2+\frac{1}{2}\rho v_2^2$

$P_1-P_2=\frac{1}{2}\rho (v_2^2-v_1^2)$

Using continuity ($v_2=v_1\cdot \frac{S_1}{S_2}$):

$\Delta P=\frac{1}{2}\rho v_1^2\left[{\left(\frac{S_1}{S_2}\right)}^2-1\right]>0$

$\therefore P_1>P_2$

Venturi Effect

Reducing the cross-section of a conduit increases fluid velocity and decreases pressure in that region.

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ffect: in a constriction, fluid velocity {1:increases} and pressure {1:decreases}.

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Archimedes’ Principle

Stevino’s Law (Hydrostatic Pressure Difference)

From Bernoulli at static conditions ($v_1=v_2=0$):

$P_1+\rho g{h}_1=P_2+\rho g{h}_2$

$\Delta P=\rho g\Delta h$

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: $\Delta P=\rho g\Delta h$, meaning pressure in a fluid {1:increases linearly with depth}.

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Derivation of Archimedes’ Force

Using the solidification principle (a fluid element in equilibrium):

• Pressure increases with depth (Stevino’s law)
• Pressure is perpendicular to the surface (Pascal)
• Upper surface: $P_1=\rho g{h}_1$ (downward)
• Lower surface: $P_2=\rho g{h}_2$ (upward)

$\Delta P={\rho }_{L}g(h_2-h_1)\Rightarrow F_A={\rho }_{L}Vg$

$\boxed{F_A={\rho }_{L}Vg}$

Archimedes' Principle

The buoyant force equals the weight of the displaced fluid: $F_A={\rho }_{L}Vg$

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Sinking vs. Floating

Condition	Behaviour
${\rho }_S>{\rho }_L$	Object sinks
${\rho }_S={\rho }_L$	Neutral buoyancy
${\rho }_S<{\rho }_L$	Object floats

Net force:

$\Delta F=({\rho }_C-{\rho }_L)Vg$

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orce is $F_A={\rho }_{L}Vg$. An object floats if its density is {1:less than} that of the fluid.

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Floating Equilibrium

When an object floats, it submerges until the Archimedes force equals its weight:

$\Delta F=0\Rightarrow m_{C}g=V_{IMM}{\rho }_{L}g$

$\frac{V_{IMM}}{V_C}=\frac{{\rho }_C}{{\rho }_L}$

Example

An iceberg with ${\rho }_{ice}\approx 917kg/{\mathrm{m}}^3$ in seawater (${\rho }_{sw}\approx 1025kg/{\mathrm{m}}^3$):
about 89% of its volume is submerged.

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g object, the fraction of volume submerged is $\frac{V_{IMM}}{V_C}=\frac{{\rho }_C}{{\rho }_L}$, showing that a denser object {1:sinks deeper}.

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Exercises

Pressure at the Bottom of a Pool

Problem: A pool of $24\mathrm{m}\times 24\mathrm{m}$, depth $2.4\mathrm{m}$, filled completely with water. What is the pressure at the bottom? What if filled with mercury (${\rho }_{Hg}=1.36\times 10^{4}kg/{\mathrm{m}}^3$)?

Solution:

$P_{bottom}=P_{surface}+{\rho }_{w}gh$

$P_{bottom}=10^5+10^3\times 9.8\times 2.4\approx 1.24\times 10^5\mathrm{Pa}$

For mercury:

$P_{bottom}=10^5+1.36\times 10^4\times 9.8\times 2.4\approx 4.2\times 10^5\mathrm{Pa}$

The pool dimensions ( $24\mathrm{m}\times 24\mathrm{m}$) are irrelevant — pressure at the bottom depends only on depth $h$ and fluid density $\rho$.

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at the bottom of a fluid column depends on {1:depth and fluid density}, not on the {1:surface area} of the container.

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TLDR

Summary — Fluid Mechanics: Ideal Fluids

• Ideal fluid: incompressible, non-viscous; real fluids are partly compressible and viscous
• Density: $\rho =m/V$, SI unit $kg/{\mathrm{m}}^3$
• Pressure: $P=F_n/S$, SI unit Pa ($N/{\mathrm{m}}^2$); increases with depth as $P=\rho gh$
• Pascal’s principle: pressure applied to a static fluid is transmitted uniformly in all directions
• Pressure units: $1\mathrm{atm}=760\mathrm{mmHg}=1.013\times 10^5\mathrm{Pa}$; $1\mathrm{bar}=10^5\mathrm{Pa}$
• Blood pressure: systolic ~120 mmHg, diastolic ~80 mmHg; mean cardiac pressure ~100 mmHg; hydrostatic effects shift pressure in upright position (~61 mmHg at head, ~200 mmHg at feet)
• Flow rate: $Q=S\cdot v$; continuity equation: $S_1{v}_1=S_2{v}_2$ (faster in narrower sections)
• Pump power: $W=P\cdot Q$; heart ~1.3 W at ~6 L/min
• Bernoulli’s theorem: $P+\rho gh+\frac{1}{2}\rho v^2=\text{const}$ — conservation of mechanical energy along a streamline
• Torricelli’s theorem: efflux velocity $v=\sqrt{2gh}$ (same as free-fall kinematics)
• Venturi effect: reducing cross-section → increased velocity, decreased pressure; $\Delta P=\frac{1}{2}\rho v_1^2\left[{\left(\frac{S_1}{S_2}\right)}^2-1\right]$
• Stevino’s law: $\Delta P=\rho g\Delta h$ (hydrostatic pressure difference)
• Archimedes’ principle: buoyant force $F_A={\rho }_{L}Vg$ = weight of displaced fluid; object floats if ${\rho }_C<{\rho }_L$; submerged fraction = ${\rho }_C/{\rho }_L$