Course n°: 3rd

Title:“RESPIRATORY PHYSIOLOGY, PHYSICS AND

PATHOLOGY IN RELATION TO ANAESTHESIA AND

INTENSIVE CARE”

Sub-category: PHYSICS AND PRINCIPLES OF MEASUREMENT

Topic: LAW OF PHYSICS

Date: 7TH May 2016

Language: English City: Lahore

Country: Pakistan

Speaker: Dr. Liaqat Ali.

Associate Professor. Kaul Associates / AIMC/JH, Lahore.

Objectives

• Variables of Physics Laws

• Highlights the most essentials laws of physics

• Their clinical implications

Variables of Physics Laws

• Volume: Space occupied by a substance measured in three dimensions by

cubic cm or cubic mm

• Temperature: All gases laws use Kelvin temperatures. It measures heat,

a form of energy, kinetic energy which comes from movement of the

molecules. Its Units = O OC = 32 OF = 273.15 OK.

• Absolute Temp. O OK = 273 OC.

• All gases laws are made at standard temperatura and pressure (STP)

T = O OC P = 760mmHg

• Pressure: Force per unit area.

• P = F/A.

• Units, I Barr= I Atmosphere=101 kpa=760mmHg=14.7psi = 1000cm H2O

• Gauge Pressure = Content Pressure & Atmospheric Pressure

• Absolute Pressure = Content pressure

Boyle’s Law

• At constant temperature the volume of a mass of gas varies

inversely with its absolute pressure.

• V α 1/P or VP= K 1 Constant or

• Same number of molecules at the same temperature occupying

different volumes will exert different pressures. However, their

product will be the same. So V1 P1 = V2 P2 = V3 P3

• P1 = Gauge pressure of cylinder

• V1 = Physical volume filled by water depending upon size of cylinder.

• P2 = Atmospheric pressure

• V2 = Actual amount of gas stored in the cylinder to be delivered.

Application of Boyle’s law in Anaesthetic practice.

• Medical gases are stored in large amounts in cylinders.

• The volume of gas present in a clynder can be measured by this law.

• P1 and P2 should have same units e.g. Bars, kPa

• Example; if V1 is 10 liter, P1 is 137 Bar (gauge pressure) & P2 is

atmospheric pressure (100 kPa or 1 Bar or 14.7 psi)

• P1 x V1 = P2 x V2

• V2 = P1 x V1 / P2 = 137 Bars x 10 litres / 1 Bar = 1380 L

Charles Law; At constant pressure the volume of mass of a gas varies

directly with the absolute temperature.

V α T V/T = K 2

Third Perfect Gas Law (Gay Lussacs Law)

• At constant volume the absolute pressure of a mass of gas varies directly

with the temperature

• P α T P/T = K 3 So P1/T1= P2/T2

Application ;

• As cylinders are tested to bear only 210 bar of pressure.

If external temperature rises so that P is > 210 bar cylinder will burst.

• Molybdenum steel can withstand pressures till 210 bars. Weakening of

metal in damaged cylinders are at a greater risk of explosion due to rise

in temperature

Avogadro’s Hypothesis

• Equal volumes of gases, at the same temperature and pressure

contain equal numbers of molecules.

• However, as the molecular weight of different substances is

different; their mass will be different in the same volume.

• A Mole; Is the quantity of any substance containing the same no. of

particles (molecular weight of a substance in grams).

• Avogadro’s no. 1 mole = 6.02 x 1023

• For any gas at as STP, 1 mole =22.4 liters

• So 2 g of Hydrogen, 32 g of Oxygen and 44 g of CO2 occupies

22.4 L as STP

Avogadro’s Hypothesis; Clinical implications

• Calibration of Vaporizer: If a steady stream of oxygen is made to

completely vaporize 19.7g of Halothane into a volume of 224 liters,

The mean percentage concentration of halothane can be ;

• 197 g of halothane is 1 mole and occupies 22.4 L

• 19.7 g ……..0.1 mol…….2.24 L

• 224L of O2 contains 2.24L of Halothane

1L….2.24/224 = 0.01 i.e. 1%

Content of Nitrous Oxide cylinder:

• If N2O cylinder contains 3.4kg of N2O. How much N2O gas will be

available for inhalation at room temperature?

• Mol wt of N2O is 44 so 1 mol of N2O is 44g 44 gm…..22.4 Litres

• 1gmm..22.4/44 3400g (3.4kg)….22.4 / 44 X 3400 = 1730L

• If gauge pressure of N2O cylinder 746 psi content are liquid

• If gauge pressure of N2O cylinder < 746 contents are gas

Universal Gas Constant

• The concepts of perfect Gas law, Avogadros’s hypothesis and mole

can be combined as.

• PV = k1, V / T = k2, P / T = k3

• For one mol of any perfect gas, PV / T = R constant where

• R is (Universal) Molar gas constant.

• PV = RT

• Where n is no of mol of gas, PV = nRT

Clinical implication

• In OT temp & volume of cylinder is constant

P is directly proportional to n ( no. of moles of gas).

Pressure gauge acts as a content gauge.

Dalton’s Law of partial pressures • In a mixture of gases, the pressure exerted by each gas is the same as

that which it would have exerted if it alone occupied the container.

Application: In a cylinder of Entonoux 50 % O2 and 50 % N2O

Pressure exerted by the N2O would be the same as it would exert if

it alone occupied the cylinder but in this case the available space for

N2O will increase from half a cylinder to full cylinder.

• Acc to Boyle’s Law pressure will be 100 kPa to 50 kPa (P ∝ 1 / V)

• Manufacturers use these two laws when filling cylinders with gas

mixtures.

• Cylinder of air at a pressure of 100kPa.

02 = 20.95% will exert a pressure of 100 × 20.95/100 = 20.95kpa

or 760 × 20.95/]00=159.22 mm Hg

N2=79.07% will exert a pressure of 100 × 79.07/100=79.07kpa or

760 × 79.07/100=600.932 mmHg

• Critical Temperature:

• Temperature above which a gas cannot be liquefied by pressure alone.

e.g. N2O = 36.5 C O2 = -119 C

• Critical Pressure:

Pressure required to liquefy a gas at its critical temperature.

e.g. N2O = 73 bar @ 36.5 C while N2O =52 bar@20.0 C

A Gas: a substance in the gaseous phase above its critical temperature.

A Vapour: a substance in the gaseous phase below its critical

temperature.

• Pseudo-Critical Temperature: For a mixture of gases at a specific

pressure, the specific temperature at which the individual gases may

separate from the gaseous phase.

• N2O 50% / O2 50% = - 5.5°C for cylinders (most likely at 117 bar)

N2O 50% / O2 50% = - 30° C for piped gas.

• Poynting Effect (Overpressure Effect): Critical temperature and

critical pressure of one gas may be affected by its admixture with

the other gas.

• Filling Ratio: The mass of the gas in the cylinder divided by the

mass of the water that would fill the cylinder. N2O = 0.65 (UK)

• At room temperature a full N2O cylinder contains about 9/10 of

liquid and 1/10 of gas.

Alveolar Air Equation

• Shows the amount of 02 available in the alveoli.

• It is derived from the Dalton's Law of Partial Pressures.

• Equation states; PAO2 = Fi02 X (PB-PH2O)- PaCO2 / R

• Where R is the respiratory Quotient.

• CO2 production in a normal adult is 200 ml/min,

• while 02 consumption is 250 m]/min. So R is equal to:

• R=200/250=0.8

Derivation of Alveolar Air Equation from

Dalton’s Law of Partial Pressure

According to Dalton's Law

• Pt = P1+P2+P3…………. Pn

• So PB = PAO2 + PAN2 + PAC02 + PH20

• PA02 = PB - PH20 - PAN2 - PACO2

• PAO2 = FiO2 (PB-PH2O)-PAC02/R

• PA02 = P1O2 - PACO2 /R

Effect of Hypoventilation on PAO2

According to Alveolar ventilation equation VA and Pa CO2 have

direct inverse relationship VA = VCO2 / PaCO2 x K

• 1f ventilation is halved; PaC02 is doubled i.e. 80 mmHg.

Let's put this value of PaC02 in alveolar air equation

• PAO2 = PiO2 – PaCO2 / R PAO2 = 150 – 80 / 0.8 So

= 150 – 100 = 50 mmHg

• If PaCO2 is increased to 80, PAO2 will decrease to 50 mm Hg

Calculating O2 Supplementation Requirement

Q. A patient has the PACO2 of 80 mmHg. His PAO2 is decreased

to 50 mm Hg. How much FiO2 we should provide to this patient

to bring his PAO2 to 100 mmHg?

• We have to calculate required FiO2

• For this we first calculate P1O2 Acc to Alveolar air equation

• PAO2= P1O2 - Pa CO2 / R 100 = P1O2 – 80 / 0.8

P102 = 100 + 80 / 0.8 = 100 +100 = 200 mm Hg

• Now required PiO2 is 200 mm Hg

• PiO2 = FiO2 x (PB-PH2O) FiO2 = PiO2 / (PB-PH20)

• FiO2 = 200 / 760 – 47 = 200 / 713 = 0.28

The Ohm’s Law

• The current flow between any two points is directly proportional

to the potential difference between those two points and inversely

proportional to the resistance. I = Δ V / R

• Application in Respiratory Physiology: Flow = ΔP / R

Application of Ohm’s Law in Human Physiology:

• Human physiology; CO, vascular resistance & BP

• Resistance in vessels is not exactly similar to electrical resistance

– Electrical resistance is not affected by the change in upstream

or downstream pressures

– Vascular resistance is changed if the pressure at the arterial or

venous ends changes due to dilatation and recruitment of

vessels at the capillary level

Flow

• Flow is defined as the quantity of a fluid passing a point in a unit

time. F = Q / t Where F=Flow, Q=quantity, and t=Time

Laminar Flow:

Fluid moves in a smooth and steady manner.

• Greatest in the centre (twice the mean flow) and minimum at the

sides of the tube, Parabolic profile.

Hagen Poiseuille's Formula:

• F ∝ ΔPr4 / η or F = π ΔPr4 / 8ηI

• If the radius of a tube is halved the flow will decrease by 16 times

Clinical implications

• Flow of gas : ETT size, tubing diameter & length.

• Blood flow in cannulas, CV lines, peripheral lines

.

Flow

Turbulent Flow

• Chaotic and random flow. It is characterized by the

presence of Eddie’s currents

• Reynold's Number = ρvd / η

• >2000= turbulent flow

• Here, ρ (rho) = density , v= velocity, d diameter, η= viscosity.

• When > 2000 flow changes from laminar to turbulent

FLOW

• Applications:

• Using an undersized ETT decrease the flow of gases

• Every piece of anaesthetic equipment effect FGF.

• Wide bore & curved rather than sharp angles preferred.

• In respiratory tract obstruction, oxygen – helium mixtures are

given to reduce density and improve the flow.

• Laminar flow during quiet breathing is changed to turbulent during

speaking & coughing ;

In the flow meter

• At low flows, Hagen – Poiseuille’s Law applies as flow is laminar

• At higher flows, the law applicable to turbulent flow is applicable.

Critical Flow

• Flow rates laminar flow changes to turbulent

• Numerical value for critical flow in liters per minute

for the anaesthetic gas mixture of O2 + N2O is the

same as the internal diameter in millimeters.

• For an ETT of 8mm, it is 8L/min.

• Breathing circuit it is 22L/min). At these flow rates

the flow changes to turbulent from laminar

Bernoulli’s Principle

• Flowing fluid possesses two forms of energy, kinetic

energy and potential energy.

• If there is a constriction in the tube, kinetic energy

increases and potential energy (pressure) falls, since

the total energy must remain the same.

• Application; Suction apparatus, nebulizers and fixed

performance venturi masks are such devices.

Coanda Effect

• If such a constriction occurs at bifurcation because of increase

in velocity and reduction in the pressure, fluid (air, blood)

tends to stick to one side of the branch causing maldistribution.

Applications

• Mucus plug at the branching of tracheo-bronchial tree may

cause maldistribution of respiratory gases.

• Unequal flow may result because of atherosclerotic plaques in

the vascular tree.

• Fluid logic used in ventilators employs this principle to replace

valves or mobile parts.

•

Fick's Law of Diffusion

• It states, "movement of a gas across a tissue is directly

proportional to the area, difference between partial pressure

across the tissue and diffusion constant, and is inversely

proportional to thickness of the tissue"

• Fick's Equation:

• Vgas ∝ A . D . (P1-P2)/T where

• Vgas = Volume of gas moving across

• A=Area T=Thickness of the Tissue

• D=Diffusion Constant (Diffusibility of the gas)

• P=Partial pressure of the gas

Application.

Diffusion of gases across Alveli, capillaries & placenta.

Graham's Law of Diffusion

• It states “diffusion of molecules is directly proportional to their

solubility, and inversely proportional to the square root of their

molecular weight".

The Diffusibility of a Gas: According to Graham's Law

Diffusion Constant (D) ∝ Solubility / √MW

• The dependence on inverse of mol wt has come from kinetic theory

of gasses. Which says that all gas at a particular temperature have

the same kinetic energy.

• 1/2 mv2 = K m=mass, and v=velocity. K = constant

• v = √2K/m so

• v ∝ 1 / √m

Henry’s Law

• It states "At a constant temperature, the amount of a given gas that

dissolves in a given type and volume of liquid is directly

proportional to the partial pressure of that gas in equilibrium

with that liquid.”

• Example: If we equilibrate water with air at 37° c in a closed

chamber at 150 mm Hg (PAO2),

• The amount of O2 in air would be 21 ml/dl (21%)

• While the amount of 02 in water or blood at same partial

pressure would be 0.45 ml/dl .

• Solubility = 0.45/150 = 0.003 ml/dl/mm Hg.

• By ↑ing partial pressure solubility can be ↑ ed

• Hyperbaric O2; By ↑ing external pressure dissolved O2 ↑es sig

• Divers; Use compressed air mixtures. 10meter dive= ↑ 1 atmp

• Beers Law; Absorption of radiation by a given

thickness and concentration of a solution is the same

as twice the thickness with half the concentration.

• Lamberts Law; Equal thicknesses absorb equal

amounts of radiation.

• Both laws say that the absorption of radiation

depends on the amount of a particular substance. This

fact has been utilized in pulse oximetry

The Laplace's Law

• The tension on the wall of a sphere is the product of the pressure

times the radius of the chamber and the tension is inversely related

to the thickness of the wall.

• T = P × R or P = T / R

• For Vessels : P = T / R & For spheres e.g. Bubbles P = 2T/R

For Alveoli P = 4T / R

• P= inward directed collapsing pressure

• T=surface tension & R-radius

Clinical implications:

• Aortic aneurysm is more likely to rupture than a normal segment of

the aorta because the radius is greater - increasing wall tension to

breaking point.

• Capillary will withstand a pressure of 100 mmHg better than a vein

because capillary has a tiny radius compared to vein.

Adiabatic Change

• The change of physical state of a gas, without the transfer of

heat energy to or from the surrounding environment.

• In rapid expansion, energy is required to overcome Van der

Waal’s forces of attraction.

• Energy cannot be gained from the surroundings, it is taken

from the kinetic energy of the molecules. Temp decreases.

• It is basis of the cryoprobe.

• In rapid compression, reverse occurs & the heat is produced

in large amount e.g. compression ignition in a diesel engine

(Joule - Thompson Effect or

Joule-Kelvin Principle).

QUESTIONS

• Q.1. How much O2 gas volume will be present in E

cylinder if gauge pressure is 100 bar.?

• A. 100 L

• B. 200 L

• C. 300 L

• D. 400 L

• E. 500 L

• Q.1. How much O2 gas volume will be present in E

cylinder if gauge pressure is 100 bar.?

• A. 100 L

• B. 200 L

• C. 300 L

• D. 400 L

• E. 500 L

• Q. 2 . Entonox cylinder F type is present in labour

room. Gauge showing pressure of 1600 psi. How

much gas content will be available for patient at room

temperature?

• A. 700 L

• B. 800 L

• C. 900 L

• D. 1000 L

• E. 1100 L

• Q. 2 . Entonox cylinder F type is present in labour

room. Gauge showing pressure of 1600 psi. How

much gas content will be available for patient at room

temperature?

• A. 700 L

• B. 800 L

• C. 900 L

• D. 1000 L

• E. 1100 L

• Q.3. What will be the gas content of N2O cylinder(E

type) if gauge pressure is 746psi? Cylinder has 3.5

kg liquid gas.

• A. 1250 L

• B. 1350 L

• C. 1450 L

• D. 1550 L

• E. 1650 L

• Q.3. What will be the gas content of N2O cylinder(E

type) if gauge pressure is 746psi? Cylinder has 3.5

kg liquid gas.

• A. 1250 L

• B. 1350 L

• C. 1450 L

• D. 1550 L

• E. 1650 L

• Q.5. A patient has the PACO2 of 80 mmHg. His PAO2

is decreased to 50 mm Hg. How much FiO2 we

should provide to this patient to bring his PAO2 to 100

mmHg?

• A. 0.24

• B. 0.28

• C. 0.32

• D. 0.36

• E. 0.40

• Q.5. A patient has the PACO2 of 80 mmHg. His PAO2

is decreased to 50 mm Hg. How much FiO2 we

should provide to this patient to bring his PAO2 to 100

mmHg?

• A. 0.24

• B. 0.28

• C. 0.32

• D. 0.36

• E. 0.40