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CHAPTER 1

Anatomy and physiology of the humanrespiratory system

R.L. Johnson, Jr., & C.C.W. HsiaDepartment of Internal Medicine, University of Texas SouthwesternMedical Center, Dallas, Texas, USA.

Abstract

A brief review of the anatomy and fundamental physiologic concepts of the humanrespiratory system is presented, including postnatal lung growth and development,mechanical function of the airway and lung parenchyma, alveolar gas exchange,ventilation-perfusion and diffusion-perfusion relationships, alveolar microvascularrecruitment, pulmonary circulatory function and respiratory muscle energetics. Theemphasis is placed on the close interactions among different respiratory componentsand between the heart and the lung, which are essential for optimizing oxygentransport in health and disease.

1 Postnatal growth and development

1.1 Anatomy of the adult lung

In the right lung there are 3 lobes (upper, middle and lower), which are subdividedinto 9 sublobular segments. In the left lung there are 2 lobes (upper and lower), whichare subdivided into 8 sublobular segments. There are approximately 23 generationsof airways (Z) that distribute respired air to and from the smallest gas-exchangeunits in the lung (alveoli) [1], Fig. 1.

The path length and number of generations from trachea to different acini in thelung are not uniform owing to nonspherical asymmetry. Path lengths from tracheato acini can vary widely. Alveoli are small pockets that bud off of acinar airways,shaped like the cells of a beehive with a hexagonal mouth and a diameter of about150 µm at birth.

www.witpress.com, ISSN 1755-8336 (on-line) WIT Transactions on State of the Art in Science and Engineering, Vol 24, © 2006 WIT Press

doi:10.2495/978-1-85312-944-5/01

2 Human Respiration

Figure 1: Diagram of the branching airways. Z designates the order of generationswith the trachea as zero, followed by bronchi (BR) and bronchioles (BL).On average the first 16 generations are conducting airways that do notparticipate in gas exchange. The last conducting airway is a terminalbronchiole (TBL), which empties into an acinus. In the acinus there are3 generations of partially alveolated respiratory bronchioles (RBL) fol-lowed by alveolar ducts (AD) that are completely alveolated in adults andempty into an alveolar sac (AS). From Weibel [1].

1.2 Postnatal development

According to Hislop et al [2] conducting airways at birth are miniature versions ofthose in the adult. No new generations are added after birth; only size and lengthchange during postnatal growth. This is not true of the acinus. Alveoli multiplyby adding more alveolar ducts as well as by lengthening of existing ducts andincreasing alveolization [3]. The number of alveoli continues to increase after birth(Fig. 2) [4–6], most rapidly in the first 2 years but continues up to age 5 to 8 years.

Based on these measurements about 2/3 of the increase in acinar volume duringpostnatal growth is from an increase in number of alveoli within the first 5 to 8 yearsand the remaining 1/3 from a steady increase in alveolar size that continues untilthe lung reaches adult size. Rates of increase in size of the thorax and of the lung


Anatomy and Physiology of the Human Respiratory System 3

0

100

200

300

400

Num

ber

of A

lveo

li (

mill

ions

)

0 5 10 15 20 25

Years since birth

Adult

Figure 2: Number of alveoli increase about 15-fold after birth. Alveolar diameteralso increases linearly with time after birth until growth is complete [4];alveolar diameters approximately double, causing an approximate 8-foldincrease in volume.

must remain exactly matched throughout postnatal growth and there probably existmechanisms of feedback control between the lungs and thorax controlling this rate.

2 Conducting airways structure and function

2.1 Functional anatomy

2.1.1 Distribution of inspired airConducting air passages distribute respired air to and from lung acini where gasexchange occurs; they constitute what is called anatomical dead space that must bedisplaced during inspiration before fresh air reaches the acinus where gas exchangeoccurs. Upper airways consist of the nasopharynx, oropharynx and glottis. Belowthe glottis, Fig. 1, there is the trachea followed by an average of 16 generations ofdichotomously branching conducting airways down to terminal bronchioles, whichaverage approximately 0.06 cm in diameter. Terminal bronchioles empty into acinarairways where gas exchange occurs in alveolated airways. In each acinus there areapproximately 3 generations of respiratory bronchioles that are partially alveolated,and additional generations of alveolar ducts that are completely alveolated, finallyending in alveolar sacs, Fig. 2. Based on Weibel’s model of regular dichotomy[1] there are 65,000 terminal bronchioles in a 3/4 inflated lung to a volume of4800 ml minus the volume of conducting airways (150 ml); hence, the average sizeof individual acini would be about 0.072 cm3. These structures are made up of veryfine alveolar septa where circulating red blood cells are separated from alveolar airby a distance of less than a µm [1].


4 Human Respiration

2.1.2 Air conditioning and filtering inspired airAnother major function of conducting air passages is to humidify the inspiredair [7] and filter out small particulate matter to protect this thin gas-exchangesurface from dehydration and accumulation of inspired atmospheric debris [8, 9].Particulates with an aerodynamic diameter larger than 10 µm are filtered out byinertial impact at airway branches and seldom reach an acinus; they are removedby retrograde ciliary transport in the conducting airways to be coughed out or swal-lowed. A fraction of smaller particulates may sediment out in the central portionof the acinus and be removed by scavenging macrophages or breathed out againwithout being deposited. There appears to be little difference between the adult andchild’s lung in the efficiency of humidification and filtration of particulate matter.If anything, however, the child’s lung may be a little more efficient than the adultlung [2].

2.2 Viscous resistance to flow in upper airways, lungs and thorax

In 1915 Fritz Rohrer [10] published a classic paper describing the quantitativeanatomy of the irregular dichotomous branching of airways down to subsegmen-tal airways 1 mm in diameter. These emptied into what he designated as a lobule,which, based on Weibel’s model [10], would be about 5 generations above the aci-nus. Between the carina of the trachea and lobules in different regions of the lunghe measured the numbers of generations, airway diameters and lengths, angles ofbranching and changes in cross-sectional area at each branch point.Airways dimen-sions in the lobules were generated by an equation for regular dichotomy derived byvon Recklinghausen [11]. Based on the anatomy of the branching airways systemand physics of flow in tubes, Rohrer derived the pressure–flow relationships anddistribution of ventilation in human airways. Distribution was not uniform becausethe dichotomous branching was not uniform. Path length from the carina to thelobules varied from 8 cm in the central regions of the lung to 14 cm in the moreperipheral regions causing uneven distribution of resistance. The overall pressuredrop from nasopharynx to alveoli required to generate increasing flow could bereduced to a single comprehensive equation of the following form:

�P = K1V + K2(V )2, (1)

where �P is the pressure drop in cmH2O across the lung; V is gas flow in l/s; K1is a constant derived from the summation of laminar-flow resistances in branchingairways in cmH2O/(l/s) and K2 is a constant derived from changing resistancefrom turbulence and from inertance at airway branch points and at changes incross-sectional area of the airways; units are in cmH2O/(l/s)2. Results reported byRohrer for resistances across the upper airways, lower airways and total are givenin Table 1 [10].

Fifty years later K1 and K2 in eqn. (1) derived by Rohrer have been determinedexperimentally to describe viscous properties of lung airways, lung tissues andthorax, Table 2.



Table 1: Pressure–flow relationships derived by Rohrer.

Upper airways �P = 0.426 · V + 0.714 · (V )2

Lower airwaysBronchial �P = 0.106 · V + 0.080 · (V )2

Lobular �P = 0.258 · V + 0.006 · (V )2

Total �P = 0.790 · V + 0.800 · (V )2

Table 2: Resistance to airflow in the respiratory system of an average normal personbreathing through the mouth or nose [12].

Mouth breathing Nasal breathing

Flow Laminar Turbulent Laminar Turbulent

Resistance K1 K2 K1 K2

Units cmH2O/(l/s) cmH2O/(l/s)2 cmH2O/(l/s) cmH2O/(l/s)2

Air flow 1.2 0.3 1.8 3.0Lung tissue 0.2 0 0.2 0Chest wall 1.0 0 1.0 0Resp. system 2.4 0.3 3.0 3.0

Thus, to describe the pressure in cmH2O to generate a volume flow of V l/sthrough the mouth would be �P = 2.4V + 0.3 · [V ]2 and through the nose wouldbe �P = 3.0V + 3.0·[V ]2. Results are close to those originally described by Rohrer,particularly when resistances imposed by viscoelastic properties of the lung andthoracic tissues are removed, since these are not a part of Rohrer’s estimates [10].

2.3 Alveolar ventilation is not uniform even in the normal lung

There are space-filling constraints within the lungs and thorax that prevent uniformdistribution of ventilation even in the normal lung. As pointed out above, Rohrerfound that resistance to flow is greater to peripheral lung lobules than to more centralones [10]. These regional differences in flow resistance do not necessarily mean thatventilation will be nonuniform. The rate at which a lung unit fills is not determinedsolely by flow resistance; it is in part determined by compliance of the unit (C),where C = the change in volume (�V ) divided by the change in filling pressure (�P)= �V /�P in l/cmH2O. The product RC is the time constant for filling of the unit inseconds. Even if resistance to flow in peripheral units is greater than in more centralunits ventilation may be the same if RC constants are the same, in analogy to RCconstants in an electrical circuit. If RC constants for filling peripheral lobular unitsare longer than for low-resistance central units, ventilation will become more and


6 Human Respiration

more nonuniform as the frequency of breathing increases. If RC constants matchbetween central and peripheral units, ventilation will remain uniform in spite ofunequal resistances to flow. Rohrer did not measure regional compliances althoughhe discussed this problem. Ross [13] compared measurements and calculations tothose of Rohrer in plaster casts of the dog lung between the carina and airwaysdown to 1.5 mm diameter. He also found that airway path lengths varied from 2 cmcentrally to 13.7 cm peripherally, with corresponding differences in airway dead-space volumes. Anatomical dead-space volume is the volume of air in conductingairways that must be cleared during a breath before fresh air reaches the alveoliwhere gas exchange occurs. Alveolar ventilation (VA) is related to total ventilation(VT) as follows: VA = VT − fVD, where f is frequency of ventilation VD is dead-space volume between the inlet to the alveoli. Thus, even if ventilation to peripheraland central lobules is the same, alveolar ventilation to peripheral lobules will beless because of the larger dead-space volume that must be cleared. The effects ofspace-filling constraints on airway branching and distribution of ventilation extendall the way down to the alveolar ducts, as pointed out by the work of a numberof investigators [13–16]. As pointed out by Johnson and Curtis [17], “because ofthese space-limiting constraints, irregular branching of the bronchial tree becomesnecessary to pack the largest alveolar surface area into a limited space with theleast power requirements for distribution of ventilation; the trade off is unevenventilation.”

2.4 Elastic properties of the lungs and thorax

The balance between the static elastic recoil forces in the lungs and thorax deter-mines the pressure–volume relationships of the respiratory system when respiratorymuscles are completely relaxed, Fig. 3 [18].

In Fig. 3, at the end of a normal breath when respiratory muscles are relaxed,the inspiratory pressure exerted by recoil of the chest wall exactly balances theexpiratory pressure exerted by recoil of the lungs. The lung volume at this balancepoint is referred to as the functional residual capacity (FRC). The total pressure thatis generated by the respiratory system at a given lung volume (excluding muscle) isthe algebraic sum of the recoil pressure exerted by the chest and that exerted by thelungs. If the lungs are held at a given volume with the mouth open and unobstructed,the negative intrapleural pressure (i.e., the pressure in the virtual space between theadjacent surfaces of the lung and chest wall) reflects the expiratory recoil pressureof the lungs. We have discussed the pressure required to overcome flow resistancein the airways to inspire, eqn. (1). The additional pressure (�P) required to inflatethe lungs by a volume �V against the elastic recoil pressure of the lungs and thoraxis given by

�P = �V /CRS, (2)

where CRS is compliance of the lungs and thorax as defined in Fig. 3 and is expressedin l/cmH2O.



Figure 3: Static pressure–volume curves of lungs and thorax in a normal uprightsubject. The pressure generated by the total respiratory system at a givenlung volume is represented by the heavy curved line. The two thin curvedlines represent pressures exerted by the chest wall on the left and by thelungs on the right. A positive sign indicates an expiratory pressure; anegative sign indicates an inspiratory pressure. Reproduced from Milleret al [18].

2.5 Work of breathing

Ventilation requires energy to overcome both elastic and viscous forces in lungsand thorax. Incorporating both eqns. (1) and (2) the following approximations werederived by Otis et al [19]:

P = V

CRS+ K1V + K2 · [V ]2 , (3)

where V is the volume above the functional residual capacity (FRC) and othersymbols are as defined for eqns. (1) and (2). Multiplying both sides by V we havethe power requirement at any instant:

Instantaneous Power = PV = V · V

CRS+ K1 · [V ]2 + K2 · [V ]3.


8 Human Respiration

If we assume that ventilation follows a sine-wave pattern, V = VPsin(2πft) whereVP is peak flow in l/s and we assume that energy is required only during inspirationsince during quiet breathing the elastic energy during inspiration is stored andreturned passively during expiration, we can derive the following equation forestimating energy requirements of quiet breathing:

W = Power = (1/2fCRS)[VE]2 + 60 ·[(K1/4) π2 [VE

]2 + (2K2/3) π2 · [VE]3]

,

(4)where VE has been converted from l/s in eqn. (3) to ventilation in l/min; f is respira-tory rate in min−1; 60 is s/min so that power is now expressed as cmH2O·l·min−1.

During exercise, when respiratory muscle energy is active throughout both inspi-ration and expiration, eqn. (4) can be modified as follows to approximate energyrequirements of breathing during heavy exercise:

W = 2 · 60 ·[(K1/4) π2 · [VE

]2 + (2K2/3) π2 · [VE]3]

/100. (5)

Dividing by 100 converts power from units of cmH2O·l·min−1 to more familiarengineering units of kg·m·min−1, Fig. 4.

Ventilatory capacity may be limited in the patient because respiratory musclemass is insufficient to generate the pressures required for a normal ventilatoryeffort. Another possible limitation of ventilation and exercise capacity in the patientconcerns metabolic requirements, which will be considered later.

0 50 100 150 200 250

250

200

150

100

50

0

Athlete

Patient with asthma

Ventilation (l/min)

Pow

er R

equi

rem

ents

(kg

m/m

in)

Figure 4: Power requirements are plotted in an athlete assuming normal valuesof K1 and K2 for lungs and thorax from Table 2 compared to powerrequirements in a hypothetical patient with asthma assuming about a5-fold increase of K1 and K2 during an asthmatic attack (i.e., K1 = 10cmH2O/(l/s) and K2 = 1.5 cmH2O/(l/s)2).



3 Mechanisms of gas exchange

3.1 Convective and diffusive distribution and mixing in airways

As a front of inspired air moves distally down the airways, the pattern of oxygentransport and mixing within this front changes from a combination of turbulent andlaminar convection in the conducting airways to predominantly diffusive mixingas it moves into the acini and finally into the alveoli where alveolar capillarygas exchange occurs. Weibel’s branching model of the airways [1] provides thedimensions of each airway generation. From these dimensions there are two non-dimensional numbers that can define critical change in flow patterns from laminarto turbulent, or where diffusive transport becomes faster than convective flow.These flow patterns are important in how inspired air is mixed during inspiration.

3.1.1 Relative importance of laminar and turbulent flowThe relative importance of laminar and turbulent flow patterns at different gen-erations in the conducting airways can be derived from the Reynolds number,which is a nondimensional ratio of inertial to viscous forces. The Reynolds number= (y · v)/(µ/ρ) where y = tube diameter in cm, v = linear velocity in cm/s, µ =fluid viscosity in poise and ρ = fluid density in g/cm3 (µ/ρ is called the kinematicviscosity). Turbulence develops normally at corners, sudden changes in direction orairway diameter as at branch points. Once developed, inertial forces tend to main-tain turbulence, while viscosity tends to damp it out, depending on the balancebetween these two forces. Even in unbranched smooth tubes turbulence tends tobecome sustained when the Reynolds number exceeds 2000. Resistance to laminarflow in airways, K1 in eqn. (3), is directly proportional to viscosity (µ) and vol-ume velocity (V ). Resistance to turbulent flow, K2, is directly proportional to fluiddensity and the square of the volume velocity, [V ]2, as in eqn. (3). Although tur-bulence increases the power requirements of breathing it is an important source ofgas mixing, heat exchange, humidification, filtration of inspired particulate matterand gas mixing in the lung before it reaches the alveoli.

3.1.2 Relative importance of convection and diffusion in transportThe relative importance of convective and diffusive transport in different genera-tions of airways can be estimated from the Peclet number (Pe). Pe is a dimensionlessratio of rate of transport by convection down a concentration gradient to the rate ofdiffusion down the same concentration gradient, i.e., (v · x)/d where v = velocityof convective flow down the tube in cm/s, x = length of the tube in cm and d isthe diffusion coefficient of oxygen in respired air given in cm2/s. As Pe falls below1.0 the rate of transport by diffusion begins to exceed that by convection. In thelower conducting airways and acini the combination of diffusion and convectionprovides an important source for mixing of gases not only within each acinus butalso between acini. Table 3 summarizes these relationships. During quiet breath-ing turbulence is minimal except at airway branch points and diffusive transport is


10 Human Respiration

Tabl

e3:

Con

vect

ive

and

diff

usiv

ega

str

ansp

orta

ndm

ixin

gin

the

airw

ays

betw

een

trac

hea

and

alve

oli.

Res

ting

vent

ilatio

n=

6l/m

inE

xerc

ise

vent

ilatio

n=

100

l/min

Peak

flow

‡Pe

akflo

w‡

Gen

erat

ion

Air

way

∗cm

/sPe

clet

num

ber

Rey

nold

snu

mbe

rcm

/sPe

clet

num

ber

Rey

nold

snu

mbe

r

0T

rach

ea13

1.23

6151

.514

08.5

1968

.592

273.

621

128.

86

Bro

nchu

s84

.17

295.

914

0.5

1262

.644

38.9

221

08.1

16T

B1.

852

1.19

40.

663

27.7

7817

.904

9.93

817

RB

10.

967

0.53

30.

311

14.5

097.

991

4.67

218

RB

20.

467

0.21

40.

139

7.01

13.

204

2.09

019

RB

30.

211

0.08

20.

059

3.16

61.

224

0.88

720

AD

10.

078

0.02

50.

021

1.17

60.

381

0.31

621

AD

20.

035

0.01

00.

009

0.53

00.

145

0.13

622

AD

30.

016

0.00

40.

004

0.24

00.

055

0.05

923

AS

0.00

80.

002

0.00

20.

120

0.02

30.

029

∗ TB

=te

rmin

albr

onch

iole

ente

ring

the

acin

us;R

B=

resp

irat

ory

bron

chio

le;A

D=

alve

olar

duct

and

AS

=al

veol

arsa

c.‡Pe

akflo

wdu

ring

are

spir

ator

ycy

cle

isab

out3

times

the

aver

age

vent

ilatio

n.Sh

aded

area

sun

der

Pecl

etnu

mbe

rsde

sign

ate

airw

ays

atth

egi

ven

vent

ilatio

nw

here

rate

ofdi

ffus

ive

tran

spor

tex

ceed

sth

era

teco

nvec

tive

tran

spor

t.Sh

aded

area

sun

der

the

colu

mns

ofR

eyno

lds

num

bers

indi

cate

thos

eai

rway

sat

the

give

nve

ntila

tion

whe

retu

rbul

ence

tend

sto

beco

me

sust

aine

d.



much faster that convection in the acinus. During heavy exercise sustained turbu-lence extends down to subsegmental bronchi and diffusive transport is still fasterthan convective transport in the alveolar ducts.

3.2 Alveolar gas exchange at equilibrium

At steady state the alveolar CO2 and O2 tensions in the lungs are determined by theventilation relative to the respective rates of CO2 output and O2 uptake.

3.2.1 CO2 equilibriumThus, the alveolar equation for CO2 states

PaCO2 = VCO2

V a

· 863, (6)

where PaCO2 = alveolar CO2 tension, VCO2 = CO2 output, V a = Alveolar venti-lation and 863 = 760 (body temp. in Kelvin)/273.

The difference between end-capillary CO2 content in mM/ml leaving the lung(Cc′

CO2 ) and mixed venous CO2 content (CvCO2 ) entering the lung multiplied byblood flow (Q) in ml/min equals the CO2 output (VCO2) in ml/min. This is the Fickequation for CO2:

VCO2 := Q(CvCO2 − Cc′CO2

) · 22.2, (7)

where 22.2 = ml/mM.Combining eqns. (6) and (7) and rearranging we have

Cc′CO2

= CvCO2 −[

V a/Q

19.2

]· PaCO2 , (8)

which, for a given ratio of ventilation to blood flow, V a/Q, and the mixed venousCO2 content (CvCO2 ) of blood entering the lung tells us the trajectory along whichblood CO2 content must fall toward equilibrium with the alveolar CO2 tension(PaCO2 ). The point of equilibrium can be derived graphically from the intersectionof this trajectory with the CO2 dissociation curve, Fig. 5. Thus, if we know the ratioof ventilation to perfusion, V a/Q, in any region of lung and the mixed venous CO2content we can estimate the CO2 content in blood leaving that region at equilibriumwith the PaCO2 .

3.2.2 O2 equilibriumThe same set of equations can be derived for oxygen. The alveolar equation for O2states that

PaO2 = (Pb − PH2O) · FiO2 − PaCO2 ·(

FiO2 + FiN2

R

), (9)



Endcapillary

blood

Figure 5: CO2 equilibrium between alveolar air and blood leaving lung capillaries ata known mixed venous CO2 content and a ratio of ventilation to perfusion(V a/Q) = 3. The solid straight line is eqn. (8) and the curved line is theCO2 dissociation curve. The point of intersection indicates equilibrium.

where PaCO2 = alveolar O2 tension; Pb = barometric pressure; PH2O = watervapor tension at body temperature (47 mmHg at 37◦C); FiO2 = inspired O2 fraction(approximately 0.21 breathing room air at sea level); FiN2 = inspired N2 fraction(normally taken as 0.78 in room air at sea level).

Furthermore, there is also a Fick equation for oxygen:

VO2 = Q(Sc′O2

− SvO2 ) · Cap, (10)

where Sc′O2

and SvO2 are fractional oxyhemoglobin saturation in the end-capillaryand mixed venous blood, respectively, and Cap is the O2 capacity of blood in mlof O2/ml blood.

Then, combining eqns. (6), (9) and (10),

[Sc′

O2− SvO2

] = [PiO2 − PaO2 · FiO2

] ·⌊

V a/Q

863 · FiN2 · Cap

⌋

− ·⌊

V a/Q

863 · FiN2 · Cap

⌋· PaO2 . (11)

Equation (11) determines the trajectory along which oxygen saturation rises fromthat in mixed venous blood toward equilibration with the alveolar O2 tension (PaO2 )for a given ratio of V a/Q. The point of equilibrium can be derived graphicallyfrom the intersection of this trajectory with the oxyhemoglobin dissociation curve,Fig. 6.



0

25

50

75

100

0 25 50 75 100 125 150

Oxygen Tension (mmHg)

VA/Q = 6.0

VA/Q = 0.5

PAO2PaO2

SaO2

b

0

25

50

75

100O

xyge

n Sa

tura

tion

(%

)

0 25 50 75 100 125 150

Oxygen Tension (mmHg)

mixed venous

PIO2- KPAO2

Equilibrium VA/Q = 3

a Oxyhemoglobindissociation curve

Figure 6: (Panel a) O2 equilibrium between alveolar air and blood leaving lungcapillaries at a known inspired O2 tension (PiO2 ) and mixed venous O2saturation. The solid straight line is eqn. (11) at a ventilation perfusionratio (V a/Q) = 3. The curved line is the O2 dissociation curve. Equi-librium is at the point of intersection. (Panel b) The effect of unevenV a/Q. This shows the effect of two regions of lung with equal bloodflows but different V a/Q ratios of 6 and 0.5. A mixture of blood from thetwo regions will have an O2 saturation and PaCO2 approximately half-way between and a resulting increase in PaO2 − PaO2 as indicated in thediagram. After Hsia and Johnson [20].

3.2.3 Effect of uneven V a/Q ratios in the lungThe most efficient gas exchange is achieved when V a/Q ratios are uniform andthe alveolar-arterial O2 tension and CO2 tension differences (PaO2 − PaO2 andPaCO2 − PaCO2 ) are zero. Increasing nonuniformity of ventilation with respect toblood flow causes these gradients to increase. Ventilation has to be increased inorder to bring the arterial CO2 and O2 tensions back toward normal levels.

3.3 Alveolar capillary gas exchange and physiologic laws of diffusion

3.3.1 Diffusing capacity of the lung (DL)DL is defined as the rate of transfer of a respired gas between alveolar air andbinding sites on hemoglobin in capillary red cells divided by the driving pressurein mmHg and expressed in units of ml·(min·mmHg)−1. Thus, for oxygen

DLO2= VO2

PaO2 − PcO2

, (12)

where VO2 = Oxygen uptake in ml/min, PaO2 = Alveolar O2 tension and PcO2 =mean oxygen tension of capillary hemoglobin, both in mmHg.



The difficulty with measuring DLO2arises from the fact that PcO2 changes as

blood passes through the lung. Although PaO2 and VO2 can be measured PcO2

cannot because it changes during transit until equilibrium is reached or leaves thecapillary bed without being accessible to measurement. On the other hand, underconditions used for measuring DLCO changes in PcCO are negligible during transit.CO binds to hemoglobin so tightly that multiple measurements of DLCO can bemade before PcCO rises significantly. DLCO can be measured essentially neglectingcapillary CO tension as follows:

DLCO = VCO

PaCO. (13)

To make this measurement during breath–holding, a test gas is made up containinga small concentration of CO and of an inert insoluble tracer gas such as He orCH4 mixed with a balance a balance of air or oxygen. A measured volume of thismixture is inspired. The breath is held for a measured time interval of about 10 s.Then an expired sample is collected to determine how much CO has disappearedwith respect to the inert, insoluble tracer for calculation of DLCO. This is the breath-holding technique and is the most common clinical measurement. The inspiredbreath can also be breathed back and forth in a reservoir bag while the disappearanceof CO with respect to the tracer is followed with each breath by a rapid analyzerfor calculation of DLCO. The latter is called the rebreathing technique and hasthe advantage of providing measurements during exercise. A small concentrationof acetylene can also be added to the test gas for simultaneous measurement ofcardiac output. Nitric oxide also reacts extremely rapidly with hemoglobin and istaken up much more rapidly by red cells than CO. DLNO can be measured by eitherthe breath-holding or rebreathing method simultaneously with measurements ofDLCO. However, before going further it is important to review the physical lawsinvolved in these measurements.

3.3.2 Fick’s law of diffusion defines how diffusive transport moves down aconcentration gradient

The rate of diffusive transport of a gas in a uniform media such as plasma or thealveolar wall is described by Fick’s law of diffusion as follows:

dQ

dt= Ad

∂C

∂x, (14)

where

dQ

dt= volume transfer of the gas in cm3/s in the x direction;

A = area in cm2 of the barrier through which diffusion is occurring.

d = the diffusion coefficient in cm2/s, and∂C

∂x= concentration gradient through the barrier in the x direction in cm−1



3.3.3 Krogh’s diffusion constant defines how a concentration gradient istranslated into a more convenient pressure gradient [21]

When a gas diffuses from one media into another where gas solubilities differ, e.g.,from alveolar air into alveolar wall, the standard Fick equation does not work well;hence Krogh modified the standard Fick diffusion equation as follows:

dQ

dt= A ·

⌊6 · α · d

760

⌋· ∂P

∂x, (15)

where

dQ

dt= volume transfer in cm3/min from air through the alveolar wall

A = surface area of the alveolar wall in cm2

α = Bunsen solubility coefficient in atm−1

d = diffusion coefficient in cm2/s⌊6 · α · d

760

⌋= Krogh’s diffusion constant in cm2·min−1·mmHg−1,

∂P

∂x= pressure gradient in mmHg/cm across the wall.

Thus, using Krogh’s constant [21], diffusion is described as being driven by apressure gradient instead of a concentration gradient. Furthermore, it allows us totranslate a membrane-diffusing capacity measured for one gas into that for anothergas of different molecular weight and solubility.

3.3.4 Diffusing capacity of the pulmonary membrane (DM) and itsrelationship to the Krogh constant

This is given by

DM =⌊

60 · α · d

760

⌋A

�x= K · A

�x, (16)

where A is the area and �x the thickness of the barrier and DM is the membrane-diffusing capacity of the lung;α and d are assumed to be similar in tissue and plasma.Approximate values of the Krogh constant are given in Table 4 for different gasesthat are important in pulmonary physiology.

If Krogh diffusion constants are known, measurements of DM for one gas canbe translated into an estimate for another gas. For example:

DMO2

DMCO

= KO2

KCO= 3.87

3.48. (17)

3.3.5 Rate of uptake of respired gases by red cells involving diffusion andchemical binding

Diffusive uptake of respired gases is determined not only by the resistance ofthe alveolar capillary membrane to diffusion but also by the resistance imposed by



Table 4: Estimated Krogh diffusion constants (K) in cm2 min−1 · mmHg−1.

Gas Diffusing in N2¶ In lung tissue, plasma or water

Respiratory gases O2 1.79 · 10−2 3.87 · 10−8 #

CO2 1.26 · 10−2 64.5 · 10−8 ∗Gases used to measure CO 1.65 · 10−2 3.48 · 10−8 #

diffusing capacity NO 1.59 · 10−2 6.84 · 10−8 ‡

¶[22–24].#Based on measurements in lung tissue;∗Based on measurements in plasma;‡Based on measurements in water [25].

Table 5: Rate of gas uptake by red cells in a ml of normal whole blood (θ) in(ml·min−1)STPD·mmHg−1 gas tension in surrounding plasma*.

Gas θ in min−1 · mmHg−1 References

O2 3.9 [28]CO2 5.8 [29]CO 1/(0.73 + 0.0058 · PO2 ) [26]NO approaching infinity [30]

*θ is estimated for a normal hemoglobin concentration of 14.9 g/dl hence it mustbe corrected for a deviation from 14.9 by the product θ · [measured Hb/14.9].

diffusion into the red cells and rate of chemical binding to hemoglobin. Specific ratesof gas uptake (referred to as θ) have been measured for O2, CO2, CO and NO usingrapid reaction-rate techniques in red-cell suspensions and the results expressed interms of [(ml/min)gas uptake/mmHg gas tension]/ml of blood having an O2 capacityof 20 ml/dl. Representative data are given in Table 5 with references. CO competeswith O2 for binding sites on hemoglobin; hence, as oxygen tension increases θCOdecreases as shown in Table 5 and has provided a method for measuring the truemembrane-diffusing capacity for CO and pulmonary capillary blood volume by atechnique devised by Roughton and Forster [26, 27]. There is no such competitionwith oxygen for binding sites by the other gases.

3.3.6 The Roughton–Forster method for estimating membrane-diffusingcapacity (DM) and pulmonary capillary blood volume (Vc) [27]

The Roughton–Forster equation [27] states that the total resistance to oxygenuptake by blood flowing through the lung capillary bed equals the reciprocal ofthe lung-diffusing capacity of the lung (1/DLO2

), which in turn is the sum ofthe resistances in series imposed by the alveolar capillary membrane (1/DMO2

)and that imposed by red cells in pulmonary capillary blood ((1/θO2 ) · (1/Vc)) as



summarized below:

Total Resistance = Membrane Resistance + Red-Cell Resistance

1

DLO2

= 1

DMO2

+ 1

θO2

· 1

Vc. (18)

Similar equations can be set up for DLCO measured at two or more alveolar oxygentensions, eqns. (19) and (20). As oxygen tension increases 1/θCO increases andthe measured DLCO correspondingly falls, Table 5; hence at two different oxygentensions the Roughton–Forster equation [27] yields two simultaneous equationswith two unknowns that can be solved for DMCO and Vc.[

1

DLCO

]O2

=[

1

DMCO

]+[

1

θCO

]O2

· 1

Vc(19)

[1

DLCO

]Air

=[

1

DMCO

]+[

1

θCO

]Air

· 1

Vc. (20)

The Krogh constants for O2 and CO from Table 4 and θO2 from Table 5 can thenbe used to estimate DLO2

from the DMCO and Vc derived from the two simultaneousequations, again using the Roughton–Forster equation as follows:

1

DLO2

= KCO

KO2 DMCO

+ 1

θO2

· 1

Vc. (21)

More recently, methods have been developed to measure DLNO and DLCO simul-taneously. NO reacts so rapidly with hemoglobin in red cells that the resistanceimposed by the red cell to NO uptake approaches zero and DLNO is the same asDMNO. Based on this assumption DMCO and Vc can be estimated empirically fromthe following modification of the Roughton–Forster equation:

1

DLCO

= 1

2.4 · DLNO

+ 1

θCO· 1

Vc. (22)

3.3.7 Recruitment of diffusing capacity during exerciseDuring exercise all components of DLCO increase in an approximately linear rela-tionship to cardiac output and a plateau is not reached within the physiologicalrange, Fig. 7. DLO2

has been estimated indirectly from DMCO and Vc as illustratedin the previous sections and is about 50% higher at any given cardiac output thanDLCO. DLO2

is higher primarily because the rate of uptake of oxygen by red cellsis much faster than the rate of uptake of CO. DLNO is approximately 4 times DLCO

at the same cardiac output primarily because of its greater solubility in lung tissueand because the red cells offer little resistance to NO uptake.

3.3.8 The importance of the relationship between DLO2and cardiac

output [20]If DLO2

did not increase as cardiac output increases during exercise arterialoxyhemoglobin desaturation would occur at relatively low levels of exercise.



Figure 7: Diffusing capacity of the lung for CO, O2 and NO (not shown here)increase from rest to heavy exercise in a linear relationship with car-diac output and does not reach a plateau over the physiological range ofmeasurement.

From eqns. (10) and (12):

VO2 = Q(Sc′

O2− SvO2

) · Cap

VO2 = DLO2· (PaO2 − PcO2

),

it is possible to derive that the ratio of diffusing capacity to pulmonary capillaryblood flow, DLO2

/Qc bears the following relationship to oxygen saturation of bloodleaving lung capillaries (Sc′

O2) when mixed venous oxygen saturation (SvO2) and

alveolar oxygen tension (PaO2 ) are known:

DLO2

Qc= Cap

∫ Sc′O2

SvO2

d(ScO2 )

(PaO2 − PcO2 ). (23)

In order to numerically integrate the right-hand side of eqn. (23) to calculatethe relationship between Sc′

O2and DLO2

/Qc at rest and exercise the shape andposition of the oxyhemoglobin dissociation curve must be known. This can beprovided with reasonable accuracy by the Hill equation in the following formPcO2 = P50[ScO2/(1 − ScO2 )]0.37. The P50 is the oxygen tension at which thehemoglobin saturation (ScO2 ) is 50%. The P50 represents the position of the curvethat changes when body temperature, blood PCO2 , or pH change, for example withexercise. The shape of the curve remains fixed. Under normal resting conditionsthe P50 is 26.5 mmHg but increases during exercise as body temperature increasesand pH falls. We have calculated the relationship between end-capillary oxygensaturation Sc′

O2and DLO2

/Qc as illustrated in Fig. 8.

Results show that as DLO2/Qc falls during exercise oxygen saturation of blood

leaving the lung remains almost fixed until a critical level is exceeded when Sc′O2



Figure 8: As the ratio of DLO2/Qc moves below a critical point, there is a rapid fall

of oxygen saturation of blood as it leaves the lung. The fall occurs earlierat a low alveolar oxygen tension such as at high altitude or when the P50of hemoglobin falls as when the pH declines during heavy exercise orwhen body temperature rises. From Hsia and Johnson [20].

falls rapidly. Data was calculated to show the effect of changing alveolar PO2

between 110 to 80 mmHg, equivalent to that expected during exercise at sea leveland at an altitude of 10,000 feet, respectively. Data was also calculated to show theeffect of changing arterial pH from 7.4 to 7.2 as occurs from rest to heavy exercise.Now we can address the question of what would happen to arterial O2 saturationif DLO2

remained fixed from rest to exercise. At sea level, at rest, DLO2/Qc would

be about 7.5 and Sc′O2

would be in equilibrium with PaO2 at about 97.5%. Bythe time cardiac output reached 22 L/min Sc′

O2would have fallen to about 80%.

At 10,000 feet arterial saturation would fall to between 70 and 80% at a cardiacoutput of 15 l/min if DLO2

failed to rise above its resting level. Why then doesn’tthe system operate with the pulmonary capillary bed fully recruited at rest? As willbe discussed in the next section, part of the explanation arises from the fact thatthe pulmonary circulation must operate at low pressures. In the upright position thepulmonary artery pressure at rest or light exercise is too low to completely recruitthe upper lobes of the lung.

4 Pulmonary circulation

4.1 A low-pressure system

The pulmonary circulation is a low-pressure system relative to the systemic circula-tion, Fig. 9. It has to operate at a relatively low pressure level in order to sustain rapiddiffusive gas exchange between inspired air and the cardiac output as it is pumped ata high rate through thin walled capillaries. The high rate of diffusion requires a thin



0

10

20

30

40

50

Mea

n P

ulm

onar

y V

ascu

lar

Pre

ssur

es (

mm

Hg)

0 10 20 30Cardiac Output (l/min)

Capillary Wedge

Pulmonary Artery

Figure 9: Mean vascular pressures are much lower in the lung than in the systemiccirculation. This is necessary to avoid stress failure of the pulmonarycapillaries during exercise.

membrane that must be protected from pressure damage. For proper perspectiveit helps to visualize the actual dimensions within which this gas exchange occurs.The pulmonary capillary bed contains 100 to 200 ml of blood enclosed within athin membrane wall of about 1 µm average thickness and spread over a surfacearea that is the size of a tennis court. At rest it operates at a mean pulmonary artery(PA) pressure of about 15 mmHg (compared to a mean systemic arterial pressureof about 90 mmHg) and at peak exercise rises to between 30 to 40 mmHg. Pul-monary capillary wedge (PCW) pressure may approach between 20 to 25 mmHg.Capillary pressures not too much higher than that can cause pulmonary edema andhemorrhage by stress failure in pulmonary capillaries [31].

4.2 The pulmonary vascular waterfall and lung zones

The top of the upright human lung is about 25 to 30 cm above the mid-right ventricle(RV). Thus, at rest the RV must generate a PApressure head of about 18 to 22 mmHgto reach the top of the lung; however, at rest it operates at a PA pressure of onlyabout 15 mmHg. Hence, at rest the perfusion pressure to the apical 20% of the lungis zero and this region is only perfused by the bronchial circulation that operates atsystemic pressure and keeps the region viable. Because the pulmonary circulationoperates at such low pressures it is considered to be functionally divided into 3pressure zones. In a region where PA pressure is lower than arterial hydrostaticpressure, capillary perfusion pressure is zero and the region operates under zone-1conditions. In a region where RV pressure exceeds both arterial hydrostatic andalveolar pressures but venous hydrostatic pressure is lower than alveolar pressure,a waterfall condition occurs. Capillary blood exits the alveoli down a waterfall tothe lower venous pressure. The capillary perfusion pressure becomes arterial minus



Figure 10: Uneven distribution of blood flow (Qc/VA) and diffusing capacity(DL/VA) from lung apex to base owing to the high hydrostatic pres-sure relative to right ventricular pressure at rest. The upper 20% of thelung is not perfused at rest except by the high-pressure bronchial circu-lation (i.e., Qc/VA = 0). From Michaelson et al [36].

alveolar pressure (Part – Palv); this is a zone-2 condition. In a region where botharterial and venous pressures exceed hydrostatic and alveolar pressure, perfusionpressure becomes arterial minus venous pressure (Part – Pven); this is a zone-3condition. These kinds of zones are unique to the lung where blood vessels arecollapsible and changes in alveolar pressure can collapse or dilate the capillaries tochange their resistance or where at zero alveolar pressure capillaries can collapse ina region where hydrostatic pressure is less than zero; blood will be sucked out as iffalling down a waterfall [32, 33]. Regions operating under zone-1 conditions do notparticipate in gas exchange. The capillary bed is only partially open under zone-2conditions but may be completely open in zone-3. These zone conditions create anuneven distribution of blood flow and diffusing capacity imposed by gravitationalstresses from the top to the base of the lung [34, 35] even in normal subjects, Fig. 10.

During exercise, zone 3 moves up as pulmonary arterial and capillary wedgepressure increase. More capillaries open and this is one source of capillary recruit-ment and increase in diffusing capacity but probably not the major source. Theremay be even greater recruitment, however, within isogravitational planes [37].

5 Respiratory muscles

5.1 Innervation and muscle mass of respiratory muscles

Inspiratory and expiratory muscle groups are shown in Fig. 11 along with their seg-mental innervation. Inspiratory muscles receive most of their segmental innervation



Figure 11: Respiratory muscles and their segmental innervation. The intercartilage-nous portions of the internal intercostals supplied by T1 to T4 (paraster-nals) are inspiratory muscles. After Miller et al [18].

from cervical and thoracic segments, expiratory muscles from thoracic and lum-bar segments. Systematic measures of respiratory muscle mass in humans is onlyavailable for the diaphragm at post mortem [38]. For humans with sedentary occu-pations, diaphragm weight (g) = 4.18 · body weight (kg) – 21.8. Muscle mass andcapability for generating power of all respiratory muscles have been measured indogs where the diaphragm weight (g) = 5 · body weight (kg), on average about 28%higher than in sedentary humans [39, 40]. However, in humans with occupationsrequiring heavy labor, the weight of the diaphragm is only about 10% lower than inthe dog. Segmental innervations are the same in humans and dogs. Hence studieson muscle mass and power capabilities of respiratory muscles in dogs probablyprovide a good first approximation to respiratory muscle function in humans.

The diaphragm is the most important respiratory muscle in both species. In thedog, the diaphragm constitutes only 9% of the respiratory muscle mass but supplies30 to 40 per cent of the ventilatory power. Inspiratory muscles in general constituteonly 40% of the respiratory muscle mass but deliver more than 70% of the totalpower for breathing during exercise.

5.2 Oxygen requirements of breathing

Oxidation of carbohydrates yields approximately 5 cal/mlO2 and one (kg·m) ofmechanical energy is equivalent to 2.33 cal of heat energy, hence 0.5 ml of oxygenwould be needed to support one (kg·m) kg m of mechanical work if all of the energygenerated by burning carbohydrate could be utilized (100% efficiency). Mechanicalefficiency of the respiratory muscles has been measured at 20 to 25% in both the dogand human [41, 42]; this is essentially the same as other exercising skeletal muscle.Hence, assuming 25% efficiency, each (kg·m) of mechanical work generated will



require approximately 2 ml of oxygen uptake. Respiratory muscles are an essentialpart of the oxygen-transport system to provide oxygen for exercise. However,respiratory-muscle function comes at a metabolic cost measured by the oxygen costof breathing. From the data used to generate Fig. 4 describing power requirements ofventilation Tables 6 and 7 have been set up to show oxygen requirements of breath-ing and possible mechanisms whereby muscles of breathing can limit exercise.

5.3 Possible limits imposed by respiratory muscles in Olympic athletes

Calculations in Table 6 are based on measurements in an Olympic bicyclist with amaximal O2 uptake of 6.14 l/min, maximal ventilation of 200 l/min and a maximalcardiac output of about 35 l/min [43]. We used eqn. (5) to calculate the mechanicalpower required to support ventilation as bicycle workload increases, assuming nor-mal K1 and K2 for mouth breathing, Table 2. Mechanical power was converted tooxygen requirements assuming carbohydrate metabolism for which 2 ml of oxygenare required to support each 1 kg·m of mechanical work at mechanical efficiencyof 25%. Based on these calculations at maximal bicycle workload O2 requirementsof respiratory muscles would be 0.334 l/min, which would be 5.4% of the total O2uptake. Blood-flow requirements of respiratory muscles necessary to supply oxygenneeds were calculated assuming an arterial O2 saturation of 95% with 95% extrac-tion of oxygen delivered. Based on these calculations blood flow to respiratorymuscles at maximal O2 uptake would be 1.76 l/min, 5% of the cardiac output. If thediaphragm were generating 30% of the mechanical work of breathing, as has beenmeasured in the dog, diaphragm blood flow would be 528 ml/min. If the diaphragmweight is 5 gram/kg body weight, blood flow to the diaphragm of this 75-kgathlete would be 1.4 ml/min/g of muscle. The highest blood flow that we have seenin the dog diaphragm under extreme conditions of inspiratory loading averaged2 ml/min/g of muscle [39]; hence this estimated diaphragm blood flow at peakexercise is a reasonable magnitude in the athlete but might be approaching anupper limit. More than likely in the Olympic athlete all transport steps, maximalcardiac output, maximal power development by the respiratory muscles and max-imal power from exercising skeletal muscles are approaching upper limits at thesame time.

5.4 Possible limits imposed by respiratory muscles in a patient with asthma

The situation is different in the asthmatic patient described in Table 7 where similarcalculations have been made to those in Table 6, except that K1 and K2 are assumedto be 5-fold higher for estimating the greater power requirements of breathing inasthma and mechanical efficiency is less. Note in the last column of the table thatin this patient the oxygen requirements of the respiratory muscles become so greatthat any attempt to increase ventilation above 100 l/min provides no further increasein oxygen for the rest of the body including locomotor muscle. Exercise could notbe sustained at this level. Total body oxygen uptake that could be sustained would



Tabl

e6:

Tota

lO

2up

take

duri

ngex

erci

sein

anat

hlet

epl

uspe

rce

ntof

the

tota

lre

quir

edby

resp

irat

ory

mus

cles

(RM

)fo

ren

ergy

requ

irem

ents

ofbr

eath

ing

and

that

left

over

for

sust

aini

ngex

erci

se*.

Tota

lO2

upta

keV

entil

atio

nC

ardi

acou

tput

RM

O2

upta

keR

MO

2up

take

RM

bloo

dflo

wR

Mbl

ood

flow

O2

avai

labl

efo

r(l

/min

)(l

/min

)(l

/min

)(l

/min

)(%

ofto

tal)

(l/m

in)

(%of

tota

l)ex

erci

se(l

/min

)

1.94

6012

.60.

019

1.0

0.1

0.8

1.82

2.45

8015

.70.

036

1.5

0.19

21.

22.

413.

0710

018

.90.

061

2.0

0.32

31.

73.

013.

6812

022

.10.

095

2.6

0.49

92.

33.

594.

2914

025

.30.

138

3.2

0.72

42.

94.

154.

9116

028

.50.

191

3.9

1.00

03.

94.

725.

5218

031

.70.

256

4.6

1.35

04.

35.

266.

1420

034

.90.

334

5.4

1.76

05.

05.

81

*Ass

umin

gK

1=

2.4

cmH

2O

/(l/s

)an

dK

2=

0.3

cmH

2O

/(l/s

)2in

eqn.

(5),

anal

veol

arve

ntila

tion

80%

ofth

eto

tals

oth

atea

ch20

l/min

incr

ease

inve

ntila

tion

prov

ides

0.61

l/min

utili

zabl

eO

2,2

5%m

echa

nica

leffi

cien

cyfo

rmus

cles

ofbr

eath

ing,

O2

capa

city

of0.

2lO

2/l

ofbl

ood

and

95%

O2

extr

actio

nby

exer

cisi

ngm

uscl

es.



Tabl

e7:

Tota

lO2

upta

kedu

ring

exer

cise

inan

asth

ma

patie

ntpl

uspe

rcen

toft

heto

talr

equi

red

byre

spir

ator

ym

uscl

es(R

M)f

oren

ergy

requ

irem

ents

ofbr

eath

ing

and

that

left

over

for

sust

aini

ngex

erci

se*.

Tota

lO2

upta

keV

entil

atio

nC

ardi

acou

tput

RM

O2

upta

keR

MO

2up

take

RM

bloo

dflo

wR

Mbl

ood

flow

O2

avai

labl

efo

r(l

/min

)(l

/min

)(l

/min

)(l

/min

)(%

ofto

tal)

(l/m

in)

(%of

tota

l)ex

erci

se(l

/min

)

0.38

205.

00.

018

4.7

0.10

2.0

0.36

0.77

407.

90.

083

10.8

0.44

6.3

0.69

1.15

609.

00.

207

18.0

1.09

12.1

0.94

1.53

8011

.00.

403

26.3

2.12

19.3

1.13

1.92

100

13.0

0.68

535

.73.

6127

.81.

242.

3012

015

.01.

066

46.3

5.61

37.4

1.23

2.68

140

16.9

1.55

858

.18.

2048

.51.

123.

0716

019

.02.

175

70.8

11.4

560

.30.

893.

4518

020

.92.

9384

.915

.42

73.8

0.52

*Ass

umin

gK

1=

10cm

H2O

/(l/s

)an

dK

2=

1.5

cmH

2O

/(l/s

)2in

eqn.

(5),

alve

olar

vent

ilatio

n=

50%

ofto

tal

vent

ilatio

nso

that

each

incr

emen

t20

l/min

vent

ilatio

npr

ovid

es0.

35l/m

inof

utili

zabl

eO

2,1

0%m

echa

nica

leffi

cien

cyfo

rmus

cles

ofbr

eath

ing,

anO

2ca

paci

tyof

0.2

lO2/l

ofbl

ood

and

95%

O2

extr

actio

nby

exer

cisi

ngm

uscl

es.



probably be less than 1 l/min. Exercise is clearly limited by the respiratory musclesand high energy requirements of breathing.

6 Interactions of heart and lungs

The lungs, thorax and heart constitute two interactive pumps, one pumping air(lungs and respiratory muscles of the thorax) and the other pumping blood (the rightventricle pumping blood through the lung for adding oxygen and removing carbondioxide from blood returning from the periphery and the left ventricle pumpingblood into peripheral organs to provide substrate and oxygen. The two pumps mustinteract efficiently within the gas-exchange region of the lungs to maintain normalarterial blood gases leaving the lung. Furthermore, since the heart pump lies entirelywithin the thoracic pump they must interact mechanically in a way that does notimpede the function of the other pump, despite the fact that they pump at differentrates and stroke volumes in a limited amount of space [44]. In fact, the interactionin many instances is probably beneficial. The pumping action of the heart causespressure fluctuations in conducting airways and gas-exchange regions of the lungthat may enhance uniformity of gas mixing in the lung. The diaphragm is the majorinspiratory pump in the thorax and operates like an inspiratory piston that movesdownward in upright humans to pull air into the lung during inspiration. In humansthe diaphragm has a firm connection with the heart, which holds the diaphragm upagainst the pull of gravity at the end of a normal breath; at the end of a breath it is inposition to exert its piston-like function for the next inspiration. Quadrupeds whonormally run in a horizontal orientation do not have this connection between heartand diaphragm; consequently when they assume an upright position the diaphragmis not held up in a position necessary for an effective inspiratory stroke. Thesemechanical interactions between heart, lungs and thorax are not well understood inhealth or disease and need further study.

Acknowledgement

Figure 1 is reprinted with kind permission from Springer Science and BusinessMedia ([1], Figure 82, page 111). Figure 10 is reprinted with permission fromThe American Society for Clinical Investigation ([36], Figure 5, page 364).Figures 3 and 11 [18], and Figures 6 and 8 [20] are reprinted with permissionfrom Elsevier. The authors wish to acknowledge the assistance of D. Merrill Daneand Jeanne-Marie Quevedo on this chapter.

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[3] Hislop,A. & Reid, L., Development of the acinus in the human lung. Thorax,29(1), pp. 90–94, 1974.

[4] Dunnill, M.S., Postnatal growth of the lung. Thorax, 17, pp. 329–333, 1962.[5] Davies, G. & Reid, L., Growth of the alveoli and pulmonary arteries in

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[10] Rohrer, F., Der Stromungswiderstand in den mensachlichens AtemwegenundVerzeigung des Bronchialsystems auf denAtmungsverlauf in verschiede-nen Lungenbezwirken. Pfluger’s Archiv fur die gesamte Physiologie, 162,pp. 225–299, 1915.

[11] von Recklinghausen, H., Uber dieAtemgrosse des Neugebhorenen. Pfluger’sArchiv fur die gesante Physiologie, 62, pp. 451–493, 1894.

[12] Mead, J. & Agostoni, E., Dynamics of breathing. Handbook of Physiology,Section 3 Respiration, eds. A. Wallace, O. Fenn & H. Rahn, American Phys-iological Society: Washington, p. 411–428, 1964.

[13] Ross, B.B., Influence of bronchial tree structure on ventilation in the dog’slung as inferred from measurements of a plastic cast. Journal of AppliedPhysiology, 10(1), pp. 1–14, 1957.

[14] Fung, Y.C., A model of the lung structure and its validation. Journal ofApplied Physiology, 64(5), pp. 2132–2141, 1988.

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[17] Johnson, R.L. Jr. & Curtis, S.E., Experimental approaches to the study of gasexchange. Lung Biology in Health Disease, vol. 148: Pulmonary and Periph-eral Gas Exchange in Health and Disease, eds. J. Roca, R. Rodriguez-Roisin& P.D. Wagner, Marcel Dekker: New York and Basel, pp. 45–119, 2000.

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