CHAPTER 1 Anatomy and Physiology of the Human Respiratory System

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CHAPTER 1

Anatomy and physiology of the human respiratory system

R.L. Johnson, Jr., & C.C.W. Hsia Department of Internal Medicine, University of Texas Southwestern Medical Center, Dallas, Texas, USA.

Abstract

A brief review of the anatomy and fundamental physiologic concepts of the human respiratory 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 microvascular recruitment, pulmonary circulatory function and respiratory muscle energetics. The emphasis is placed on the close interactions among different respiratory components and between the heart and the lung, which are essential for optimizing oxygen transport 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 subdivided into 9 sublobular segments. In the left lung there are 2 lobes (upper and lower), which are subdivided into 8 sublobular segments. There are approximately 23 generations of airways (Z) that distribute respired air to and from the smallest gas-exchange units in the lung (alveoli) [1], Fig. 1. The path length and number of generations from trachea to different acini in the lung are not uniform owing to nonspherical asymmetry. Path lengths from trachea to 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 about 150 µm at birth.

Figure 1: Diagram of the branching airways. Z designates the order of generations with the trachea as zero, followed by bronchi (BR) and bronchioles (BL). On average the ﬁrst 16 generations are conducting airways that do not participate in gas exchange. The last conducting airway is a terminal bronchiole (TBL), which empties into an acinus. In the acinus there are 3 generations of partially alveolated respiratory bronchioles (RBL) followed by alveolar ducts (AD) that are completely alveolated in adults and empty 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 of those in the adult. No new generations are added after birth; only size and length change during postnatal growth. This is not true of the acinus. Alveoli multiply by adding more alveolar ducts as well as by lengthening of existing ducts and increasing 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 during postnatal growth is from an increase in number of alveoli within the ﬁrst 5 to 8 years and the remaining 1/3 from a steady increase in alveolar size that continues until the lung reaches adult size. Rates of increase in size of the thorax and of the lung

400 Adult

300

200

100 Number of Alveoli (millions)

0 0 5 10 15 20 25 Years since birth

Figure 2: Number of alveoli increase about 15-fold after birth. Alveolar diameter also increases linearly with time after birth until growth is complete [4]; alveolar diameters approximately double, causing an approximate 8-fold increase in volume. must remain exactly matched throughout postnatal growth and there probably exist mechanisms 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 air Conducting air passages distribute respired air to and from lung acini where gas exchange occurs; they constitute what is called anatomical dead space that must be displaced during inspiration before fresh air reaches the acinus where gas exchange occurs. Upper airways consist of the nasopharynx, oropharynx and glottis. Below the glottis, Fig. 1, there is the trachea followed by an average of 16 generations of dichotomously branching conducting airways down to terminal bronchioles, which average approximately 0.06 cm in diameter. Terminal bronchioles empty into acinar airways where gas exchange occurs in alveolated airways. In each acinus there are approximately 3 generations of respiratory bronchioles that are partially alveolated, and additional generations of alveolar ducts that are completely alveolated, finally ending 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 of 4800 ml minus the volume of conducting airways (150 ml); hence, the average size of individual acini would be about 0.072 cm3. These structures are made up of very fine alveolar septa where circulating red blood cells are separated from alveolar air by a distance of less than a µm [1].

2.1.2 Air conditioning and filtering inspired air Another major function of conducting air passages is to humidify the inspired air [7] and filter out small particulate matter to protect this thin gas-exchange surface from dehydration and accumulation of inspired atmospheric debris [8, 9]. Particulates with an aerodynamic diameter larger than 10 µm are filtered out by inertial impact at airway branches and seldom reach an acinus; they are removed by 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 portion of the acinus and be removed by scavenging macrophages or breathed out again without being deposited. There appears to be little difference between the adult and child’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 adult lung [2].

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

In 1915 Fritz Rohrer [10] published a classic paper describing the quantitative anatomy of the irregular dichotomous branching of airways down to subsegmental 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 acinus. Between the carina of the trachea and lobules in different regions of the lung he measured the numbers of generations, airway diameters and lengths, angles of branching and changes in cross-sectional area at each branch point.Airways dimensions in the lobules were generated by an equation for regular dichotomy derived by von Recklinghausen [11]. Based on the anatomy of the branching airways system and physics of flow in tubes, Rohrer derived the pressure–flow relationships and distribution of ventilation in human airways. Distribution was not uniform because the dichotomous branching was not uniform. Path length from the carina to the lobules varied from 8 cm in the central regions of the lung to 14 cm in the more peripheral regions causing uneven distribution of resistance. The overall pressure drop from nasopharynx to alveoli required to generate increasing flow could be reduced to a single comprehensive equation of the following form:

˙ ˙ 2 P = K1V + K2(V ) , (1) ˙ where P is the pressure drop in cmH2O across the lung; V is gas ﬂow in l/s; K1 is a constant derived from the summation of laminar-ﬂow resistances in branching airways in cmH2O/(l/s) and K2 is a constant derived from changing resistance from turbulence and from inertance at airway branch points and at changes in 2 cross-sectional area of the airways; units are in cmH2O/(l/s) . Results reported by Rohrer for resistances across the upper airways, lower airways and total are given in Table 1 [10]. Fifty years later K1 and K2 in eqn. (1) derived by Rohrer have been determined experimentally to describe viscous properties of lung airways, lung tissues and thorax, Table 2.

Table 1: Pressure–ﬂow relationships derived by Rohrer.

Upper airways P = 0.426 · V˙ + 0.714 · (V˙ )2 Lower airways Bronchial 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 person breathing through the mouth or nose [12].

Mouth breathing Nasal breathing

Flow Laminar Turbulent Laminar Turbulent

Resistance K1 K2 K1 K2 2 2 Units cmH2O/(l/s) cmH2O/(l/s) cmH2O/(l/s) cmH2O/(l/s) Air ﬂow 1.2 0.3 1.8 3.0 Lung tissue 0.2 0 0.2 0 Chest wall 1.0 0 1.0 0 Resp. system 2.4 0.3 3.0 3.0

˙ Thus, to describe the pressure in cmH2O to generate a volume ﬂow of V l/s through the mouth would be P = 2.4V˙ + 0.3 · [V˙ ]2 and through the nose would be 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 and thoracic 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 uniform distribution of ventilation even in the normal lung. As pointed out above, Rohrer found that resistance to flow is greater to peripheral lung lobules than to more central ones [10]. These regional differences in flow resistance do not necessarily mean that ventilation will be nonuniform. The rate at which a lung unit fills is not determined solely 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 in seconds. Even if resistance to flow in peripheral units is greater than in more central units ventilation may be the same if RC constants are the same, in analogy to RC constants in an electrical circuit. If RC constants for filling peripheral lobular units are longer than for low-resistance central units, ventilation will become more and

WIT Transactions on State of the Art in Science and Eng ineering, Vol 24, © 2006 WIT Press www.witpress.com, ISSN 1755-8336 (on-line) 6 Human Respiration more nonuniform as the frequency of breathing increases. If RC constants match between central and peripheral units, ventilation will remain uniform in spite of unequal resistances to ﬂow. Rohrer did not measure regional compliances although he discussed this problem. Ross [13] compared measurements and calculations to those of Rohrer in plaster casts of the dog lung between the carina and airways down to 1.5 mm diameter. He also found that airway path lengths varied from 2 cm centrally to 13.7 cm peripherally, with corresponding differences in airway dead- space volumes. Anatomical dead-space volume is the volume of air in conducting airways that must be cleared during a breath before fresh air reaches the alveoli ˙ where 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 peripheral and central lobules is the same, alveolar ventilation to peripheral lobules will be less because of the larger dead-space volume that must be cleared. The effects of space-ﬁlling constraints on airway branching and distribution of ventilation extend all the way down to the alveolar ducts, as pointed out by the work of a number of investigators [13–16]. As pointed out by Johnson and Curtis [17], “because of these space-limiting constraints, irregular branching of the bronchial tree becomes necessary to pack the largest alveolar surface area into a limited space with the least power requirements for distribution of ventilation; the trade off is uneven ventilation.”

2.4 Elastic properties of the lungs and thorax

The balance between the static elastic recoil forces in the lungs and thorax determines the pressure–volume relationships of the respiratory system when respiratory muscles 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 the expiratory pressure exerted by recoil of the lungs. The lung volume at this balance point is referred to as the functional residual capacity (FRC). The total pressure that is generated by the respiratory system at a given lung volume (excluding muscle) is the algebraic sum of the recoil pressure exerted by the chest and that exerted by the lungs. 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 the adjacent surfaces of the lung and chest wall) reflects the expiratory recoil pressure of the lungs. We have discussed the pressure required to overcome flow resistance in the airways to inspire, eqn. (1). The additional pressure (P) required to inflate the lungs by a volume V against the elastic recoil pressure of the lungs and thorax is given by

P = V /CRS, (2) where CRS is compliance of the lungs and thorax as deﬁned in Fig. 3 and is expressed in l/cmH2O.

Figure 3: Static pressure–volume curves of lungs and thorax in a normal upright subject. The pressure generated by the total respiratory system at a given lung volume is represented by the heavy curved line. The two thin curved lines represent pressures exerted by the chest wall on the left and by the lungs on the right. A positive sign indicates an expiratory pressure; a negative sign indicates an inspiratory pressure. Reproduced from Miller et al [18].

2.5 Work of breathing

Ventilation requires energy to overcome both elastic and viscous forces in lungs and thorax. Incorporating both eqns. (1) and (2) the following approximations were derived by Otis et al [19]: V ˙ ˙ 2 P = + K1V + K2 · V , (3) CRS where V is the volume above the functional residual capacity (FRC) and other symbols are as deﬁned for eqns. (1) and (2). Multiplying both sides by V˙ we have the power requirement at any instant: · ˙ ˙ V V ˙ 2 ˙ 3 Instantaneous Power = PV = + K1 · V + K2 · [V ] . CRS

WIT Transactions on State of the Art in Science and Eng ineering, Vol 24, © 2006 WIT Press www.witpress.com, ISSN 1755-8336 (on-line) 8 Human Respiration ˙ ˙ If we assume that ventilation follows a sine-wave pattern, V = VPsin(2πft) where ˙ VP is peak ﬂow in l/s and we assume that energy is required only during inspiration since during quiet breathing the elastic energy during inspiration is stored and returned passively during expiration, we can derive the following equation for estimating energy requirements of quiet breathing: ˙ = = ˙ 2 + · π2 ˙ 2 + π2 · ˙ 3 W Power (1/2fCRS)[VE] 60 (K1/4) VE (2K2/3) VE , (4) ˙ where VE has been converted from l/s in eqn. (3) to ventilation in l/min; f is respira- −1 −1 tory rate in min ; 60 is s/min so that power is now expressed as cmH2O·l·min . During exercise, when respiratory muscle energy is active throughout both inspiration and expiration, eqn. (4) can be modiﬁed as follows to approximate energy requirements of breathing during heavy exercise: ˙ 2 ˙ 2 2 ˙ 3 W = 2 · 60 · (K1/4) π · VE + (2K2/3) π · VE /100. (5)

−1 Dividing by 100 converts power from units of cmH2O·l·min to more familiar engineering units of kg·m·min−1, Fig. 4. Ventilatory capacity may be limited in the patient because respiratory muscle mass is insufﬁcient to generate the pressures required for a normal ventilatory effort.Another possible limitation of ventilation and exercise capacity in the patient concerns metabolic requirements, which will be considered later.

250 Athlete Patient with asthma 200

150

100

50 Power Requirements (kg Requirements Power m/min) 0 0 50 100 150 200 250 Ventilation (l/min)

Figure 4: Power requirements are plotted in an athlete assuming normal values of K1 and K2 for lungs and thorax from Table 2 compared to power requirements in a hypothetical patient with asthma assuming about a 5-fold increase of K1 and K2 during an asthmatic attack (i.e., K1 =10 2 cmH2O/(l/s) and K2 = 1.5 cmH2O/(l/s) ).

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 oxygen transport and mixing within this front changes from a combination of turbulent and laminar convection in the conducting airways to predominantly diffusive mixing as it moves into the acini and finally into the alveoli where alveolar capillary gas exchange occurs. Weibel’s branching model of the airways [1] provides the dimensions of each airway generation. From these dimensions there are two nondimensional numbers that can define critical change in flow patterns from laminar to 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 flow The relative importance of laminar and turbulent flow patterns at different generations 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 kinematic viscosity). Turbulence develops normally at corners, sudden changes in direction or airway diameter as at branch points. Once developed, inertial forces tend to maintain turbulence, while viscosity tends to damp it out, depending on the balance between these two forces. Even in unbranched smooth tubes turbulence tends to become sustained when the Reynolds number exceeds 2000. Resistance to laminar flow 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 fluid density and the square of the volume velocity, [V˙ ]2, as in eqn. (3). Although turbulence increases the power requirements of breathing it is an important source of gas mixing, heat exchange, humidification, filtration of inspired particulate matter and gas mixing in the lung before it reaches the alveoli.

3.1.2 Relative importance of convection and diffusion in transport The relative importance of convective and diffusive transport in different generations of airways can be estimated from the Peclet number (Pe). Pe is a dimensionless ratio of rate of transport by convection down a concentration gradient to the rate of diffusion down the same concentration gradient, i.e., (v · x)/d where v = velocity of convective ﬂow down the tube in cm/s, x = length of the tube in cm and d is the diffusion coefﬁcient of oxygen in respired air given in cm2/s. As Pe falls below 1.0 the rate of transport by diffusion begins to exceed that by convection. In the lower conducting airways and acini the combination of diffusion and convection provides an important source for mixing of gases not only within each acinus but also between acini. Table 3 summarizes these relationships. During quiet breathing turbulence is minimal except at airway branch points and diffusive transport is

WIT Transactions on State of the Art in Science and Eng ineering, Vol 24, © 2006 WIT Press www.witpress.com, ISSN 1755-8336 (on-line) 10 Human Respiration 0.887 9.938 0.316 0.136 0.059 0.029 2.090 4.672 2108.1 21128.8 Reynolds number 0.381 0.145 0.055 0.023 ‡ Exercise ventilation = 100 l/min between trachea and alveoli. 7.011 3.204 3.1661.176 0.530 0.240 1.224 0.120 Peak flow Shaded areas under Peclet numbers designate airways AD = alveolar duct and AS = alveolar sac. rate convective transport. Shaded areas under the columns of 0.311 14.509 7.991 0.139 0.059 0.021 0.009 0.004 0.002 0.663 27.778 17.904 140.5 1262.6 4438.92 where turbulence tends to become sustained. 1408.5 1968.5 92273.6 0.533 0.214 0.082 0.025 0.010 0.004 0.002 Resting ventilation = 6 l/min ‡ 0.211 1.8520.967 1.194 0.035 0.016 0.008 0.467 0.078 cm/s Peclet number Reynolds number cm/s Peclet number Peak flow ∗ Table 3: Convective and diffusive gas transport and mixing in the airways 6 Bronchus 84.17 295.9 0 Trachea 131.23 6151.5 Peak flow during a respiratory cycle is about 3 times the average ventilation. TB = terminal bronchiole entering the acinus; RB = respiratory bronchiole; 1617 TB RB1 212223 AD2 AD3 AS 19 RB3 18 RB2 20 AD1 Generation Airway ∗ ‡ at the given ventilationReynolds where numbers rate indicate of those airways diffusive at transport the exceeds given the ventilation

much faster that convection in the acinus. During heavy exercise sustained turbulence extends down to subsegmental bronchi and diffusive transport is still faster than 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 the ventilation relative to the respective rates of CO2 output and O2 uptake.

3.2.1 CO2 equilibrium Thus, the alveolar equation for CO2 states ˙ VCO2 PaCO = · 863, (6) 2 V˙ a a = ˙ = ˙ a = where P CO2 alveolar CO2 tension, VCO2 CO2 output, V Alveolar ventilation and 863 = 760 (body temp. in Kelvin)/273. The difference between end-capillary CO2 content in mM/ml leaving the lung (Cc CO ) and mixed venous CO2 content (Cv¯CO ) entering the lung multiplied by 2 ˙ 2 ˙ blood ﬂow (Q) in ml/min equals the CO2 output (VCO2 ) in ml/min. This is the Fick equation for CO2:

V˙ := Q˙ (Cv¯ − Cc ) · 22.2, (7) CO2 CO2 CO2 where 22.2 = ml/mM. Combining eqns. (6) and (7) and rearranging we have ˙ ˙ V a/Q Cc = Cv¯CO − · PaCO , (8) CO2 2 19.2 2

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

3.2.2 O2 equilibrium The same set of equations can be derived for oxygen. The alveolar equation for O2 states that Fi Pa = (Pb − P ) · Fi − Pa · Fi + N2 , (9) O2 H2O O2 CO2 O2 R

End capillary blood

Figure 5: CO2 equilibrium between alveolar air and blood leaving lung capillaries at a 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 the CO2 dissociation curve. The point of intersection indicates equilibrium.

a = = = where P CO2 alveolar O2 tension; Pb barometric pressure; PH2O water ◦ i = vapor tension at body temperature (47 mmHg at 37 C); F O2 inspired O2 fraction i = (approximately 0.21 breathing room air at sea level); F N2 inspired N2 fraction (normally taken as 0.78 in room air at sea level). Furthermore, there is also a Fick equation for oxygen:

V˙ = Q˙ (Sc − Sv¯ ) · Cap, (10) O2 O2 O2 where Sc and Sv¯ are fractional oxyhemoglobin saturation in the end-capillary O2 O2 and mixed venous blood, respectively, and Cap is the O2 capacity of blood in ml of O2/ml blood. Then, combining eqns. (6), (9) and (10), ˙ ˙ V a/Q Sc − Sv¯O = PiO − PaO · FiO · O2 2 2 2 2 · i · 863 F N2 Cap V˙ a/Q˙ −· · Pa . (11) · i · O2 863 F N2 Cap

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

a Oxyhemoglobin Equilibrium b VA/Q = 6.0 100 dissociation curve VA/Q = 3 100 SaO2

75 75

50 mixed 50 VA/Q = 0.5 venous

25 25 Oxygen Saturation (%)

PAO2 PIO2-K PaO2 PA 0 0 O2 0 25 50 75 100 125 150 0 25 50 75 100 125 150 Oxygen Tension (mmHg) Oxygen Tension (mmHg)

Figure 6: (Panel a)O2 equilibrium between alveolar air and blood leaving lung i capillaries at a known inspired O2 tension (P O2 ) and mixed venous O2 saturation. The solid straight line is eqn. (11) at a ventilation perfusion ˙ ˙ ratio (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 uneven V˙ a/Q˙ . This shows the effect of two regions of lung with equal blood ﬂows but different V˙ a/Q˙ ratios of 6 and 0.5. A mixture of blood from the

two regions will have an O2 saturation and PaCO2 approximately half- a − way between and a resulting increase in P O2 PaO2 as indicated in the diagram. After Hsia and Johnson [20].

3.2.3 Effect of uneven V˙ a/Q˙ ratios in the lung The most efﬁcient gas exchange is achieved when V˙ a/Q˙ ratios are uniform and a − the alveolar-arterial O2 tension and CO2 tension differences (P O2 PaO2 and a − P CO2 PaCO2 ) are zero. Increasing nonuniformity of ventilation with respect to blood ﬂow causes these gradients to increase. Ventilation has to be increased in order 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 and binding sites on hemoglobin in capillary red cells divided by the driving pressure in mmHg and expressed in units of ml·(min·mmHg)−1. Thus, for oxygen ˙ VO2 DL = , (12) O2 a − P O2 PcO2 ˙ a where VO2 = Oxygen uptake in ml/min, P O2 = Alveolar O2 tension and PcO2 = mean oxygen tension of capillary hemoglobin, both in mmHg.

The difﬁculty with measuring D arises from the fact that Pc changes as LO2 O2 a ˙ blood passes through the lung. Although P O2 and VO2 can be measured PcO2 cannot because it changes during transit until equilibrium is reached or leaves the capillary bed without being accessible to measurement. On the other hand, under conditions used for measuring DLCO changes in PcCO are negligible during transit. CO binds to hemoglobin so tightly that multiple measurements of DLCO can be made before PcCO rises signiﬁcantly. DLCO can be measured essentially neglecting capillary CO tension as follows: ˙ = VCO DLCO . (13) PaCO To make this measurement during breath–holding, a test gas is made up containing a small concentration of CO and of an inert insoluble tracer gas such as He or CH4 mixed with a balance a balance of air or oxygen. A measured volume of this mixture 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 disappeared with respect to the inert, insoluble tracer for calculation of DLCO . This is the breath- holding technique and is the most common clinical measurement. The inspired breath can also be breathed back and forth in a reservoir bag while the disappearance of CO with respect to the tracer is followed with each breath by a rapid analyzer for calculation of DLCO . The latter is called the rebreathing technique and has the advantage of providing measurements during exercise. A small concentration of acetylene can also be added to the test gas for simultaneous measurement of cardiac output. Nitric oxide also reacts extremely rapidly with hemoglobin and is taken up much more rapidly by red cells than CO. DLNO can be measured by either the breath-holding or rebreathing method simultaneously with measurements of

DLCO . However, before going further it is important to review the physical laws involved in these measurements.

3.3.2 Fick’s law of diffusion deﬁnes how diffusive transport moves down a concentration gradient The rate of diffusive transport of a gas in a uniform media such as plasma or the alveolar wall is described by Fick’s law of diffusion as follows:

dQ ∂C = Ad , (14) dt ∂x where dQ = volume transfer of the gas in cm3/s in the x direction; dt A = area in cm2 of the barrier through which diffusion is occurring. d = the diffusion coefﬁcient in cm2/s, and ∂C − = concentration gradient through the barrier in the x direction in cm 1 ∂x

3.3.3 Krogh’s diffusion constant defines how a concentration gradient is translated 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 6 · α · d ∂P = A · · , (15) dt 760 ∂x where dQ = volume transfer in cm3/min from air through the alveolar wall dt A = surface area of the alveolar wall in cm2 − α = Bunsen solubility coefficient in atm 1 = 2 d diffusion coefficient in cm /s 6 · α · d − − = Krogh’s diffusion constant in cm2·min 1·mmHg 1, 760 ∂P = pressure gradient in mmHg/cm across the wall. ∂x Thus, using Krogh’s constant [21], diffusion is described as being driven by a pressure gradient instead of a concentration gradient. Furthermore, it allows us to translate a membrane-diffusing capacity measured for one gas into that for another gas of different molecular weight and solubility.

3.3.4 Diffusing capacity of the pulmonary membrane (DM) and its relationship to the Krogh constant This is given by 60 · α · d A A D = = K · , (16) M 760 x x

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 gases that are important in pulmonary physiology. If Krogh diffusion constants are known, measurements of DM for one gas can be translated into an estimate for another gas. For example:

DM K 3.87 O2 = O2 = . (17) DMCO KCO 3.48

3.3.5 Rate of uptake of respired gases by red cells involving diffusion and chemical binding Diffusive uptake of respired gases is determined not only by the resistance of the 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

−2 −8# Respiratory gases O2 1.79 · 10 3.87 · 10 −2 −8 ∗ CO2 1.26 · 10 64.5 · 10 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 −1 −1 (ml·min )STPD·mmHg 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 must be 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 rates of gas uptake (referred to as θ) have been measured for O2,CO2, CO and NO using rapid reaction-rate techniques in red-cell suspensions and the results expressed in terms of [(ml/min)gas uptake/mmHg gas tension]/ml of blood having an O2 capacity of 20 ml/dl. Representative data are given in Table 5 with references. CO competes with O2 for binding sites on hemoglobin; hence, as oxygen tension increases θCO decreases as shown in Table 5 and has provided a method for measuring the true membrane-diffusing capacity for CO and pulmonary capillary blood volume by a technique devised by Roughton and Forster [26, 27]. There is no such competition with oxygen for binding sites by the other gases.

3.3.6 The Roughton–Forster method for estimating membrane-diffusing capacity (DM) and pulmonary capillary blood volume (Vc) [27] The Roughton–Forster equation [27] states that the total resistance to oxygen uptake by blood ﬂowing through the lung capillary bed equals the reciprocal of the lung-diffusing capacity of the lung (1/D ), which in turn is the sum of LO2 the resistances in series imposed by the alveolar capillary membrane (1/DMO ) · 2 and that imposed by red cells in pulmonary capillary blood ((1/θO2 ) (1/Vc)) as

Similar equations can be set up for DLCO measured at two or more alveolar oxygen tensions, eqns. (19) and (20). As oxygen tension increases 1/θCO increases and the measured DLCO correspondingly falls, Table 5; hence at two different oxygen tensions the Roughton–Forster equation [27] yields two simultaneous equations with two unknowns that can be solved for DM and Vc. CO 1 1 1 1 = + · (19) DLCO DMCO θCO O Vc O2 2 1 1 1 1 = + · . (20) DLCO Air DMCO θCO Air Vc

The Krogh constants for O2 and CO from Table 4 and θO2 from Table 5 can then be used to estimate D from the D and V derived from the two simultaneous LO2 MCO c equations, again using the Roughton–Forster equation as follows: 1 K 1 1 = CO + · . (21) D K D θ V LO2 O2 MCO O2 c

More recently, methods have been developed to measure DLNO and DLCO simultaneously. NO reacts so rapidly with hemoglobin in red cells that the resistance imposed by the red cell to NO uptake approaches zero and DLNO is the same as

DMNO . Based on this assumption DMCO and Vc can be estimated empirically from the following modiﬁcation of the Roughton–Forster equation: 1 1 1 1 = + · . · (22) DLCO 2.4 DLNO θCO Vc

3.3.7 Recruitment of diffusing capacity during exercise

During exercise all components of DLCO increase in an approximately linear relationship to cardiac output and a plateau is not reached within the physiological range, Fig. 7. D has been estimated indirectly from D and V as illustrated LO2 MCO c in the previous sections and is about 50% higher at any given cardiac output than D . D is higher primarily because the rate of uptake of oxygen by red cells LCO LO2 is 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 tissue and because the red cells offer little resistance to NO uptake.

3.3.8 The importance of the relationship between D and cardiac LO2 output [20] If D did not increase as cardiac output increases during exercise arterial LO2 oxyhemoglobin 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 cardiac output and does not reach a plateau over the physiological range of measurement.

From eqns. (10) and (12): ˙ = ˙ − ¯ · VO2 Q ScO SvO2 Cap 2 V˙ = D · Pa − Pc , O2 LO2 O2 O2 it is possible to derive that the ratio of diffusing capacity to pulmonary capillary blood flow, D /Q˙ bears the following relationship to oxygen saturation of blood LO2 c ¯ leaving lung capillaries (ScO ) when mixed venous oxygen saturation (SvO2 ) and a 2 alveolar oxygen tension (P O2 ) are known: Sc DL O2 d(Sc ) O2 = O2 ˙ Cap . (23) Q ¯ (PaO − PcO ) c SvO2 2 2 In order to numerically integrate the right-hand side of eqn. (23) to calculate the relationship between Sc and D /Q˙ at rest and exercise the shape and O2 LO2 c position of the oxyhemoglobin dissociation curve must be known. This can be provided with reasonable accuracy by the Hill equation in the following form = − 0.37 PcO2 P50[ScO2 /(1 ScO2 )] . The P50 is the oxygen tension at which the hemoglobin saturation (ScO2 ) is 50%. The P50 represents the position of the curve that changes when body temperature, blood PCO2 , or pH change, for example with exercise. The shape of the curve remains fixed. Under normal resting conditions the P50 is 26.5 mmHg but increases during exercise as body temperature increases and pH falls. We have calculated the relationship between end-capillary oxygen saturation Sc and D /Q˙ as illustrated in Fig. 8. O2 LO2 c Results show that as D /Q˙ falls during exercise oxygen saturation of blood LO2 c leaving the lung remains almost fixed until a critical level is exceeded when Sc O2

Figure 8: As the ratio of D /Q˙ moves below a critical point, there is a rapid fall LO2 c of oxygen saturation of blood as it leaves the lung. The fall occurs earlier at a low alveolar oxygen tension such as at high altitude or when the P50 of hemoglobin falls as when the pH declines during heavy exercise or when 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 level and at an altitude of 10,000 feet, respectively. Data was also calculated to show the effect 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 saturation if D remained ﬁxed from rest to exercise. At sea level, at rest, D /Q˙ would LO2 LO2 c be about 7.5 and Sc would be in equilibrium with Pa at about 97.5%. By O2 O2 the time cardiac output reached 22 L/min Sc would have fallen to about 80%. O2 At 10,000 feet arterial saturation would fall to between 70 and 80% at a cardiac output of 15 l/min if D failed to rise above its resting level. Why then doesn’t LO2 the system operate with the pulmonary capillary bed fully recruited at rest? As will be discussed in the next section, part of the explanation arises from the fact that the pulmonary circulation must operate at low pressures. In the upright position the pulmonary artery pressure at rest or light exercise is too low to completely recruit the 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 circulation, Fig. 9. It has to operate at a relatively low pressure level in order to sustain rapid diffusive gas exchange between inspired air and the cardiac output as it is pumped at a high rate through thin walled capillaries. The high rate of diffusion requires a thin

50 Pulmonary Artery 40 Capillary Wedge

Pressures (mmHg) 10 Mean Pulmonary Vascular

0 0102030 Cardiac Output (l/min)

Figure 9: Mean vascular pressures are much lower in the lung than in the systemic circulation. This is necessary to avoid stress failure of the pulmonary capillaries during exercise. membrane that must be protected from pressure damage. For proper perspective it 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 a thin membrane wall of about 1 µm average thickness and spread over a surface area 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 pressure of 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 and hemorrhage 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 mmHg to reach the top of the lung; however, at rest it operates at a PA pressure of only about 15 mmHg. Hence, at rest the perfusion pressure to the apical 20% of the lung is zero and this region is only perfused by the bronchial circulation that operates at systemic pressure and keeps the region viable. Because the pulmonary circulation operates at such low pressures it is considered to be functionally divided into 3 pressure zones. In a region where PA pressure is lower than arterial hydrostatic pressure, capillary perfusion pressure is zero and the region operates under zone-1 conditions. In a region where RV pressure exceeds both arterial hydrostatic and alveolar pressures but venous hydrostatic pressure is lower than alveolar pressure, a waterfall condition occurs. Capillary blood exits the alveoli down a waterfall to the 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 pressure relative to right ventricular pressure at rest. The upper 20% of the lung 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 both arterial and venous pressures exceed hydrostatic and alveolar pressure, perfusion pressure becomes arterial minus venous pressure (Part –Pven); this is a zone-3 condition. These kinds of zones are unique to the lung where blood vessels are collapsible and changes in alveolar pressure can collapse or dilate the capillaries to change their resistance or where at zero alveolar pressure capillaries can collapse in a region where hydrostatic pressure is less than zero; blood will be sucked out as if falling down a waterfall [32, 33]. Regions operating under zone-1 conditions do not participate in gas exchange. The capillary bed is only partially open under zone-2 conditions but may be completely open in zone-3. These zone conditions create an uneven distribution of blood ﬂow and diffusing capacity imposed by gravitational stresses 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 wedge pressure increase. More capillaries open and this is one source of capillary recruitment and increase in diffusing capacity but probably not the major source. There may 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 segmental 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 only available for the diaphragm at post mortem [38]. For humans with sedentary occupations, diaphragm weight (g) = 4.18 · body weight (kg) – 21.8. Muscle mass and capability for generating power of all respiratory muscles have been measured in dogs where the diaphragm weight (g) = 5 · body weight (kg), on average about 28% higher than in sedentary humans [39, 40]. However, in humans with occupations requiring heavy labor, the weight of the diaphragm is only about 10% lower than in the dog. Segmental innervations are the same in humans and dogs. Hence studies on muscle mass and power capabilities of respiratory muscles in dogs probably provide a good ﬁrst approximation to respiratory muscle function in humans. The diaphragm is the most important respiratory muscle in both species. In the dog, the diaphragm constitutes only 9% of the respiratory muscle mass but supplies 30 to 40 per cent of the ventilatory power. Inspiratory muscles in general constitute only 40% of the respiratory muscle mass but deliver more than 70% of the total power for breathing during exercise.

5.2 Oxygen requirements of breathing

Oxidation of carbohydrates yields approximately 5 cal/mlO2 and one (kg·m) of mechanical energy is equivalent to 2.33 cal of heat energy, hence 0.5 ml of oxygen would be needed to support one (kg·m) kg m of mechanical work if all of the energy generated by burning carbohydrate could be utilized (100% efficiency). Mechanical efficiency of the respiratory muscles has been measured at 20 to 25% in both the dog and 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

WIT Transactions on State of the Art in Science and Eng ineering, Vol 24, © 2006 WIT Press www.witpress.com, ISSN 1755-8336 (on-line) Anatomy and Physiology of the Human Respiratory System 23 require approximately 2 ml of oxygen uptake. Respiratory muscles are an essential part of the oxygen-transport system to provide oxygen for exercise. However, respiratory-muscle function comes at a metabolic cost measured by the oxygen cost of breathing. From the data used to generate Fig. 4 describing power requirements of ventilation Tables 6 and 7 have been set up to show oxygen requirements of breathing 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 a maximal O2 uptake of 6.14 l/min, maximal ventilation of 200 l/min and a maximal cardiac output of about 35 l/min [43]. We used eqn. (5) to calculate the mechanical power required to support ventilation as bicycle workload increases, assuming normal K1 and K2 for mouth breathing, Table 2. Mechanical power was converted to oxygen requirements assuming carbohydrate metabolism for which 2 ml of oxygen are required to support each 1 kg·m of mechanical work at mechanical efficiency of 25%. Based on these calculations at maximal bicycle workload O2 requirements of respiratory muscles would be 0.334 l/min, which would be 5.4% of the total O2 uptake. Blood-flow requirements of respiratory muscles necessary to supply oxygen needs were calculated assuming an arterial O2 saturation of 95% with 95% extraction of oxygen delivered. Based on these calculations blood flow to respiratory muscles at maximal O2 uptake would be 1.76 l/min, 5% of the cardiac output. If the diaphragm were generating 30% of the mechanical work of breathing, as has been measured in the dog, diaphragm blood flow would be 528 ml/min. If the diaphragm weight is 5 gram/kg body weight, blood flow to the diaphragm of this 75-kg athlete would be 1.4 ml/min/g of muscle. The highest blood flow that we have seen in the dog diaphragm under extreme conditions of inspiratory loading averaged 2 ml/min/g of muscle [39]; hence this estimated diaphragm blood flow at peak exercise is a reasonable magnitude in the athlete but might be approaching an upper limit. More than likely in the Olympic athlete all transport steps, maximal cardiac output, maximal power development by the respiratory muscles and maximal power from exercising skeletal muscles are approaching upper limits at the same 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 similar calculations have been made to those in Table 6, except that K1 and K2 are assumed to be 5-fold higher for estimating the greater power requirements of breathing in asthma and mechanical efﬁciency is less. Note in the last column of the table that in this patient the oxygen requirements of the respiratory muscles become so great that any attempt to increase ventilation above 100 l/min provides no further increase in oxygen for the rest of the body including locomotor muscle. Exercise could not be sustained at this level. Total body oxygen uptake that could be sustained would

WIT Transactions on State of the Art in Science and Eng ineering, Vol 24, © 2006 WIT Press www.witpress.com, ISSN 1755-8336 (on-line) 24 Human Respiration /l 2 1.82 2.41 3.01 3.59 4.15 4.72 5.26 5.81 available for 2 capacity of 0.2 l O 2 0.8 1.2 1.7 2.3 2.9 3.9 4.3 5.0 (% of total) exercise (l/min) respiratory muscles (RM) for energy 0.1 0.192 0.323 0.499 0.724 1.000 1.350 1.760 uptake RM blood flow RM blood flow O 2 in eqn. (5), an alveolar ventilation 80% of the total so that each 20 l/min 2 uptake RM O , 25% mechanical efficiency for muscles of breathing, O 2 2 O/(l/s) 2 = 0.3 cmH 2 K O/(l/s) and 2 6080 12.6 15.7 0.019 0.036 1.0 1.5 extraction by exercising muscles. 100120140160 18.9180 22.1200 25.3 28.5 0.061 31.7 0.095 34.9 0.138 0.191 0.256 2.0 0.334 2.6 3.2 3.9 4.6 5.4 2 (l/min) (l/min) (l/min) (% of total) (l/min) uptake during exercise in an athlete plus per cent of the total required by 2 = 2.4 cmH 1 K uptake Ventilation Cardiac output RM O requirements of breathing and that left over for sustaining exercise*. 2 of blood and 95% O (l/min) increase in ventilation provides 0.61 l/min utilizable O 1.94 2.45 3.07 3.68 4.29 4.91 5.52 6.14 *Assuming Table 6: Total O Total O

WIT Transactions on State of the Art in Science and Eng ineering, Vol 24, © 2006 WIT Press www.witpress.com, ISSN 1755-8336 (on-line) Anatomy and Physiology of the Human Respiratory System 25 capacity 2 0.94 1.13 1.24 1.23 1.12 0.52 0.36 0.69 0.89 available for 2 2.0 6.3 60.3 (% of total) exercise (l/min) 50% of total ventilation so that each by respiratory muscles (RM) for energy = 0.10 0.44 1.092.123.615.618.20 12.1 19.3 27.8 37.4 48.5 uptake RM blood flow RM blood flow O 4.7 2 84.9 15.42 73.8 , 10% mechanical efficiency for muscles of breathing, an O 2 in eqn. (5), alveolar ventilation 2 uptake RM O 2 O/(l/s) 2 1 . 5 cmH = 2 K extraction by exercising muscles. 2 O/(l/s) and 2 80 11.0 0.403 26.3 20 5.0 0.018 4060 7.9 9.0 0.083 0.207 10.8 18.0 100120140160 13.0180 15.0 16.9 19.0 0.685 20.9 1.066 1.558 2.175 35.7 2.93 46.3 58.1 70.8 11.45 (l/min) (l/min) (l/min) (% of total) (l/min) 10 cmH uptake during exercise in an asthma patient plus per cent of the total required 2 = 1 K /l of blood and 95% O 2 uptake Ventilation Cardiac output RM O requirements of breathing and that left over for sustaining exercise*. 2 (l/min) of 0.2 l O increment 20 l/min ventilation provides 0.35 l/min of utilizable O 1.53 1.92 2.30 2.68 3.07 3.45 0.38 *Assuming 0.77 1.15 Table 7: Total O Total O

probably be less than 1 l/min. Exercise is clearly limited by the respiratory muscles and 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 right ventricle pumping blood through the lung for adding oxygen and removing carbon dioxide from blood returning from the periphery and the left ventricle pumping blood into peripheral organs to provide substrate and oxygen. The two pumps must interact efficiently within the gas-exchange region of the lungs to maintain normal arterial blood gases leaving the lung. Furthermore, since the heart pump lies entirely within the thoracic pump they must interact mechanically in a way that does not impede the function of the other pump, despite the fact that they pump at different rates and stroke volumes in a limited amount of space [44]. In fact, the interaction in many instances is probably beneficial. The pumping action of the heart causes pressure fluctuations in conducting airways and gas-exchange regions of the lung that may enhance uniformity of gas mixing in the lung. The diaphragm is the major inspiratory pump in the thorax and operates like an inspiratory piston that moves downward in upright humans to pull air into the lung during inspiration. In humans the diaphragm has a firm connection with the heart, which holds the diaphragm up against the pull of gravity at the end of a normal breath; at the end of a breath it is in position to exert its piston-like function for the next inspiration. Quadrupeds who normally run in a horizontal orientation do not have this connection between heart and diaphragm; consequently when they assume an upright position the diaphragm is not held up in a position necessary for an effective inspiratory stroke. These mechanical interactions between heart, lungs and thorax are not well understood in health or disease and need further study.

Acknowledgement

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

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