Download CHAPTER 1 Anatomy and physiology of the human respiratory system

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.
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2 Human Respiration
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
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Number of Alveoli (millions)
400
Adult
300
200
100
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].
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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 swallowed. 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:
P = K1 V̇ + K2 (V̇ )2 ,
(1)
where P is the pressure drop in cmH2 O across the lung; V̇ is gas flow in l/s; K1
is a constant derived from the summation of laminar-flow resistances in branching
airways in cmH2 O/(l/s) and K2 is a constant derived from changing resistance
from turbulence and from inertance at airway branch points and at changes in
cross-sectional area of the airways; units are in cmH2 O/(l/s)2 . 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.
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Table 1: Pressure–flow relationships derived by Rohrer.
P = 0.426 · V̇ + 0.714 · (V̇ )2
Upper airways
Lower airways
Bronchial
Lobular
P = 0.106 · V̇ + 0.080 · (V̇ )2
P = 0.258 · V̇ + 0.006 · (V̇ )2
Total
P = 0.790 · V̇ + 0.800 · (V̇ )2
Table 2: Resistance to airflow in the respiratory system of an average normal person
breathing through the mouth or nose [12].
Mouth breathing
Flow
Resistance
Units
Air flow
Lung tissue
Chest wall
Resp. system
Nasal breathing
Laminar
Turbulent
Laminar
Turbulent
K1
cmH2 O/(l/s)
1.2
0.2
1.0
2.4
K2
cmH2 O/(l/s)2
0.3
0
0
0.3
K1
cmH2 O/(l/s)
1.8
0.2
1.0
3.0
K2
cmH2 O/(l/s)2
3.0
0
0
3.0
Thus, to describe the pressure in cmH2 O to generate a volume flow of V̇ l/s
through the mouth would be P = 2.4V̇ + 0.3 · [V̇ ]2 and through the nose would
be P = 3.0V̇ + 3.0·[V̇ ]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/cmH2 O. 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
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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 deadspace 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 (V̇A ) is related to total ventilation
(V̇T ) as follows: V̇A = V̇T − fVD , where f is frequency of ventilation VD is deadspace 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/cmH2 O.
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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]:
P=
2
V
+ K1 V̇ + K2 · V̇ ,
CRS
(3)
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 V̇ we have
the power requirement at any instant:
Instantaneous Power = P V̇ =
2
V · V̇
+ K1 · V̇ + K2 · [V̇ ]3 .
CRS
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If we assume that ventilation follows a sine-wave pattern, V̇ = V̇P sin(2πft) where
V̇P 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
3 Ẇ = Power = (1/2fC RS )[V̇E ]2 + 60 · (K1 /4) π2 V̇E + (2K2 /3) π2 · V̇E
,
(4)
where V̇E has been converted from l/s in eqn. (3) to ventilation in l/min; f is respiratory rate in min−1 ; 60 is s/min so that power is now expressed as cmH2 O·l·min−1 .
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
3 Ẇ = 2 · 60 · (K1 /4) π2 · V̇E + (2K2 /3) π2 · V̇E
/100.
(5)
Dividing by 100 converts power from units of cmH2 O·l·min−1 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.
Athlete
Power Requirements (kg m/min)
250
Patient with asthma
200
150
100
50
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
cmH2 O/(l/s) and K2 = 1.5 cmH2 O/(l/s)2 ).
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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 volume velocity (V̇ ). Resistance to turbulent flow, K2 , is directly proportional to fluid
density and the square of the volume velocity, [V̇ ]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
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Trachea
Bronchus
TB
RB1
RB2
RB3
AD1
AD2
AD3
AS
Generation
0
6
16
17
18
19
20
21
22
23
131.23
84.17
1.852
0.967
0.467
0.211
0.078
0.035
0.016
0.008
Peak flow‡
cm/s
6151.5
295.9
1.194
0.533
0.214
0.082
0.025
0.010
0.004
0.002
Peclet number
1408.5
140.5
0.663
0.311
0.139
0.059
0.021
0.009
0.004
0.002
Reynolds number
1968.5
1262.6
27.778
14.509
7.011
3.166
1.176
0.530
0.240
0.120
Peak flow‡
cm/s
92273.6
4438.92
17.904
7.991
3.204
1.224
0.381
0.145
0.055
0.023
Peclet number
21128.8
2108.1
9.938
4.672
2.090
0.887
0.316
0.136
0.059
0.029
Reynolds number
Exercise ventilation = 100 l/min
‡ Peak
= terminal bronchiole entering the acinus; RB = respiratory bronchiole; AD = alveolar duct and AS = alveolar sac.
flow during a respiratory cycle is about 3 times the average ventilation. Shaded areas under Peclet numbers designate airways
at the given ventilation where rate of diffusive transport exceeds the rate convective transport. Shaded areas under the columns of
Reynolds numbers indicate those airways at the given ventilation where turbulence tends to become sustained.
∗ TB
Airway∗
Resting ventilation = 6 l/min
Table 3: Convective and diffusive gas transport and mixing in the airways between trachea and alveoli.
10 Human Respiration
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11
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
PaCO2 =
V̇CO2
· 863,
V̇ a
(6)
where PaCO2 = alveolar CO2 tension, V̇CO2 = CO2 output, V̇ a = Alveolar ventilation 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 (C v̄CO2 ) entering the lung multiplied by
blood flow (Q̇) in ml/min equals the CO2 output (V̇CO2 ) in ml/min. This is the Fick
equation for CO2 :
V̇CO2 := Q̇(C v̄CO2 − CcCO2 ) · 22.2,
where 22.2 = ml/mM.
Combining eqns. (6) and (7) and rearranging we have
V̇ a/Q̇
CcCO2 = C v̄CO2 −
· PaCO2 ,
19.2
(7)
(8)
which, for a given ratio of ventilation to blood flow, V̇ a/Q̇, and the mixed venous
CO2 content (C v̄CO2 ) of blood entering the lung tells us the trajectory along which
blood CO2 content must fall toward equilibrium with the alveolar CO2 tension
(PaCO2 ). 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, V̇ a/Q̇, in any region of lung and the mixed venous CO2
content we can estimate the CO2 content in blood leaving that region at equilibrium
with the PaCO2 .
3.2.2 O2 equilibrium
The same set of equations can be derived for oxygen. The alveolar equation for O2
states that
FiN2
PaO2 = (Pb − PH2 O ) · FiO2 − PaCO2 · FiO2 +
,
(9)
R
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12 Human Respiration
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
(V̇ a/Q̇) = 3. The solid straight line is eqn. (8) and the curved line is the
CO2 dissociation curve. The point of intersection indicates equilibrium.
where PaCO2 = alveolar O2 tension; Pb = barometric pressure; PH2 O = water
vapor 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:
V̇O2 = Q̇(ScO2 − S v̄O2 ) · Cap,
(10)
where ScO2 and S v̄O2 are fractional oxyhemoglobin saturation in the end-capillary
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),
ScO2
− S v̄O2 = PiO2 − PaO2 · FiO2 ·
V̇ a/Q̇
−·
863 · FiN2 · Cap
V̇ a/Q̇
863 · FiN2 · Cap
· PaO2 .
(11)
Equation (11) determines the trajectory along which oxygen saturation rises from
that in mixed venous blood toward equilibration with the alveolar O2 tension (PaO2 )
for a given ratio of V̇ a/Q̇. The point of equilibrium can be derived graphically
from the intersection of this trajectory with the oxyhemoglobin dissociation curve,
Fig. 6.
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a
Oxygen Saturation (%)
100
Oxyhemoglobin
dissociation curve
Equilibrium
VA/Q = 3
VA /Q = 6.0
b
100
13
SaO2
75
75
50
50
mixed
venous
VA/Q = 0.5
25
25
PAO2 PI O2 - K
PaO2
0
0
0
25
50
75
100
125
Oxygen Tension (mmHg)
150
0
25
50
P A O2
75
100
125
150
Oxygen Tension (mmHg)
Figure 6: (Panel a) O2 equilibrium between alveolar air and blood leaving lung
capillaries at a known inspired O2 tension (PiO2 ) and mixed venous O2
saturation. The solid straight line is eqn. (11) at a ventilation perfusion
ratio (V̇ a/Q̇) = 3. The curved line is the O2 dissociation curve. Equilibrium is at the point of intersection. (Panel b) The effect of uneven
V̇ a/Q̇. This shows the effect of two regions of lung with equal blood
flows but different V̇ a/Q̇ ratios of 6 and 0.5. A mixture of blood from the
two regions will have an O2 saturation and PaCO2 approximately halfway between and a resulting increase in PaO2 − PaO2 as indicated in the
diagram. After Hsia and Johnson [20].
3.2.3 Effect of uneven V̇ a/Q̇ ratios in the lung
The most efficient gas exchange is achieved when V̇ a/Q̇ ratios are uniform and
the alveolar-arterial O2 tension and CO2 tension differences (PaO2 − PaO2 and
PaCO2 − 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
DLO2 =
V̇O2
,
PaO2 − PcO2
(12)
where V̇O2 = Oxygen uptake in ml/min, PaO2 = Alveolar O2 tension and PcO2 =
mean oxygen tension of capillary hemoglobin, both in mmHg.
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The difficulty with measuring DLO2 arises from the fact that PcO2 changes as
blood passes through the lung. Although PaO2 and V̇O2 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:
DLCO =
V̇CO
.
PaCO
(13)
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 breathholding 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
,
dt
∂x
(14)
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
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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
d = diffusion coefficient in cm2 /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
A
60 · α · d A
DM =
=K·
,
760
x
x
(16)
where A is the area and x the thickness of the barrier and DM is the membranediffusing 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:
DMO2
DMCO
=
KO2
3.87
.
=
KCO
3.48
(17)
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
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16 Human Respiration
Table 4: Estimated Krogh diffusion constants (K) in cm2 min−1 · mmHg−1 .
Respiratory gases
Gases used to measure
diffusing capacity
Gas
Diffusing in N2 ¶
In lung tissue, plasma or water
O2
CO2
CO
NO
1.79 · 10−2
1.26 · 10−2
1.65 · 10−2
1.59 · 10−2
3.87 · 10−8 #
64.5 · 10−8 ∗
3.48 · 10−8 #
6.84 · 10−8 ‡
¶ [22–24].
# Based
on measurements in lung tissue;
on measurements in plasma;
‡ Based on measurements in water [25].
∗ Based
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
O2
CO2
CO
NO
3.9
5.8
1/(0.73 + 0.0058 · PO2 )
approaching infinity
References
[28]
[29]
[26]
[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/DLO2 ), which in turn is the sum of
the 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
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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 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 DMCO and Vc .
1
1
1
1
+
=
·
(19)
DLCO O2
DMCO
θCO O2 Vc
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 DLO2 from the DMCO and Vc derived from the two simultaneous
equations, again using the Roughton–Forster equation as follows:
1
DLO2
=
KCO
1
1
+
· .
KO2 DMCO
θO2 Vc
(21)
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
=
+
· .
DLCO
2.4 · DLNO
θCO Vc
(22)
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. DLO2 has been estimated indirectly from DMCO and Vc as illustrated
in the previous sections and is about 50% higher at any given cardiac output than
DLCO . DLO2 is higher primarily because the rate of uptake of oxygen by red cells
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 DLO2 and cardiac
output [20]
If DLO2 did not increase as cardiac output increases during exercise arterial
oxyhemoglobin desaturation would occur at relatively low levels of exercise.
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18 Human Respiration
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):
V̇O2 = Q̇ ScO2 − S v̄O2 · Cap
V̇O2 = DLO2 · PaO2 − PcO2 ,
it is possible to derive that the ratio of diffusing capacity to pulmonary capillary
blood flow, DLO2 /Q̇c bears the following relationship to oxygen saturation of blood
leaving lung capillaries (ScO2 ) when mixed venous oxygen saturation (S v̄O2 ) and
alveolar oxygen tension (PaO2 ) are known:
DLO2
Q̇c
= Cap
ScO
S v̄O2
2
d(ScO2 )
.
(PaO2 − PcO2 )
(23)
In order to numerically integrate the right-hand side of eqn. (23) to calculate
the relationship between ScO2 and DLO2 /Q̇c at rest and exercise the shape and
position of the oxyhemoglobin dissociation curve must be known. This can be
provided with reasonable accuracy by the Hill equation in the following form
PcO2 = P50 [ScO2 /(1 − ScO2 )]0.37 . 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 ScO2 and DLO2 /Q̇c as illustrated in Fig. 8.
Results show that as DLO2 /Q̇c falls during exercise oxygen saturation of blood
leaving the lung remains almost fixed until a critical level is exceeded when ScO2
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Figure 8: As the ratio of DLO2 /Q̇c moves below a critical point, there is a rapid fall
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 DLO2 remained fixed from rest to exercise. At sea level, at rest, DLO2 /Q̇c would
be about 7.5 and ScO2 would be in equilibrium with PaO2 at about 97.5%. By
the time cardiac output reached 22 L/min ScO2 would have fallen to about 80%.
At 10,000 feet arterial saturation would fall to between 70 and 80% at a cardiac
output of 15 l/min if DLO2 failed to rise above its resting level. Why then doesn’t
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
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20 Human Respiration
50
Mean Pulmonary Vascular
Pressures (mmHg)
Pulmonary Artery
Capillary Wedge
40
30
20
10
0
0
10
20
30
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. Pulmonary 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 PA pressure 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
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Figure 10: Uneven distribution of blood flow (Q̇c /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 circulation (i.e., Q̇c /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
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22 Human Respiration
Figure 11: Respiratory muscles and their segmental innervation. The intercartilagenous portions of the internal intercostals supplied by T1 to T4 (parasternals) are inspiratory muscles. After Miller et al [18].
from cervical and thoracic segments, expiratory muscles from thoracic and lumbar 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
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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
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60
80
100
120
140
160
180
200
12.6
15.7
18.9
22.1
25.3
28.5
31.7
34.9
0.019
0.036
0.061
0.095
0.138
0.191
0.256
0.334
1.0
1.5
2.0
2.6
3.2
3.9
4.6
5.4
0.1
0.192
0.323
0.499
0.724
1.000
1.350
1.760
0.8
1.2
1.7
2.3
2.9
3.9
4.3
5.0
1.82
2.41
3.01
3.59
4.15
4.72
5.26
5.81
*Assuming K1 = 2.4 cmH2 O/(l/s) and K2 = 0.3 cmH2 O/(l/s)2 in eqn. (5), an alveolar ventilation 80% of the total so that each 20 l/min
increase in ventilation provides 0.61 l/min utilizable O2 , 25% mechanical efficiency for muscles of breathing, O2 capacity of 0.2 l O2 /l
of blood and 95% O2 extraction by exercising muscles.
1.94
2.45
3.07
3.68
4.29
4.91
5.52
6.14
Total O2 uptake Ventilation Cardiac output RM O2 uptake RM O2 uptake RM blood flow RM blood flow O2 available for
(l/min)
(l/min)
(l/min)
(l/min)
(% of total)
(l/min)
(% of total)
exercise (l/min)
Table 6: Total O2 uptake during exercise in an athlete plus per cent of the total required by respiratory muscles (RM) for energy
requirements of breathing and that left over for sustaining exercise*.
24 Human Respiration
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20
40
60
80
100
120
140
160
180
5.0
7.9
9.0
11.0
13.0
15.0
16.9
19.0
20.9
0.018
0.083
0.207
0.403
0.685
1.066
1.558
2.175
2.93
4.7
10.8
18.0
26.3
35.7
46.3
58.1
70.8
84.9
0.10
0.44
1.09
2.12
3.61
5.61
8.20
11.45
15.42
2.0
6.3
12.1
19.3
27.8
37.4
48.5
60.3
73.8
0.36
0.69
0.94
1.13
1.24
1.23
1.12
0.89
0.52
*Assuming K1 = 10 cmH2 O/(l/s) and K2 = 1.5 cmH2 O/(l/s)2 in eqn. (5), alveolar ventilation = 50% of total ventilation so that each
increment 20 l/min ventilation provides 0.35 l/min of utilizable O2 , 10% mechanical efficiency for muscles of breathing, an O2 capacity
of 0.2 l O2 /l of blood and 95% O2 extraction by exercising muscles.
0.38
0.77
1.15
1.53
1.92
2.30
2.68
3.07
3.45
Total O2 uptake Ventilation Cardiac output RM O2 uptake RM O2 uptake RM blood flow RM blood flow O2 available for
(l/min)
(l/min)
(l/min)
(l/min)
(% of total)
(l/min)
(% of total)
exercise (l/min)
Table 7: Total O2 uptake during exercise in an asthma patient plus per cent of the total required by respiratory muscles (RM) for energy
requirements of breathing and that left over for sustaining exercise*.
Anatomy and Physiology of the Human Respiratory System
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25
26 Human Respiration
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.
References
[1] Weibel, E.R., Morphometry of the Human Lung. Springer-Verlag: Berlin,
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