Lungs and Breathing

6 L/min of air, which is the ventilation or inhalation rate. (This is comparable to the $$\sim $$5–6 L of blood pumped per min in the pulmonary circulation through the lungs, which is largely by coincidence.) Because air is $$\sim $$20% oxygen, we inhale (inspire) $$\sim $$1.2 L oxygen/min. The breathing rate is typically 12/min for men, 20/min for women, and 60/min for infants. The air we inhale has $$\sim $$80% N$$_{2}/20\%$$ O$$_{2}$$ (or more precisely $$78.084\%$$ N$$_{2}/20.947\%$$ O$$_{2}/0.934\%$$ Ar$$/0.04\%$$ CO$$_{2}$$ for dry air in 2015), and the air we exhale (expire) has $$\sim $$80% N$$_{2}/16\%$$ O$$_{2}/4\%$$ CO$$_{2}$$, excluding water vapor and argon. (If the air we exhaled had little or no oxygen, we could not use it for mouth-to-mouth resuscitation.) We breathe in roughly 10 kg air/day, with $$\sim $$2 kg O$$_{2} /$$day. The lungs absorb about 0.5 kg O$$_{2}/$$day (400 L). We exhale air with $$ \sim $$0.5 kg water vapor/day.


Because we inspire and expire air at the same rate (if not where would the difference go?), it is clear the body uses $$\sim $$0.2–0.3 L oxygen/min during usual sedentary activity, delivered by the $$\sim $$5–6 L of blood pumped per min. We have called this rate of oxygen consumption in the body $$\mathrm{d}V_{\mathrm {O}_{2}}/\mathrm{d}t$$ in Chaps. 6 and 8. During aerobic exercise $$\mathrm{d}V_{\mathrm {O}_{2}}/\mathrm{d}t$$ increases linearly with cardiac output $$Q_{\mathrm {t}}$$ (see Fig. 8.​26, and (6.​18) and (8.​28)). The maximum rate of oxygen usage is $$\sim $$2.8 L/min for a person of average fitness and $$\sim $$ $$4$$ L/min for a highly fit person. This assumes the lungs bring in air at a rate fast enough to maintain the needed oxygenation of arterial blood. With advancing age, the main factor in decreasing athletic performance is decreasing oxygen intake. At maximal exercise in endurance-trained men, oxygen consumption (divided by body mass) decreases by $$\sim $$28% from $$\sim $$68 mL/kg-min at 28 years of age to $$\sim $$49 mL/kg-min at 60 years of age [36].

Gauge pressures, relative to atmosphere, are usually used in discussing breathing. Two roughly equal types of units are commonly used, mmHg and cmH$$_{2}$$O, with 1 mmHg = 1.36 cmH$$_{2}$$O.

A114622_2_En_9_Fig1_HTML.gif


Fig. 9.1
Diagram of parts of the respiratory system. (These components are also important in voice production (Chap. 10). The vocal cords (or vocal folds) used in speaking are in the larynx.) (From [6]. Used with permission)


9.1 Structure of the Lungs


Air is inhaled through the nose or mouth and then through the pharynx, larynx, and the trachea (windpipe) (Fig. 9.1). The trachea divides into the right and left bronchus (Fig. 9.2), each of which continues to bifurcate into smaller and smaller bronchi and bronchioles over 23 levels of bifurcation ($$2^{24} = 1.6 \times 10^{8}$$) (Table 9.1, Figs. 9.2 and 9.3) until they form alveoli (which is the plural of alveolus) (al-vee-oh’-lie (lus)), which are the actual operating units of the lungs. The average diameter of the airways decreases with generation z, as $$d(z)=2^{-z/3}d(0)$$ until generation 16. This relation is the optimal design of a branched system of tubes in hydrodynamics. There are about $$3 \times 10^{8}$$ alveoli, each $$\sim $$0.2–0.3 mm in diameter, with walls that are $$\sim $$ $$0.4 \;\upmu $$m thick. They are in contact with blood in the pulmonary capillaries (Fig. 9.4), which themselves form after subdividing in 17 branches (Table 9.2, Figs. 9.2 and 9.5). Oxygen diffuses from the alveoli to the red blood cells, while carbon dioxide diffuses from the blood into the air in the alveoli. The total surface area of the alveoli is $$\sim $$80 m$$^{2}$$ (ranging from 50–100 m$$^{2}$$). The total external surface area of the lungs is only $$\sim $$0.1 m$$^{2}$$, so subdividing into alveoli results in a tremendous increase in the surface area in contact with the blood, by a factor of almost 1,000. This is also the factor by which the oxygen intake increases. Without this, we would never even come close to meeting our metabolic needs for oxygen. Our chests expand when we breathe because incoming air filling the alveoli makes each one bigger, just as with ordinary bubbles. Models of O$$_{2}$$ and CO$$_{2}$$ diffusion across the alveoli and pulmonary capillary walls and how this progresses along the capillaries can be found in [13, 14, 28], along with models on how these gases are transported between the alveoli and the rest of the lungs (ventilation) and how the capillaries deliver blood to tissue (perfusion).

A114622_2_En_9_Fig2_HTML.gif


Fig. 9.2
The relationship between the lung and heart is shown. The first few generations of the branching of the air vessels in the lungs, pulmonary arteries, and the pulmonary veins are shown. These three systems can be called the three “trees” of the lung. Note that the pulmonary arteries are close to the bronchi, while the pulmonary veins stand alone (From [9])



Table 9.1
Approximate quantification of the bronchial system




























































































































































































































Pulmonary branch
Generation z

Branch diameter (mm)
Branch length (mm)
Total cross-sectional area (cm$$^{2}$$)
Volume (cm/s)
Air speed (cm$$^{3}$$)

Trachea
0
18.0
120.0
2.5
31
393

Main bronchus
1
12.2
47.6
2.3
11
427

Lobar bronchus
2
8.3
19.0
2.1
4.0
462
 
3
5.6
7.6
2.0
1.5
507

Segmental bronchus
4
4.5
12.7
2.5
3.5
392
 
5
3.5
10.7
3.1
3.3
325

Bronchi
6
2.8
9.0
4.0
3.5
254

w/cartilage in wall
7
2.3
7.6
5.1
3.8
188
 
8
1.86
6.4
7.0
4.4
144
 
9
1.54
5.4
9.6
5.2
105
 
10
1.30
4.6
13
6.2
73.6

Terminal bronchus
11
1.09
3.9
20
7.6
52.3
 
12
0.95
3.3
29
9.8
34.4

Bronchioles
13
0.82
2.7
44
12
23.1

w/muscle in wall
14
0.74
2.3
69
16
14.1
 
15
0.66
2.0
113
22
8.92

Terminal bronchiole
16
0.60
1.65
180
30
5.40

Respiratory bronchiole
17
0.54
1.41
300
42
3.33

Respiratory bronchiole
18
0.50
1.17
534
61
1.94

Respiratory bronchiole
19
0.47
0.99
944
93
1.10

Alveolar duct
20
0.45
0.83
1,600
139
0.60

Alveolar duct
21
0.43
0.70
3,200
224
0.32

Alveolar duct
22
0.41
0.59
5,900
350
0.18

Alveolar sac
23
0.41
0.50
12,000
591
0.09

Alveoli, 21 per duct
  
0.28
0.23
  
3,200
  


Using data from [8, 38]. Also see [39, 40]

The air speed is assumed to be 1 L/s. The data include that for both lungs. The number in each generation is $$2^{z}$$ (for generations $$z=0$$–23), and $$300\times 10^{6}$$ for the alveoli

The circulatory system is the conduit for the transfer of O$$_{2}$$ and CO$$_{2}$$ between the alveoli and tissues, and so we should track the partial pressure in each system. Within the alveoli the partial pressure of O$$_{2}$$ is $$\simeq $$105 mmHg, which is smaller than that in the atmosphere (159 mmHg = 21% of 760 mmHg) because of the dead volume in the respiratory system. The partial pressure of O$$_{2}$$ in blood in the pulmonary capillaries increases from 40 to $$\simeq $$100 mmHg after O$$_{2}$$ is transferred from the alveoli, and this is the partial pressure in the pulmonary veins and systemic arteries. The partial pressure of O$$_{2}$$ in tissue is 40 mmHg, so that after transfer of O$$_{2}$$ from the capillaries to surrounding tissues, the partial pressure in the systemic veins and pulmonary arteries is also $$\simeq $$40 mmHg—and then it is again increased to 100 mmHg in the lungs.

A114622_2_En_9_Fig3_HTML.gif


Fig. 9.3
Bifurcations of lung airways, showing generation number z (From [30])


A114622_2_En_9_Fig4_HTML.gif


Fig. 9.4
The details of the alveolar bifurcation are shown in (a). These alveoli are sacs imbedded in capillary beds. The details of the interaction between the alveoli and capillaries are depicted in (b) and (c) (From [30])

Similarly, within thealveoli the partial pressure of CO$$_{2}$$ is $$\simeq $$40 mmHg; this is much larger than that in the atmosphere ($$\sim $$0.3 mmHg), again because of the dead volume. The partial pressure of CO$$ _{2}$$ in blood in the pulmonary capillaries decreases from 46 to $$\simeq $$40 mmHg after CO$$_{2}$$ is transferred to the alveoli, and this is the partial pressure in the pulmonary veins and systemic arteries. The partial pressure of CO$$_{2}$$ in tissue is 46 mmHg, so that after transfer of CO$$_{2}$$ into the capillaries from the tissues, the partial pressure in the systemic veins and pulmonary arteries is also $$\simeq $$46 mmHg—and then it is again decreased to 40 mmHg in the lungs. With advancing age, the difference in arterial and venous O$$_{2}$$ levels decreases, which indicates less oxygen is being used by the body. At maximal exercise in endurance-trained men, this difference decreases by $$\sim $$8% from $$\sim $$16.7 mL(O$$_{2}$$)/100 mL(blood) at 28 years of age to $$\sim $$15.2 mL/100 mL at 60 years of age [36].


Table 9.2
Branching structure of the pulmonary arterial network



































































































Pulmonary branching order
Number of branches of each order
Vessel length (mm)
Vessel diameter (mm)

1
1
90.5
30.0

2
3
32.0
14.83

3
8
10.9
8.06

4
20
20.7
5.82

5
66
17.9
3.65

6
203
10.5
2.09

7
675
6.6
1.33

8
2,290
4.69
0.85

9
5,861
3.16
0.525

10
17,560
2.10
0.351

11
52,550
1.38
0.224

12
157,400
0.91
0.138

13
471,300
0.65
0.086

14
1,411,000
0.44
0.054

15
4,226,000
0.29
0.034

16
12,660,000
0.20
0.021

17
300,000,000
0.13
0.013


Using data from [8, 33]



A114622_2_En_9_Fig5_HTML.gif


Fig. 9.5
A silicone elastomer cast of the venous tree of the lung of a cat. The venous pressure was $$-7$$ cmH$$_{2}$$O ($$= -5$$ mmHg), the airway pressure was 10 cmH$$_{2}$$O (= 7 mmHg), and the pleural pressure was 0 cmH$$_{2}$$O (From [9])


9.2 The Physics of the Alveoli


The alveoli are similar to interconnected bubbles. Inside them the pressure is $$P_{ \mathrm {in}}$$ and outside the pressure is $$P_{\mathrm {out}}$$, with $$\Delta P=P_{ \mathrm {in}}-P_{\mathrm {out}}$$, and they have a radius R. The Law of Laplace for a sphere (7.​9) is


$$\begin{aligned} \Delta P=\frac{2T}{R}, \end{aligned}$$

(9.1)
where T is the tension in the sphere walls. The main source of this tension in the alveoli is not within the walls but on the surfaces. This contribution is called the surface tension $$\gamma $$, which has the same units as T—of force/length or energy/area. In typical bubbles, such as soap bubbles, both surfaces contribute the same surface tension and so T is replaced by $$2\gamma $$. Therefore we find


$$\begin{aligned} \Delta P=P_{\mathrm {in}}-P_{\mathrm {out}}=\frac{4\gamma }{R}. \end{aligned}$$

(9.2)
For the water/air interface $$\gamma \simeq 7.2 \times 10^{-4}$$ N/m (Table 7.​2). In alveoli, however, only the surface tension of the inner surface is really important because it is a fluid/air interface and has larger surface tension than the fluid/fluid interface of the outer surface, and so


$$\begin{aligned} \Delta P_{\mathrm {alveoli}}=P_{\mathrm {in}}-P_{\mathrm {out}}=\frac{2\gamma }{R}. \end{aligned}$$

(9.3)


A114622_2_En_9_Fig6_HTML.gif


Fig. 9.6
Instability of bubbles, according to the Law of Laplace. This assumes that the surface tension does not change with bubble (or alveolus) radius. The external pressure is 0

There is an apparent instability that seemingly leads to an unreasonable situation in interconnected bubbles or alveoli. Consider two bubbles that are initially not interconnected, as in Fig. 9.6, because there is a plug between them. Bubble #1 has an internal pressure $$P_{1}$$ and radius $$R_{1}$$, and Bubble #2 has an internal pressure $$P_{2}$$ and radius $$R_{2}$$. (Because the difference between the pressure inside and outside the bubble is what is significant, the external pressure is equal to zero.) In equilibrium, the internal pressure $$P_{\mathrm {internal}}=4\gamma /R$$ for each bubble. (Whether this factor is 4 or 2 is not significant here.) Say Bubble #2 is the smaller bubble. Because $$R_{2}<R_{1}$$, in equilibrium $$P_{2 }>P_{1}$$” src=”http://basicmedicalkey.com/wp-content/uploads/2017/06/A114622_2_En_9_Chapter_IEq93.gif”></SPAN>; the smaller bubble has the higher internal pressure. If the plug is opened, air will flow from higher pressure to lower pressure, and therefore from the smaller bubble to the larger bubble. The loss of air in Bubble #2 makes it smaller. With this smaller radius, the equilibrium internal pressure increases. Because this pressure is still higher than in Bubble #1, air continues to flow from the smaller bubble to the larger bubble, until it collapses.</DIV><br /> <DIV id=Par12 class=Para>This implies that the largest of the hundreds of millions of alveoli would get ever larger at the expense of all of the smaller ones and the system of alveoli we have described for the lungs could not be stable. What is wrong with this scenario? There is no error in our reasoning; however, we have made one assumption that is not accurate for alveoli. We implicitly assumed that the surface tension is not a function of radius <SPAN class=EmphasisTypeItalic>R</SPAN>. There is a surfactant on the surfaces of the alveoli of healthy people, containing dipalmitoyl phosphatidycholine or DPPC, that causes <SPAN id=IEq94 class=InlineEquation><IMG alt= to be independent of R from 0 to a critical size ($$R_{0}$$). With $$\Delta P=2\gamma (R)/R$$, some bubbles become bigger and others smaller as in Fig. 9.6. However, eventually $$\gamma (R)$$ increases with larger R faster than R does itself, so $$\Delta P$$ begins to increase with even larger R. Such a system of interconnected alveoli is stable.

A114622_2_En_9_Fig7_HTML.gif


Fig. 9.7
Surface tension on alveoli walls (lung extract) in (b), as measured by the surface balance in (a) which measures surface tension versus area. Similar measurements for detergent and water are also shown (Based on [12, 42])

We can see how such a dependence of $$\gamma (R)$$ can occur with the following model. The surface of an alveolus can be covered either with a lipoprotein or by water; the surface tension of the lipoprotein is much lower ($$\gamma _{\mathrm {lung}} = 1\times 10^{-3}$$ N/m) than that of water ($$\gamma _{\mathrm {water}}= 7.2\times 10^{-2}$$ N/m (= 72 dynes/cm)). Assuming the alveolus is spherical, for one particular radius $$R_{0}$$ there is exactly one monolayer of lipoprotein on the whole surface and at that radius the surface tension is $$\gamma _{\mathrm {lung}}$$ over the $$4 \pi R_{0}^{2}$$ surface area. If this alveolus becomes smaller, so $$R<R_{0}$$, it has several monolayers of lipoprotein on its surface and its surface tension is still $$\gamma _{\mathrm {lung}}$$, and so


$$\begin{aligned} \gamma (R)=\gamma _{\mathrm {lung}}\mathrm { \ \ \ for \;}R\le R_{0}. \end{aligned}$$

(9.4)
If this same alveolus instead becomes larger, so $$R>R_{0}$$” src=”http://basicmedicalkey.com/wp-content/uploads/2017/06/A114622_2_En_9_Chapter_IEq107.gif”></SPAN>, it has a monolayer of lipoprotein over only a portion of its surface (of surface area <SPAN id=IEq108 class=InlineEquation><IMG alt= because the layer cannot become smaller than a monolayer) and water over the rest of the surface (of area $$4\pi R^{2}- 4\pi R_{0}^{2}$$). So the average surface tension is


$$\begin{aligned} \gamma (R)=\frac{4\pi R_{0}^{2}\gamma _{\mathrm {lung}}+(4\pi R^{2}-4\pi R_{0}^{2})\gamma _{\mathrm {water}}}{4\pi R^{2}}\mathrm { \ \ \ for\;}R>R_{0} \end{aligned}$$” src=”http://basicmedicalkey.com/wp-content/uploads/2017/06/A114622_2_En_9_Chapter_Equ5.gif”></DIV></DIV><br /> <DIV class=EquationNumber>(9.5)</DIV></DIV>or<br /> <DIV id=Equ6 class=Equation><br /> <DIV class=EquationContent><br /> <DIV class=MediaObject><IMG alt=R_{0}. \end{aligned}$$” src=”http://basicmedicalkey.com/wp-content/uploads/2017/06/A114622_2_En_9_Chapter_Equ6.gif”>

(9.6)
This approaches the much larger $$\gamma _{\mathrm {water}}$$ for $$R\gg R_{0}$$ at a rate that is faster than R, so the alveoli will be stable.

Because this lipoprotein is only on one of the surfaces, the stability condition is $$\Delta P=2\gamma /R$$. For $$R>R_{0}$$” src=”http://basicmedicalkey.com/wp-content/uploads/2017/06/A114622_2_En_9_Chapter_IEq113.gif”></SPAN>, there is a stable equilibrium when <SPAN id=IEq114 class=InlineEquation><IMG alt= or


$$\begin{aligned} \frac{\mathrm{d}(\Delta P)}{\mathrm{d}R}=-\frac{2\gamma _{\mathrm {water}}}{R^{2}}+6\frac{ R_{0}^{2}}{R^{4}}(\gamma _{\mathrm {water}}-\gamma _{\mathrm {lung}})=0 \end{aligned}$$

(9.7)
or


$$\begin{aligned} R_{\mathrm {eq}}=\sqrt{3\frac{\gamma _{\mathrm {water}}-\gamma _{\mathrm {lung}}}{ \gamma _{\mathrm {water}}}}\;R_{0}. \end{aligned}$$

(9.8)
Because $$\gamma _{\mathrm {water}}\gg \gamma _{\mathrm {lung}}$$, the equilibrium radius $$R_{\mathrm {eq}}\simeq \sqrt{3}R_{0}$$.

Figure 9.7 shows that this surface tension of the surfactant in the lung decreases from $$5\times 10^{-2}$$ N/m (50 dynes/cm) to zero as the area of the film gets smaller. Alveoli are typically stable at approximately 1/4 of their maximum size.

One function of the surfactant is to provide alveolus stability. Another function is to lower the amount of force needed to be supplied by the diaphragm to inflate the alveoli. With $$\gamma _{\mathrm {water}}= 7.2\times 10^{-2}$$ N/m and $$R= 0.05$$ mm of the alveoli when they are collapsed (and need to be inflated), (9.3) gives $$\Delta P_{\mathrm {alveoli}} = 2.9\times 10^{3}$$ N/m$$^{2} = 22$$ mmHg. The area of an adult diaphragm muscle is about 500 cm$$^{2}$$, so the force it needs to exert to expand the alveoli for breathing is $$\sim $$150 N—which corresponds to a weight with mass 15 kg. With the lower surface tension of the lung surfactant, this force is over an order of magnitude smaller and breathing is easier, especially for infants. This explains why people with insufficient surfactant—with hyaline membrane disease—have difficulty breathing.


9.3 Physics of Breathing


Each lung is surrounded by a sac membrane within the thoracic cavity. We can picture the pleural sac as a balloon, as in Fig. 9.8, filled with intrapleural fluid. The inside wall of this sac, the visceral pleura (membrane), attaches to the outer lung wall. The outside wall of this sac, the parietal pleura (membrane), attaches to the thoracic wall. It is the springiness of the lung that pulls the two pleural membranes apart, and this causes a slight decrease of pressure of the pleural sac relative to atmospheric pressure of $$-4$$ mmHg to $$-6$$ mmHg. This pressure difference is what keeps the lungs expanded, and keeps them from collapsing. The mechanical “driving force” in controlling lung volume is the transpulmonary pressure , which is the difference in pressure in the alveoli in the lungs and that around the lung in the pleural sac, which is called the intrapleural (or pleural) pressure. (The alveolar and pleural pressures are gauge pressures, referenced to atmospheric pressure.)

A114622_2_En_9_Fig8_HTML.gif


Fig. 9.8
a Pushing a fist into a balloon is analogous to the lungs in the pleural cavity. b Schematic of the lungs in the pleural cavities (Based on [34])

The lungs are expanded and contracted by the motion of structures surrounding them by way of inspiratory and expiratory muscles. This occurs in two ways (Fig. 9.9), of which only the first one is used during quiet breathing. (1) The diaphragm moves downward to lengthen the chest cavity (by pulling the bottom of the lungs downward) during inspiration. During quiet breathing, the lungs contract by the natural elastic recoil of the lungs and chest wall, with the diaphragm relaxed, while in heavier breathing this contraction is accelerated by the contraction of the abdominal muscles that push the abdominal contents and then the diaphragm upward to shorten the chest cavity. (2) The ribs are elevated by the neck muscles to increase the anteroposterior (front-to-back) diameter of the chest cavity and are depressed (lowered) by the abdominal recti to decrease it. This causes chest cavity expansion and contraction, respectively, because the ribs slant outward and have larger transverse cross-sectional areas in the lower sections; this can increase the anterior–posterior chest thickness by about 20% during inspiration.

A114622_2_En_9_Fig9_HTML.gif


Fig. 9.9
Expansion and contraction of the thoracic cage during expiration and inspiration, showing the ribs, lungs and heart, the external intercostal muscles (that contract during inspiration to elevate the rib cage and widen it laterally so the cage increases in all three dimensions), and the diaphragm (that contracts to increase the vertical dimension of the cage during inspiration) (From [31])


A114622_2_En_9_Fig10_HTML.gif


Fig. 9.10
Force balance of the visceral pleura/outer lung wall during preinspiration. Note that the forces are really normal to the wall everywhere, not just at the bottom as depicted (Based on [34])

How does this help bring air into the lungs? Before inspiration, there is atmospheric pressure in the lungs. The attractive force of the visceral pleura for the parietal pleura and the outward force of the outer lung wall due to the lower-than-atmospheric pressure in the pleural sac ($$\sim $$–4 mmHg) cause each lung to expand. In equilibrium their sum is balanced by the tendency of the lungs to contract due to their springiness. This preinspiration force balance is shown in Fig. 9.10. They are no longer in balance during inspiration.

A114622_2_En_9_Fig11_HTML.gif


Fig. 9.11
a Force imbalance at the onset of inspiratory muscle contraction (and expansion of the thoracic wall/parietal pleura) leads to a (b) subatmospheric pressure in the lungs and flow of air into the lungs during inspiration. (Compare this to the preinspiration force balance in Fig. 9.10) (Based on [34])

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Jun 11, 2017 | Posted by in GENERAL & FAMILY MEDICINE | Comments Off on Lungs and Breathing

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