Ventilation and Perfusion of the Lungs

About 30% of total ventilation in a respiratory cycle is wasted ventilating anatomical dead space (i.e., conducting airways)

Total ventilation ( ) is the volume of air moved out of the lungs per unit of time:

Here V is the volume of air exiting the lungs during a series of breaths. Note that we are using
differently than in Chapter 27 , where
represented flow through an airway at a particular instant in time. A practical definition is that
is the product of tidal volume (TV or V _T ) and the respiratory frequency (f). Thus, for someone with a tidal volume of 0.5 L, breathing 12 breaths/min,

Because total ventilation usually is reported in L/min, it is sometimes called minute ventilation.

Before an inspiration, the conducting airways are filled with “stale” air having the same composition as alveolar air ( Fig. 31-1 , step 1); we will see why shortly. During inspiration, ~500 mL of “fresh” atmospheric air (high /low ) enters the body (step 2). However, only the first 350 mL reaches the alveoli; the final 150 mL remains in the conducting airways (i.e., nose, pharynx, larynx, trachea, and other airways without alveoli)—that is, the anatomical dead space. These figures are typical for a 70-kg person; V _T and V _D are roughly proportional to body size. During inspiration, ~500 mL of air also enters the alveoli. However, the first 150 mL is stale air previously in the conducting airways; only the final 350 mL is fresh air. By the end of inspiration, the 500 mL of air that entered the alveoli (150 mL of stale air plus 350 mL of fresh air) has mixed by diffusion with the pre-existing alveolar air (see Fig. 31-1 , step 3). During expiration (step 4), the first 150 mL of air emerging from the body is the fresh air left in the conducting airways from the previous inspiration. As the expiration continues, 350 mL of stale alveolar air sequentially moves into the conducting airways and then exits the body—for a total of 500 mL of air leaving the body. Simultaneously, 500 mL of air leaves the alveoli. The first 350 mL is the same 350 mL that exited the body. The final 150 mL of stale air to exit the alveoli remains in the conducting airways, as we are ready to begin the next inspiration.

Thus, with each 500-mL inspiration, only the initial 350 mL of fresh air entering the body reaches the alveoli. With each 500-mL expiration, only the final 350 mL of air exiting the body comes from the alveoli. One 150-mL bolus of fresh air shuttles back and forth between the atmosphere and conducting airways. Another 150-mL bolus of stale air shuttles back and forth between the conducting airways and alveoli. Dead-space ventilation ( ) is the volume of the stale air so shuttled per minute. Alveolar ventilation ( ) is the volume of fresh air per minute that actually reaches the alveoli, or the volume of stale alveolar air that reaches the atmosphere. N31-1 Thus, total ventilation—a reflection of the work invested in breathing—is the sum of the wasted dead-space ventilation and the useful alveolar ventilation. In our example,

so that the dead-space ventilation is 30% of the total ventilation.

The inset of Figure 31-1 illustrates how inspiration and expiration lead to small fluctuations in the alveolar partial pressures for O ₂ and CO ₂ , noted above in Figure 30-3 B . Throughout the respiratory cycle, blood flowing through the pulmonary capillaries continuously draws O ₂ out of the alveolar air and adds CO ₂ . Just before an inspiration, alveolar has fallen to its lowest point, and alveolar has risen to its highest. During inspiration, a new bolus of inspired fresh air mixes with pre-existing alveolar air, which causes alveolar to rise and alveolar to fall. During expiration and until the next inspiration, alveolar and gradually drift to the values that we saw at the start of the respiratory cycle. Assuming that functional residual capacity (FRC) is 3 L and that each breath adds 350 mL of fresh air, one can calculate that alveolar oscillates with an amplitude of ~5 mm Hg, whereas alveolar oscillates with an amplitude of ~4 mm Hg. N31-2

N31-2

Oscillations in
and
During Breathing

We will assume that the volume of alveolar air after a quiet expiration is 3000 mL (the FRC), and that this air has a of 98.4 mm Hg—the nadir of during our hypothetical respiratory cycle. The subsequent inspiration delivers to the alveoli 350 mL of fresh air (at 500 mL tidal volume less 150 mL of anatomical dead space) that has a of 149 mm Hg. Thus, after the 350 mL of fresh air mixes with the 3000 mL of pre-existing alveolar air, the alveolar will be

This value is the zenith of during our hypothetical respiratory cycle. During the ensuing several seconds, the alveolar drifts back down to 98.4 mm Hg as O ₂ diffuses from the alveolar air into the pulmonary-capillary blood. Thus, during a respiratory cycle, alveolar oscillates from a low of 98.4 mm Hg to a high of 103.6 mm Hg, with a mean of 101 mm Hg. In other words, alveolar oscillates around a mean of 101 mm Hg, with the peak and nadir deviating from the mean by ~2.6 mm Hg—an amplitude of ~5 mm Hg.

In the case of CO ₂ , the analysis is similar. After a quiet expiration, the 3000 mL of alveolar air has a of 42.2 mm Hg—the zenith of during our hypothetical respiratory cycle. The subsequent inspiration delivers to the alveoli 350 mL of fresh air, which has a of ~0 mm Hg. Thus, after the 350 mL of fresh air mixes with the 3000 mL of pre-existing alveolar air, the alveolar will be

This value is the nadir of during our hypothetical respiratory cycle. During the ensuing several seconds, the alveolar drifts back up to 42.2 mm Hg as CO ₂ diffuses from the pulmonary-capillary blood to the alveolar air. Thus, during a respiratory cycle, alveolar oscillates from a high of 42.2 mm Hg to a low of 37.8 mm Hg, with a mean of 40 mm Hg. In other words, alveolar oscillates around a mean of 40 mm Hg, with the peak and nadir deviating from the mean by ~2.2 mm Hg—an amplitude of ~4 mm Hg.

The Fowler single-breath N ₂ -washout technique estimates anatomical dead space

In 1948, Ward Fowler introduced an approach for estimating the anatomical dead space, based on the washout of N ₂ from the lungs. The key concept is that N ₂ is physiologically inert. After the subject has been breathing room air, the alveolar air is ~75% N ₂ . After a quiet expiration, when lung volume is FRC ( Fig. 31-2 A , step 1), the subject takes a single normal-sized breath (~500 mL). The inspired air is 100% O ₂ , although, in principle we could use any nontoxic gas mixture lacking N ₂ . The first portion of inspired O ₂ enters the alveolar spaces (step 2), where it rapidly mixes by diffusion and dilutes the N ₂ and other gases remaining after the previous breaths of room air (step 3). The last portion of the inspired O ₂ (~150 mL) remains in the conducting airways, which have a of zero.

The subject now exhales ~500 mL of air (see Fig. 31-2 A , step 4). If no mixing occurred between the N ₂ -free air in the most distal conducting airways and the N ₂ -containing air in the most proximal alveolar spaces, then the first ~150 mL of air emerging from the body would have an [N ₂ ] of zero (see Fig. 31-2 B , red lines). After this would come a sharp transition to a much higher [N ₂ ] for the final ~350 mL of expired air. Thus, the expired volume with an [N ₂ ] of zero would represent air from the conducting airways (anatomical dead space), whereas the remainder would represent air from the alveoli.

In reality, some mixing occurs between the air in the conducting airways and alveoli, so that the transition is S-shaped (see Fig. 31-2 C , red curve). A vertical line drawn through the S-shaped curve N31-3 so that area a is the same as area b marks the idealized transition between air from conducing and alveolar airways—as in Figure 31-2 B . The expired lung volume at the point of this vertical line is thus the anatomical dead space. In Figure 31-2 C , the part of the S-shaped curve with an expired [N ₂ ] of zero represents pure dead-space air, the part where [N ₂ ] gradually rises represents a mixture of dead-space and alveolar air, and the part where [N ₂ ] is high and flat represents pure alveolar air. This plateau is important, because it is during this plateau that one obtains an alveolar gas sample. N31-4

The Bohr expired-[CO ₂ ] approach estimates physiological dead space

In principle, we could compute the dead space using any gas whose expiration profile looks like that of N ₂ . Nitrogen is useful because we can easily create an artificial situation in which the subject makes a single inhalation of N ₂ -free air (e.g., a single breath of 100% O ₂ ). Another possibility is CO ₂ . Its profile during expiration is similar to that of N ₂ . Moreover, we do not need to use any special tricks to get it to work because room air has practically no CO ₂ . Yet plenty of CO ₂ is in the alveoli, where it evolves from the incoming mixed-venous blood. After a quiet expiration ( Fig. 31-3 A , step 1), the of the alveolar air is virtually the same as the of the arterial blood (see p. 673 ), ~40 mm Hg. The subject now inhales a normal tidal volume (~500 mL) of room air, although any CO ₂ -free gas mixture would do. The first portion enters the alveoli (step 2), where it rapidly dilutes the CO ₂ and other gases remaining after the previous breath (step 3). The rest (~150 mL) remains in the conducting airways, which now have a of ~0. When the subject now expires (step 4), the first air that exits the body is the CO ₂ -free gas that had filled the conducting airways, followed by the CO ₂ -containing alveolar air. Thus, the idealized profile of expired [CO ₂ ] (see Fig. 31-3 B , red lines) is similar to the idealized [N ₂ ] profile (see Fig. 31-2 B ). In particular, the volume of expired air at the vertical line in Figure 31-3 B is the estimated anatomical dead space.

One could use a CO ₂ probe to record the expired [CO ₂ ] profile during a single-breath CO ₂ washout, rather than the [N ₂ ] profile that Fowler used to measure anatomical dead space. However, because CO ₂ probes did not exist in his day, Christian Bohr N31-5 used a single-breath CO ₂ washout but analyzed the average in the mixed -expired air (i.e., averaged over the dead space plus expired alveolar air).

The principle of the Bohr approach is that the amount of CO ₂ present in the volume of mixed-expired air (V _E ) is the sum of the CO ₂ contributed by the volume of air from the dead space (V _D ) plus the CO ₂ contributed by the volume of air coming from the alveoli (V _E − V _D ). As summarized in Figure 31-3 B :

• 1

  The amount of CO ₂ coming from the dead space is the product of V _D and [CO ₂ ] in this dead-space air. Because [CO ₂ ] _D is zero, the area beneath V _D is also zero.

• 2

  The amount of CO ₂ coming from alveolar air is the product of (V _E − V _D ) and alveolar [CO ₂ ] and is represented by the rose area in Figure 31-3 B .

• 3

  The total amount of CO ₂ in the mixed-expired air is the product of V _E and the average [CO ₂ ] in this air and is represented by the hatched area in Figure 31-3 B .

Because the rose and hatched areas in Figure 31-3 B must be equal, and because the alveolar and expired [CO ₂ ] values are proportional to their respective values, it is possible to show that N31-6

This is the Bohr equation. Typically, V _D /V _E ranges between 0.20 and 0.35. For a V _D of 150 mL and a V _E of 500 mL, V _D /V _E would be 0.30. For example, if the alveolar is 40 mm Hg and the mixed-expired is 28 mm Hg, then

Equation 31-4 makes good intuitive sense. In an imaginary case in which we reduced V _D to zero, the expired air would be entirely from the alveoli, so that

On the other hand, if we reduced the tidal volume to a value at or below the dead-space volume, then all of the expired air would be dead-space air. In this case,
would be zero, and

N31-6

The Bohr Equation

The principle of the Bohr approach is that the CO ₂ concentration of the expired air is the CO ₂ concentration of the alveolar air —as diluted by the CO ₂ in the dead space, which contains no CO ₂ . Imagine that we have a volume of expired air (V _E ). The total amount of CO ₂ present in this expired air is the sum of the CO ₂ contributed by the volume of air from the dead space (V _D ) and the CO ₂ contributed by the volume of air coming from the alveoli (V _E − V _D ):

Knowing that the amount of the gas is simply the product of the volume and the concentration of the gas in that volume, we can write

Note that the volume of air coming from the alveoli is not the total volume of alveolar air, which might be a couple of liters, but rather that part of the expired volume that came from the alveoli. For example, if the expired volume were 500 mL and the dead space were 150 mL, the air coming from the alveoli (V _E − V _D ) would be 350 mL.

Because [CO ₂ ] _D is virtually zero, we can drop the “conducting airway” term from Equation NE 31-4 . In other words, the hatched area beneath the dashed line in Figure 31-3 B is equal to the red area under the solid red line. The simplified version of Equation NE 31-4 is thus

Furthermore, because the CO ₂ concentration is proportional to the CO ₂ partial pressure, and because the proportionality constant is the same for the expired air and the alveolar air,

We can rearrange this equation and solve for the ratio V _D /V _E , which is the fraction of the expired volume that came from the dead space:

This is the Bohr equation. Typically, V _D /V _E ranges between 0.20 and 0.35. For a V _D of 150 mL and a V _E of 500 mL, V _D /V _E would be 0.30. Because the partial pressure of CO ₂ in the dead-space air is virtually zero, must be less than . For example, imagine that the alveolar were 40 mm Hg and the average in the expired air were 28 mm Hg:

Equation NE 31-7 makes good intuitive sense. In an extreme hypothetical case in which we reduced V _D to zero, the expired air would be entirely from the alveoli, so that

On the other hand, if the tidal volume and the dead space were both 150 mL, then all of the expired air would be dead-space air. Thus, the would have to be zero, and

If the tidal volume were increased from 500 mL to 600 mL and V _D remained at 150 mL, the ratio V _D /V _E would fall from 0.30 to 0.25. With a greater V _E , the alveolar air would make a greater contribution to the mixed expired air, so that the mean expired would be 30 mm Hg (i.e., closer to ) instead of only 28 mm Hg. Thus, if you wish to compute V _D using the Bohr approach, it is best to use an expired volume that is larger than V _D , but not too much larger.

Two examples of this principle are of practical importance. During panting, the respiratory frequency is very high, but the tidal volume is only slightly greater than the anatomical dead space. Thus, most of the total ventilation is wasted as dead-space ventilation. If we reduced tidal volume below V _D , then in principle there would be no alveolar ventilation at all! During snorkeling, a swimmer breathes through a tube that increases V _D . If the snorkeling tube had a volume of 350 mL and the dead space within the body of the swimmer were 150 mL, then a tidal volume of 500 mL would in principle produce no alveolar ventilation. Consequently, the swimmer would suffocate, even though total ventilation was normal! Thus, the fractional dead space (V _D /V _E ) depends critically on tidal volume.

It is important to recognize that although the Fowler and Bohr methods yield about the same estimate for V _D in healthy individuals, the two techniques actually measure somewhat different things. The Fowler approach measures anatomical dead space —the volume of the conducting airways from the mouth/nose up to the point where N ₂ in the alveolar gas rapidly dilutes inspired 100% O ₂ . The Bohr approach, on the other hand, measures the physiological dead space —the volume of airways not receiving CO ₂ from the pulmonary circulation, and, therefore, not engaging in gas exchange. In a healthy person, the anatomical and physiological dead spaces are identical—the volume of the conducting airways. However, if some alveoli are ventilated but not perfused by pulmonary-capillary blood, these unperfused alveoli, like conducting airways, do not contain CO ₂ . The air in such unperfused alveoli, known as alveolar dead space, contributes to the physiological dead space:

The Fowler and Bohr methods could yield very different results in a patient with a pulmonary embolism, a condition in which a mass such as a blood clot wedges into and obstructs part or all of the pulmonary circulation. Alveoli downstream from the embolus are ventilated but not perfused; that is, they are alveolar dead space (see Fig. 31-3 C ). Thus, the Bohr method—but not the Fowler method—could detect an increase in the physiological dead space caused by a pulmonary embolism.

Alveolar ventilation is the ratio of CO ₂ production rate to CO ₂ mole fraction in alveolar air

One way of computing alveolar ventilation is to subtract the dead space from the tidal volume and multiply the difference by the respiratory frequency (see Equation 31-3 ). We can also calculate from alveolar . The body produces CO ₂ via oxidative metabolism at a rate of ~200 mL/min. In the steady state, this rate of CO ₂ production ( ) must equal the rate at which the CO ₂ enters the alveoli, and the rate at which we exhale the CO ₂ . Of course, this 200 mL/min of exhaled CO ₂ is part of the ~4200 mL of total alveolar air that we exhale each minute. Therefore, the exhaled 200 mL of CO ₂ is ~5% of the exhaled 4200 mL of alveolar air:

Rearranging the foregoing equation and solving for
yields

The equation above the brace is true only when we measure all parameters under the same conditions. This obvious point would hardly be worth noting if respiratory physiologists had not managed, by historical accident, to measure the two volume terms under different conditions:

• 1

  Body temperature and pressure, saturated with water vapor (BTPS; see p. 594 ) for .

• 2

  Standard temperature and pressure/dry (STPD; see p. 594 ) for .

Thus, in Equation 31-10 we introduce a constant k that not only indicates that alveolar ( ; measured at 37°C) is proportional to the mole fraction of CO ₂ in alveolar air, but also accounts for the different conditions for measuring the parameters. N31-7

N31-7

The Conversion Factor 0.863

Unfortunately for the student of respiratory physiology, the way customs have evolved for measuring respiratory volumes and partial pressures is a case study in mixed conventions. In a rational world, all parameters would be measured under a consistent set of conditions. In the real world of respiratory science, however, the parameters you (and clinicians) will need to compute alveolar ventilation, alveolar , and alveolar are measured under at least three very different sets of conditions. As a result, the student is faced with “correction factors” such as 0.863. Our advice to the student, when working numerical problems, is to insert the laboratory data directly into the correct equation—which you ought to intuitively understand—and use the proper correction factor. You will get the right answer. Before explaining the origin of 0.863 in Equation 31-11 , we will examine the system of units in which clinical laboratories report each of the three terms in that equation.

First, , the alveolar ventilation, is reported in the same units as (the rate at which air is expired from the lungs). If you collect all the warm, moist air that a subject exhales over a period of 1 minute, the laboratory will report that volume BTPS (body temperature and pressure, saturated with water vapor), precisely the same volume that the sample of expired air would have occupied in the warm, moist confines of the lung (see Box 26-3 ).

Second, , the rate at which the body produces CO ₂ , is measured at STPD (standard temperature and pressure/dry). Thus, if you collect a sample of warm, moist air expired over a period of 1 minute and give it to a clinical laboratory for analysis, the laboratory will report the value as the volume that the dry CO ₂ in the sample would occupy at 0°C. The rationale is that this is the way chemists treat gases.

Third, , the partial pressure of CO ₂ in the alveoli, is reported in BTPD (body temperature and pressure/dry). If you obtain a sample of alveolar air and send it to a clinical laboratory, it will take the warm, moist air you send it and keep it at 37°C but remove the H ₂ O. Of course, if the sample is in a rigid container and if P b is 760 mm Hg, this removal of water vapor will lower the total pressure of the sample by 47 mm Hg (i.e., the vapor pressure of water at 37°C) to 713 mm Hg. To keep the pressure constant, the laboratory will allow the sample volume to decrease to a volume that is 713/760 of the original volume. This, however, means that the mole fraction of CO ₂ increases by the fraction 760/713. Thus, the in the warm, moist alveolar air is actually lower than the reported BTPD; the BTPS value of is the reported value multiplied by (713/760) or ~0.983. The real tragedy of the convention for reporting at BTPD is that the same laboratory will report arterial as BTPS.

In deriving the factor 0.863, we begin with a rearrangement of Equation 31-9 , using a consistent set of units (i.e., BTPS):

Because is actually reported in STPD, our first task is to convert (STPD) to (BTPS). Because standard temperature is 0°C and body temperature is 37°C, we must multiply (STPD) by the ratio of temperatures in degrees Kelvin, the factor [(273 + 37)/273]. Furthermore, as noted in the previous paragraph, when we add water vapor to a dry sample of a gas and then expand the volume to keep the total pressure fixed at P b , the partial pressure of that gas will decrease to (713/760) of its dry partial pressure. Thus,

Our final task is to replace (%CO ₂ BTPS) _A with (BTPD). Because (BTPS) = (%CO ₂ BTPS) _A × P b ,

Because is reported in BTPD, we must multiply it by the factor (713/760). Thus,

Substituting the expression for (%CO ₂ BTPS) _A in Equation NE 31-14 into Equation NE 31-12 , we have

In the above equation, both and are reported in mL/min. Because it is customary to report in L/min, the factor 863 becomes 0.863:

The above is the alveolar ventilation equation.

For example, if is 200 mL/min and is 40 mm Hg, then

This is the alveolar ventilation equation, which we can use to compute . We determine by collecting a known volume of expired air over a fixed time period and analyzing its CO ₂ content. For a 70-kg human, is ~200 mL/min. We can determine by sampling the expired air at the end of an expiration—an alveolar gas sample (see pp. 676–677 ). N31-4 This end-tidal (see Fig. 31-2 C ) is ~40 mm Hg. In practice, clinicians generally measure arterial ( ), and assume that alveolar and arterial are identical (see p. 673 ). Inserting these values into the alveolar ventilation equation, we have

For the sake of simplicity—and consistency with our example in Equation 31-3 —we round this figure to 4.2 L/min.

Although it is usually safe to regard as being con­stant for a person at rest, a clinical example in which can increase markedly is malignant hyperthermia (see Box 9-2 ), which is associated with increased oxidative metabolism. One hallmark of this clinical catastrophe is that the increase in leads to an increase in even though is normal.

Alveolar and arterial are inversely proportional to alveolar ventilation

Viewing Equation 31-11 from a different perspective illustrates one of the most important concepts in respiratory physiology: Other things being equal, alveolar is inversely proportional to alveolar ventilation. This conclusion makes intuitive sense because the greater the , the more the fresh, inspired air dilutes the alveolar CO ₂ . Rearranging Equation 31-11 yields

In other words, if CO ₂ production is fixed, then doubling causes to fall to half of its initial value. Conversely, halving causes to double. Because, arterial is virtually the same as alveolar (see p. 673 ), changes in affect both and .

The blue curve in Figure 31-4 helps illustrate the principle. Imagine that your tissues are producing 200 mL/min of CO ₂ . In a steady state, your lungs must blow off 200 mL of CO ₂ each minute. Also, imagine that your lungs are exhaling 4200 mL/min of alveolar air. Because the 200 mL of expired CO ₂ must dissolve in the 4200 mL of exhaled alveolar air (center red point in Fig. 31-4 ), Equation 31-13 tells us that your (and thus ) must be ~40 mm Hg.

What would happen if the excitement of reading about respiratory physiology caused your alveolar ventilation to double, to 8400 mL/min? This is an example of hyperventilation. You could double either by doubling respiratory frequency or by doubling the difference between tidal volume and dead space, or a combination of the two (see Equation 31-3 ). Immediately on doubling , you would be blowing off at twice the previous rate not only alveolar air (i.e., 8400 mL/min) but also CO ₂ (i.e., 400 mL/min). Because your body would continue to produce only 200 mL/min of CO ₂ —assuming that doubling does not increase —you would initially blow off CO ₂ faster than you made it, causing CO ₂ levels throughout your body to fall. However, the falling of mixed-venous blood would cause alveolar to fall as well, and thus the rate at which you blow off CO ₂ would gradually fall. Eventually, you would reach a new steady state in which the rate at which you blow off CO ₂ would exactly match the rate at which you produce CO ₂ (i.e., 200 mL/min).

When you reach a new steady state, the values in your mixed-venous blood, arterial blood, and alveolar air would be stable. But what would be the ? Because each minute you now are blowing off 8400 mL of alveolar air (i.e., twice normal) but still only 200 mL of CO ₂ (right red point in Fig. 31-4 ), your must be half normal, or ~20 mm Hg. Not only does the hyperventilation cause alveolar to fall by half, it also causes arterial to fall by half. Thus, hyperventilation leads to respiratory alkalosis (see p. 634 ). This respiratory alkalosis causes the arterioles in the brain to constrict (see p. 559 ), reducing blood flow to the brain, which causes dizziness.

What would happen if, instead of doubling alveolar ventilation, you halved it from 4200 to 2100 mL/min? This is an example of hypoventilation. At the instant you began hypoventilating, the volume of CO ₂ expired per unit time would fall by half, to 100 mL/min, even though CO ₂ production by the tissues would remain at 200 mL/min. Thus, CO ₂ would build up throughout the body, causing to rise. To what value would have to increase before you would reach a new steady state? Because each minute you must exhale 200 mL of CO ₂ but this can be diluted in only 2100 mL, or half the usual amount of alveolar air (left red point in Fig. 31-4 ), the alveolar [CO ₂ ] must double from ~40 to 80 mm Hg. Of course, this doubling of alveolar is paralleled by a doubling of arterial , leading to a respiratory acidosis (see p. 633 ).

Therefore, the steady-state alveolar is inversely proportional to alveolar ventilation. The higher the , the lower the . If were infinitely high, then would theoretically fall to zero, the of inspired air.

Alveolar and arterial rise with increased alveolar ventilation

As illustrated by the red curve in Fig. 31-4 , increases in alveolar ventilation cause alveolar to rise and—at an infinite —approach the inspired of ~149 mm Hg.

Although alveolar obviously depends critically on , it is also influenced to a lesser extent by alveolar gases other than O ₂ —namely, H ₂ O, N ₂ , and CO ₂ . The partial pressure of H ₂ O at 37°C is 47 mm Hg, and will not change unless body temperature changes. The partial pressures of all of the other gases “fit” into what remains of barometric pressure (P b ) after has claimed its mandatory 47 mm Hg. We can think of N ₂ as a “spectator molecule,” because it is not metabolized; is whatever it has to be to keep the total pressure of the dry alveolar air at 760 − 47 = 713 mm Hg. That leaves us to deal with CO ₂ .

Because inspired air contains virtually no CO ₂ , we can think of the alveolar CO ₂ as coming exclusively from metabolism. However, depends not only on how fast the tissues burn O ₂ but also on the kind of fuel they burn. If the fuel is carbohydrate, then the tissues produce one molecule of CO ₂ for each O ₂ consumed (see p. 1188 ). This ratio is termed the respiratory quotient (RQ):

In this example the RQ is 1, which is a good place to start when considering how alveolar affects alveolar . If we consider only the dry part of the inspired air that enters the alveoli (see Table 26-1 ), then is 713 × 0.78 = 557, is 713 × 0.2095 = 149 mm Hg, and is ~0. As pulmonary-capillary blood takes up incoming O ₂ , it replaces the O ₂ with an equal number of CO ₂ molecules in the steady state (RQ = 1). Because the exchange of O ₂ for CO ₂ is precisely 1 for 1, alveolar is what is left of the inspired after metabolism replaces some alveolar O ₂ with CO ₂ ( = 40 mm Hg):

A typical fat-containing Western-pattern diet produces an RQ of ~0.8 (see p. 1188 ), so that 8 molecules of CO ₂ replace 10 molecules of O ₂ in the alveolar air. This 8-for-10 replacement has two consequences. First, the volume of alveolar air falls slightly during gas exchange. Because the non-H ₂ O pressure remains at 713 mm Hg, this volume contraction concentrates the N ₂ and dilutes the O ₂ . Second, the volume of expired alveolar air is slightly less than the volume of inspired air.

The alveolar gas equation N31-8 describes how alveolar depends on RQ:

N31-8

The Alveolar Gas Equation

The alveolar gas equation allows one to compute the ideal alveolar ( ) from one's knowledge of the alveolar ( ) as well as two other parameters, the mole fraction of O ₂ in the dry inspired air ( ) and the RQ. We begin by deriving an expression for RQ, which, as described on page 1188 , is defined as the ratio of the rate of metabolic CO ₂ production ( ) to the rate of O ₂ consumption ( ):

For RQ values less than unity, the body consumes more O ₂ than it produces CO ₂ , so that the inspired alveolar ventilation ( ) must be greater than the expired alveolar ventilation ( ). Because the metabolically produced CO ₂ must all appear in the expired alveolar gas, is the fraction of the expired alveolar ventilation that is CO ₂ gas ( ):

Similarly, the consumed O ₂ must all enter the body via the lungs; is the difference between (1) the amount of O ₂ entering the alveoli during inspirations over a certain period of time, and (2) the amount of O ₂ exiting the alveoli during expirations over that same time interval:

Here, is the mole fraction of O ₂ in the inspired air, and is the mole fraction of O ₂ in the expired alveolar air. Substituting the expressions for (see Equation NE 31-19 ) and (see Equation NE 31-20 ) into the definition of RQ (see Equation NE 31-18 ), we have

We now need an expression for the ratio . In the derivation of this expression, the fundamental assumption is that the N ₂ gas in the lungs is not metabolized. Thus, in the steady state, the amount of N ₂ entering the alveoli with inspirations over a certain period of time is the same as the amount of N ₂ leaving the alveoli with the expirations over the same period. In other words,

Here, is the mole fraction of N ₂ in this inspired gas, and is the mole fraction of N ₂ in the expired alveolar air. For the rest of this derivation, we will consider only dry gases; the laboratory reports all partial pressure (P) and mole fraction (F) values in terms of the “dry” gas (see Box 26-1 ). Thus, considering only the dry inspired air, the mole fractions of N ₂ , O ₂ , and CO ₂ must sum to 1:

Because the inspired air contains virtually no CO ₂ (i.e., ≅ 0),

Similarly, considering only the dry expired air, the mole fractions of N ₂ , O ₂ , and CO ₂ must likewise sum to 1:

If we now substitute the expressions for (see Equation NE 31-24 ) and (see Equation NE 31-25 ) into Equation NE 31-22 , we have

Solving for the ratio of inspired to expired alveolar ventilation yields

We now have the ratio ( ) that we needed for Equation NE 31-21 . Substituting Equation NE 31-27 into Equation NE 31-21 , we get

If we now solve Equation NE 31-28 for , we have

Realizing that the “dry” partial pressure (P) is the product of mole fraction (the above F terms) and (P b − 47), we can multiply Equation NE 31-12 through by (P b − 47) to arrive at our final equation, which expresses quantities in terms of partial pressures:

Remember, all the partial-pressure and mole-fraction values in the above equation refer to dry gases. Thus, is × (P b − 47).

On page 681 , we provide a simplified version of Equation NE 31-30 :

Why is this equation a reasonable approximation? Examination of Equations NE 31-30 and NE 31-31 shows that they are identical except that the term in brackets in Equation NE 31-30 is replaced by 1/RQ in Equation NE 31-31 . For Equation NE 31-30 , we can rearrange the term in brackets as follows:

Under physiological conditions—when is 0.21 and RQ is 0.8—the above expression evaluates to 0.958/RQ, which is quite close to 1/RQ … which is to say that Equation NE 31-31 is a reasonable approximation of Equation 31-30 under these conditions.

When RQ is unity, the term in brackets in Equation 31-30 evaluates to unity, and Equation NE 31-30 simplifies to Equation 31-16 in the text:

is the fraction of inspired dry air that is O ₂ , which is 0.21 for room air (see Table 26-1 ). Note that when RQ is 1, the term in parentheses becomes unity, and Equation 31-17 reduces to Equation 31-16 . The term in parentheses also becomes unity, regardless of RQ, if is 100% (i.e., the subject breathes pure O ₂ )—in this case, no N ₂ is present to dilute the O ₂ .

A simplified version of Equation 31-17 is nearly as accurate:

The concepts developed in the last two sections allow us to compute both alveolar and alveolar . The approach is first to use Equation 31-13 to calculate from and , and then use Equation 31-17 to calculate from and RQ. Imagine that we first found that is 40 mm Hg and that we know that RQ is 0.8. What is ?

By default, the partial pressure of N ₂ and other gases (e.g., argon) is P b − − or 713 − 40 − 101 = 572 mm Hg. For simplicity, we round down this to 100 mm Hg in our examples.

Because of the action of gravity on the lung, regional ventilation in an upright subject is normally greater at the base than the apex

Until now, we have assumed that all alveoli are ventilated to the same extent. We could test this hypothesis by using an imaging technique for assessing the uniformity of ventilation. Imagine that a subject who is standing up breathes air containing ¹³³ Xe. Because Xe has very low water solubility, it (like He and N ₂ ) has a very low diffusing capacity (see pp. 661–663 ) and—over a short period—remains almost entirely within the alveoli.

Imaging the ¹³³ Xe radioactivity immediately after a single breath of ¹³³ Xe provides an index of absolute regional ventilation ( Fig. 31-5 A ). However, [ ¹³³ Xe] might be low in a particular region either because the alveoli truly receive little ventilation or because the region has relatively little tissue. Therefore, we normalize the absolute data to the maximal regional alveolar volume. The subject continues to breathe the ¹³³ Xe until [ ¹³³ Xe] values stabilize throughout the lungs. When the subject now makes a maximal inspiratory effort (V _L = total lung capacity [TLC]), the level of radioactivity detected over any region reflects that region's maximal volume. Dividing the single-breath image by the steady-state image at TLC yields a ratio that describes regional ventilation per unit volume.

This sort of analysis shows that alveolar ventilation in a standing person gradually falls from the base to the apex of the lung (see Fig. 31-5 B ). Why? The answers are posture and gravity. In Chapter 27 we saw that, because of the lung's weight, intrapleural pressure (P _IP ) is more negative at the apex than at the base when the subject is upright (see Fig. 31-5 C ). The practical consequences of this P _IP gradient become clear when we examine a static pressure-volume diagram not for the lungs as a whole (see Fig. 27-5 ), but rather for a small piece of lung (see Fig. 31-5 D ). We assume that the intrinsic mechanical properties of the airways are the same, regardless of whether the tissue is at the base or at the apex. At the base, where P _IP might be only −2.5 cm H ₂ O at FRC, the alveoli are relatively underinflated compared to tissues at the apex, where P _IP might be −10 cm H ₂ O and the alveoli are relatively overinflated. However, because the base of the lung is underinflated at FRC, it is on a steeper part of the pressure-volume curve (i.e., it has a greater static compliance ) than the overinflated apex. Thus, during an inspiration, the same ΔP _IP (e.g., 2.5 cm H ₂ O) produces a larger ΔV _L near the base than near the apex. Keep in mind that it is the change in volume per unit time, not the initial volume, that defines ventilation.

The relationship between ventilation and basal versus apical location in the lung would be reversed if the subject hung by the knees from a trapeze. A person reclining on the right side would ventilate the dependent lung tissue on the right side better than the elevated lung tissue on the left. Of course, the right-to-left P _IP gradient in the reclining subject would be smaller than the apex-to-base P _IP gradient in the standing subject, reflecting the smaller distance (i.e., smaller hydrostatic pressure difference). Subjects under microgravity conditions (see p. 1233 ), such as astronauts aboard the International Space Station, experience no P _IP gradients, and thus no gravity-dependent regional differences in ventilation.

Restrictive and obstructive pulmonary diseases can exacerbate the nonuniformity of ventilation

Even in microgravity, where we would expect no regional dif­ferences in ventilation, ventilation would still be nonuniform at the microscopic or local level because of seemingly random differences in local static compliance (C) and airway resistance (R). In fact, such local differences in the ventilation of alveolar units are probably more impressive than gravity-dependent regional differences. Moreover, pathological changes in compliance and resistance can substantially increase the local differences and thus the nonuniformity of ventilation. N31-9

Restrictive Pulmonary Disease

As discussed in Box 27-1 , restrictive pulmonary diseases include disorders that decrease the static compliance of alveoli (e.g., fibrosis) as well as disorders that limit the expansion of the lung (e.g., pulmonary effusion). Figure 31-6 A shows a hypothetical example in which R is normal and disease has halved the static compliance of one lung but left the other unaffected. N31-10 Thus, for the usual change in P _IP , the final volume change (ΔV) of the diseased lung is only half normal, so that its ventilation is also halved. Because the ventilation of the other unit is normal, decreased local compliance has increased the nonuniformity of ventilation.

N31-10

Effect of Changes in Compliance on the Time Constant Governing Changes in Lung Volume

On page 622 , we saw that, during inspiration and expiration, the time course of lung volume (V _L ) is approximately exponential. The time constant (τ) for the change in V _L (ΔV _L ) is the time required for the change in V _L to be ~63% complete. Moreover, N27-10 explains why τ is the product of airway resistance (R) and alveolar compliance (C):

$\tau =\text{R}\times \text{C}$

Decreased Compliance

In the example discussed under restrictive pulmonary disease on page 683 , we decreased the compliance of one lung by half. Of course, other things being equal, the ΔV of the affected lung will be half normal, as illustrated in Figure 31-6 A . What is not so obvious is that τ of the affected lung will also be half normal. In other words, the lung with half-normal compliance will achieve its final volume twice as fast as normal. The reason is that, with a normal airway resistance, the inhaled air—at any instant in time—will flow at a more-or-less normal rate, so that the affected lung achieves its half-normal ΔV earlier than the normal lung.

Although it might seem that the decreased τ in this example is a good thing, the problem is that the reduced ΔV translates to less ventilation, and that is not a good thing. As described in the text, this reduced ventilation increases the nonuniformity of ventilation, which in turn—as discussed below in this chapter—tends to lead to hypoxia and respiratory acidosis.

Increased Compliance

What would be the effect of increasing the static compliance of one lung? A disease that increases the static compliance is emphysema (see Fig. 27-5 ). Other things being equal, an increase in compliance will cause the ΔV of the affected lung to be greater than normal. At the same time, the τ of the affected lung will also be increased. The reason is that, with a normal airway resistance, the inhaled air—at any instant in time—will flow at a more-or-less normal rate, so that the affected lung achieves its greater-than-normal ΔV later than a normal lung. If τ is too large, then the time that the person allows for inspiration may not be long enough for the affected lung to increase its volume to the level that it could achieve if the inspiration were infinitely long. As a result, the inspiration will be truncated, and the true ΔV for the affected lung will be less than normal, thereby exacerbating the unevenness of ventilation. Moreover, the greater the respiratory frequency, the greater the truncation of the inspiration, and thus the greater the exacerbation of the unevenness of ventilation.

Figure 27-15 A illustrates the effect of increased airway resistance on the time course of V _L , and Figure 27-15 B illustrates the effect on the change in lung volume, which is proportional to dynamic compliance. In the example in Figure 27-15 , the static compliance was normal and the final ΔV was the same for the normal and affected lung. The example we are discussing in this webnote is just the opposite (i.e., a normal airway resistance but altered compliance). Nevertheless, the principle of how increases in τ affect ventilation is the same: the greater the τ, the greater the chance that increased respiratory frequency will lead to a truncated inspiration and thus a greater unevenness of ventilation.

Although we have focused on inspiration in these examples, the same principles apply to expiration, namely, the greater the τ, the greater the chance of a truncated expiration.

The pulmonary circulation has low pressure and resistance but high compliance

The pulmonary circulatory system handles the same cardiac output as the systemic circulation, but in a very different way. The systemic circulation is a high-pressure system. This high pressure is necessary to pump blood to the top of the brain while the individual is standing, or even to a maximally elevated fingertip. The systemic circulation also needs to be a high-pressure system because it is a high- resistance system. It uses this high resistance to control the distribution of blood flow. Thus, at rest, a substantial fraction of the systemic capillaries are closed, which gives the system the flexibility to redistribute large amounts of blood (e.g., to muscle during exercise). The mean pressure of the aorta is ~95 mm Hg ( Table 31-1 ). At the opposite end of the circuit is the right atrium, which has a mean pressure of ~2 mm Hg. Thus, the driving pressure for blood flow through the systemic circulation is ~93 mm Hg. Given a cardiac output ( ) of 5 L/min or 83 mL/s, we can compute the resistance of the systemic system using an equation like Ohm's law (see Equation 17-10 ):

PRU is a peripheral resistance unit (see p. 415 ), which has the dimension mm Hg/(mL/s).

In contrast, the pulmonary circulation is a low-pressure system. It can afford to be a low-pressure system because it needs to pump blood only to the top of the lung. Moreover, it must be a low-pressure system to avoid the consequences of Starling forces (see pp. 467–468 ), which would otherwise flood the lung with edema fluid. The mean pressure in the pulmonary artery is only ~15 mm Hg. Because the mean pressure of the left atrium, at the other end of the circuit, is ~8 mm Hg, and because the cardiac output of the right heart is the same as for the left, we have

Thus, the total resistance of the pulmonary circulation is less than one tenth that of the systemic system, which explains how the pulmonary circulation accomplishes its mission at such low pressures. Unlike in the systemic cir­culation, where most of the pressure drop occurs in the arterioles (i.e., between the terminal arteries and beginning of the capillaries), in the pulmonary circulation almost the entire pressure drop occurs rather uniformly between the pulmonary artery and the end of the capillaries. In particular, the arterioles make a much smaller contribution to resistance in the pulmonary circulation than in the systemic circulation.

What are the properties of the pulmonary vasculature that give it such a low resistance? First, let us examine the complete circuit. The pulmonary artery arises from the right ventricle, bifurcates, and carries relatively deoxygenated blood to each lung. The two main branches of the pulmonary artery follow the two mainstem bronchi into the lungs and bifurcate along with the bronchi and bronchioles. A single pulmonary arteriole supplies all of the capillaries of a terminal respiratory unit (see p. 597 ). Together, the two lungs have ~300 million alveoli. However, they may have as many as 280 billion highly anastomosing capillary segments (each looking like the edge of a hexagon in a piece of chicken wire), or nearly 1000 such capillary segments per alveolus—creating a surface for gas exchange of ~100 m ² . It is easy to imagine why some have described the pulmonary capillary bed as a nearly continuous flowing sheet of blood surrounding the alveoli. At rest, the erythrocytes spend ~0.75 second navi­gating this capillary bed, which contains ~75 mL of blood. During exercise, capillary blood volume may increase to ~200 mL. Pulmonary venules collect the oxygenated blood from the capillary network, converge, course between the lobules, converge some more, and eventually enter the left atrium via the pulmonary veins. The total circulation time through the pulmonary system is 4 to 5 seconds.

Pulmonary blood vessels are generally shorter and wider than their counterparts on the systemic side. Arterioles are also present in much higher numbers in the pulmonary circulation. Although the pulmonary arterioles contain smooth muscle and can constrict, these vessels are far less muscular than their systemic counterparts, and their resting tone is low. These properties combine to produce a system with an unusually low resistance.

The walls of pulmonary vessels have another key property: thinness, like the walls of veins elsewhere in the body. The thin walls and paucity of smooth muscle give the pulmonary vessels a high compliance, which has three consequences. First, pulmonary vessels can accept the relatively large amounts of blood that shift from the legs to the lungs when a person changes from a standing to a recumbent position. Second, as we discuss below, the high compliance also allows the vessels to dilate in response to modest increases in pulmonary arterial pressure. Third, the pulse pressure in the pulmonary system is rather low (on an absolute scale). The systolic/diastolic pressures in the pulmonary artery are typically 25/8 mm Hg, yielding a pulse pressure of ~17 mm Hg. In contrast, the systolic/diastolic pressures in the aorta are ~120/80 mm Hg, for a pulse pressure of ~40 mm Hg. Nonetheless, relative to the mean pulmonary artery pressure of 15 mm Hg, the pulmonary pulse pressure of 17 mm Hg is quite high.

Overall pulmonary vascular resistance is minimal at FRC

Because pulmonary blood vessels are so compliant, they are especially susceptible to deformation by external forces. These forces are very different for vessels that are surrounded by alveoli (i.e., “alveolar” vessels) compared with those that are not (i.e., “extra-alveolar” vessels). In both types, the key consideration is whether these external forces pull vessels open or crush them.

Alveolar Vessels

Alveolar vessels include the capillaries, as well as slightly larger vessels that are also surrounded on all sides by alveoli ( Fig. 31-7 A ). The resistance of these alveolar vessels depends on both the transmural pressure gradient and lung volume.

We have already introduced the transmural pressure gradient (P _TM ) in our discussions of systemic blood vessels (see p. 414 ) and conducting airways (see pp. 624–625 ). For alveolar vessels, P _TM is the difference between the pressure in the vessel lumen and that in the surrounding alveoli (P a ). For simplicity, we consider the factors affecting P _TM at a fixed lung volume.

The pressure inside these vessels varies with the cardiac cycle; indeed, the pulmonary capillary bed is one of the few in which flow is pulsatile (see p. 513 ). The pressure inside the alveolar vessels also depends greatly on their vertical position relative to the left atrium: the higher the vessel, the lower the pressure.

The pressure in the alveoli varies with the respiratory cycle. With no airflow and the glottis open, P a is the same as P b (i.e., 0 cm H ₂ O). On the other hand, P a is negative during inspiration and positive during expiration. A combination of a high intravascular pressure and a negative P a tends to dilate the compliant alveolar vessels, lowering their resistance. But a combination of a low intravascular pressure and a positive P a crushes these vessels, raising their resistance.

Changes in lung volume (V _L ) have characteristic effects on alveolar vessels. For simplicity, here we assume that each time we examine a new V _L , airflow has stopped, so that P a is zero. As V _L increases, the alveolar walls become more stretched out. Consequently, the alveolar vessels become stretched along their longitudinal axis but crushed when viewed in cross section. Both of these effects tend to raise vessel resistance. Thus, as V _L increases, the resistance of the alveolar vessels also increases (see Fig. 31-7 B , red curve).

Increases in pulmonary arterial pressure reduce pulmonary vascular resistance by recruiting and distending pulmonary capillaries

Although the pulmonary circulation is normally a low-resistance system under resting conditions, it has a remarkable ability to lower its resistance even further. During exercise, 2-fold to 3-fold increases in cardiac output may elicit only a minor increase in mean pulmonary arterial pressure. In other words, a slight increase in pulmonary arterial pressure is somehow able to markedly decrease resistance ( Fig. 31-8 A ) and thus markedly increase flow (see Fig. 31-8 B ). This behavior is a general property of a passive/elastic vascular bed (see Fig. 19-7 A , red curve). Two “passive” mechanisms—that is, mechanisms not related to “active” changes in the tone of vascular smooth muscle—are at work here: the recruitment and distention of pulmonary capillaries. However, before we can understand either change, we must more completely describe the pulmonary capillaries at “rest.”

Under “resting” conditions (i.e., at relatively low values of pulmonary arterial pressure), some pulmonary capillaries are open and conducting blood, others are open but not conducting substantial amounts of blood, and still others are closed (see Fig. 31-8 C ). Why should some capillaries be open but have no flow? In a highly anastomosing capillary network, tiny differences in driving pressure might exist. In addition, seemingly random differences in the dimensions of parallel capillaries may lead to differences in resistance. In low-pressure systems, such slight differences in absolute resistance allow pathways with relatively low resistances to steal flow from neighbors with slightly higher resistances, leaving some “open” pathways heavily underutilized. A familiar example is a type of garden hose used to drizzle water on a flower bed; this hose is closed at its distal end, but perforated with hundreds of tiny holes. If water pressure is low, only some of the holes conduct water.

Why should some parallel vessels be closed ? Popping open a previously closed vessel requires that the perfusion pressure overcome the tone of the vascular smooth muscle and reach that vessel's critical closing pressure (see p. 454 ), which varies from vessel to vessel. As we discuss below, alveolar vessels also may be closed because the alveolar pressure exceeds intravascular pressure, thereby crushing the vessel.