Respiratory Aids - Introduction to Biomechatronics - page 479

FIGURE 9-8
Graph showing the
lung volume as a
function of the
relaxation pressure
of the lung, the
chest cage, and the
combined pressure.
The slope of the
curves gives the
compliance
(133.34 Pa per
mmHg and
1.36 cmH 2 O per
mmHg).

opposite. At this point, lung volume corresponds to functional residual capacity (FRC).
With inspiration, the lung is stretched farther and exhibits a greater recoil pressure. At the
same time, the chest cage is less compressed, so its negative recoil pressure diminishes
as it approaches its equilibrium volume. The chest cage reaches its equilibrium volume
(point B), and the lung and lung-chest cage relaxation curves intersect (point C). At this
lung volume, all measured relaxation pressure for the lung-chest cage system is from the
lung alone. If an even greater air volume is inhaled (point D), both the lung and chest cage
are stretched beyond their equilibrium volumes.
Note that the compliance curve for the combined lung-chest cage becomes more
flattened (less compliant) at this point because the lung and chest cage are both tending to
recoil toward smaller equilibrium volumes.
If the total lung-chest cage system is returned to resting end expiration (point A) and
then more air is expelled, a negative relaxation pressure results for both the chest cage
and the combined lung-chest cage (point E). At this point, the chest cage is compressed
as more and more air is expelled, with the negative recoil pressure reflecting its tendency
to expand toward its equilibrium volume (point B). At the same time, the lung contributes
little positive relaxation pressure because it is close to its equilibrium volume because it
is stretched very little.
During normal breathing, the tidal volume range is between points A and B. During
inspiration, the individual moves up the lung-chest cage compliance curve from point A
toward point C and in the opposite direction during expiration. Note that over the normal
tidal volume range total pulmonary compliance, C tot , given by the slope of the curves is
less than the compliance of either the lung, C lung , or chest cage, C chest , alone, as can be
seen in Figure 9-9. This occurs because the lung and chest cage are physically arranged
in a series and therefore must be added as reciprocals

1
C tot =

1
C lung +

1
C chest

(9.1)