Page 129 - Color_Atlas_of_Physiology_5th_Ed._-_A._Despopoulos_2003

Pressure–Volume Curve, Respiratory determine the peak expiratory and inspiratory
Work pressures (! A, red and green curves). Only a
very small pressure can be generated from a
Resting position (RP) is the position to which position of near-maximum expiration (V pulm #
the lung—chest system returns at the end of 0; ! A7) compared to a peak pressure of about
normal quiet expiration; this lung volume 15 kPa (! 110 mmHg) at V pulm $ 0 (Valsalva’s
equals the functional residual capacity (FRC, maneuver; ! A5). Likewise, the greatest
! p. 114). Its value is set at zero (V pulm = 0) in negative pressure (suction) (ca. – 10 kPa =
A–C. RP (! A1) is a stable central position 75 mmHg) can be generated from a position of
characterized by the reciprocal cancellation of maximum expiration (Müller’s maneuver;
two passive forces: chest expansion force (CEF) ! A6), but not from an inspiratory position
and lung contraction force (LCF). As we (! A4).
breathe in and out, the lung – chest system A dynamic PV curve is obtained during res-
makes excursions from resting position; thus, piration (! C). The result is a loop consisting of
LCF ! CEF during inspiration, and CEF ! LCF the opposing inspiratory (red) and expiratory
during expiration. The difference between LCF (green) curves transected by the resting curve
and CEF, i.e. the net force (! blue arrows in A2, (blue) because airway flow resistance (RL)
3, 5, 6), is equal to the alveolar pressure (PA
Respiration ing a stopcock, as in A1–3, 5, 6) after a known middle airways) while inhaling in the one
must be overcome (mainly in the upper and
! p. 108) if the airway is closed (e.g., by turn-
direction and exhaling in the other. The driving
air volume has been inhaled (V pulm ! 0, ! A2)
pressure gradients (∆P) also oppose each other
! A3). (In the resting position, CEF = LCF, and
5 from a spirometer or expelled into it (V pulm " 0, (inspiratory PA " 0; expiratory PA ! 0; ! p.
109 B). As in Ohm’s law, ∆P = RL ! respiratory
.
PA = 0). Therefore, the relationship between flow rate (V). Therefore, ∆P must increase if
V pulm and PA in the lung—chest system can be the bronchial tubes narrow and/or if the respi-
determined as illustrated by the static resting ratory flow rate increases (! C).
pressure—volume (PV) curve (! blue curve
in A–C) (“static” = measured while holding the In asthma, the airway radius (r) decreases and a very
4
breath; “resting” = with the respiratory high ∆P is needed for normal ventilation (RL ! 1/r !).
During exspiration, a high ∆P decreases the transpul-
muscles relaxed). monary pressure (= PA % P pl) and thereby squeezes
(Compression and expansion of V pulm by a positive or the airways (R L"). The high RL results in a pressure
negative PA during measurement has to be taken decrease along the expiratory airway (P airway#) until
into account; ! A, dark-gray areas). P airway % P pl " 0. At this point, the airway will collapse.
This is called dynamic airway compression, which
The slope of the static resting PV curve, ∆V pulm/ often results in a life-threatening vicious cycle: r# !
∆PA, represents the (static) compliance of the ∆P"! r##! ∆P""....
lung–chest system (! B). The steepest part of Respiratory work. The colored areas within the
the curve (range of greatest compliance; ca. loop (AR insp and AR exp; ! C) represent the inspiratory
and expiratory PV work (! p. 374) exerted to over-
1 L/kPa in an adult) lies between RP and V pulm = come flow resistance. The cross-hatched area (! C)
1 L. This is in the normal respiratory range. The is the work required to overcome the intrinsic elastic
curve loses its steepness, i.e. compliance force of the lungs and chest (A elast). Inspiratory work is
decreases, in old age or in the presence of lung defined as AR insp + A elast. The inspiratory muscles
disease. As a result, greater effort is needed to (! p. 108) must overcome the elastic force, whereas
breathe the same tidal volume. the same elastic force provides the (passive) driving
force for expiration at rest (sign reverses for A elast).
The above statements apply to lung-and-chest com- Thus, expiratory work is AR exp – A elast. Expiration can
pliance. It is also possible to calculate compliance for also require muscle energy if AR exp becomes larger
the chest wall alone (∆VA/∆P pl = 2 L/kPa) or for the than A elast—e.g., during forced respiration or if RL is
lung alone (∆V/∆[PA – P pl] = 2 L/kPa) when the elevated.
pleural pressure (P pl) is known (! p. 108).
PV relationships can also be plotted during
116
maximum expiratory and inspiratory effort to
Despopoulos, Color Atlas of Physiology © 2003 Thieme
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